NetNews Usenet Archive 1992 #30/NN_1992_30.iso/spool/sci/math/stat/2649

Text File | 1992-12-21 | 2.0 KB | 47 lines

Newsgroups: sci.math.stat
Path: sparky!uunet!spool.mu.edu!agate!usenet.ins.cwru.edu!news.csuohio.edu!vmcms.csuohio.edu!R0264
From: [e-mail]
Subject: Re: Simple Odds on dice.
Message-ID: <[e-mail]>
Sender: [e-mail] (USENET News System)
Organization: CSU
References: <[e-mail]> <[e-mail]>
Date: Fri, 18 Dec 1992 20:08:13 GMT
Lines: 35

In article <[e-mail]>
[e-mail] writes:

>
>In article <[e-mail]>
>[e-mail] (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.