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I understood the formula as:

Given there are 5 letters {A,B,C,D,E}, pick 3 letters randomly(putting them back) and the combination formula is C(5, 3). r=3 means how many I choose from the 5 to make the combinations.

However, using the same logic and I couldn't comprehend the answer for the following question.

Suppose we roll eight fair six-sided dice. What is the probability that the sum of the eight dice is equal to 9

The answer is $\frac{8!}{1!7!}\times6^{-8}$

I don't get why it is C(8, 1), from my understand of r above, this means I am picking r=1 from 8 dices.

I do understand the only possible combination to this question is 2111 1111 and the order of 2 doesn't matter.

Chung
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    The dice have at least the value $1$, so the sum is at least $8$. To get a sum $9$, there must be exactly one die with value $2$, all the others with $1$. And yes the order does matter (this won't change the probability, and the answer is much simpler as the the outcomes are equiprobable and the denominator is $6^8$). – Jean-Claude Arbaut Jul 06 '23 at 23:05
  • Do you mean the order doesn't matter? – Chung Jul 06 '23 at 23:26
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    I mean it's up to you, but if you decide the order doesn't matter, then you have to account for the fact that (for instance) $12345678$ and $11112222$ don't have the same probability. That is, you can't just divide the number of favorable cases by the total number of cases (which assumes equiprobability), and you will have to consider patterns of repeated digits. It's a mess, and it's absolutely equivalent to consider that order does matter, and then everything is simple. – Jean-Claude Arbaut Jul 06 '23 at 23:34
  • Typo above, $12345678$ is of course not possible with $6$-sided dice, but you get the idea, and you can consider $12345666$ instead. – Jean-Claude Arbaut Jul 06 '23 at 23:45

1 Answers1

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As you observed in order to get a total of $9$ you must roll a $2$ and seven $1$'s. The probability of obtaining such combination equals the number of favorable cases divided by the number of all cases, under the tacit assumption that all outcomes are equally probable.

To count the favourable cases you only need to decide which dice will show the $2$. That means choosing a dice out of $8$, i.e. $C(8,1)=8$ choices.

Since the number of all possible combinations is $6^8$, the probability is $$ p=8\cdot\frac1{6^8}. $$

In the same way the probability to obtain $10$ is $$ p=(8+C(8,2))\frac1{6^8}=36\cdot\frac1{6^8}=\frac1{6^6} $$ and so on.

Andrea Mori
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  • "and so on" is a bit optimistic: for larger sums, you may have more than two different values, and combinations are not going to be enough (or at least not as straightforward). – Jean-Claude Arbaut Jul 06 '23 at 23:18
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    @Jean-ClaudeArbaut, of course you're right! But in principle one would solve the problem in the same way: enumerate the combinations that give a certain total $T$, for each of these combinations compute the number of occurrences, add up all these numbers and divide by $6^8$. – Andrea Mori Jul 06 '23 at 23:28