Simple Probability Question.

[ajizzle]

You are playing with a 52 card deck. Assuming a random shuffle, you deal cards face up from the top until the first Ace is dealt. Which has a higher probability of being the next card off the deck, the Ace of Spades or the Duece of Clubs.

[maurile]

The deuce of clubs.

After the first ace has been dealt, there is a 1/4 chance that the ace of spades is gone, so a 3/4 chance that it's still available to be dealt out.

The first ace will come up, on average, before 1/4 of the deck has been dealt out. (There's about a 50% chance that the first ace will be dealt within the first eight cards.) So at the time that the first ace is dealt, there's a greater-than-3/4 chance that the two of clubs remains in the deck to be dealt out.

[pzhon]

[ QUOTE ]
You are playing with a 52 card deck. Assuming a random shuffle, you deal cards face up from the top until the first Ace is dealt. Which has a higher probability of being the next card off the deck, the Ace of Spades or the Duece of Clubs.

[/ QUOTE ]
Interesting question. I hadn't seen that one.

Let's ignore the differences within the set of aces and within the set of nonaces. What is the probability that the card after the first ace is a second ace?

There are 52C4 ways the aces can be placed among the 52 cards. In 51C3 of them, the second ace is immediately after the first ace. You can get all such arrangements uniquely by placing 3 aces in 51 slots, then replacing the first ace with a pair of aces.

Since each order is equally likely, the probability that the card following the first ace is another ace is 51C3/52C4 = 1/13. The probability this card is the ace of spades is 1/4 of 1/13, 1/52. The probability the card is not an ace is 12/13. The probability the card is the deuce of clubs is 1/48 of that, 1/52.

It's not obvious to me why these are equal.

[SumZero]

An intuitive reason why the probability is the same follows:

Call the first card of the deck the 1st card, the second the 2nd, etc. Then:

What is the probability that the duce of clubs is the nth card for any given n? 1/52.
What is the probability that the ace of spades is the nth card for any given n? 1/52.

Now consider the randomly shuffled deck to be circular. That is when you count past the end of the deck you return to the beginning section. Now, reindex the deck so that each card that was value n before is value n+k mod 52. (this is the circular deck part). Now for any given n from above, and any given k what is the probability that card 1 now is the duce of clubs? Still 1/52. What is the probability that this card 1 is the ace of spades for any given card? Still 1/52.

Now how do we pick a k? Well as long as our card that we are peaking past is not: definitely the duce of clubs, definitely the ace of spades, and definitely the bottom card on the original deck you should be fine (you want to avoid the 1 * 0 that may occur if you dealt down to definitely the ace of spades). But you'd be fine if you dealt down to either the ace of spades or the duce of clubs. The probability the next card was either of those cards is still going to be 1/52.

But I'd rather see the math as the above is the type of intuitive that sometimes goes astray in probability trick questions.

[BruceZ]

[ QUOTE ]
It's not obvious to me why these are equal.

[/ QUOTE ]

Perhaps this will make it intuitive. There are 52! ways to order the deck. Now glue the 2c to the back of one of the aces so that these two cards stay together like one card. Now there are 51! ways to order the deck. 1/4 of these will have our glued card as the first ace, but we could have picked any of the 4 aces to glue, so there are (1/4)*51!*4 = 51! ways to order the deck with the 2c after the first ace, so the probability is 51!/52! = 1/52. Now instead of gluing the 2c to the ace, glue the As to an ace. Now our glued card will be the first ace 1/3 of the time, and there are 3 aces we can pick to glue, so this time there are (1/3)*51!*3 = 51! ways to order the deck with As after the first ace, same as before with the 2c. In either case the probability is 51!/52! = 1/52.

[turbo the man]

appears to be a trick question...for some.

1 in 4 the first ace IS the ace of spades and cannot be the next card off the deck. 3 in 4 the first ace is not the ace of spades and the probability of the next card being either the ace of spades or the deuce of clubs. therefore the ace of spades will have a 25% less chance of being the next card compared to the deuce of clubs.

[BruceZ]

[ QUOTE ]
appears to be a trick question...for some.

[/ QUOTE ]

Agreed!


[ QUOTE ]

1 in 4 the first ace IS the ace of spades and cannot be the next card off the deck. 3 in 4 the first ace is not the ace of spades and the probability of the next card being either the ace of spades or the deuce of clubs. therefore the ace of spades will have a 25% less chance of being the next card compared to the deuce of clubs.

[/ QUOTE ]

As near as I can make sense of this, you are stating without proof that as long as the first ace is not the ace of spades, then the next card has equal probability of being either the As or the 2c. Of course this is nonsense since the 2c could have already been dealt, while we are guaranteed that the As has not been dealt.

You'll want to refer to the posts by Pzhon and myself for independent correct proofs that the probabilities are both 1/52. Then come back with any specific questions you may have about those proofs. If you still want to bet against both Pzhon and me, I'm sure you can find plenty of good action. /images/graemlins/wink.gif

[pzhon]

[ QUOTE ]

Perhaps this will make it intuitive. There are 52! ways to order the deck. Now glue the 2c to the back of one of the aces so that these two cards stay together like one card. Now there are 51! ways to order the deck. 1/4 of these will have our glued card as the first ace, but we could have picked any of the 4 aces to glue, so there are (1/4)*51!*4 = 51! ways to order the deck with the 2c after the first ace, so the probability is 51!/52! = 1/52. Now instead of gluing the 2c to the ace, glue the As to an ace. Now our glued card will be the first ace 1/3 of the time, and there are 3 aces we can pick to glue, so this time there are (1/3)*51!*3 = 51! ways to order the deck with As after the first ace, same as before with the 2c. In either case the probability is 51!/52! = 1/52.

[/ QUOTE ]
Nice argument.

I think there is something deeper going on. I don't like SumZero's assertion that if you can't rule out a card and it isn't forced, the probability is 1/52. However, that holds in many more situations, so maybe it should be called a conjecture. For example, suppose you deal until either the first ace or 20 nonaces, then take the next card. Each card has probability 1/52 of coming next.

[Siegmund]

Equal.

I *really* didn't believe this until I sat down and worked it out. Intuition screams the two has to be more likely.

Here is my argument:

Let's start with a deck of only 5 cards, the C2, the SA, and the other three aces. This deck can be shuffled into 120 arrangements -- 20 distinct ones, since we don't care which of the other 3 aces is which.

2SAAA: neither key card is under the first ace.
2ASAA: spade comes up.
2AASA,2AAAS: neither.
S2AAA: deuce comes up.
SA2AA, SAA2A, SAAA2: neither.
AS2AA, ASA2A, ASAA2: spade comes up.
A2SAA, A2ASA, A2AAS: deuece comes up.
AAxxx: neither.

4 ways apiece for the spade and the deuce to be under the first ace, and 12 ways for neither one to have a chance to win. It is equally likely that you will throw the deuce away before reaching an ace vs. that the spade will be the first ace exposed.

Into any of those 20 arrangements, you can insert the 47 other cards the same number of ways, so the probabilities will remain equal no matter how many blanks you add to the deck.

[RiverTheNuts]

nm

[BillsChips]

This is a good question. It seems both overly simple and overly complex at the same time. Here's my take...

After the first Ace is dealt, the possibility that the next card is the 2C is 1 out of 51. The same goes for the Ace of Clubs, EXCEPT that 1 out of every 4 times, the Ace of Spades will be the first Ace dealt, and so it cannot be the next card.

Ace of Spades after the first Ace: 3/4 * 1/51
Two of Clubs after the first Ace: 1/51

The Two of Clubs will win this race 25% more than the Ace of Spades will.

[tylerdurden]

[ QUOTE ]
Ace of Spades after the first Ace: 3/4 * 1/51
Two of Clubs after the first Ace: 1/51

[/ QUOTE ]

This isn't quite correct.

