Player $A$ rolls one die. Player $B$ rolls two dice. If $A$ rolls a number greater or equal to the largest number rolled by $B$, then $A$ wins, otherwise $B$ wins. What is the probability that $B$ wins?
I calculated the probability to be $$\frac 1 {36} + \frac 3 {36} \frac 5 6 + \frac 5 {36} \frac 4 6 + \frac 7 {36} \frac 3 6 + \frac 9 {36} \frac 2 6 + \frac {11} {36} \frac 1 6 = 1 -\frac {125} {216}$$
I then noticed that this is the same as $(5/6)^3$, and assumed that there might be a quicker way of obtaining this directly. Is this the case?
Best Answer
Update: Sorry I didn't read carefully. You want the probability that $B$ wins, so it is indeed $\frac{125}{216}$, even though your equation gives $\frac{91}{216}$. Except for that, my analysis is correct. And yes you get your answer faster via $1-\frac{\sum_{j=1}^6 j^2}{216}=\frac{125}{216}$, but not from $\left(\frac 56\right)^3$.
Not sure how you arrived at $125$ since your equation gives $91$.
But, interestingly, $216-91=125$. Is it just a coincidence?
If Player $A$ throws a $j$, then there are $j^2$ cases that Player $B$ throws twice less than or equal to that, therefore the probability you want is $\frac{\sum_j j^2}{6^3} = \frac{91}{216}$, and you happen to have $6^3-\sum_{j=1}^6 j^2=125$. If you change 6 to some other integers you don't always get a perfect cube.
$3^3-\sum_{j=1}^3 j^2=13;$
$4^3-\sum_{j=1}^4 j^2=34;$
$5^3-\sum_{j=1}^3 j^2=70$, etc.
In general $n^3-\frac 16 n(n+1)(2n+1)=\frac 16 n(4n+1)(n-1)$ and you are lucky to have $4\cdot 6+1=25=(6-1)^2$.
(BTW if $n=14$ you get 1729, the taxicab number.)