🚀 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.

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.

In article <168C1D231.R0264@vmcms.csuohio.edu> R0264@vmcms.csuohio.edu writes:
>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.

NO PROBABILITY THAT I KNOW OF CAN BE GREATER THAN 1 !!.

Instead consider:

assuming your dice are fair we have p = 1/6 you get any number from ONE die.
Now, if you are using 3 dices the probability that comes with any combination
of outcomes of the THREE dices follows what is known as the binomial distribu-
tion, that is:
        n  x     n-x          n        n!
p(x) = ( )p (1-p)    , where ( ) = ----------
        x                     x     x!(n-x)!

p = 1/6 , n = number of trials (3 in this case), and x = number of posible out-
comes (in this case x=0, x=1, x=2, x=3)   x=0 => your number did not come up
                                          x=1 => your number came up once
                                          x=2 => your number came up twice
                                          x=3 => your number came up in all dice

if you work this up you get p(x=0) = 0.57870
                            p(x=1) = 0.34722
                            p(x=2) = 0.06944
                            p(x=3) = 0.00463
of course you might view these as percentages by multiplying by a hundred ( but
be careful with the interpretation!!)

If your friend is not satisfied with just the formula then work out the numbers
for, say, x=1. Then:
           1   5  2
p(x=1) = 3---(---)
           6   6

that is if one of the die has your number this happens with p=1/6 .Since your
number came up only ONCE the other two dice must have any of the other 5
possible outcomes with p= 5/6 each . Now, since we require that these two events
(one die has your number, the other two don't) happen simultaneously we must
multiply their probs. hence (1/6)(5/6)(5/6) the multiplier 3 is, of course,
because the event I mentioned could happen in three different ways (first with
die A, then with die B, and finally with die C).

The other probabilities are found using simmilar reasoning.
-- 
 jorge david rivas (rivas@sfu.ca)    __@ !!  /  "It has taken mankind a million
  undergrad (Math & Stats)         _(\<,_   /  years to alter irreversibly the
SIMON FRASER UNIV, BC. CANADA     (X)/ (X) / planet, something whales could not
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ do in ten times as long"  :(