[Greeksquared]

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