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For example if i roll 2 dice the probability of getting exactly doubles is 6/36.

similarly if i roll 3 dice the probability of getting exactly doubles not considering triples is 90/216.

How can i calculate probability for N dice i.e 1,2,3,4,5,6,7,8,9,10....N

is there a general formula for this?

Thank you in Advance.

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  • $\begingroup$ Would a sequence 1,1,2,2 (two doubles) count as a success? Would 1,1,2,2,2 (one double, one triple)? $\endgroup$ – Stephan Kolassa Aug 20 '20 at 7:45
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    $\begingroup$ You need to say more about what is defined as "success". Is it only one double? Are two or more identical numbers considered a success? $\endgroup$ – osmoc Aug 20 '20 at 8:56
  • $\begingroup$ 1,1,,2,2 cannot be considered as doubles. For Eg: if i roll 4 dice the probability of getting doubles can be considered as . The numbers apart from doubles have to be different. i.e (1,1,2,3) and (2,3,4,2) is considered as doubles not(1,1,2,2) $\endgroup$ – Sandeep Aug 20 '20 at 9:00
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    $\begingroup$ Then, you can't have success after $N\geq 8$, and you don't need a general formula for $N$. $\endgroup$ – gunes Aug 20 '20 at 10:01
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2 dices:

number of possible outcomes: $6^2$
number of possible pairs: $1$
probability: $6 / 6^2 = 1/6$.

3 dices:

number of possible outcomes: $6^3$
number of possible doubles: $\{(a,a,b),(a,b,a),(b,a,a)\}$ with $a,b\in\{1,2,3,4,5,6\}$ ($b \neq a$). For each possibe $b$. In total, you have $3*6*5$ possible pairs. probability = $3*6*5/6^3 = 90/216$.

4 dices:

number of possible outcomes: $6^4$
number of possible pairs:
number of possible $N$-tuples with $a,b,c$ outcomes $\{(a,a,b,c), (a,b,a,c), (a,b,c,a), \ldots \}$, with $a,b,c \in \{1, \ldots, 6\}$ times the possible values for $a,b,c=6*5*4$ ($b \neq c\neq a$).
The number of possible permutations of 4 elements with 2 repetitions is $4!/2!=12$, that must be multiplied by the number of possible different faces $6*5*4$.
probability: $12*6*5*4/6^4$

In general:

the number of possible permutations of $k$ with $2$ replicates is given by $k!/2!$, the number of possible distinct $k-1$ values is $6*5*\ldots*(6-(k-1)+1)$, and given that all possible permutations are $6^k$:

$$ P(k)=\frac{k!/2! * 6 * 5 * \ldots * (6-(k-1)+1)}{6^k}=\frac{\frac{k!}{2!} \frac{6!}{(6-(k-1))!}}{6^k} $$

From the formula, it's clear that with $k = 8$ the probability becomes 0.

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