I agree with your assessment of the 2/images/graemlins/club.gif, but not the A/images/graemlins/spade.gif.

The problem is that the A/images/graemlins/spade.gif cannot be before the first ace. It is either after the first ace, or it IS the first ace.

Here's a little chart I came up with to show the number of ways we can come up with A,2/images/graemlins/club.gif or A,A/images/graemlins/spade.gif for each possible position of the first ace.

<font class="small">Code:</font><hr /><pre>
position of A,2:club: A,A:spade:
first ace
1 4 3
2 47*4 48*3
3 47*46*4 48*47*3
4 47*46*45*4 48*47*46*3
...
13 47*...*36*4 48*...*37*3
14 47*...*35*4 48*...*36*3
...
49 47*...*0*4 48*...*1*3</pre><hr />

There are four ways you can have A,2/images/graemlins/club.gif at the very beginning of the deck - one for each possible suit of A. There are only 3 ways you can have A,A/images/graemlins/spade.gif - one for each of the non-spade suits. If the 1st ace is the 2nd card, there are 47*4 ways to make A,2/images/graemlins/club.gif since there are 47 cards other than the four aces and the 2/images/graemlins/club.gif. For A,A/images/graemlins/spade.gif, there are 48*3 ways, since there are 48 cards other than the four aces. So on down the chart.

Note that if the first ace is the 13th card, there are an equal number of ways to get A,2/images/graemlins/club.gif and A,A/images/graemlins/spade.gif (47*...*36*4 = 48*...*37*3 - everything between 47 and 37 cancels out, so you get 36*4 = 48*3).

If the first Ace is before the 13th card, then A,2/images/graemlins/club.gif is more likely. If the first ace is after the 13th card, then A,A/images/graemlins/spade.gif is more likely.

Note we can collapse that table down some more by cancelling out and find the relative probability of each.

<font class="small">Code:</font><hr /><pre>
position of A,2:club: A,A:spade:
first ace
1 4 3
2 47*4 48*3
3 46*4 48*3
4 45*4 48*3
...
13 36*4 48*3
14 35*4 48*3
...
48 1*4 48*3
49 0*4 48*3</pre><hr />

Sum up those and we get 4516*(some number) of possibilities for A,2:club and 7056*(the same number) of possibilities for A,A/images/graemlins/spade.gif.

A,A/images/graemlins/spade.gif is more likely.

[llamaoo7]

I founds errors in my logic. Ignore post.

[tylerdurden]

[ QUOTE ]

Sum up those and we get 4516*(some number) of possibilities for A,2:club and 7056*(the same number) of possibilities for A,A/images/graemlins/spade.gif.

[/ QUOTE ]

Well, this is wrong. You can't sum up all these simplified terms because the terms I'm cancelling out aren't the same for each case. you *can* compare A,2/images/graemlins/club.gif and A,A/images/graemlins/spade.gif at the same position of the first ace since in that case it is the same term that is cancelled out, but you can't compare A,2/images/graemlins/club.gif where the first ace is the 5th card and A,2/images/graemlins/club.gif where the first ace is the 6th card because different terms were dropped.

However, I believe the conclusion is correct. I'll probably get around to re-summing those correctly later today.

[BillsChips]

They say you should always go with your instincts...I think they're right. At first glance I thought that the Deuce Clubs and Ace Spades had an equal chance of coming immediately after the first ace. The more I thought about it, the more complicated it got. Now after doing the math, I see that my original thought was correct.

I reduced the set to 4 cards: The 2 of clubs, the Ace of Spades, and 2 more Aces. There are 24 possible combinations (4 * 3 * 2). The 2C comes out after the 1st ace exactly 6 times. The Ace of Spades comes out after the 1st ace exactly 6 times. Each has a probability of 6 out of 24 or 1 out of 4, or 1 out of N. For a 52 card deck, the probability is 1/52.

Essentially the problem is saying "What is the probability that a certain card will be found in a chosen position. The fact that that position is chosen by the location of the 1st ace is irrelevant, and only adds to the confusion. It's no different than choosing a position randomly.

[tylerdurden]

[ QUOTE ]
The fact that that position is chosen by the location of the 1st ace is irrelevant, and only adds to the confusion. It's no different than choosing a position randomly.

[/ QUOTE ]

By this logic the first card after the Ace of Spades is as likely to be the 2/images/graemlins/club.gif as the A/images/graemlins/spade.gif. After all, it's no different than randomly selecting, right? Using the first ace as the selector cleary DOES have an impact since one of the cards you are looking for is an ace.

[reubenf]

[ QUOTE ]
By this logic the first card after the Ace of Spades is as likely to be the 2/images/graemlins/club.gif as the A/images/graemlins/spade.gif. After all, it's no different than randomly selecting, right? Using the first ace as the selector cleary DOES have an impact since one of the cards you are looking for is an ace.

[/ QUOTE ]

I thought that, too, when I read his post.

So I did this:

It should be clear that the probability that the next card is 2/images/graemlins/club.gif is 1/51.

When the A/images/graemlins/spade.gif is the first ace in the deck, the probability that the next card is A/images/graemlins/spade.gif is 0. This happens 1/4 of the time.

When the the A/images/graemlins/spade.gif is not the first ace in the deck, the probability that the next card is A/images/graemlins/spade.gif is 1/(52-N), where N is the position of the first ace. The average value for N is 51/4 + 1 = 55/4. The probability that the next card is A/images/graemlins/spade.gif given that the A/images/graemlins/spade.gif is not the first ace is 1/(52-55/4) = 4/153. This happens 3/4 of the time.

The probability that the card after the first ace is A/images/graemlins/spade.gif is 4/153 * 3/4 = 1/51.

[BillsChips]

Let's beat this dead horse one more time.
With a 4 card deck consisting of 2 of Clubs, Ace of Spades, Ace of Diamonds, and Ace of Hearts, here are all of the possible combinations: (2 is the deuce, 1 = Ace of Spades, H = Ace of Hearts, D = Ace of Diamonds)

The Ace wins 6 times, the Deuce wins 6 times.

12DH = Deuce
12HD = Deuce
1D2H
1DH2
1H2D
1HD2
21DH
21HD
2D1H = Ace of Spades
2DH1
2H1D = Ace of Spades
2HD1
D12H = Ace of Spades
D1H2 = Ace of Spades
D21H = Deuce
D2H1 = Deuce
DH12
DH21
H12D = Ace of Spades
H1D2 = Ace of Spades
H21D = Deuce
H2D1 = Deuce
HD12
HD21

[pzhon]

[ QUOTE ]
It should be clear that the probability that the next card is 2/images/graemlins/club.gif is 1/51.

[/ QUOTE ]
No, there have been multiple correct calculations on this thread that say it is 1/52.

Why do so many people post incorrect answers days after fully justified, correct answers have been posted? Do people not bother to read the thread before responding?

[reubenf]

[ QUOTE ]
Why do so many people post incorrect answers days after fully justified, correct answers have been posted? Do people not bother to read the thread before responding?

[/ QUOTE ]

I wanted to do the math on my own. Where is the error in my math? It must be in my calculation of the average position of the first ace?

[tylerdurden]

[ QUOTE ]
Let's beat this dead horse one more time.

[/ QUOTE ]

Beat away. Your trivial example is interesting, but it doesn't scale up. When you add some cards that don't matter (i.e. a card that is neither a 2/images/graemlins/club.gif nor an Ace) it breaks down.

[elitegimp]

[ QUOTE ]

Sum up those and we get 4516*(some number) of possibilities for A,2:club and 7056*(the same number) of possibilities for A,A/images/graemlins/spade.gif.

