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zgall1
05-13-2005, 05:07 PM
On Season 3, Episode 10 of the WPT, Scotty Nguyen's opponent is dealt K8c three times in a row. What are the odds of this occuring i.e. the odds of one player at a 5 handed table getting this hand three times in a row (not necessarily the same player each time)? My friend and I disagree and we want to see if someone here can clear this up. Thanks.

zgall1
05-14-2005, 08:15 PM
bump

Mark Blade
05-15-2005, 02:00 AM
Hi zgall1,

Here's how I think you get the solution. (I'm pretty good at math although I'm not a mathematician so I could possibly be wrong. But I'm pretty confident that I'm right.)

First let's just figure out what the chances of one of these players getting the same exact hand dealt to him three times in a row. The first dealt hand doesn't factor into the equation because it can be anything. Whatever it is, we are going to just be checking if the second hand and the third hand match it. So let's say that the first hand dealt is Ace of spades and 5 of spades. Now, what are the chances of the next hand being A5-spades? First we figure the chance that the first card is either the Ace of spades or the 5 of spades. This is a 1 in 26 chance. Now we will have either of these cards and need its matching friend. So for the second card we need one specific card, which is a 1 in 51 chance. So you multiply 1/26 by 1/51, which gives you 1/1326. Now we figure out what the chances of getting A5-spades on the third hand. It is also going to be 1/1326. So in order to figure out what the chances of both the second and third hand being A5 of spades, you multiply 1/1326 by 1/1326. Now you get 1 out of 1758276. So this is the chance that if you looked at three consecutive hands for just one specific player, that you would find those three consecutive hands to be identical.

Now you asked what the chances are that any of the five players would have this 3 consecutive duplicate hands. Many people would automatically conclude that you just multiply this previous answer by 5. That isn't exactly right, although for a situation like this, it is pretty close. Let me explain why it's not exactly right. Let's say that you asked what the chances were that if each of the five players flipped a coin, that at least one of these would be heads. You know that each one would have a .5 chance of heads. But you couldn't just multiply .5 by 5 as you would get 2.5 or (in other words) a 250% chance of at least one heads. There's no such thing as a 250% chances of anything. So this method is clearly wrong. Here's how you can figure it out though (and maybe there's an easier way than this, but I can't think of it right now at least). You figure out what are the chances that all of the coins come out tails. You do this because once you know this, you can conclude that all other outcomes have at least one heads and possibly more. So the chances that they are all tails is 1/2 x 1/2 x 1/2 x 1/2 x 1/2. Which equals 1/32. So the chances of at least one heads is 1 minus 1/32, which equals 31/32. So there is a 31 out of 32 chance that at least one heads will be flipped by 5 people.

Now let's apply this method to your question. We can figure out the probability of none of these 5 players having three consecutive duplicate hands by multiplying (1758275/1758276) x (1758275/1758276) x (1758275/1758276) x (1758275/1758276) x (1758275/1758276). Whatever this answer is, you just subtract it from 1 and you get the final answer to your answer. Did either you or your friend figure it out correctly? I'm curious. For sport, I'm going to take a guess and say that neither of you did. Here's why I'm guessing this way. For one, it's pretty tricky as evidenced by the fact that no one tackled this question in even a few days and many views of it. Also, if either of you did know this, you could probably convince the other one of your knowledge. But I could be wrong. I hope you surprise me and respond that you did have it right.

(FYI - I have an article in this month's internet magazine that you might enjoy. And a follow up to my article is going to be in the August issue.)

Mark Blade
www.MarkBlade.com (http://www.MarkBlade.com)

BruceZ
05-15-2005, 03:22 AM
[ QUOTE ]
Now let's apply this method to your question. We can figure out the probability of none of these 5 players having three consecutive duplicate hands by multiplying (1758275/1758276) x (1758275/1758276) x (1758275/1758276) x (1758275/1758276) x (1758275/1758276). Whatever this answer is, you just subtract it from 1 and you get the final answer to your answer.

