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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?

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  • $\begingroup$ It should be $\frac{91}{216}$, not $\frac{125}{216}$. $\endgroup$
    – Neat Math
    Commented Dec 1, 2020 at 22:13
  • $\begingroup$ The question asks for the probability that $B$ wins. That is $125/216$. $\endgroup$
    – mjqxxxx
    Commented Dec 1, 2020 at 23:34
  • $\begingroup$ ah, yes. I forgot to subtract it from 1. I'm still curious if there is a direct way of seeing that the answer is 1 - (5/6)^3 $\endgroup$
    – user1488
    Commented Dec 1, 2020 at 23:47
  • $\begingroup$ Read my answer below. It is a coincidence. $\endgroup$
    – Neat Math
    Commented Dec 1, 2020 at 23:57

3 Answers 3

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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.)

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The probability is given by $$ \sum_{k=1}^6P[\max(B_1,B_2)>k\mid A=k]\cdot P[A=k] = \tfrac16\Bigl(\tfrac{35}{36} +\tfrac{34}{36} +\tfrac{27}{36} +\tfrac{20}{36} +\tfrac{11}{36} +\tfrac{0}{36}\Bigr) =\tfrac{127}{216} \approx0.587963. $$

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Three dice are rolled. $A$ wins if his roll is the largest (or tied for largest). There are three cases: (a) all three rolls are different, (b) two rolls are the same and the third is different, and (c) all three rolls are the same. Case (a) can happen in $6\cdot 5\cdot 4$ ways, and $A$ wins with probability $1/3$ in that case. Case (c) can happen in $6$ ways, and $A$ definitely wins in that case. Finally, case (b) can happen in $6\cdot 5\cdot 3$ ways... in half of these, the pair is larger than the singleton, and $A$ wins with probability $2/3$; in the rest, the pair is smaller than the singleton, and $A$ wins with probability $1/3$... so overall, $A$ wins with probability $1/2$ in case (b).

The result is

$$ P_A=\frac{(6\cdot 5\cdot 4) \cdot \frac{1}{3}+(6\cdot 5\cdot 3)\cdot\frac{1}{2}+6}{6^3}=\frac{40+45+6}{6^3}=\frac{91}{216}. $$

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