# Two dice-related games, do the outcomes have the same distribution?

In this codegolf question, it is effectively asked to find the distribution for $X$, defined below:

• Throw $2$ dice; is the sum $7$ then $X=2$, else continue.
• Throw $4$ dice; is the sum $14$ then $X=4$, else continue.
• ...
• Throw $2n$ dice; is the sum $7n$ then $X=2n$, else continue.
• ...

A few answers attempt this by calculating $Y$, defined below:

• Throw $2$ dice; is the sum $7$ then $Y=2$, else continue.
• Keep the $2$ dice on the table, throw $2$ additional dice; is the sum $14$ then $Y=4$, else continue.
• ...
• Keep the previous $2n-2$ dice on the table, throw $2$ additional dice; is the sum $7n$ then $Y=2n$, else continue.
• ...

I have the feeling that $X$ and $Y$ have (slightly) different distributions, because for $Y$ you know that the first $2n-2$ dice did not sum to zero, while for $X$ that could be the case. However, I don't remember how to approach this problem...

$X$ and $Y$ have different distributions. To see this, consider the following:

$$P[X=2] = P[Y=2] = \frac{6}{36} = \frac{1}{6}$$

For $X$ to equal 4, we first must not hit 7 with the first two dice, and then hit 14 with the four new dice. To throw these last four, we can first throw two new dice and then another two dice. If the first two dice hit 2, we need 12 with the other two; if they hit 3, we need 11 etc. As such, we get:

$$P[X=4] = (1-P[X=2]) \cdot \frac{1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 4 \cdot 4 + 5 \cdot 5 + 6 \cdot 6 + 5 \cdot 5 + 4 \cdot 4 + 3 \cdot 3 + 2 \cdot 2 + 1 \cdot 1}{6^4} = \frac{5}{6} \cdot \frac{146}{1296} \approx 0.09388$$

For $Y$ to equal 4, we need the first two dice to not hit 7, and given this information, two new dice to complete the sum. As such, we get:

$$P[Y=4] = \frac{1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 4 \cdot 4 + 5 \cdot 5 + 5 \cdot 5 + 4 \cdot 4 + 3 \cdot 3 + 2 \cdot 2 + 1 \cdot 1}{6^4} = \frac{110}{1296} \approx 0.08488$$

• Yes, this is the way! I got stuck trying to prove the general case, but it is sufficient and much smarter to just do it for $X=4$. – user193810 Jun 13 '17 at 18:29