🚀 go-pugleaf

In article <Russel_Mcauliffe.5w90@equinox.gen.nz>
Russel_Mcauliffe@equinox.gen.nz (Russel Mcauliffe) writes:
 
>
>I have just been playing drinking games with a friend of mine and we had an
>argument about the odds of wining [losing??!!], and I was wondering if
>anyone could clarify the situation.
>
>The game was a very simple one (you don't need to be at all sober to play)
>and it basically goes like this:-
>
>There are 3 die(dice). You call a number before throwing the die and however
>many times that number comes up (out of the three die) you have to 'skull'
>that many glasses. My friend reckoned that there was a 1 in 6 chance of
>'skulling' but obviously that was completely wrong. No matter how I tried to
>point out he was wrong, he wouldn't beleive me and as I was [am] quite drunk
>I couldn't come up with a suitable arguement. What I would like to know is
>the odds of throwing a single, a double and a triple of the number.
>
>Any written proof in my argument would be much appreciated.
>Cheers!!!!!
>RJ.
Well, unless there is some trick that I don't see, P(just once) =
(1/6)(5/6)(5/6); P(twice) = (1/6)(1/6)(5/6); P(thrice) = (1/6)(1/6)(1/6).
These are mutually exclusive, so P(once or more) = 31/16.
Phil, Cleveland State University.