🚀 go-pugleaf

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"  :(