[Texas Hold'em] I'm writing a non-GUI intensive console-based poker analysis game and am having a difficult time "getting" how to apply some very basic formulas to a series of related questions. I am trying to find the probabilities for whole families of problems. Below I have enumerated three questions that sum up all the problems I am having, and answering any one of them will unlock clues to many of the rest for me. Thanks for any help here... *NOTE* I am completely comfortable with the theory of probabilistic AND and OR, meaning: Pr{A and B} = Pr{A} * Pr{B; given A} = Pr{A}Pr{B|A} Pr{A or B} = Pr{A} plus Pr{B} - Pr{A and B} = Pr{A} plus Pr{B}-Pr{A}Pr{B|A} It is in the "math modeling" aspect (translating these formulae into meaningful representations of my problems at hand) where I am falling short. 1. [The Warm-Up] In a game with 6 players, what is the probability of 2 or more players getting 3-of-a-kind? I model this as: Let "trip(x)" = "x players get a 3-of-a-kind" Pr{trip(x >=2)} = Pr{trip(2) or trip(3) or trip(4) or trip(5) or trip(6)} = Pr{trip(2)} plus Pr{trip(3)} plus ... Pr{trip(6)} - Pr{trip(2) and trip(3)and ... and trip(6)} = Pr{trip(2)} plus Pr{trip(3)} plus ... Pr{trip(6)} - 0 (mutually exclusive) But I fear this model is incorrect because the answer doesn't "add up." 2. [The Head-Scratcher] In a game with 3 players, what is the probability that each of the three players will be dealt exactly one Ace in their pockets cards? (Example: Player 1 has Ace of Clubs, Player 2 has Ace of Diamonds and Player 3 has Ace of Spades, leaving the Aces of Hearts remaining somewhere in the deck or muck.) 3. [The Brain Strainer] I have a *hunch* that (i) the number of players and (ii) your position at the table directly affects your raw probs/odds (regardless fo strategy). For instance in a 10-player hand the likelihood of one of the players having a particular hand will be greater than the likelihood of the same hand showing up at a finals table between two players. And in regards to table position, if I am in a game with three other people, and I am sitting to the immediate left of the current dealer, then the pocket cards will be allocated as such: 52 shuffled cards to start with... DEALER: 4th card of deck -> 48 remain; 8th card of deck -> 44 remain ME: 1st card of deck -> 51 remain; 5th card of deck -> 47 remain PLAYER 3: 2nd card of deck -> 50 remain; 6th card of deck -> 46 remain PLAYER 4: 3rd card of deck -> 49 remain; 7th card of deck -> 45 remain I have a feeling that depending on where you sit at the table, that because this affects how many cards are currently left in the pile at the time your card is dealt to you, that this affects the probability of you getting a particular hand. In the above example, the probability of me getting an Ace as my first pocket card should (I think) be 4/52. But the probability for the next player (again, I think) should be out of 51, because now there are only 51 cards that remain. This spawned an whole second train of thought, because now I have no way to model the numerator in Player 3's Ace-on-first-pocket-card-probability. I can't say that his probability is 4/51, because the numerator is contingent on whether or not I got an Ace on the opening pocket card. If I did, I would think his prob is 3/51, else 4/51... So basically, there are 2 major sub-questions to this one question: (a) Can you elaborate - specifically - as to how #players and table position affect one's probabilities? This includes any known formulae or "rules of thumb," or just common knowledge. (b) Can you elaborate on that second train of thought posted right above, regarding how to model the numerator for each players' probabilities, taking into consideration the incomplete information that cannot be gathered from previous (those sitting closer to the dealer) players' pocket cards? Any help here *at all* is enormously appreciated. Also if anybody is big into poker and knows some good sources (sites or books) on heavy duty poker math (not just the standard "probabilities and odds") please let me know. -ply

I’m a bit rusty with probability theory... I hope the following make sense...

Quote:Original post by plywood
2. [The Head-Scratcher] In a game with 3 players, what is the probability that each of the three players will be dealt exactly one Ace in their pockets cards? (Example: Player 1 has Ace of Clubs, Player 2 has Ace of Diamonds and Player 3 has Ace of Spades, leaving the Aces of Hearts remaining somewhere in the deck or muck.)

If I remember correctly the rules of texas hold'em every player have two cards and the total number of cards is 52.
The total numbers of combinations of cards for 3 players are:
choose(52,2)*choose(50,2)*choose(48,2) = 1832266800
The total numbers of combination where each player have one ace is:
(4*48)*(3*47)*(2*46) = 2490624
So the probability should be 2490624/1832266800 = 0.0013593

Quote:Original post by plywood
3. [The Brain Strainer] I have a *hunch* that (i) the number of players and (ii) your position at the table directly affects your raw probs/odds (regardless fo strategy).