[/ QUOTE ]

I really liked this method, but I think you want to do a weighted sum -- i.e. P(A is first card)*combinations + P(A is second card)*combinations + ...

hope that made sense, I'm trying to keep this short /images/graemlins/smile.gif

[pzhon]

[ QUOTE ]
[ QUOTE ]
Why do so many people post incorrect answers days after fully justified, correct answers have been posted? Do people not bother to read the thread before responding?

[/ QUOTE ]

[ QUOTE ]
I wanted to do the math on my own. Where is the error in my math?

[/ QUOTE ]
Do you want to figure that out on your own? Answer in white:
<font color="white">First, you simply assumed an incorrect answer for the probability that the next card was the 2 of clubs, calling the incorrect value "clear." Second, you miscalculated the average position of the first ace, 53/5. (Do you see why? Imagine 5 aces in 53 cards, one labeled "cut.") Third, you want average(f(position)) not f(average(position)). They are not always the same, and they are different here.</font>
Working on the problem on your own is fine, but why not read what others write before you post?

[BruceZ]

I'm wondering if my post and Pzhon's post did not state the final conclusion clearly enough. Some of you seem to be spending a lot of time on complicated ways to do this problem which has already been solved very simply and conclusively. To be clear, the 2c and the As will immediately follow the first ace with equal probability, and that probability is 1/52. Let me restate my simple proof more clearly:

1. The number of ways to order the deck so that the 2c immediately follows any A is 4*51!. That is 51! for each of the 4 aces.

2. Out of these 4*51! arrangements containing Ax2c, the number of these which have the Ax as the first A is 51!.

3. The number of ways to order the deck so that the As immediately follows any A is 3*51!. That is 51! for each of the 3 remaining aces.

4. Out of these 3*51! arrangements containing AxAs, the number of these which have the Ax as the first A is 51!.

5. Therefore the number of arrangements of the deck containing Ax2c and AxAs, where Ax is the first ace, are both 51!, so the probabilities of both of these are the same, and they are both equal to 51!/52! = 1/52.

End of proof. QED


I'll be happy to entertain questions about any of these steps.

Note: Many of you seem to be convinced that the Ax2c must happen more often than AxAs simply because of the fact that when the As is the first ace, then it cannot be the second ace. What you must also realize is that when one of the other aces is the first ace, the As will always be a candidate to be the next card. This cannot be said for the 2c because the 2c may have already been dealt. This means that when the first ace is anything other than the As (3/4 of the time) the As will follow the first ace more often than the 2c, and it turns out that this effect exactly cancels the 1/4 of the time that the first ace is the As.

Another way to look at it is that even though the Ax2c can be made 4 ways, while the AxAs can only be made 3 ways, this is exactly canceled by the fact that it is 4/3 times more likely that the AxAs will be the first ace, since there are only 3 aces competing to be first instead of 4.

The discussion in the last two paragraphs is not part of the proof. The proof consists only of steps 1-5 above.

So to make it clear once again, [b]the As and the 2c will follow the first ace equally often[b].

[tylerdurden]

[ QUOTE ]
[ QUOTE ]
Let's beat this dead horse one more time.

[/ QUOTE ]

Beat away. Your trivial example is interesting, but it doesn't scale up.

[/ QUOTE ]

Ignore this. I did my test case incorrectly. /images/graemlins/blush.gif

[tylerdurden]

[ QUOTE ]
2. Out of these 4*51! arrangements containing Ax2c, the number of these which have the Ax as the first A is 51!.

[/ QUOTE ]

I think this is probably correct, but it's not obvious (at least, not to me).

[ QUOTE ]
4. Out of these 3*51! arrangements containing AxAs, the number of these which have the Ax as the first A is 51!.

[/ QUOTE ]

Again, can you expand this a bit?

Other than that (which may be a nitpick), I agree that this proof looks correct. That leads me to wonder why my count is incorrect. I found that the number of deck configurations with AxAs (where Ax is the first Ace) is two orders of magnitude bigger than Ax2c (where Ax is the first Ace). Obviously I must be counting some cases multiple times or making some other mistake.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
2. Out of these 4*51! arrangements containing Ax2c, the number of these which have the Ax as the first A is 51!.

[/ QUOTE ]

I think this is probably correct, but it's not obvious (at least, not to me).

[/ QUOTE ]

I'll try to spell it out some more. We have 51 objects, 4 of which are called As, Ac, Ah, and Ad2c. We are going to form all 51! permutations of these 51 objects, and we observe by symmetry that of these 4 objects, the Ad2c will occur first as often as As, Ac, or Ah. Therefore it will occur first in 1/4 of the 51! permutations. Then we will repeat this whole process 3 more times, binding the 2c to each of the other aces. Each time we will form all 51! permutations, and each time we observe that the object with the ace bound to the 2c will occur first in 1/4 of the permutations. So all together we have formed 4*51! permutations which represent all of the permutations of the deck for which the 2c can immediately follow an A, and we have found that the 2c has followed the first ace in 1/4 of these 4*51! permutations, or 51! permutations.


[ QUOTE ]
[ QUOTE ]
4. Out of these 3*51! arrangements containing AxAs, the number of these which have the Ax as the first A is 51!.

[/ QUOTE ]

Again, can you expand this a bit?

[/ QUOTE ]

Argue this exactly the same way. We have 51 objects, 3 of which are called Ac, Ah, and AdAs. We are going to form all 51! permutations of these 51 objects, and we observe by symmetry that of these 3 objects, the AdAs will occur first as often as Ac, or Ah. Therefore it will occur first in 1/3 of the 51! permutations. Then we will repeat this whole process 2 more times, binding the As to each of the other 2 aces. Each time we will form all 51! permutations, and each time we observe that the object with the ace bound to the As will occur first in 1/3 of the permutations. So all together we have formed 3*51! permutations which represent all of the permutations of the deck for which the As can immediately follow an ace, and we have found that the As has followed the first ace in 1/3 of these 3*51! permutations, or 51! permutations.


[ QUOTE ]
Other than that (which may be a nitpick), I agree that this proof looks correct. That leads me to wonder why my count is incorrect. I found that the number of deck configurations with AxAs (where Ax is the first Ace) is two orders of magnitude bigger than Ax2c (where Ax is the first Ace). Obviously I must be counting some cases multiple times or making some other mistake.

[/ QUOTE ]

I believe you already found that you could not add these terms directly since each one represents a different number of permutations. You are only considering the number of permutations of the first N cards, but to count all the permutations you must also consider the cards that occur after N since there will be a different number of these for each value of N. I believe you were going to try a weighted average, but I have no interest in pursuing something that messy when we have the above simple and elegant solution.

Let me know if it is clear now. You can also refer to my original post (with the light bulb icon) where I go into more detail about gluing the cards together, etc. Pzhon's solution is also correct, but frankly if you had trouble understanding mine, you will find his much more terse, IMO.

[tylerdurden]

[ QUOTE ]
we observe by symmetry that of these 4 objects, the Ad2c will occur first as often as As, Ac, or Ah. Therefore it will occur first in 1/4 of the 51! permutations.

[/ QUOTE ]

I don't see how these two follow. I definitely agree that the 2c will follow each of the four aces with the same frequency, but that alone isn't enough to prove that it will follow the first ace with the same frequency that it will follow the 2nd, 3rd, or 4th ace.

Don't worry about expanding it, I already convinced myself that it is true, I just don't think your particular argument is the best way to show it. No biggie.

[ QUOTE ]
You are only considering the number of permutations of the first N cards, but to count all the permutations you must also consider the cards that occur after N since there will be a different number of these for each value of N.

[/ QUOTE ]

Yes, that's it. OK.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
we observe by symmetry that of these 4 objects, the Ad2c will occur first as often as As, Ac, or Ah. Therefore it will occur first in 1/4 of the 51! permutations.

