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Hello everyone I wish to know the formula for calculating how

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Hello everyone

I wish to know the formula for calculating how much probability do I have of getting same numbers on some d10s.

For example, how much probable is to get a double (1,1 or 2,2 or 3,3 etc) on 3d10?
And a triple (1,1,1 or 2,2,2 etc)?

I just need the formula and some explanation (bonus), I think i can figure out the rest.

Thank you very much
>>
>>378336
>to get a double
Any double.
1* 1/10. The first roll counts only for the number you are aiming for, anything 0-9. Therefore the probability is 1. The second roll (regardless of it being the second or third dice) needs to repeat the first roll with a probability of one out of ten. Therefore 1/10. You didn't put restrictions on which double, nor you didn't put restrictions for Only double. You wanted only any double.

>And a triple
Any triple. 1 * 1/10 * 1/10. Same thing, the first roll is allowed to be anything, therefore 1. The second and third roll need both to repeat the first roll with a probability of one out of them for both, so you multiply those. You didn't put restrictions on which triple.
>>
>>378472
>aiming for
*getting
>>
I fucked >>378667 up
It's just the number of possibilities for each roll, multiplied together.

For double anything: 1 (possibility for the first roll) * 10 (possibilities for the second roll) = 10 possibilities, so 1 in 10

For triple anything: 1 (possibility for the first roll) * 10 (possibilities for the second roll) * 10 (possibilities for the third roll) = 100 possibilities, so 1 in 100

For double something, e.g. double 7s: 10 (possibilities for the first roll) * 10 (possibilities for the second roll) = 100 possibilities, so 1 in 100

For triple something, e.g. triple 7s: 10 (possibilities for the first roll) * 10 (possibilities for the second roll) * 10 (possibilities for the third roll) = 1,000 possibilities, so 1 in 1,000
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