If I roll 12 dice, what are the probabilities that a) exactly 9 are the same (any number), and b) at least 9 are the same?
Please provide an equation so I can calculate the answer for different numbers, e.g. 10 of 14.
Thanks
Too many variables. Please provide more information.
Six sided dice? Could be. But we gamers have d4,d6,d8,d10,d12,d20 and d100.
You can use the binomial probability distribution to find the probability of getting a certain number X of “successes,” out of a certain number n of trials, where there is a certain fixed probability p of success on each trial.
In the example given in the OP, X=9, n=12, and, assuming fair six-sided dice, p = 1/6 = 0.1666666…, so the probability of rolling exactly 9 of any one particular number is 0.00001263333 using this online calculator; and the probability of rolling at least 9 of any one particular number is 0.00001341935.
For the probabilities of rolling 9 of any particular number, multiply these by 6:
0.00007579998 probability of exactly 9 the same
0.00001341935 probability of at least 9 the same
(By contrast, the probability of rolling a “natural Yahtzee”—5 out of 5 the same—is 0.00077160492.)
That’s great - thank you.
And yes, six-sided dice. Actually, it is not dice at all - just something that has a random 1 in 6 probability so I thought I would call it dice to be more understandable than the actual situation.
And, looking back, I see that this somehow didn’t get multiplied by 6 like I intended. It should be 0.0000805161.
If it’s not confidential trade secrets or something, then you might want to tell us the original situation anyway: Probabilities often depend on easily-overlooked subtleties, and it’s quite common to come up with something (like dice) that looks like it should accurately model your situation, but turns out not to.
(a) Starting at https://www.wolframalpha.com , enter6C(12,9)(6-1)^(12-9)/6^12and get 0.00007579995…
Except for the constant 1, you can see that this depends only on your parameters 9, 12, along with 6 for 6-sided dice.
(b) Repeat (a), substituting 10, 11, then 12 for 9. Sum the four results
.00007579995
.00000454797
.00000016514
.00000000276
to get an answer close to Mr. Boink’s answer.
Did you reverse those inadvertently? This would literally mean it is likelier to roll exactly 9 than to roll at least 9 which, obviously, is impossible. It had to be MORE likely to roll at least 9 than to roll exactly 9.
See my correction in Post #5.
You can have WolframAlpha do the summing for you as well.
This…
sum(C(12, 9+n) * 6 * 5^(3-n)/(6^12)) 0 to 3
Gives us…
9737/120932352 ~= 0.000080516088862639502785821944486782163965520161222035936…
Do not play Russian roulette!
Not many people lose at Russian Roulette more than once 