[/ QUOTE ]

I don't see how these two follow. I definitely agree that the 2c will follow each of the four aces with the same frequency, but that alone isn't enough to prove that it will follow the first ace with the same frequency that it will follow the 2nd, 3rd, or 4th ace.

[/ QUOTE ]

NOOOOO, YOU'RE NOT GETTING THIS AT ALL! I'm saying start by BINDING the 2c to the ace of diamonds so that they cannot be separated. Staple them, glue them, whatever, so that this Ad2c HYBRID ACTS LIKE A SINGLE CARD, so we really effectively only have 51 cards now instead of 52. So instead of having 52! possible permutations of the deck, there are now only 51! permutations (I'm assuming you know that there are 52! permutations of a normal deck) since no matter how we shuffle, the Ad2c stick together as one card, so that all of these permutations have the 2c immediately after the Ad, and this will generate ALL of the permutations for which the 2c immediately follows the Ad. Now all I'm doing is observing that there is no fundamental difference between this stapled hybrid card that we made and the other 3 aces. They are all just cards, and they will each take turns being the FIRST ACE. That is, the As, Ac, Ah, and Ad2c hybrid will each be the first ace 1/4 of the 51! permutations. This much alone proves that out of all the times that the 2c follows the Ad, the Ad will be first 1/4 of the time. Now do the same thing with each of the other 3 possible hybrids (Ah2c, As2c, and Ac2c), and show in the same way that in each case the hybrid will be the first ace 1/4 of the permutations. Now we have generated ALL of the permutations for which the 2c immediately follows any ace, and we have shown that the Ax2c combination will be the first ace exactly 1/4 of the time.


[ QUOTE ]
Don't worry about expanding it, I already convinced myself that it is true, I just don't think your particular argument is the best way to show it.

[/ QUOTE ]

That's just nonsense. It is hard to imagine how this proof can be any more direct, and if you don't see that, then I guarantee you it is because you have not yet grasped the mental construction being used here, and not because the proof is not completely solid. When you have understood this proof, you will not make this statement. The kind of mental construction I've made here is absolutely typical of problems in combinatorics.

[tylerdurden]

[ QUOTE ]
The kind of mental construction I've made here is absolutely typical of problems in combinatorics.

[/ QUOTE ]

I'm sure it is. I just wasn't familiar with it. I get it now, though. /images/graemlins/smile.gif

[motorholdem]

I have been viewing this thread and watching with interest the different ways that people have tried to solve this problem. The most ironic part is the very first response (by Maurille) provided the correct answer. Nevertheless, BruceZ, Pzhon and others seem to think “they” have the right answer, and that the discussion is closed.

The fact is the 2 of clubs will come more frequently than the Ace of spades. Period.

For proof, you only need to look at what kind of scenario would cause equal probability, keeping in mind the constant fact that once the 1st ace hits there will ALWAYS be a 75% chance that the A f spades is still to come.


Scenario #1 – Equal Probability

So, if the first Ace always appeared as the 13th card, the there would be a 75% chance the A of spades is still to come. There would also a 75% chance that the 2 of clubs is still in play, as exactly ¼ of the deck (including an A), has been exposed. This is the point of equal probability.

Scenario #2 – A of spades more likely

What happens if you move the 1st Ace to the 26th card? If you did this, the Ace of spades would more likely to come up because it STILL AND ALWAYS has a 75% chance of being in the deck, but the chance that the 2 of clubs is still in play is now only 50% (actually it’s less, because all the Aces are hogging otherwise random spots in the last half of the deck).

Scenario #3 – 2 clubs more Likely

What happens, then, if the first A is the 4th card out? Clearly, the 2 of clubs will be more likely to come next, as it has over a 90% chance of still being in the deck, while the A of spades STILL has on a 75% chance.


So, in order to accept the mathematical argument by Brucez , pzhon and others that the chances of the A of spades and 2 of clubs following the first Ace is equal, you would (IMO) therefore have to accept the argument that (on average) the 1st Ace will appear as the 13th card over 1000’s of random simulations.

You don’t need to be an MIT grad to know that with aces making up 1/13th of the deck, that the chances of the first Aces average position being 13th is not possible in any study with a sufficient random sample size!!! Over a 1,000, 000 hands the Ace will be 1st card x amount of times, 2nd card x amount of times, and so on. It will even appear in the 30’s and 40’s sometimes. Over the long-term, though, it could NEVER average 13th place.

Try even a small sample of hundred simulations and see if the average position of the first Ace is not anywhere near 13th position. You’ll probably find it to be in the 7th to 11th position, on average. And in this position the 2 of clubs is more likely to still be in the deck, thus more likely to come next.

I did 100 random shuffles and started peeling until I hit the first A. Only 14 out of 100 times did the 2clubs appear before the first Ace. So, the 2 of clubs was still in the deck 86% of the time after this small sample. This is not surprising, as the largest cluster of first aces came in the 7th to 11th range (though I had 5 aces in the 30s, I also had 9 aces in the 1st and 2nd positions combined!!!).

If after reading this if anyone wants to place a 1000 bets that the odds are even, I’ll take the 2 of clubs and you can have the A of spades. Bring on you money and your complex math formulas. I’ll count my profits while you are scratching your had looking for the flaw in the formula.

And remember, Maurille was the first to respond, and he had it right all

[SumZero]

I am amused by the same thing, especially since your "correct" answer is in fact incorrect.

[ QUOTE ]
So, in order to accept the mathematical argument by Brucez , pzhon and others that the chances of the A of spades and 2 of clubs following the first Ace is equal, you would (IMO) therefore have to accept the argument that (on average) the 1st Ace will appear as the 13th card over 1000’s of random simulations.

[/ QUOTE ]

Wrong. Just because in the mean position of the first Ace the 2 is more likely does NOT mean that. For instance the distribution around this position is NOT symetric. And the probability doesn't need to follow it as well.

[ QUOTE ]
If after reading this if anyone wants to place a 1000 bets that the odds are even, I’ll take the 2 of clubs and you can have the A of spades. Bring on you money and your complex math formulas. I’ll count my profits while you are scratching your had looking for the flaw in the formula.

[/ QUOTE ]

What a good idea to simulate this to show you how wrong you are. But why stop at 100 or 1000 and why do your silly is the 2 of C before the 1st A. Why not simulate the real answer as well? Here's what I got:

Overall the first ace was at 10.5908226923077 position (1-indexed).
Overall there were 50362 ace of spades and 49672 2C.
This was with a sample size of 2 600 000. Guess what 2.6 million/52 is? 50,000.

Hmm, could it be the correct answer was 1/52, equal probability? Seems likely.

Here is the perl code I used (this was Q&amp;D code, don't shoot me b/c I'm not a real perl hacker):

#!/usr/local/bin/perl
$pos = 0;
$num = 0;
$ace = 0;
$two = 0;
@a = ();
for($num=0;$num&lt;2600000;$num++) {
for($i=0;$i&lt;52;$i++) {
$a[$i] = $i;
}

#shuffle

for($i=0;$i&lt;52;$i++) {
$j = int((52-$i)*rand());
$tmp = $a[$i];
$a[$i] = $a[$i+$j];
$a[$i+$j] = $tmp;
}
$k=1;
for($i=0; ($k==1) and ($i&lt;52); $i++) {
if(($a[$i] % 13) == 0) {
#print "First Ace: " . ($i + 1) . " position. ";
#print "Next card: " . $a[$i+1] . " (note AS == 0, 2C == 14)\n";
if($a[$i+1] == 0) {
$ace++;
}
if($a[$i+1] == 14) {
$two++;
}
$pos += $i;
$k = 0;
}
}
}

print "Overall the first ace was at " . (($pos/$num) + 1) . "
position.\n";
print "Overall there were $ace ace of spades and $two 2C.\n";

[motorholdem]

[ QUOTE ]
I am amused by the same thing, especially since your "correct" answer is in fact incorrect.

