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Telamon

Hold'em Odds Puzzle

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My friends and I are arguing about this. Maybe you can help us. Situation: You are playing Texas Hold'em poker with a 52 card deck. You deal out 7 two card hands. Then you deal a 3 card flop. The flop pairs (so it looks like 2-2-3 or 5-5-7 for example). What are the odds that SOMEONE has trips at this point? (None of the hands were allowed to fold preflop no matter what they were). My reasoning: In an N person game like this, the odds of someone having trips after the flop is 1 - (47/49)^(2*N). After the flop, there are 49 cards left unseen. Either they are in the deck or they have been dealt to a player. For the sake of lucidity, say the flop was (5-5-3). There are two 5's left unseen in 49 cards, so the chance of any unseen card being a 5 is 2/49. Conversely, the chance of any one card NOT being a 5 is 47/49. Thus the chance that none of the 14 cards you dealt to the players is a 5 is (47/49)^14. So the chance that someone does have a 5 is 1 - (47/49)^14. My friends are giving me crap about conditionally probability, which I claim doesn't apply in this situation because the 14 cards are dealt independently and we are assuming the flop has already paired.

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Also I realize this calculation isn't super useful in predicting what to do in a real game because people tend to fold low cards, while holding high cards. Also people who get dealt a pocket pair will likely play it.

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Your formula is assuming there's replacement, but there's no replacement. If you deal out 1 card, then yes the probability that card is not a 5 is 47/49. But if you deal out 2 cards, the probability that the second card is not a 5 is 46/48. So for N players, each taking 2 cards, the prob that none of them have a 5 looks like:

(47/49)(46/48)(45/47)(....)(47-2n / 49-2n)

which I think would look like this:

47! (49-2n)!
-----------
(47-2n)! 49!

and the prob of someone having a 5 is 1 minus that

But I think you're right that conditional probability doesn't really apply.

[Edited by - pinacolada on August 22, 2005 3:35:27 PM]

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Guest Anonymous Poster
Quote:
Original post by Telamon
In an N person game like this, the odds of someone having trips after the flop is 1 - (47/49)^(2*N).

You can very easily see that yor solution is incorrect by setting N=24. Now, someone MUST get one of the remaining 5's, since all cards but one have been dealt, but your solution gives lower than 100% probability.

The correct solution I believe is

P = 1 - (49-2N)(48-2N)/(49*48)

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So I was trying to think about this problem a little different. Essentially there are 17 cards dealt out of 52, and what are the chances that three of the same number were within these 17. However, that is not the correct approach since this problem requires 2 of the cards to be placed on the table (known), with a remainder in someone's hand. So it's conditionally dependent in that regard. You require that an event of a pair is dealt on the flop (which has its own percentage), and NOW what are the chances of someone having trips. Of course both of your approaches are taking this into consideration, but this might be what your friends were referring to as conditional.

As for an answer.. There are 49 choose 14 for the total number of possibilities across everyone's hands (which equates to ~675 billion combinations). Now figure out how many these possibilities have the required card to determine if someone has AT LEAST trips. To do that, reconfigure a list of possiblities. Remove the desired 'trips' card from the 49 cards and consider it 'dealt' and count the possibilities. Now you have one less choice, so it's 48 choose 13 possibilities. Divide the number of possibilities with this card, by the total to see the answer.

(#combos with card & 13 others) / (#combos of 14 cards)
(48)nCr(13) / (49)nCr(14)

And you get about 29%. The problem with this is I'm not throwing in the chance of the 'other' trips card getting into play. You can't simply double this, since there is an 'intersection' of probability (because 7.7% of the time, both cards are dealt - two people have trips or one has 4 of a kind) And I haven't done probability in a while, so hopefully someone will chime in and fix my equation. Though, I think you can add both probabilities, and subtract the intersection 29% + 29% - 7.7% and get roughly 50%. Though, I put no guarantee on that :). But I can at least say its over 30%!

edit: Yes, what I assumed was correct. I just checked the previous posters 'simpler' equation of P = 1 - (49-2N)(48-2N)/(49*48) and its equal to doubling my above equation and subtracting the 7.7% intersection.

Final Answer: 49.40%

Which is surprisingly high. Going to have to keep that in mind on a 7 person table :) And of course, all of this has to be renumbered if you burn a card first.


[Edited by - Gluc0se on August 22, 2005 3:28:33 PM]

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