Re: probability homework question
Hi Bruce - I didn't solve it for the specific example, but for the general case when hero has trips with one card higher and one card lower. For trip eights, it wouldn't only be 88897, but
88896
88895
88894
88893
88892
8889A
and then
888T7
888T6
888T5
888T4
888T3
888T2
888TA
etc.
and then there would be trip sevens, and they're a bit different from trip eights, and then trip sixes which are also a bit different,
etc.
I did it for all of them. For all cases.
It always depends on how you look at things, I guess.
I'll get out of your hair now.
Thanks.
Buzz
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