[ QUOTE ]
So, in order to accept the mathematical argument by Brucez , pzhon and others that the chances of the A of spades and 2 of clubs following the first Ace is equal, you would (IMO) therefore have to accept the argument that (on average) the 1st Ace will appear as the 13th card over 1000’s of random simulations.

[/ QUOTE ]

Wrong. Just because in the mean position of the first Ace the 2 is more likely does NOT mean that. For instance the distribution around this position is NOT symetric. And the probability doesn't need to follow it as well.

[ QUOTE ]
If after reading this if anyone wants to place a 1000 bets that the odds are even, I’ll take the 2 of clubs and you can have the A of spades. Bring on you money and your complex math formulas. I’ll count my profits while you are scratching your had looking for the flaw in the formula.

[/ QUOTE ]

What a good idea to simulate this to show you how wrong you are. But why stop at 100 or 1000 and why do your silly is the 2 of C before the 1st A. Why not simulate the real answer as well? Here's what I got:

Overall the first ace was at 10.5908226923077 position (1-indexed).
Overall there were 50362 ace of spades and 49672 2C.
This was with a sample size of 2 600 000. Guess what 2.6 million/52 is? 50,000.

Hmm, could it be the correct answer was 1/52, equal probability? Seems likely.

Here is the perl code I used (this was Q&amp;D code, don't shoot me b/c I'm not a real perl hacker):

#!/usr/local/bin/perl
$pos = 0;
$num = 0;
$ace = 0;
$two = 0;
@a = ();
for($num=0;$num&lt;2600000;$num++) {
for($i=0;$i&lt;52;$i++) {
$a[$i] = $i;
}

#shuffle

for($i=0;$i&lt;52;$i++) {
$j = int((52-$i)*rand());
$tmp = $a[$i];
$a[$i] = $a[$i+$j];
$a[$i+$j] = $tmp;
}
$k=1;
for($i=0; ($k==1) and ($i&lt;52); $i++) {
if(($a[$i] % 13) == 0) {
#print "First Ace: " . ($i + 1) . " position. ";
#print "Next card: " . $a[$i+1] . " (note AS == 0, 2C == 14)\n";
if($a[$i+1] == 0) {
$ace++;
}
if($a[$i+1] == 14) {
$two++;
}
$pos += $i;
$k = 0;
}
}
}

print "Overall the first ace was at " . (($pos/$num) + 1) . "
position.\n";
print "Overall there were $ace ace of spades and $two 2C.\n";

[/ QUOTE ]


Sorry, but I'm afraid you response has done nothing but reaffrim my position. If you want to address it, please refute the folowing:

The first ace was at 10.59 position (your info- yes/no)
Can we use the 11th position to round off? (yes/no)
One or more 2s could occur before the first Ace (yes/no)

The 1st Ace has a 25% chance of being Ace of spades, so there is a 75% chance that the Ace of spades is still in the deck after the 11th position (yes/no)

Of the 10 cards preceding the first Ace, there is an 83.33% (1/12 x 10) chance that a 2 is there, and there is 20.83% (83.33% /4)chance that the 2 of clubs is there. (yes/no)

So, when the cards come after the first Ace, there is a 75% that the Ace of spades is in the deck , and a 79.17% chance that the 2 of clubs is still in the deck (yes/no).

Every card remaining in the deck (all 41 of them) has an equal chance of being the next card (Yes/no.)

If the Ace of spades has a 4% less chance of still being in the deck than the 2 of clubs, then it CANNOT occur as the NEXT card with equal probability.(correct/incorrect)

Sorry, your argument does not hold up. The 2 of clubs will occur more frequently after the first Ace appears.

[pzhon]

[ QUOTE ]
So, in order to accept the mathematical argument by Brucez , pzhon and others that the chances of the A of spades and 2 of clubs following the first Ace is equal, you would (IMO) therefore have to accept the argument that (on average) the 1st Ace will appear as the 13th card over 1000’s of random simulations.

[/ QUOTE ]
No, our arguments were completely different. You have injected a false argument into the discussion that assumes E(f(X)) = f(E(X)), where X is the position of the first ace, and f(n) is the probability the next card is the deuce of clubs if the first ace is in position n. f isn't linear, and E(f(X)) is not equal to f(E(X)), so there is no reason E(X) would have to be 13 for the As and 2c to be equally likely after the first ace. In fact, as I have pointed out elsewhere on this thread, E(X) = 53/5, which agrees with Sumzero's simulation.

Our arguments do not rely on this false step. Our arguments are correct, and the As and 2c both have probability 1/52 to occur immediately after the first ace.

[ QUOTE ]
If after reading this if anyone wants to place a 1000 bets that the odds are even, I’ll take the 2 of clubs and you can have the A of spades. Bring on you money and your complex math formulas. I’ll count my profits while you are scratching your had looking for the flaw in the formula.

[/ QUOTE ]
That would be a 0 EV bet for us, while you think you would profit. That's not the fair way to resolve a disagreement. Since you believe the 2c is more likely, please say how much more likely you think it is, and we can split the difference between those odds and even odds. That way, both of us will see it as a +EV wager. For example, if you believe the 2c is a 6:5 favorite, let's bet as though it is an 11:10 favorite. Let's not do it just 1000 times, but enough that whichever one of us is right expects to be ahead by a few standard deviations. Let's put some real money on it. See my PM.

[SumZero]

[ QUOTE ]

Sorry, but I'm afraid you response has done nothing but reaffrim my position.


[/ QUOTE ]

It's good you don't let the facts get in the way of your opinion!

[ QUOTE ]
If you want to address it, please refute the folowing:

The first ace was at 10.59 position (your info- yes/no)


[/ QUOTE ]

YES

[ QUOTE ]
Can we use the 11th position to round off? (yes/no)

[/ QUOTE ]

NO We can't make that assumption, especially since it is wrong. Fortunately I can still point the critical flaw out to you (as this mistake is a mistake, and is wrong, but it isn't the critical reason why you are incorrect).

[ QUOTE ]

One or more 2s could occur before the first Ace (yes/no)


[/ QUOTE ]

YES

[ QUOTE ]
The 1st Ace has a 25% chance of being Ace of spades

[/ QUOTE ]

YES

[ QUOTE ]
The 1st Ace has a 25% chance of being Ace of spades, so there is a 75% chance that the Ace of spades is still in the deck after the 11th position (yes/no)

[/ QUOTE ]

NO, not necessarily. The sentence that is clearly true is this: The 1st Ace has a 25% chance of being Ace of spades, so there is a 75% chance that the Ace of spades is still in the deck after the first Ace. You seem to think that because the expected position of the first Ace is X that then you can reason about cards before and after X in the general case. You can't. There is a greater than 75% chance that the Ace of spades is after the 11th position. How do I know this? Let's imagine the following game with a standard 52 card deck. We shuffle the cards. You get the first 11 cards. I get the next 41 cards. If you get the Ace of spades I owe you $3. If I get the Ace of spades you owe me $1. If your above conjecture about the 11th position were correct, then this would be a fair game. It isn't. I expect to win $41 and lose (3*11)=$33 every 52 games we play. The per game expectation is 8/52 for me. Do you see this? This is the critical mistake you make.

The rest of your questions are irrelevent as your "proof" falls apart here and your questions have invalid implicit assumptions. Like:

[ QUOTE ]
Of the 10 cards preceding the first Ace, there is an 83.33% (1/12 x 10) chance that a 2 is there, and there is 20.83% (83.33% /4)chance that the 2 of clubs is there. (yes/no)

[/ QUOTE ]

What if the first Ace is the first card in the deck? Why do there have to be 10 cards in front of it? There don't.

