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

Dear Sietse,

On 27 май, 17:27, sbbrouwer.s...@xxxxxxxxx wrote:
Indeed, the number of equations ( (12-2+1)-1 = 10, "-1" since the
equation for 6+6 is depends fully from the rest equations +
relations for probabilities)

I don't think I understand you completely, although I understand
that you're striking the dependent equations. I think I understand
the "-1" in (12-2+1)-1=10, but I don't understand the "-2" or the
"+1", or why you start with 12 instead of 13. Could you perhaps
explain it to me?

We have the sequence of numbers (2,3,...12), which is the set of
possible values of the sum of two dices, and we want to count them.
First, we reduce it to sequence (2-2,3-2,...,12-2)=(0,1,...,10) and
then we know that this sequence has 10+1=11 elements.

Well, a criterion for correctness could be 'the best-fitting model',
where we define 'best-fitting' as 'having the smallest sum of
squares'.

The question: why this criterion will be correct in this task and what
statistical meaning will the results have?

at the origin we have only the data (d1[1],d2[1]), (d1[2],d2[2]),
... and estimation of probabilities P{d1+d2=x}. We can represent
d1=e1+e2, where, for example, e2=0..5, then use the same procedure
for estimation of probabilities of P{e1=x} and P{e2=z} and should
get finally the same result (in some statistical meaning).

I don't quite get what you're describing here, sorry. Is it an
extension of this method to three dice? Oh, do you mean that if we
have 2 dice, d1 and d2; or we have 3 dice (e1, e2, and d2);
then if e1+e2 has the same probability distribution as d1, then
the 2-dice and the 3-dice model are the same? That sounds about
right, yes. In which case it will be easier to compute the 2-dice
solution than the three-dice solution.

I meant the following. After we built the procedure we would have at
least three ways to estimate propability distributions for the *same*
input data:
(1) To estimate P{d1=x}, P{d2=y} for two dices d1+d2=r, d1,d2 lie in
1..6.
(2) To estimate P{d1=x}, P{d2=y} for two dices d1+d2=r, d1,d2 lie in
1..6, but then represent d1=e1+e2, where e1 lies in 1..6, e2 lies in
0..5 and apply the same procedure for e1+e2=d1. Finally we will have
propability distributions P{e1=x}, P{e2=y}, P{d2=z}.
(3) To estimate P{e1=x}, P{e2=y}, P{d2=y} for three dices e1+e2+d2=r,
e1,d2 lie in 1..6, and e2 lies in 0..5.

All these solutions should be almost the same in some statistical
meaning since all of them was estimated from almost the same input
data (in order to the input data would be the same in (1), (2) and (3)
we can provide input numbers (d1[1],d2[1]), (d1[2],d2[2]) as
(e1[1],e2[1],d2[1]) and take d1[i]=e1[i]+e2[i] in (1) and (2)).

Thank you for your interest.

Regards,

-Andrey

.

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