Originally Posted by statmanhal
Working out probabilities like this with more than one opponent is often difficult. To overcome this I wrote a simulation program that looks at these types of problems.
With you holding a pair and with 5 other opponents, the chance at least one player has a pair is 26.4%.
With a pair of sevens, the chance at least one opponent has a higher pair is 16.1%
The chance for a lower pair is 12.1%.
These results were based on 5 million simulations.
I would be interested to see an analytical solution or another simulation result to verify these. Although I have made hundreds of checks on the program, an undiscovered bug is always possible.
For pairs, the independence approximation is so good that it makes exact calculations with the inclusion-exclusion principle
or simulated results unnecessary for all practical purposes. The approximate results are generally within a tenth of a percent.
By this approximation, when you hold 77, the probability that one of your 5 opponents holds a pair is
1 - [1 - 73/C(50,2)]5
Inclusion-exclusion gives 26.41%.
For a higher pair it is:
1 - [1 - 42/C(50,2)]5
Inclusion-exclusion gives 16.05%.
For a lower pair it is:
1 - [1 - 30/C(50,2)]5
Inclusion-exclusion gives 11.71%. Since this is the only one that differs from yours, here are the first 3 terms of inclusion-exclusion which is good to much better than 0.01%:
C(5,3)*30*(24*20 + 1*24)/C(50,2)/C(48,2)/C(46,2)
You should check that last one or run it some more. You should be closer than 12.1% after 5 million.