[ QUOTE ]

So, when the cards come after the first Ace, there is a 75% that the Ace of spades is in the deck , and a 79.17% chance that the 2 of clubs is still in the deck (yes/no).

[/ QUOTE ]

Up to the and YES, after the and, NO (see above)

[ QUOTE ]

Every card remaining in the deck (all 41 of them) has an equal chance of being the next card (Yes/no.)

[/ QUOTE ]

The non parenthesized part is YES. But there doesn't have to be exactly 41 of them.

[ QUOTE ]

If the Ace of spades has a 4% less chance of still being in the deck than the 2 of clubs, then it CANNOT occur as the NEXT card with equal probability.(correct/incorrect)

[/ QUOTE ]

This one is a tricky answer. The answer is correct. The conclusion is wrong because the "if" clause is false. But if the if clause were true (which it isn't), then the reasoning is correct, hence the correct statement even though your final answer is wrong.

[ QUOTE ]
Sorry, your argument does not hold up. The 2 of clubs will occur more frequently after the first Ace appears.

[/ QUOTE ]

You are very, very wrong. How much more frequently do you think? I'm willing to wager on this or on the game I describe above that exploits your claim that the Ace of spades has only a 75% chance to be in the final 41 cards of a shuffled deck.

Hopefully though after my post, and pzon's post that makes the same point but uses math equations instead of prose, you'll have realized you are mistaken and will be willing to post a retraction.

[pudley4]

[ QUOTE ]
Nevertheless, BruceZ, Pzhon and others seem to think “they” have the right answer, and that the discussion is closed.

[/ QUOTE ]

I don't ever recall seeing BruceZ or pzhon come up with a blatantly incorrect answer (I know Bruce has posted some slightly incorrect answers before, but he's always come back and corrected them on his own). When they both agree on an answer, it is as correct as saying 2+2=4.

[Peter_rus]

Both cards have the same probability. If only not an ace of clubs was the first ace we encountered in a deck.

[yellowjack]

I argue that they have no significance at all. For all blanks (47 other cards) before the deck, we deal through them and they do not affect the experiment. For all blanks two cards and further from the initial ace, we never get to them. For the blank directly after the initial ace, the probability of having the 2c and the As is equal, 0. We move on.

1 = 2c
2 = As
3 = Ah
4 = Ac
5 = Ad

In my mind, there are 5! (120) combinations of deck arrangements that this problem can have. They are as follows:

12345
12354
12435
12453
etc.

I am too lazy/mentally challenged to work on all the combinations and find their outcomes to weigh the chance of As and 2c against each other.

Some input will be greatly appreciated from regulars. I will check on this thread again in about 10 hours.

[Siegmund]

motorholdem got off to a good start here:

[ QUOTE ]
For proof, you only need to look at what kind of scenario would cause equal probability, keeping in mind the constant fact that once the 1st ace hits there will ALWAYS be a 75% chance that the A f spades is still to come.


Scenario #1 – Equal Probability

So, if the first Ace always appeared as the 13th card, the there would be a 75% chance the A of spades is still to come. There would also a 75% chance that the 2 of clubs is still in play, as exactly ¼ of the deck (including an A), has been exposed. This is the point of equal probability.

Scenario #2 – A of spades more likely

What happens if you move the 1st Ace to the 26th card? If you did this, the Ace of spades would more likely to come up because it STILL AND ALWAYS has a 75% chance of being in the deck, but the chance that the 2 of clubs is still in play is now only 50% (actually it’s less, because all the Aces are hogging otherwise random spots in the last half of the deck).

Scenario #3 – 2 clubs more Likely

What happens, then, if the first A is the 4th card out? Clearly, the 2 of clubs will be more likely to come next, as it has over a 90% chance of still being in the deck, while the A of spades STILL has on a 75% chance.


[/ QUOTE ]

So far so good. The farther down we deal before we come to an ace, the more likely we are to have thrown away the C2 already. The chance that we throw away the SA (by having it be the first ace) is always 1/4.
[ QUOTE ]

So, in order to accept the mathematical argument by Brucez , pzhon and others that the chances of the A of spades and 2 of clubs following the first Ace is equal, you would (IMO) therefore have to accept the argument that (on average) the 1st Ace will appear as the 13th card over 1000’s of random simulations.


[/ QUOTE ]

Almost, but not quite. We have to accept that Scenario 2 and Scenario 3 occur with such frequency as to cause the advantages to exactly cancel out. And that happens to be exactly what they do.

We could deal off anywhere from 0 to 48 cards before we hit an ace. Let's work our way down, case by case:

0 cards dealt before we hit an ace: This happens with probability 1/13, and, when it does, the C2 has a 4:3 advantage: 3/4 of the time the SA is in the deck, all the time the C2 is in the deck.

1 card dealt off before we hit an ace: This happens with probability 16/221, and when it does, the C2 has a 47:36 advantage: 3/4 of the time the SA is in the deck, 47/48 of the time the C2 is in the deck.

2 cards dealt off before we hit an ace: This happens with probability 376/5525, and when it does, the C2 has a 23:18 advantage.

N cards dealt off before we hit an ace: This happens with probability 4* 48! * (51-N)! / (48-N)! / 52!, and when it does, the relative advantage is 48-N:36, in favor of C2 for N&lt;12, even for N=12, in favor of SA for N&gt;13.

Your "Scenario #3" happens almost 2/3rds of the time, with the C2 having a small edge. "Scenario #1" happens the ~3% of the time that the first ace is the 13th card. "Scenario #2" happens the remaining 1/3rd of the time, but when you deal down a very long way, the SA can acquire a very large edge.

If you are reallllly bored, you can write out all 49 cases and add up the probabilites:

[4 * 48! (51-N)! / (48-k)! / 52!] * 3/4 * 1/(51-N) for the probability the card under the first ace is the SA;

[4 * 48! (51-N)! / (48-k)! / 52!] * (48-N)/48 * 1/(51-N) for the probability the card under the first ace is the C2.

You will find they both add up to 1/52. (Yes, I checked them, with a calculator, just now.)

[ QUOTE ]

You don’t need to be an MIT grad to know that with aces making up 1/13th of the deck, that the chances of the first Aces average position being 13th is not possible in any


[/ QUOTE ]

True but not the point, as others have noted; the end result is a weighted combination of where the first ace is and what the C2 and SA's chances are when it's in that position.

So are you going offer us a fair bet or not? /images/graemlins/smile.gif

[SumZero]

Ok, I'll bite. There are 5! arrangements of the deck. 20% of them have the 2C first and then the question is what fraction of these have the AS in the 3rd position? Answer 25%. What percentage have the 2C in the third position? Answer 0% (it was the 1st card!). Of the 5! deck arrangements 80% have an A in the first position, what percentage of these have the 2C in the 2nd position? Answer 25%. Of the 80% with an A in the first position 75% of the time it is not the A of spades (25% of the time it is, and when it is it can't also be in the second position). In these cases the 2nd card will be the A of spades 25% of the time.

Therefore I get:

2 Clubs:
(5!*20%*0%+5!*80%*25%)/5! = 1/5
A of Spades:
(5!*20%*25%+5!*80%*75%*25%)/5! = 1/5

Which is not a surprise given what we know about the true correct answer with the 52 card deck.

[motorholdem]

[ QUOTE ]
motorholdem got off to a good start here:

[ QUOTE ]
For proof, you only need to look at what kind of scenario would cause equal probability, keeping in mind the constant fact that once the 1st ace hits there will ALWAYS be a 75% chance that the A f spades is still to come.


Scenario #1 – Equal Probability

So, if the first Ace always appeared as the 13th card, the there would be a 75% chance the A of spades is still to come. There would also a 75% chance that the 2 of clubs is still in play, as exactly ¼ of the deck (including an A), has been exposed. This is the point of equal probability.

