what are the odds

[carsten]

I'll assume you want me to clarify why (44 choose 5) = (44*43*42*41*40)/(5*4*3*2*1).

You are correct in stating that (44 choose 5) = 44!/(39!*5!). Now observe that you can write 44! as 44*43*42*41*40*(39!), since 39! is the product of all integers from 1 to 39, and the term 44*43*42*41*40 completes the expression to the product of all integers from 1 to 44, which is 44!.

With this in mind, you get (44 choose 5) = 44!/(39!*5!) = (44*43*42*41*40*39!)/(39!*5!), which yields my original expression by writing 5! as 5*4*3*2*1 and cancelling out the 39!. The end result is an expression that's much more practical to compute.

-Carsten

[TonyBlair]

[ QUOTE ]
The probability of getting to the river with a pair or better is.

=1-(44c5)/(50c5)= 48.74%

[/ QUOTE ]

I'm pretty sure you'll end up with a pair or better more times than this when all-in with AKs.