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In article <168C1D231.R0264@vmcms.csuohio.edu>
R0264@vmcms.csuohio.edu writes:
 
>
>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.
OOPS! I missed a trick.  It should be p(once) = 3(1/6)(5/6)(5/6);
p(twice) = 3(1/6)(1/6)(5/6); and p(thrice) was ok.  So P(once or more)
= 91/216, quite a difference.  Also, and easier, p(none) = (5/6)(5/6)(5/6)
and p(one or more) = 1 - p(none).
Phil again.