Probability Question - give it a try.

[LinusKS]

The other day I was in a live game of seven card stud. I'd just been dealt my cards, and I happened to notice my up-card was an ace.

I looked at the first of the two down cards I'd been dealt, and I saw that it was an ace. I then shuffled my two down cards, and drew one at random. It was an ace.

I hadn't seen any of the other players' cards, and I couldn't remember the suit of the first down card.

What are the odds the card that's still face down (we'll call it card #3) is an ace?

[tubbyspencer]

I think you'd better work on your memory. That may even be more important than the math.

[uuDevil]

You've seen two aces, what is the probability a 3rd card is an ace?

There are 2 aces left in the 50 cards you haven't seen (or don't remember-- either way you don't have the information), so the probability is 2/50=1/25, or odds of 24:1 against.

[LinusKS]

[ QUOTE ]
I think you'd better work on your memory. That may even be more important than the math.

[/ QUOTE ]

This is true.

But at least I'm not tubby.

[LinusKS]

[ QUOTE ]
You've seen two aces, what is the probability a 3rd card is an ace?

There are 2 aces left in the 50 cards you haven't seen (or don't remember-- either way you don't have the information), so the probability is 2/50=1/25, or odds of 24:1 against.

[/ QUOTE ]

24:1 is not correct, given the information in the question.

[LinusKS]

Note: I edited the question slightly, to make it clearer what I'm asking.

[uuDevil]

Ahh, so this is one of those "Let's Make a Deal" deals?
Well this is what I came up with:

Given that your upcard is an A, before you look at your hole cards there are 3 possible cases:

Case 1: AA, occurs with probability P1=(3/51)(2/50)
Case 2: AX, occurs with probability P2=2*(3/51)(48/50)
Case 3: XX, occurs with probability P3=(48/51)(47/50)

When you look, you know Case 3 is not possible. So now

P(AA)=P1/(P1+P2)=.0204, or odds of 48:1

How's that?

[topspin]

[ QUOTE ]
The other day I was in a live game of seven card stud. I'd just been dealt my cards, and I happened to notice my up-card was an ace.

I looked at the first of the two down cards I'd been dealt, and I saw that it was an ace. I then shuffled my two down cards, and drew one at random. It was an ace.

I hadn't seen any of the other players' cards, and I couldn't remember the suit of the first down card.

What are the odds the card that's still face down (we'll call it card #3) is an ace?

[/ QUOTE ]

Condition on whether you looked at both down cards or the same down card twice. Either happens with probability 1/2.

If you looked at both down cards, P(third card is ace) = 1. If you looked at the same card twice, P(third card is ace) = 2/50.

P(third card is ace) = 0.5*1 + 0.5*(2/50) = 0.52.

[uuDevil]

[ QUOTE ]
Given that your upcard is an A, before you look at your hole cards there are 3 possible cases:

Case 1: AA, occurs with probability P1=(3/51)(2/50)
Case 2: AX, occurs with probability P2=2*(3/51)(48/50)
Case 3: XX, occurs with probability P3=(48/51)(47/50)

When you look, you know Case 3 is not possible. So now

P(AA)=P1/(P1+P2)=.0204, or odds of 48:1


[/ QUOTE ]

No, still not right. Going for 3/3:

Since you sampled your hole cards twice, you have one more piece of information.

For Case 1, P(observing A two times)=1.0
For Case 2, P(observing A two times)=(1/2)*(1/2)=1/4

So the overall probability is Po=P(AA)/{P(AA)+(1/4)*P(AX)}

and Po=.07692, or odds of 12:1 against.

I have a headache. /images/graemlins/frown.gif

[topspin]

You'll find theTheorem of Total Probability (http://mathworld.wolfram.com/TotalProbabilityTheorem.html) useful for this particular problem. The two mutually exclusive events of interest here are (1) that you actually saw both your hole cards, or (2) that you saw the same hole card twice (and hence the 2nd hole card is still unknown).

[BruceZ]

[ QUOTE ]
The other day I was in a live game of seven card stud. I'd just been dealt my cards, and I happened to notice my up-card was an ace.

I looked at the first of the two down cards I'd been dealt, and I saw that it was an ace. I then shuffled my two down cards, and drew one at random. It was an ace.

I hadn't seen any of the other players' cards, and I couldn't remember the suit of the first down card.

What are the odds the card that's still face down (we'll call it card #3) is an ace?

[/ QUOTE ]

Let Obs be the observations as stated in the problem, that we have at least AA, and we observed the randomly selected card which could either be the 3rd A, or else an A already observed.

From the definition of the conditional probability of AAA given Obs:

P(AAA | Obs) = P(Obs and AAA) / P(Obs)

P(AAA | Obs) = P(Obs | AAA)*P(AAA) / P(Obs)

Since AAA will always produce Obs:

P(AAA | Obs) = 1*P(AAA) / P(Obs)

Now Bayes' theorem:

P(AAA | Obs) = P(AAA) / [ P(Obs | AAA)*P(AAA) + P(Obs | AAx)*P(AAx) ]

P(AAA | Obs) = 4/C(52,3) / [ 1*4/C(52,3) + 0.5 * C(4,2)*48/C(52,3) ]

P(AAA | Obs) = 1/(1 + 36) = 1/37 = odds of 36-to-1.

[topspin]

Ugh, there was a glitch in my math too. Thanks for your post, which prompted me to go back over my reasoning.