Scenario #2 – A of spades more likely

What happens if you move the 1st Ace to the 26th card? If you did this, the Ace of spades would more likely to come up because it STILL AND ALWAYS has a 75% chance of being in the deck, but the chance that the 2 of clubs is still in play is now only 50% (actually it’s less, because all the Aces are hogging otherwise random spots in the last half of the deck).

Scenario #3 – 2 clubs more Likely

What happens, then, if the first A is the 4th card out? Clearly, the 2 of clubs will be more likely to come next, as it has over a 90% chance of still being in the deck, while the A of spades STILL has on a 75% chance.


[/ QUOTE ]

So far so good. The farther down we deal before we come to an ace, the more likely we are to have thrown away the C2 already. The chance that we throw away the SA (by having it be the first ace) is always 1/4.
[ QUOTE ]

So, in order to accept the mathematical argument by Brucez , pzhon and others that the chances of the A of spades and 2 of clubs following the first Ace is equal, you would (IMO) therefore have to accept the argument that (on average) the 1st Ace will appear as the 13th card over 1000’s of random simulations.


[/ QUOTE ]

Almost, but not quite. We have to accept that Scenario 2 and Scenario 3 occur with such frequency as to cause the advantages to exactly cancel out. And that happens to be exactly what they do.

We could deal off anywhere from 0 to 48 cards before we hit an ace. Let's work our way down, case by case:

0 cards dealt before we hit an ace: This happens with probability 1/13, and, when it does, the C2 has a 4:3 advantage: 3/4 of the time the SA is in the deck, all the time the C2 is in the deck.

1 card dealt off before we hit an ace: This happens with probability 16/221, and when it does, the C2 has a 47:36 advantage: 3/4 of the time the SA is in the deck, 47/48 of the time the C2 is in the deck.

2 cards dealt off before we hit an ace: This happens with probability 376/5525, and when it does, the C2 has a 23:18 advantage.

N cards dealt off before we hit an ace: This happens with probability 4* 48! * (51-N)! / (48-N)! / 52!, and when it does, the relative advantage is 48-N:36, in favor of C2 for N&lt;12, even for N=12, in favor of SA for N&gt;13.

Your "Scenario #3" happens almost 2/3rds of the time, with the C2 having a small edge. "Scenario #1" happens the ~3% of the time that the first ace is the 13th card. "Scenario #2" happens the remaining 1/3rd of the time, but when you deal down a very long way, the SA can acquire a very large edge.

If you are reallllly bored, you can write out all 49 cases and add up the probabilites:

[4 * 48! (51-N)! / (48-k)! / 52!] * 3/4 * 1/(51-N) for the probability the card under the first ace is the SA;

[4 * 48! (51-N)! / (48-k)! / 52!] * (48-N)/48 * 1/(51-N) for the probability the card under the first ace is the C2.

You will find they both add up to 1/52. (Yes, I checked them, with a calculator, just now.)

[ QUOTE ]

You don’t need to be an MIT grad to know that with aces making up 1/13th of the deck, that the chances of the first Aces average position being 13th is not possible in any


[/ QUOTE ]

True but not the point, as others have noted; the end result is a weighted combination of where the first ace is and what the C2 and SA's chances are when it's in that position.

So are you going offer us a fair bet or not? /images/graemlins/smile.gif

[/ QUOTE ]


Thanks for the explanation. If I understand you correctly, the flaw in my 11th (or 13th card) mid-point position argument is that while a minority of 1st Aces will occur beyond the 26th position (+13 and beyond), none can occur before the 1st position (-13). Thus, even though the number of times the A can occur very late in the deck is rare, the much higher chance that the 2c is gone compensates for the lack of frequency (and offets for the probability of higher frequencies for Aces in earlier positions)?

Is that it? If so, I better start typing my humble pie post to a few folks here. I'll wait to hear if I understand the above correctly.



It all seems so counter-intuitive....as in my 1st manual run on this (real cards) I only had the 2c precede the 1st A 13 times in 100 tries, which left it in the deck 87% ( I know, too small a sample size).

PS. In one deal it took me until the 41st card to pull an Ace

[yellowjack]

1 = 2c; 2 = As; 3 = Ah; 4 = Ac; 5 = Ad

132xx, 142xx, 152xx (6 combinations): 6 As and 0 2c
21xxx (6 combinations): 0 As and 6 2c
31xxx (6 combinations): 0 As and 6 2c
32xxx (6 combinations): 6 As and 0 2c
41xxx (6 combinations): 0 As and 6 2c
42xxx (6 combinations): 6 As and 0 2c
51xxx (6 combinations): 0 As and 6 2c
52xxx (6 combinations): 6 As and 0 2c

Only 48 of the 120 combinations have one of the two cards coming up; neither of them come up for the remainder of them. 48/120 = 40% of them, which is exactly equal to the proportion of cards (2 out of 5).

Of these combinations where there is an outcome, for 24 of them the As will come up, and for 24 of them the 2c will come up.

So the 2c will come up as often as the As. I think this proves what was intuitive for many people.

[MickeyHoldem]

For more proof that the probabilities are equal and 1/52:

Using excel functions
Let p be the position of the 2 of clubs.
The 2c cannot follow the first ace in positions 1,50,51,52
For all other p the number of unique decks that the 2c follows the first ace is = 4*PERMUT(47,p-2)*FACT(52-p)
Sum these for p=2 to 49

Now the Ace of spades cannot follow the first ace in positions 1,51,52
For all other p, the number of unique decks with the As following the first ace = 3*PERMUT(48,p-2)*FACT(52-p)
Sum these for p=2 to 50

Not surprising is the fact that these are equal and also equal to 51!

Dividing the sums by 52!(number of unique decks) indeed gives you 1/52.

Use a line chart to chart the data and watch the 2 data lines intersect at p=14, the position where it is equally likely that the 2c or As will follow the first ace. Before this position it is more likely for the 2c to be the next card, and after this position it is more likely that the As will follow.

[Mike Haven]

You are playing with a 52 card deck. Assuming a random shuffle, you deal cards face up from the top until the first Ace is dealt. Which has a higher probability of being the next card off the deck, the Ace of Spades or the Two of Clubs?

i think that BruceZ and Pzhon are wrong here

here is my simple solution:

it is only the order of the five cards AcAdAhAs2c which matters: the rest are standing stones

there are 120 permutations in ordering 5 things: 5x4x3x2x1

of these 120 perms 24 begin with As, so 2c is certainly the next relevant card after the first Ace, As, is dealt

also, 24 perms begin with 2c, but six of these are neutral, because the second card is As, so neither the As nor the 2c can be next after the first Ace, As, is dealt

the other 18 perms starting with 2c mean As will be before 2c, because the 2c is already out before the first Ace, Ac/Ad/Ah, shows

all the other 72 perms have an equal chance of As before or after 2c, the first Ace being one of the other Aces, so 36 to each side

summary:

6 are neutral
24 + 36 = 60 times: 2c before As
18 + 36 = 54 times: As before 2c

therefore, the 2c has a higher probability of being the next card off the deck after the first Ace is dealt

(a priori of course - if you have actually seen all the cards dealt you would know the situation... which was first ace?... has 2 gone? ... then the chances would be obvious)

if this is not the correct answer, please tell me where i am wrong

[pzhon]

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of these 120 perms 24 begin with As, so 2c is certainly the next relevant card after the first Ace, As, is dealt

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In 1/4 of those 24, 6 times, the 2c is next.

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also, 24 perms begin with 2c, but six of these are neutral, because the second card is As, so neither the As nor the 2c can be next after the first Ace, As, is dealt

the other 18 perms starting with 2c mean As will be before 2c, because the 2c is already out before the first Ace,

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In 1/3 of those 18, 6 times, the As is next.