The position doesn’t affect the probability because all the permutations of the cards have equal probability.

Quote:Original post by ToohrVyk

Quote:Original post by plywood
I have a feeling that depending on where you sit at the table, that because this affects how many cards are currently left in the pile at the time your card is dealt to you, that this affects the probability of you getting a particular hand.

No. All permutations are equiprobable when dealing cards. Meaning that swapping places with another player does not change the odds. So no, position does not affect the probability of being dealt something.

Can you explain the reason for this please as I feel this is incorrect. If you have a card and your second card is being dealt to you, then you have a different probability of getting a (insert type) compared to the person next to you due to that there is one more card in the pack. As you do not know other peoples cards, this effects the probability as you do not know which cards have already been dealt to other people.

[Edited by - CmpDev on July 7, 2008 7:33:03 AM]

Quote:Original post by CmpDev

Quote:Original post by ToohrVyk

Quote:Original post by plywood
I have a feeling that depending on where you sit at the table, that because this affects how many cards are currently left in the pile at the time your card is dealt to you, that this affects the probability of you getting a particular hand.

No. All permutations are equiprobable when dealing cards. Meaning that swapping places with another player does not change the odds. So no, position does not affect the probability of being dealt something.

Can you explain the reason for this please as I feel this is incorrect. If you have a card and your second card is being dealt to you, then you have a different probability of getting a (insert type) compared to the person next to you due to that there is one more card in the pack. As you do not know other peoples cards, this effects the probability as you do not know which cards have already been dealt to other people.

The probability to have a card as the first card in the deck is equal to the probability to have the same card in the third, fourth or every other position in the deck. So there aren’t differences if you are the first or the last player, all have the same probability to get a specific card.

Edit: I have make it clearer

[Edited by - apatriarca on July 8, 2008 1:29:35 AM]

Quote:Original post by apatriarca
The probability to have a card as the first card in the deck is equal to the probability to have the same card in the third, fourth or every other position in the deck. So there aren’t differences if you are the first or the last player, all have the same probability to get a specific card.

I agree all cards have the same probability when talking about a single card(ie not ace followed by ace); yet disagree with your last statement as your own calculations in thread contradict.

Quote:The total numbers of combinations of cards for 3 players are:
choose(52,2)*choose(50,2)*choose(48,2) = 1832266800
The total numbers of combination where each player have one ace is:
(4*48)*(3*47)*(2*46) = 2490624

Why do you decrease the number of cards in the pack if this has no effect?


Anyway, your position in respect to the dealer does have an effect in a game of poker, as it effects the pot odds and how you may play a hand ie you may check when the pot odds say raise due to being the small blind and setting a trap, you may raise the minimum when the odds say raise larger.

Quote:Original post by CmpDev
I agree all cards have the same probability when talking about a single card(ie not ace followed by ace); yet disagree with your last statement as your own calculations in thread contradict.

Quote:The total numbers of combinations of cards for 3 players are:
choose(52,2)*choose(50,2)*choose(48,2) = 1832266800
The total numbers of combination where each player have one ace is:
(4*48)*(3*47)*(2*46) = 2490624

Why do you decrease the number of cards in the pack if this has no effect?

Because I have “simulated” the extraction of the cards to make it easier to follow. I see no contradiction, it’s just a method to calculate it. You can also calculate the number of permutations of the deck with an ace in the first two positions, another ace in the third or fourth position and a third in the fifth or sixth position.
Total number of permutations of the deck: 52!
Number of permutation that satisfy the property: (4*48*2)*(3*47*2)*(2*46*2)*46!
Result: 368/270725 = 0.0013593
Here I’m not simulating a game, I’m simply using combinatorics. If you have asked for the aces to be in other positions (for example in last positions of the deck) I have done exactly the same calculations.

Quote:Original post by CmpDev
Anyway, your position in respect to the dealer does have an effect in a game of poker, as it effects the pot odds and how you may play a hand ie you may check when the pot odds say raise due to being the small blind and setting a trap, you may raise the minimum when the odds say raise larger.

Yes, I agree. If you are the last player you have seen how the others players have played and you can probably make more assumptions about the game, but the last player continue to have the same probability to have something in his hand than the other players.

[Edited by - apatriarca on July 8, 2008 3:12:51 PM]