Quads (Little Content)

[Chairman Wood]

I've played close to 2600 hands today. I have hit quads 6 times 4 of them being the kind where you your card hits with 3 on the board. Is this possible?

[jason_t]

Clearly the answer is yes.

I think that the probability that you hit quads 4 times in a session of 2600 hands where one card in the hole hits 3 on the board is .12.

Do you see why? I'll let others elaborate.

[blank frank]

I'm assuming this is Texas Hold'em, although it wasn't specified.

The odds that you get four of a kind in hold'em are the same as in seven stud. There are 224,848 hands with four of a kind out of 133,784,560 possible seven card hands. That's a 0.1681% chance. Using the binomial formula for 6 successes out of 2600 trials with a probability of 0.1681%, you get a 12.24% chance of getting exactly 6 four of a kinds in 2600 hands, and a 27.49% chance of getting at least 6 four of a kinds in 2600 hands.

Now, There are 35 ways to arrange those four cards among the seven cards in the hand. There are two ways to put one in the hole and ten ways to put three of them on the board, so there is a 20/35 = 57.14% chance any given four of a kind has one in the hole and three on the board. That makes it a 0.096% chance of getting such a hand. Again using the binomial density function, we get a 13.34% chance of getting exactly four such hands in 2600 hands. Using the density function, we get a 24.17% chance of getting at least four such hands out of 2600.

[jason_t]

I agree with all of your calculations, which are the same ones that I made, up until you use the binomial distribution to calculate the probability of getting exactly four such hands in 2600 hands. The problem is that you are not first computing the expected number of hands you are dealt that are NOT a pocket pair. The answer to this is 2600 - 2600/17 = 2447. Now use the binomial distriubtion over 2477 hand to obtain the result I provited before.

[blank frank]

[ QUOTE ]
I agree with all of your calculations, which are the same ones that I made, up until you use the binomial distribution to calculate the probability of getting exactly four such hands in 2600 hands. The problem is that you are not first computing the expected number of hands you are dealt that are NOT a pocket pair. The answer to this is 2600 - 2600/17 = 2447. Now use the binomial distriubtion over 2477 hand to obtain the result I provited before.

[/ QUOTE ]

Why would I need to do that? The 0.096% chance already takes that into account. Any hand that is a pocket pair is automatically part of the complement of 99.904%, as it cannot be a four of a kind with exactly one of the four in the hole.

[jason_t]

Here's how I computed the answer. If there is an error, please let me know.

Say I am dealt two cards, and they are different. I want to compute the probability that I hit quads in this situation.

There are (50 choose 5) = 2118760 different boards. How many of them have three of one of the cards in my hand? That would be 46 * 45 = 2070. So the probability that I hit quads is 2070/2118760 = .000 976 986 539.

That last number is the probability that I hit quads given that I was not dealt a pocket pair.

The probability that I am not dealt a pocket pair is 48/51 = .941 176 470 588. By Bayes' theorem, the probability that I hit quads and am not dealt a pocket pair is the product of the two probabilities I computed above and I obtain that this probability is p = .000 919 516 743. Now we use the binomial distribution

(2600 choose 4)* p^4 * (1-p)^2596 = .124 667 075.

This is my answer, and is the one I produced last night via a slighly different method.

[jason_t]

I found an error. Here is a corrected version, and you'll see we now have the same answer. The error was in computing the number of boards giving quads. Before I computed 46 * 45 and it should be 47 * 46. Off by one: the classic computer science mistake. So my method was correct, but not the details.

Say I am dealt two cards, and they are different. I want to compute the probability that I hit quads in this situation.

There are (50 choose 5) = 2118760 different boards. How many of them have three of one of the cards in my hand? That would be 47 * 46 = 2162. So the probability that I hit quads is 2070/2118760 = .001 020 408 16.

That last number is the probability that I hit quads given that I was not dealt a pocket pair.

The probability that I am not dealt a pocket pair is 48/51 = .941 176 470 588. By Bayes' theorem, the probability that I hit quads and am not dealt a pocket pair is the product of the two probabilities I computed above and I obtain that this probability is p = .000 960 384 154. Now we use the binomial distribution

(2600 choose 4)* p^4 * (1-p)^2596 = .133 405 801.

Final answer.