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What is the expected number of rolls for any combination of sums including the same sum multiple times if rolling two six sided dice. For example, what is the expected number of rolls to achieve the sum of 7 six times, the sum of 6 four times, and the sum of 8 four times. This leads to the question of what is the most efficient chosen set of sums for any given number of rolled sums. Suppose I wanted to get to a set of 14 numbers (as in the previous example) with the least number of rolls. What sums would I select and how many of each would I select? This problem arises in a game we play with our elementary students where they do experimentation and then predict what 11 numbers they should select (duplicates allowed) to be the first to have all their numbers rolled. Please see the M&M game on the following site: http://www.mathwire.com/data/dicetoss2.html

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  • $\begingroup$ I would personally select sums of 6,7,8, because there are most likely. Note that for these number no matter what the first dice is, we still have chance to reach the sum. So the probability is $\frac 16$, while for any other sum the probability is lower. $\endgroup$ – Stefan4024 Sep 13 '13 at 0:37
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Here is how I would go about calculating the optimal choice. We know that if the number of rolls were very high, tending to infinity, the proportion of the number of rolls for each sum will converge to the probability distribution of the sums over the numbers 2 - 12 in a single roll.

Now what I would do is calculate the probability of each number occurring

$$P(2) = \frac{1}{36};\ \ \ \ \ P(3) = \frac{2}{36};\ \ \ \ \ P(4) = \frac{3}{36};\ \ \ \ \ P(5) = \frac{4}{36}\\ P(6) = \frac{5}{36};\ \ \ \ \ P(7) = \frac{6}{36};\ \ \ \ \ ...\ \ \ \ \ P(11) = \frac{2}{36};\ \ \ \ \ P(12) = \frac{1}{36}$$

Now, it is easy to see that if the two dice were rolled 36 times, I would have picked the numbers so that 2 occurs once, 3 occurs twice .. 7 occurs 6 times etc...

Since they are rolled 11 times, however, you will have to do some rounding to get the distribution that closest approximates this. My guess: 4, 5, 6, 6, 7, 7, 7, 8, 8, 9, 10.

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As I said in the comments I would stick to selecting the sum 7 because the probability is $\frac 16$. because no matter what the first dice have rolled, we can still rech the sum.

Let's say we choose 14 sums, as I said we would stick just to 6,7 and 8. For ex: The sum of six 5 times, the sum of seven 5 times, the sum of eight 4 times.

Expected number of rolls for a particular sum is $\frac 1p$, where $p$ is the probability for that sum to be reached.

So the expected number of rolls for reaching all 14 sums will be:

$$E(T) = E(t_1) + E(t_2) + ... + E(t_4)$$

Because for every sum te probability is $\frac 16$, the expected number of rolls will be $6$, so the expected number of total throws will be:

$$E(T) = 6 + 6 + ... + 6$$ $$E(T) = 14 \cdot 6$$ $$E(T) = 84$$

So the expected number of rolls will be 84.

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    $\begingroup$ I don't quite follow why $p = \frac 16$ for all sums. It seems to me that $p_6 = p_8 = \frac{5}{36}$ while $p_7 = \frac{1}{6}$. Am I missing something? (Sorry. Just realize this was almost a year old.) $\endgroup$ – Tunococ Aug 15 '14 at 13:50
  • $\begingroup$ @Tunococ Yes, you are right. As you've mentioned this was almost year ago so I can't exactly recall what was I thinking then. $\endgroup$ – Stefan4024 Aug 17 '14 at 2:21

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