What is the probability that, when $4$ dices are rolled, we can choose two of them such that the sum of the numbers on the upper faces is $7$?

I get 139/216 but I'm not sure I'm correct.

  • 1
    $\begingroup$ Yes, that is correct. (check numerically with excel) $\endgroup$ – turkeyhundt Nov 30 '14 at 23:06

It is convenient to assume that the dice were rolled one at a time, and that we recorded the results as they came in. So there are $6^4$ equally likely outcomes. We count the favourables.

First we count the number of strings that have $6$ and a $1$. It is easier to count the strings that don't. There are $5^4$ strings that don't have a $1$, and $5^4$ that don't have a $6$, and $4^4$ that are missing both. So the number of strings that have a $6$ and a $1$ is $6^4-2\cdot 5^4+4^4$. Call this number $a$.

There are also $a$ strings with a $2$ and a $5$, and $a$ with a $3$ and a $4$. So our first estimate for the number of favourables is $3a$.

However, we have double-counted the strings that have, for example, both $1$ and $6$ and $2$ and $5$ (or the other $2$ possibilities).

For both $1$ and $6$ and $2$ and $5$, we have $\binom{4}{2}\binom{2}{1}$. Multiply by $3$. We conclude that the number of favourables is $3a-3\binom{4}{2}\binom{2}{1}$.

We leave it to you to put the pieces together.


I think you're right (and it will probably line up with André's results in the end), but let's check by implicitly counting.

We can, without loss of generality, assume that the first die is a 1 - because whatever roll it is just changes what other roll will get us to 7, and every roll is equally likely so we would wind up counting 6 times as many successful rolls and 6 times as many possible rolls, which would cancel out.

Simplest case, then, is if the 2nd die rolls a 6. There is one way to do this, multiplied by 36 possible results on the last two dice, all of which we consider a success. So we have 36 successes out of 36 rolls for this.

Next simplest is if the 2nd die rolls another 1. Then if the third die is a 6, we don't care what the 4th is, so that's another 6 successes out of 6 rolls. If the third die is a 1, then there are 6 possible rolls for the fourth die but only one success - which is a 6. So let's add another 1 success out of 6 rolls. Finally, for each other possible roll of the 3rd die (2, 3, 4 or 5), there are two possible successes on the 4th - {5, 4, 3, 2} or {6, 6, 6, 6}. So out of those 24 possible rolls (4 ways to roll on the 3rd die and 6 on the 4th), there are a total of 12 successes. So, that's a total of 1+6+12=19 successes, out of 6+6+24=36 rolls.

The final case we need to consider is if the 2nd die is one of 2, 3, 4 or 5. Again, it doesn't matter which one of these we choose, so we'll total up the counts for a single choice - let's pick 2 - then multiply both successes and total rolls by 4.

So we've rolled 1, 2. If the 3rd die is a 5 or 6, we're home. So out of 12 rolls, there are 12 successes. If the 3rd die is another 1 or 2, then the 4th must be a 5 or 6, so that's 4 successes from 12 rolls. Finally, if the 3rd die is a 3 or 4, then the 4th die needs to be a 5, 6 or whatever the complement of the 3rd die is. So in those last 12 rolls, there are 6 successes.

Adding it up, that's 12+4+6=22 successes in 36 rolls. Multiplying those both by 4 gives us 88 successes for 144 rolls.

Finally, we add them all together. So we have 36+19+88=139 successes, out of 36+36+144=216 rolls, which matches your result.


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