# Probability question involving three dice

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?

• It should be $\frac{91}{216}$, not $\frac{125}{216}$. Commented Dec 1, 2020 at 22:13
• The question asks for the probability that $B$ wins. That is $125/216$. Commented Dec 1, 2020 at 23:34
• 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 Commented Dec 1, 2020 at 23:47
• Read my answer below. It is a coincidence. Commented Dec 1, 2020 at 23:57

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.)

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

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

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