If I have nine 100 sided dice & need them to land on numbers 1-15, what is...?

Rigs

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I feel like I may be overthinking this, but I can't seem to come to the right answer

If I have nine 100 sided dice and I need them to land on the numbers 1-15, what is the probability that
1,2,3...9 of them would all land on the numbers 1-15.
I know the chance for a single die throw to land on 15 is 15% or 0.15. I figure 0.15^9 will not give me the correct probability to roll all dice between 1-15.

Can someone help me find the 9 different probabilities I'm looking for. The probability for 1/9 dice to be 1-15, 2/9 to be 1-15, and so on
 
Dice throw probability

I'm working on a game and I'm curious about the actual odds.
There's a jackpot that would require you to roll nine 100 sided dice and get a 1-15 on every one of them.
Is the probability for that to occur 0.000000038443359375, or about 1 in 40 million?
I'd like the know the probability to for 1,2,3...9 of the dice to all land on 1-15 if anyone could help.
 
I feel like I may be overthinking this, but I can't seem to come to the right answer

If I have nine 100 sided dice and I need them to land on the numbers 1-15, what is the probability that
1,2,3...9 of them would all land on the numbers 1-15.
I know the chance for a single die throw to land on 15 is 15% or 0.15. I figure 0.15^9 will not give me the correct probability to roll all dice between 1-15.

Can someone help me find the 9 different probabilities I'm looking for. The probability for 1/9 dice to be 1-15, 2/9 to be 1-15, and so on
Your question is unclear.

Are you asking for the probability that all nine dice land on the same number and that the identical number fall in the range of 1 through 15? The probability of that is 15 in
1,000,000,000,000,000,000, which is less than 1 in 500 quadrillion.

Or are you asking what is the probability that each of the nine dice land on any one of the numbers in the range 1 through 15? That probability is indeed about
\(\displaystyle 0.15^9 \approx\) 4 in a hundred million.
 
Can someone help me find the 9 different probabilities I'm looking for. The probability for 1/9 dice to be 1-15, 2/9 to be 1-15, and so on

I'd like the know the probability to for 1,2,3...9 of the dice to all land on 1-15 if anyone could help.

These two sentences sound as if you want, in addition to the probability that all nine dice roll 1 to 15, the individual probabilities for exactly 1, 2, 3, ... 9 of them to roll 1 to 15. This is a binomial probability. Are you familiar with that? (The last of them will be 0.15^9; the sum of all of them would give the probability that at least one rolls 1 to 15, which is 1 - 0.85^9.)
 
If each die has 100 sides, and all 100 are equally likely, the probability that a specific die comes up 1-15 is 15/100= 3/20. The probability a specific die come up 16- 100 (NOT 1-15) is (100- 15)/100= 85/100= 17/20. First imagine labeling the dice 1- 9. The probability die "1" is 1-15 and all other are not is \(\displaystyle (13/20)(17/20)^8\(\displaystyle . The probability each specific die is 1-15 is exactly the same so the probability one is 1-15 and the other 8 not is \(\displaystyle 8(13/20)(17/20)^8\).

Similarly, the probability and specific 2 are 1-15 and the others are not is \(\displaystyle (13/20)^2(17/20)^7\). But then there are \(\displaystyle \begin{pmatrix}9 \\ 2\end{pmatrix}= \frac{9!}{2!7!}\) way to allocate those "2" and "7" among the 9 dice so the probability of exactly two dice are 1- 15 and the other 7 are not is \(\displaystyle \begin{pmatrix}9 \\ 2 \end{pmatrix}(13/20)^2(17/20)^7\).

The probability exactly 3 dice are 1-15 and the other 6 are not is \(\displaystyle \begin{pmatrix}9 \\ 3\end{pmatrix}(13/20)^3(17/20)^6\), etc.\)\)
 
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