As far as I can tell (hopefully more accurately this time /images/graemlins/smile.gif), your post is essentially correct, with one minor quibble:

[ QUOTE ]
Now Bayes' theorem:

P(AAA | Obs) = P(AAA) / [ P(Obs | AAA)*P(AAA) + P(Obs | AAx)* <font color="red"> P(AAx) </font> ]

P(AAA | Obs) = 4/C(52,3) / [ 1*4/C(52,3) + 0.5 * <font color="red"> C(4,2)*48/C(52,3) </font> ]

P(AAA | Obs) = 1/(1 + 36) = 1/37 = odds of 36-to-1.

[/ QUOTE ]

In the bolded section, you've calculated the odds of being dealt AAx (where x != A), but you actually need the more specific event of the 'x' also being the 2nd down card. Otherwise you've included the events where the non-ace card is face-up, and where the non-ace card is the first hole card that Linus specifically looks at, both of which contradict the observation. With this in mind, the probability is only 1/3 as large, or (48*4*3)/(52*51*50).

With this correction, the final probability is 1/(1+12)=1/13, or 12-to-1 against AAA.

[LinusKS]

[ QUOTE ]

Po=.07692, or odds of 12:1 against.

I have a headache. /images/graemlins/frown.gif

[/ QUOTE ]

It's harder than it seems, isn't it? Here's how I did it. (I don't have your background in math.)

Before my second peek, the odds I will be looking at an ace are 52%.

That's 50% that I'll be seeing the ace I've already seen, plus another 2% that I'll be seeing a new ace.

Once I've seen the ace - therefore - I know there's a 50/52 chance it's the one I've already seen, and a 2/50 chance it's a new ace.

In the first case - when I'm looking at the same ace twice - the chance that the other card is an ace is 2/50.

So that's (50/52)(2/50) for the first case, or 100/2600, or 1/26.

The second case is (2/52)(1) -- since in the case where you're looking at a new ace, the odds are 100% that the other card is the ace you've already seen. That simplifies to 1/26.

(1/26) plus (1/26) = 1/13.

So it's 1/13, or odds of 12:1.

[BruceZ]

[ QUOTE ]
P(AAA | Obs) = P(AAA) / [ P(Obs | AAA)*P(AAA) + P(Obs | AAx)* <font color="red">P(AAx)</font> ]
P(AAA | Obs) = 4/C(52,3) / [ 1*4/C(52,3) + 0.5 * <font color="red">C(4,2)*48/C(52,3)</font> ]



In the bolded section, you've calculated the odds of being dealt AAx (where x != A), but you actually need the more specific event of the 'x' also being the 2nd down card. Otherwise you've included the events where the non-ace card is face-up, and where the non-ace card is the first hole card that Linus specifically looks at, both of which contradict the observation. With this in mind, the probability is only 1/3 as large, or (48*4*3)/(52*51*50).

[/ QUOTE ]

You're correct, and we can go a step further since we also know that the first card dealt was an A, this term should really be (1*3*48)/(52*51*50), but since this also affects the other terms the same way, the correct answer remains 1/13 or 12-to-1. So the full expression becomes:

P(AAA | Obs) = 1*3*2/(52*51*50) / [ 1*3*2/(52*51*50) + 0.5 * 1*3*2 / (52*51*50) ]

= 1/13 or odds of 12-to-1.


Here's a much easier way to do this which still uses Bayes' theorem: Before we looked at the last card, the probability of the 3rd card being an A was 2/50. We then saw an A on the random card, but the probability of that was 1/2 + (1/2)*(2/50) = 26/50. So now the probability that the 3rd card is an A becomes 2/50 out of 26/50 or

(2/50) / (26/50) = 2/26 = 1/13.


This is still Bayes' theorem since we are doing:

P(3rd card = A | random card = A) = P(random card = A | 3rd card = A)*P(3rd card = A) / P(random card = A)

Since P(random card = A | 3rd card = A) = 1, this simplifies to:

P(3rd card = A | random card = A) = P(3rd card = A) / P(random card = A)

= (2/50) / (26/50) = 1/13.

It is an important special case of Bayes' theorem to remember, that when what you have observed will ALWAYS happen when the event occurs for which you are seeking the probabilty, then the solution will reduce to the probability of the event divided by the total probability of your observations.

[uuDevil]

[ QUOTE ]
It's harder than it seems, isn't it? Here's how I did it.

[/ QUOTE ]

I think David Sklansky would prefer your method over even BruceZ's (elegant though it may be).

Thanks for the ...uh... entertainment. /images/graemlins/grin.gif

[BruceZ]

[ QUOTE ]
I think David Sklansky would prefer your method over even BruceZ's (elegant though it may be).

[/ QUOTE ]

See the second half of my post marked "Simple answer by Bayes' theorem". This is actually the method David Sklansky uses when explaining Bayes' theorem. It's really a special case, but easy to understand.

[BruceZ]

[ QUOTE ]
You're correct, and we can go a step further since we also know that the first card dealt was an A, this term should really be (1*3*48)/(52*51*50), but since this also affects the other terms the same way, the correct answer remains 1/13 or 12-to-1. So the full expression becomes:

P(AAA | Obs) = 1*3*2/(52*51*50) / [ 1*3*2/(52*51*50) + 0.5 * 1*3*2 / (52*51*50) ]

= 1/13 or odds of 12-to-1.

[/ QUOTE ]

That should have been:

P(AAA | Obs) = 1*3*2/(52*51*50) / [ 1*3*2/(52*51*50) + 0.5 * 1*3*48 / (52*51*50) ].

Also, the reason for this is that he knows the suit of the hole card, not just that it is an A. Of course he didn't tell us, and it doesn't change the probability, just the expression.

[Gordon Scott]

[ QUOTE ]
I hadn't seen any of the other players' cards, and I couldn't remember the suit of the first down card.


[/ QUOTE ]

Could you recall the color of the ace in the hole?