Re: How to evaluate the propability distribution of two dices?

Hully Andrey,

A few questions about the dice:
* Do they both have the same number of sides? Do we know how many
sides?
* Do they have the same probability distribution, or are they
different?
* Are all possible outcomes given in the data, or can we not assume
that?

I'm not quite sure yet how to solve this, but this might help you on
your way,
at least if the dice have equal amounts of faces, and equal
probability
distributions. Start out by drawing the grid for two identical six-
sided dice:

1 2 3 4 5 6
+------------
1 | 2 3 4 5 6 7
2 | 3 4 5 .
3 | 4 5 6 .
4 | 5 . .
5 | 6 . .
6 | 7 . . . .12 ,

and realise that the probability for any single cell
p(cell) = p(row) * p(col).
And, since the dice are identical,
p(row=n) = p(col=n),
wich means that on the diagonal
p(cell) = p(row)^2 = p(col)^2.

That means we can solve p(sum=2) and p(sum=12), because they each
consist of
only one cell on the diagonal.
p(sum=2) = p(row=1) * p(col=1) = p(row=1)^2
p(row=1) = sqrt( p(sum=2) ) = p(col=1)
We then know the probability that a die throws 1, i.e. p(row=1), and
p(col=1).
Using that knowledge, we can calculate p(row=2) = p(col=2) by solving
p(sum=3) = p(row=1)*p(col=2) + p(row=2)*p(col=1)
= p(row=1)*p(col=2) * 2

We then know p(row=1), p(row=2), and the same for the columns; from
that, we can
calculate the probabilities of p(row=3). And so on, until we have done
all rows.

Of course, p(sum=2) = number_obs(sum=2) / number_obs_total.

I don't think I've explained this very clearly, but I hope it helps.
Oh, and if
there's a more statistical solution, I'd very much like to know it.

Good luck!

Cheers,

Sietse
Sietse Brouwer

.

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