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Sum of three dice (six faced) throws is $15$. What is the probability that first throw was $4$?

The way I thought of solving this was... - Given - sum of second and third throw is $11$ - Probability of getting first throw = $4$ is $1$ out of $6$, that is $\frac{1}{6}$

Is this correct?

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  • $\begingroup$ (ways to get 15 with 4 as first throw)/(ways to get 15) $\endgroup$ – Joffan May 8 '15 at 16:00
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Joffan pretty much pointed how to find the answer.

$P(D_{1}=4\mid Sum=15) = \frac {ways\ to\ get\ 11\ with\ 2\ dices}{ways\ to\ get\ 15\ with\ 3\ dices}$ (Note that getting 11 with 2 dices is like getting 4 with the first one).

$ways\ to\ get\ 11\ with\ 2\ dices = 2$ (both $(5,6)$ and $(6,5)$)

$ways\ to\ get\ 15\ with\ 3\ dices =10$

You can have 15 with $(3,6,6), (4,5,6),(5,5,5)\ and\ its\ permutations: (6,3,6), (6,6,3), (4,6,5), (5,4,6), (5,6,4), (6,4,5)\ and\ (6,5,4)$

All together gives us that $P(D_{1}=4\mid Sum=15) =\frac {2}{10}=0.2$

$0.2$, is the probability you were looking for.

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  • $\begingroup$ When calculating the "ways to get 11 with 2 dice", you should weight it by the number of possible rolls with 2 dice. Likewise with "ways to get 15 with 3 dice". Otherwise it doesn't calculate the probability correctly. $\endgroup$ – aepound May 11 '15 at 12:26
  • $\begingroup$ @aepound Even Note that I'm looking for the conditioned probability. Therefore, the possible events are the ways to get 15 with 3 dices. It doesn't have to be weighted by anything. And since the favorable cases are those where I get a 4, and I know that I got a 15 with 3 dices, since that's what I'm coditioning), It's the same as looking for the ways to get 11 with 2 dices. $\endgroup$ – Masclins May 11 '15 at 12:28
  • $\begingroup$ @aepound In other words, what you propose, is to divide both "ways to get 11 with 2 dices" and "ways to get 15 with 3 dices" by the possible ways to throww 3 dices. Doing that, would end up with the exact same factor. $\endgroup$ – Masclins May 11 '15 at 12:29
  • $\begingroup$ Oh! I should read more carefully. I derived the full probability of the roll, not the probability given that the sum was already 15. I understand your answer now. $\endgroup$ – aepound May 11 '15 at 12:35
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You can state this very nicely using conditional probabilities:

Following your start, I'd look at $$ \begin{split} P\left(\sum_{i=1}^3 x_i = 15, x_1 = 4\right) &= P\left(\sum_{i=1}^3 x_i = 15|x_1 = 4 \right)P\left(x_1 = 4\right) \\ &= P\left(\sum_{i=2}^3 x_i = 11\right)P\left(x_1 = 4\right), \end{split} $$ where by independence (unless a roll depends on previous rolls) the second line is derived from the first.

You have the second term on the right side, and now need to calculate the first term.

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