[BruceZ]

[ QUOTE ]
Intuitively each player has a 1/20 chance at getting to the final table. But since, they are all playing the same tournament the chances of a second player getting to a final table is less because one spot is already taken.

The chances of 0 getting to the final table is the chance that they all don't make the final table.
(19/20)(189/199)(188/198)(187/197)(186/196)= .7717

The chance of the team getting one person to the final table of the first tourney is just

(1/20)(190/199)(189/198)(188/197)(187/196) but there are 5 combinations of this so we multiply this by 5 and get .2031

Similarly

for 2 (1/20)(9/199)(190/198)(189/197)(188/196)*10= .0193
for 3 (1/20)(9/199)(8/198)(190/197)(189/196)*10= .00085
for 4 (1/20)(9/199)(8/198)(7/197)(190/196)*5= .0000157
for 5 (1/20)(9/199)(8/198)(7/197)(6/196)= .0000000993


To find the chance that there will be 2 of the friends at atleast 2 or more of the final tables we see that there is .0208 chance at 2 or more members getting to the final table.

So the chance that this happens 0 or 1 times out of ten is just a binomial distribution. P(0)=.8104, P(1)=.1721.

So the probability of this happening 2 or more times is 1-(P(0)+P(1))=.0175

[/ QUOTE ]

You are interpreting the problem as asking for the probability of 2 or more friends facing each other at the same final table. I agree with your numbers. Here is my solution for that interpretation:

Probability of 2 or more friends making the final table at a particular tournament:

P(2) = 1 - [C(195,10) + C(5,1)*C(195,9)] / C(200,10) = 2.08%

The value in brackets is the probability of 0 or 1 of the friends making the final table, so 1 minus this is the probability of 2 or more making it. Now the probability that this happens at least once in 10 tournaments is 1 minus the probabilty that it does not occur for 10 tournaments, which is:

1 - [1 - P(2)]^10 = 19.0%

The probability that this happens in 2 or more tournaments is 1 minus the probabilty that it happens 0 or 1 time. This is:

1 - [ (1-P(2))^10 + 10*P(2)*(1-P(2))^9 ] = 1.75%.