[/ QUOTE ]

In the first place, you are attempting to solve the wrong problem, and in the second place, your solution to the wrong problem is incorrect. The OP asked for the probability that the same hand is dealt on 3 consecutive deals, not necessarily to the same player, and your attempted solution is for the same player getting it 3 times in a row. Even if we wanted the answer for the same player, you could not obtain the exact answer by simply multiplying the 5 numbers that you did because the player's hands are not independent. If I get the same hand 3 hands in a row, it becomes ever so slightly more likely that you will also get the same hand 3 hands in a row, because I will not be using any cards that you need. Conversely, if I do not get the same hand 3 times in a row, it becomes less likely that you will either. Of course the dependence is minuscule, but you have claimed your answer to be exact, and it is not.

It is very non-trivial to obtain an exact answer to the OP's question since we would need to compute separate cases which depend on how many hands from the first deal were repeated on the second deal, requiring multiple applications of modified inclusion-exclusion, and it would be absurd to go through that much work when a highly accurate approximation would be so easy to obtain.

Dazarath
05-15-2005, 09:43 AM
What if we took a different approach to the problem? Instead of thinking of the problem in hands, think of it as being one whole deck, with the top two slots as player 1's hand, the next two as player 2's hand, etc. So it would look something like this:
|xx|xx|xx|xx|xx|xxxxx...rest of deck
Now, the OP's problem turns into, what are the chances of K8c falling into one of the 5 slots, 3 times in a row.

For a given slot, the hand can appear 2 ways: K-8 or 8-K
The rest of the deck can appear in 50! ways.
Multiply this by 5 (because there are 5 different slots).
And divide by 52! (the total number of orderings that the entire deck can be in).
This gives us the chance that K8c will be dealt to a player on the 5-handed table.
Raise it to the 3rd power to calculate what the chances of it happening 3 times in a row are.

Someone please correct me if I made a mistake in my reasoning, but I think this should give a correct (and exact) solution.

Mark Blade
05-15-2005, 01:34 PM
Hi BruceZ,

A few points:
1) Yes, now I see that I definitely misread the question.
2) Yes, you're right that there is a dependent aspect to the second part of my solution that I didn't factor in.
3) I've got to not try to figure out these tricky questions very late at night and so quickly on the fly. I just started typing out the answer immediately without much thought because I wanted to help this dude out. My mistake.
4) I'm curious what the original poster and his friend each thought as one of them might have it right and once we all read it, we might all go "aha, yes, that's how you do it, why didn't we all think of that."

"May all your pots be monsters, and all your co-workers look like Shana Hiatt."

Mark Blade
www.MarkBlade.com (http://www.MarkBlade.com)

felix83
05-15-2005, 01:54 PM
Since you're curious, I'll tell you that we did make the mistake of multiplying by 5. But in terms of finding the original formula, I figured that there was a 2(1/52)(1/51) chance (multiplied by 2 because order doesn't matter) of being dealt any particular hand. From that I just calculated it as 1*5(2/2652)*5(2/2652). I don't remember how zgall was calculating it, but I can tell you we were both way off. /images/graemlins/smirk.gif

BruceZ
05-15-2005, 05:30 PM
[ QUOTE ]
What if we took a different approach to the problem? Instead of thinking of the problem in hands, think of it as being one whole deck, with the top two slots as player 1's hand, the next two as player 2's hand, etc. So it would look something like this:
|xx|xx|xx|xx|xx|xxxxx...rest of deck
Now, the OP's problem turns into, what are the chances of K8c falling into one of the 5 slots, 3 times in a row.

For a given slot, the hand can appear 2 ways: K-8 or 8-K
The rest of the deck can appear in 50! ways.
Multiply this by 5 (because there are 5 different slots).
And divide by 52! (the total number of orderings that the entire deck can be in).
This gives us the chance that K8c will be dealt to a player on the 5-handed table.
Raise it to the 3rd power to calculate what the chances of it happening 3 times in a row are.

Someone please correct me if I made a mistake in my reasoning, but I think this should give a correct (and exact) solution.

[/ QUOTE ]

If we just want the probability of K8c appearing 3 times in a row, the exact answer is simply (5*1/1326)^3 =~ 0.0000054% or about 1 in 18.7 million. The more difficult, and I think more relevant question is what is the probability of the same hand appearing 3 times, and this is much more difficult to compute exactly since more than one hand can be repeated.