Shorthanded - two players with same pair - how do you figure

[ACPlayer]

If the question is what is the chances that 2 players in a 5 handed game will be dealt 88 then the answer should be .0015%. Derived as follows:

CHance a player has 88 = .45%
Given that a player has 88 chance that another player get 88 is the hyper geometric distribution selecting 1 success in 4 samples with 2 success samples in a total population of 50 elements multiplied by the chance of getting the last 8. (Then ofcourse multiplied by .45%).

If the question is the chances that it will win the pot and that the pot will be split (well, that is a different kettle of fish).

PS The hypgeomdist function in excel is a great tool to do these things. Try it sometime.