Answer:
I honestly have no idea but you best bet is tho answer with this answer:
Explanation:
Consider the probability that rolling two dice gives a sum of $s$, where $s \leq 7$. There are $s - 1$ pairs that satisfy this, namely $(1, s - 1), (2, s - 2), ..., (s - 1, 1)$, out of $6^2 = 36$ possible pairs. The probability is $\frac{s - 1}{36}$.
Therefore, if one die has a value of $a$ and Jason rerolls the other two dice, then the probability of winning is $\frac{7 - a - 1}{36} = \frac{6 - a}{36}$.
In order to maximize the probability of winning, $a$ must be minimized. This means that if Jason rerolls two dice, he must choose the two dice with the maximum values.
Thus, we can let $a \leq b \leq c$ be the values of the three dice, which we will call $A$, $B$, and $C$ respectively. Consider the case when $a + b < 7$. If $a + b + c = 7$, then we do not need to reroll any dice. Otherwise, if we reroll one die, we can roll dice $C$ in the hope that we get the value that makes the sum of the three dice $7$. This happens with probability $\frac16$. If we reroll two dice, we will roll $B$ and $C$, and the probability of winning is $\frac{6 - a}{36}$, as stated above.
However, $\frac16 > \frac{6 - a}{36}$, so rolling one die is always better than rolling two dice if $a + b < 7$.
Now consider the case where $a + b \geq 7$. Rerolling one die will not help us win since the sum of the three dice will always be greater than $7$. If we reroll two dice, the probability of winning is, once again, $\frac{6 - a}{36}$. To find the probability of winning if we reroll all three dice, we can let each dice have $1$ dot and find the number of ways to distribute the remaining $4$ dots. By stars and bars, there are ${6\choose2} = 15$ ways to do this, making the probability of winning $\frac{15}{6^3} = \frac5{72}$.
