Quote:
Originally Posted by arlecchino76
I tried to do the math in my head and just got confused , but I came up with something like 8% , close enough.
A decent approximate method can be used.
There are 8 remaining spades, 3 of which are ten, jack, queen. The number of ways one player can have 2 spades, and at least one of which is T, J, or Q is
C(3,1)C(5,1)+C(3,2)C(5,0)=18
The total number of hands for a player from the deck of 47 is C(47,2) = 1081. Therefore the probability one player has a higher flush is 18/1081 = 0.01665. An upper bound for at least one of four players having a higher flush is 4*0.01665 = 0.0666, or 6.66%.
For a lower bound, no higher flush for one player has probability 1-0.01665 = 0.98335. Raising that to the fourth power is an upper bound for all four players not having a higher flush. Subtracting from 1 gives a lower bound to the higher flush chances, and the value is 6.496%.
Note the simulation result of 6.58% is in between the two bounds.