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Munga30
11-05-2003, 11:57 AM
I'm observing an 8 handed card game. I know the number of ways a particular hand (or hands) can be dealt and I know the number of total possible hands. How do I calculate the probability that at least one of the hands in question will be dealt on a particular deal.

As a contrived example, take holdem. I know there are 1326 possible hands and 78 of them are pairs. What is the probability that at least one of the 8 players is dealt a pair in a single deal?

I know this has been covered before but I couldn't find it. I should be able to follow quick explanations or (even better) a link to a detailed one. Thanks.

BruceZ
11-05-2003, 02:01 PM
This is a basic question which comes up often, but obtaining the exact answer is rather advanced. It involves the inclusion-exclusion principle again. You can do a search for my explanations and examples of that principle. The most recent one was from yesterday for the odds that some player has a flush draw (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=395738&amp;page=0&amp;view=ex panded&amp;sb=5&amp;o=14&amp;fpart=).

For your problem, we start by saying that the probability of one particular player having a pair is 78/1326 = 1/17. For the probability that at least 1 of 8 players has a pair, a quick and dirty first approximation is 8/17 = 47%. This is the first step in the inclusion-exclusion method, and this is higher than the exact answer because it double counts all the times exactly 2 players have it, triple counts the times 3 players have it, etc. In most problems, this will actually be within 1%, but not in this case, because the probability of a pair is fairly high. To make it more accurate, we subtract off the probability that 2 people have pairs, and we approximate that probability the same way as C(8,2)* 78/C(52,2) * 73/C(50,2) = 9.8%. C(8,2) is the number of ways to pick the 2 players. There are 73 pairs remaining after the first player chooses a pair. Subtracting this result from 47% gives 37.2%. We can continue in this manner until the answer is as accurate as we like. If we carry it out to 8 terms, the answer will be exact. Note that you alternate adding and subtracting, and each successive term will be smaller than the last, so you know when your answer is sufficiently accurate. Normally this only requires 2 or 3 terms, but in this case it will converge more slowly. Also, the next term will be a little more complicated, because the number of pairs remaining will depend on which pair the second guy chose. This is not as clean as the flush example, because in that case we could consider all combinations of 2n cards dealt to n players, so the answer was very clean and easy to carry out to all the terms. Be sure to understand that example first. I'll leave this one as an exercise to the student, as they say. /images/graemlins/grin.gif I'll be glad to help you if you run into trouble.

PS - I computed the next term, and it is 1.2%, bringing the answer to 38.4%.

Munga30
11-05-2003, 02:29 PM
Thanks, Bruce. I can see how pairs are more complicated than flushes, but I will work on a few terms, at least.

BruceZ
11-05-2003, 02:36 PM
It looks like my final answer 38.4% is accurate to all 3 significant digits after 3 terms. To see that you have to compute the 4th term, which starts to be a brain-bender because you have a tree of possibilities. This part isn't really germane to the inclusion-exclusion method in particular.

Qui-Gon
11-13-2003, 11:17 AM
Hey there!

This is my first post ever. I felt I had to register to comment on this question (and the answer given).

The calculation is actually relatively easy to conduct. The probability that at least one out of 8 players have a pocket pair is the same as (1 - "the probability that none has a pocket pair")!

That probability is a lot easier to calculate. The probability that a single player doesn't have a pocket pair is simply 1-(1/17) = 16/17.

The probability that none has a pocket pair is then (16/17 * 16/17 * 16/17 * 16/17 * 16/17 * 16/17 * 16/17 * 16/17) = 0,6157.

This means that the probability that at least one player has a pocket pair is (1-0,6157) = 0,3843!

There you go, a much easier way for the calculation.

Good luck with the maths!

BruceZ
11-13-2003, 12:04 PM
Hi Qui-Gon,

Welcome to the forum.

The probability that none has a pocket pair is then (16/17 * 16/17 * 16/17 * 16/17 * 16/17 * 16/17 * 16/17 * 16/17) = 0,6157.

This means that the probability that at least one player has a pocket pair is (1-0,6157) = 0,3843!

There you go, a much easier way for the calculation.

It is much easier, but it is only an approximation, though often a good one. It is only approximate because the event of one player not having a pair is not independent of another player not having a pair, so you cannot simply multiply these probabilities to get an exact answer. For example, if the first player does not have a pair, the probability of the second player not having a pair is no longer 16/17 = .941176, but instead it changes to 1 - (11*6+3+3)/(50*49/2) = .941224. Since these are close, the approximation is fairly good.

Another approximation which is sometimes good is to simply multiply the number of players by the probability for 1 player. This approximation assumes that the events can be approximated as mutually exclusive rather than independent, but it is not accurate in this problem. Often one or the other of these approximations is good, or both, so it is good to mention them.

The poster wanted a solution which would work for all such problems, and that required the inclusion-exclusion method. This gives an exact answer. If the elementary methods were exact, I certainly would have used them.

-Bruce

Qui-Gon
11-13-2003, 12:23 PM
Thanks Bruce.

I will enjoy the forum.

As a matter of fact I came to think about the things you stated in your reply and realized my mistake.

Have you calculated the final result? How far away is the approximation?

BruceZ
11-13-2003, 12:41 PM
0.3836 is accurate to at least +/- 0.0007, and the error is actually closer to 0.00007. This is after 4 terms, and the change was 0.000723 on the 4th term, so the error should be an order of magnitude below this. I have no desire to calculate it to any greater precision.

southerndog
11-14-2003, 01:07 PM
Bruce, what is your Prob. background? You seem to be the authority around here.

BruceZ
11-14-2003, 01:49 PM
Bruce, what is your Prob. background?

I'm an electrical engineer in the field of digital signal processing and telecommunications which are heavily based on probability. I also have degrees in physics and mathematics with an emphasis in probability and statistics.

RydenStoompala
11-17-2003, 08:59 AM
Sounds like you've been to school? Just a guess.

RydenStoompala
11-17-2003, 09:01 AM
FYI, I liked your post. For us mathidiots, it works.

BruceZ
11-25-2003, 07:55 PM
I just noticed that in the series of Poker Digest articles, I'm In...No I'm Out (http://www.math.sfu.ca/~alspach/pokerdigest.html), Brian Alspach has solved this problem using inclusion-exclusion for 1-11 players. He solves it over the course of two separate articles, and and his results appear at the bottom of article 28. His line for 8 players agrees with the value I got here (1 minus the value for 0 players or add the other columns).

The method I described here can also be easily modified to generate data for different numbers of players. His method is much more complicated in that he is considering all arrangements of hands among distinguishable players, and he handles the problem of different players having the same pair in a closed form way. His method may become easier for the cases where you want the tiny probabilities of having 4-5 players holding a pair.

This is an advanced problem, and should not be the first example of the inclusion-exclusion principle that you study. See this post for a simpler explanation and examples of the inclusion-exclusion principle (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=417383&amp;page=0&amp;view=ex panded&amp;sb=5&amp;o=14&amp;fpart=).