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all the other 72 perms have an equal chance of As before or after 2c, the first Ace being one of the other Aces, so 36 to each side

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In 1/4 of these 72, 18 times, the 2c is next. Similarly, As is next 18 times.

So, in 6+18 of the orders, the card after the first ace is the 2c. In 6+18 of the orders, the card after the first ace is the As. The probabilities are both 24/120 = 1/5.

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if this is not the correct answer, please tell me where i am wrong

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You gave equal weight to situations in which the 2c would show up 1/4 of the time and the situations in which the AS would show up 1/3 of the time. They should have been weighted by the relevant probabilities.

If we were wrong, you should have been able to find an error in our arguments.

[Mike Haven]

Which has a higher probability of being the next card off the deck, the Ace of Spades or the Two of Clubs?

i interpreted this as meaning "Which has a higher probability of being the next card you see, out of the As and the 2c, if you keep dealing off the deck?"

if my interpretation is correct, my answer is correct - i agree that your answer is correct if "next card off the deck" means ... "next card off the deck" /images/graemlins/tongue.gif

[barongreenback]

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Which has a higher probability of being the next card off the deck, the Ace of Spades or the Two of Clubs?

i interpreted this as meaning "Which has a higher probability of being the next card you see, out of the As and the 2c, if you keep dealing off the deck?"

if my interpretation is correct, my answer is correct - i agree that your answer is correct if "next card off the deck" means ... "next card off the deck" /images/graemlins/tongue.gif

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Yes you're right. I thought about it the same way as you. This may have led to some of the trouble in the thread but not most of it.

The interesting point for me is comparing two situations at the point when the first ace is dealt:
A. 2C is out. AS is still in.
B. 2C is still in. AS is out.

B occurs more often than A but the AS is more likely to be the next card in A than the 2C is in B. This is because in A there are 2 non spade aces left in but in but in B there are 3. ie fewer cards left in the deck in A.

Well, it meant something to me anyway.

[icepoker]

Here is a simple induction argument that shows that the probability of these events i.e. 2/images/graemlins/club.gif following the first ace and the A/images/graemlins/spade.gif following the first ace are equivalent.

Denote the first ace by A. For a five card deck of 2/images/graemlins/club.gifA/images/graemlins/spade.gifA/images/graemlins/heart.gifA/images/graemlins/diamond.gifA/images/graemlins/club.gif we have that Pr{2/images/graemlins/club.gif follows A}=Pr{A/images/graemlins/spade.gif follows A}=1/5. Note that this is for a 5 card deck, and we are calculating the probability of the next card after the first ace. If you have problem seeing this, just count all the possible shuffles and note that the total number of shuffles is 5!.

Now denote,
S_n(2/images/graemlins/club.gif)=number of shuffles such that 2/images/graemlins/club.gif follows A for a deck with n cards, n&gt;4.
S_n(A/images/graemlins/spade.gif)=number of shuffles such that A/images/graemlins/spade.gif follows A for a deck with n cards, n&gt;4.

From our observation above S_5(2/images/graemlins/club.gif)=S_5(A/images/graemlins/spade.gif)=4!. And again, you can do a simple count to see this. We now want to show that S_n(2/images/graemlins/club.gif)=S_n(A/images/graemlins/spade.gif)=(n-1)! for n&gt;4.

Assume this is true for n. We clearly have
S_n+1(2/images/graemlins/club.gif)=n*S_n(2/images/graemlins/club.gif)=n*(n-1)!=n!
Similarily,
S_n+1(A/images/graemlins/spade.gif)=n*S_n(A/images/graemlins/spade.gif)=n*(n-1)!=n!

By the principle of mathematical induction, we have shown that the events in question have the same probability.

[d10]

This is rediculous.

20% of the time, the 2c will be placed before all 4 aces. Given this situation, you will see the 2c pass and you will know there is now a 0% chance it will come out after the first Ace.

Another 20% of the time, the As will be placed before the other 3 aces and before the 2c. Given this situation, there is now a 0% chance that the As will come out after the first Ace.

The other 60% of the time a different Ace will come out first, and you will have seen neither the As or the 2c. You will then know that the As and the 2c are both just as likely to come next.

What's wrong with my reasoning here? I suspect that I'm the retarded one here, because I know all of you knowledgeable people would not be arguing so much if it was so simple, but I fail to see the flaw in my logic.

By the way, I've read all of the other posts here, and to say that you must weight the different outcomes is incorrect I believe. I think that is like saying "Why are my chances of hitting my flush draw on the river 9/46 when some of my opponents surely hold some of my suit".

[pzhon]

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What's wrong with my reasoning here?

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What you have said is incomplete. You broke the possible deals into 3 pieces, which was fine.

Case 1: 2c comes first.
Case 2: As comes first.
Case 3: Another ace comes first.

However, you didn't say anything about the probability that the As would be the card after the first ace in Case 1. You didn't say anything about the probability that the 2c would be the card after the first ace in Case 2 (immediately following the As). You did not give an argument that these two probabilities are equal. If you don't see that this is necessary, then perhaps you misread the question.

Once again, I'd like to point out that complete solutions were given earlier in this thread.

[irchans]

ajizzle wrote:
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"You are playing with a 52 card deck. Assuming a random shuffle, you deal cards face up from the top until the first Ace is dealt. Which has a higher probability of being the next card off the deck, the Ace of Spades or the Duece of Clubs."


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Pzhon, BruceZ, and Icepoker proved that the probability of the next card being the Ace of Spades is 1/52 and the probability of the next card being the Deuce of Clubs is 1/52. Pzhon commented, "It's not obvious to me why these are equal." This comment made me say to myself, "It's not obvious to me why these probablities are equal to the probability of just drawing the Ace of Spades randomly." What a coincidence! It seems that the dealing and stopping at the first ace process does not affect the probabilities at all. Reading everyone's insiteful proofs and remarks made me think that maybe there was a way to prove the 1/52 result without any calculations. Here is the general conclusion that I came up with.

The answer to most questions of the form "Shuffle a deck of cards and deal cards until ____ is true. What is the probability that the next card is the Ace of Spades?" is 1/52. Please check my reasoning below.


Call the phrases "until ____ is true" and "until the first Ace is dealt" the stopping criteria. Let's suppose that the stopping criteria has the property that the dealer will always stop dealing before the last card is dealt. Call this property P. Examples of stopping criteria with property P include:

1) "until three Spades have been dealt",
2) "until two Aces have been dealt", and
3) "until two or more Aces, three or more Kings, and five or more Spades have been dealt".

Consider the question: "Shuffle a deck of cards and deal cards until ____ is true. What is the probability that the LAST card in the deck is the Ace of Spades?". (Assume the stopping criteria "until ___ is true" has property P.)

It seems obvious that the answer is 1/52 because dealing the cards off the top does not affect the last card on the bottom of the deck. Now it seems quite reasonable to assume that the following question has the same answer: "Shuffle a deck of cards and deal cards until ____ is true. What is the probability that the NEXT card in the deck is the Ace of Spades?"

Based on the reasoning above, I think we can conclude that the following statement is true:

Statement: If a deck is randomly shuffled and cards are dealt until a stopping criteria is met and that stopping criteria has property P (the dealing will stop before the last card is dealt regardless of how the deck was shuffled) then the probability of the next card being the Ace of Spades is exactly 1/52.


What do you think?

[pzhon]

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Consider the question: "Shuffle a deck of cards and deal cards until ____ is true. What is the probability that the LAST card in the deck is the Ace of Spades?".

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Very nice argument. I think that is worth publishing.

[blingice]

This "Simple Probability Question has gotten over 50 opinions.