So the question is: given that you roll $10$ dice, what is the probability of the sum of the total dice rolls adding up to $57$?

I know that there are three ways to do this:

  1. Seven die rolls must be $6$ with three $5$s
  2. Eight die rolls must be $6$, one die roll must be $5$ and one must be $4$
  3. Nine die rolls must be $6$, and one roll must be $3$

The solution states that the probability of the events are:

  1. $ \binom{10}{3} \cdot \frac{1}{6^{10}}$
  2. $ \binom{10}{1} \cdot \binom{9}{1} \cdot \frac{1}{6^{10}} $
  3. $ 10 \cdot 9 \cdot \frac{1}{6^{10}} $

I really don't understand why the probabilities work this way. I would really, really appreciate it if someone could perhaps explain this in a more intuitive way for me.

Edit: I am really sorry for the mistake. Edited so that the question reads sum up to 57.

Edit 2: Also, I think my solution sheet is missing the fact that you should sum all of these probabilities and set them over $6^{10}$. I apologize for the mess and I appreciate all the comments that pointed this out.

  • 2
    $\begingroup$ None of those add up to 52 can you edit your question to say 57? $\endgroup$ – David Kleiman Feb 7 '16 at 20:19
  • $\begingroup$ #3 is certainly wrong. $10 \cdot 9$ isn't a probability. $\endgroup$ – The Chaz 2.0 Feb 7 '16 at 20:21
  • $\begingroup$ The number of combinations for $3.$ should be $10$, not $90$ $\endgroup$ – Gregor de Cillia Feb 7 '16 at 20:24
  • $\begingroup$ none of them is probability $\endgroup$ – Jimmy Dur Feb 7 '16 at 20:26
  • 1
    $\begingroup$ The solution probably rather states that the probability of the events are $ \binom{10}{3}\cdot\frac1{6^3}$, $ \binom{10}{1} \cdot \binom{9}{1}\cdot\frac1{6^3}$ and $\binom{10}{1} \cdot\frac1{6^3}$ respectively. Decidedly, your question needs a thorough revising. $\endgroup$ – Did Feb 7 '16 at 20:28

For any of the three solutions, for each die, there is a $\tfrac{1}{6}$ chance of rolling the right number, so there is a $\tfrac{1}{6^{10}}$ chance of that specific combination turning up.
But it doesn't matter if the first die rolled a $6$ and the last the $3$ you needed, or the other way around. So we're going to try to figure out how many possible ways there are of rolling any of those combinations.

  1. Seven $6$s and three $5$s
    There is a total of $\binom{10}{3} = 120$ ways to roll this combination, each of which has a probability of $\tfrac{1}{6^{10}}$, so the total probability is $\binom{10}{3} \cdot \frac{1}{6^{10}} = \frac{120}{6^{10}}$
  2. Eight $6$s, a $5$, and a $4$
    This one isn't that hard. We can roll this combination in a total of $\binom{10}{1} \cdot \binom{9}{1} = 10 \cdot 9$ ways, which you can verify for yourself: there are $10$ dice that can be the $5$, after which there are $9$ dice that can be the $4$, with the rest being $6$s.
    Since each has a probability of $\tfrac{1}{6^{10}}$, the total probability here is $\binom{10}{1} \cdot \binom{9}{1} \cdot \frac{1}{6^{10}} = \frac{90}{6^{10}}$
  3. Nine $6$s and a $3$
    This one is the simplest of the three. There are just $\binom{10}{1} = 10$ ways of rolling this combination. Again, you can verify this yourself; any one of the dice can be $3$ while all others are $6$. The total probability here is $\frac{10}{6^{10}}$

Now that we have the separate probabilities, all that's left to do is add them up.

$\begin{align} P(57) & = \frac{120}{6^{10}} + \frac{90}{6^{10}} + \frac{10}{6^{10}} \\ & = \frac{220}{6^{10}} \\ & \approx 3.638\times10^{-6} \end{align}$

Note that this means that out of the $6^{10}$ possible rolls of the dice, only $220$ will sum up to $57$.

  • $\begingroup$ Ah, thank you so much! So in a sense, by $\binom{10}{3}$, we're essentially calculating the number of places or positions the 5s can be given our 10 rolls? It makes sense now. Thank you! $\endgroup$ – Nikitau Feb 10 '16 at 17:05

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