simple pocket pair odds

[CASHIZ]

can anyone tell me the odds that a pocketpair wil be delt out to any player in a ten handed game. thanks

[kleraudio]

hey i was just thinkin about this yesterday. its a fairly simple equation. the answer is 16:1 or 1 in 17. about 5.9%.

ok you get one card guaranteed every deal. so the odds of getting any one card is 100%.

the odds of matching that first card in rank is 3/51. (3 of that rank left and 51 cards left to be dealt) 3/51 = 1/17. thats the probability one in 17 or 16:1.

i dont believe that it matters how many people are in the game, those are the odds if its heads up or ten handed. im not sure if this is right, do odds for this change w/ number of ppl at table? i dont think so but someone correct me if im wrong.

Jim /images/graemlins/spade.gif

[Stork]

16:1 are the odds of YOU getting a pocket pair, but for any person in a 10-handed game to get one, you'de have to multiply by 10. You get dealt pocket pair 1/17, x 10 = 10/17, so the odds are 7:10 against, so you can expect a pocket pair alittle over 50% of the time.

This seems kind of strange, they seem much rarer, but I don't see any faults in my math. Anyone care to prove me wrong?

[garyc8]

Your math is incorrect. I give the correct method in answer to kleraudio's recent post about pp odds.

[BruceZ]

[ QUOTE ]
Your math is incorrect. I give the correct method in answer to kleraudio's recent post about pp odds.

[/ QUOTE ]

Your math (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=112 0725&Forum=,All_Forums,&Words=&Searchpage=0&Limit= 25&Main=1118214&Search=true&where=&Name=18926&date range=&newerval=&newertype=&olderval=&oldertype=&b odyprev=#Post1120725) is also an approximation since the hands are not independent as you have assumed by raising to the 9th power. Stork's method is an approximation since the events of different players holding pairs are not mutually exclusive. He assumes that only one player can hold a pair at a time when he multiplies 1/17 by 10. His approximation is much worse in this case, while yours happens to be close. To get the exact answer, you need the inclusion-exclusion principle (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383).

These posts follow a tired pattern. People continually use these two methods, independence and mutual exclusivity, without realizing that they are both approximations. Usually the first respondents use mutual exclusivity, followed by someone "correcting" them by using independence, unaware that this is also an approximation, and that what is really needed for the exact answer is inclusion-exclusion.

Here is an old thread which uses the inclusion-exclusion for pairs (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Number=396659&page=&view=&sb =5&o=&vc=1). This is for an 8-handed game, though it is easy to modify it for 10-handed. Notice how poster Qui-Gon tried to use a method almost identical to yours.

[garyc8]

I acknowledge that the odds probably change very slightly since you hold two specific cards.
It is the easiest way I know to quickly compute the odds that another player holds a pp. (or a very close approximation.)

[BruceZ]

[ QUOTE ]
I acknowledge that the odds probably change very slightly since you hold two specific cards.
It is the easiest way I know to quickly compute the odds that another player holds a pp. (or a very close approximation.)

[/ QUOTE ]

The issue is much more than just the fact that you hold specific cards. Each player that does not hold a pair affects the probability that every other player does not hold a pair, so each of the 9 probabilities need to be changed to the conditional probability that the nth player does not hold a pair given that the previous n-1 players do not hold a pair. The greatest inaccuracy occurs with the last term, since the probability that the 10th player does not hold a pair is most heavily influenced by the fact that that the other 9 do not hold pairs. This effect can be quite significant, or not very signficant, depending on the problem.

[garyc8]

That being said, how much is my estimate off by (in a ten handed game)?

[BruceZ]

[ QUOTE ]
That being said, how much is my estimate off by (in a ten handed game)?

[/ QUOTE ]

There is a table at the bottom of this article (http://www.math.sfu.ca/~alspach/mag29) which shows the probability of exactly N players having pairs for different numbers of opponents dealt in. For 10 players, the probability that one of your 9 opponents has a pair is the same as the probability that one player has a pair for 9 players, and from the table this is 1 - .581 = .419. This agrees with what I get with inclusion-exclusion. You get 1 - (16/17)^9 = .4205. This is very close.

I realized the reason this is so close is because we are considering all pairs. If the first player gets a pair, the number of pairs goes down by 5, but if he gets a non-pair, the number of pairs also goes down by 6, so the probability doesn't change much, and a non-pair actually makes it slightly more probable that the next player will get a non-pair rather than less probable. This would be different if we were only considering a few pairs such as AA-JJ. In that case, as each successive player does not draw one of these pairs, the number of these pairs does not change, but the number of total cards decreases, so the probability that the next player draws one of these pairs goes up . In these cases, the independence approximation would not be quite so accurate, as I have observed.

[garyc8]

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