Re: toy probability problem

On 2007-12-14, Nasser Abbasi <nma@xxxxxxxxx> wrote:

I am not familar with this game. So I do not know if what one person writes
down will depend or not on what the other does.

If what A and B do are independent,

Well... if A and B each code the pieces at home, totally
independent from each other, I think we can assume that what A
and B do are indeed independent :-)

this it seems to me all what you have to
do is simply count how many ways there are to code a 'move'. Which is simply
23x22x21x20=212,520 (I am assuming the same color can't show up twice in a
code).

The same color *can* show up twice, so there are 23^4 = 279841 ways to
code a piece.

So each code is any one of 212,520 possibilities. Since there is one
correct, hence there are 212,519 wrong codes for that one square.

I can agree: for that one square we are talking about, there is
one correct code and 279840 wrong codes.

Hence for any one code that person A picks out of those 212,519 ones,
person B has a probability of 1/212,519 of picking the same code.

Hmm... this is indeed a simple reasoning... i hadn't thought
about it that way yet... So the chance of person A and B
mistakingly coding the piece with the same wrong code is

1/(23^4-1) = 0.000003573470555

Pretty small :-)

Or Am I missing something here?

I don't know, that's why i asked the question :-) But your
analysis sure looks possible and correct to me.

Thanks,
Bart

--
"Share what you know. Learn what you don't."
.

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