if u have some spare time

Hey guys my math definitely isn't my best skill and have no idea how to figure this out but if you have some free time take a look at it.

I was at home game the other day playing in a tournament, the first hand when it got down to three players, everyone had a pocket pair and everyone went all in. What are the odds of having all three players get pocket pairs, or four of four players getting pocket pairs?
p.s. If you could dumb the math down to idiot mode i'd really appreciate it

[AaronBrown]

The computation for one player to get pocket pairs is easy. The first card can be anything, the second must be one of 3 out of 51 cards, so 1 chance in 17.

It's approximately true that the probability is (1/17)^n for n players to get pocket pairs, that is (1/17)^2 = 1/289 for two players, (1/17)^3 = 1/4,913 for three players and so on. But if there are, say, ten players at the table, you have to consider the number of ways the players can be split up. There are 10/1 = 10 ways to pick one player, 10*9/(2*1) = 45 ways to pick two players, 10*9*8/(3*2*1) = 120 ways to pick three and so on.

To get more accurate, you have to do two things. First, if one player gets pocket pairs, it affects the chance of the next player getting them slightly. The second player has 2 chances in 50 of getting the same rank as the first player paired, then she has 1 chance in 49 of getting the pair. 48 times out of 50 she will have 3 chances out of 49 instead. So you get (2/50)*(1/49)+(48/50)*(3/49) = 146/2,450 which is about 1% larger than 1/17.

Second, you can't just multiply the combinations by the probabilities. If each player had 1 chance in 17 of getting a pair, then the probability that no player has a pocket pair would be (16/17)^10 = 0.55. Of course this assumption is not exactly correct, as mentioned above. The probability that exactly one player out of ten gets a pair is 10*(1/17)*(16/17)^9 = 0.34. The probability for exactly two out of ten is 45*(1/17)^2*(16/17)^8 = 0.10.

[pzhon]

[ QUOTE ]

I was at home game the other day playing in a tournament, the first hand when it got down to three players, everyone had a pocket pair and everyone went all in. What are the odds of having all three players get pocket pairs, or four of four players getting pocket pairs?

[/ QUOTE ]
All deals are equally likely, so if we count the number of ways to deal pairs to each player, we can divide by the total number of deals to get the probability.

/images/graemlins/spade.gif 3 players

There are 52*51/2 ways to deal a hand to the first player, then 50*49/2 ways to deal a hand to the second player, then 48*47/2 ways to deal a hand to the third player. The total number of deals is 52*51*50*49*48*47/8 = 1,832,266,800

Case: All different pairs.
There are 13 ways to choose the rank of the first player's pair, then 12 ways to choose the rank of the second player's pair, then 11 ways to choose the rank of the third player's pair. Given a rank, there are 4*3/2=6 ways to get a pair of that rank, so the total number of ways three players can get pairs of three different ranks is 13*12*11*6*6*6 = 370,656.

Case: Two players get pairs of the same rank.
There are 3 choices of which player gets the odd pair, 13 choices for that rank, and 6 choices for the suits. There are 12 choices for the rank of the pairs the other two get, and 6 choices for the ways to divide the cards of that rank among the two. The total number of ways 3 players can get pairs of two ranks is 3*13*6*12*6 = 16,848.

In total, the number of ways 3 players can get dealt pairs is 387,504 / 1,832,266,800 ~ 1/4728.

As a sanity check, 17^3 = 4913.

/images/graemlins/spade.gif 4 players

There are 52*51*50*49*48*47*46*45/16 deals.

Case: 4 different ranks.
13*12*11*10*6^4.

Case: 3 different ranks.
There are 6 ways to choose the pair of people with pairs of the same rank, then 13 choices for that rank, then 6 ways to divide up the suits. There are 12*11*6*6 ways to choose the other two pairs.
6*13*6*12*11*6*6.

Case: 2 different ranks.
There are 3 ways to divide up the people into two pairs. There are 13 choices for the rank for the first pair, then 6 ways to divide up the suits. There are 12 choices for the rank of the second pair, then 6 ways to divide up the suits.
3*13*6*12*6.

Total: The probability that 4 people will each be dealt pocket pairs is (13*12*11*10*6^4 + 6*13*6*12*11*6*6 + 3*13*6*12*6)/(52*51*50*49*48*47*46*45/16) ~ 1/77,467.

As a sanity check, 17^4 = 83,521.