brain-teaser - probability of maximum payoff for 3 rolls of a dice

Question

You allow three throws of a fair dice. The payoff is the maximum digit from all 3 throws. That is, if the throws are <1, 4, 3>, the payoff is 4.

I have been given the probability for each payoff $= [m^3 - (m-1)^3]/216$ Is that correct, and how would one work it out?

My simple estimation is,

1. for the case of payoff = 6,
2. the probability = (1/6) + (5/6) * 1/6 + (5/6)^2 + 1/6

I think my (2) is fine... how would I work it out?

Kind regards

Answer 1

That is correct.

$m^3$ is the number of ways to have all values $\le m$.

$(m-1)^3$ is the number of ways to have all values $\le m-1$.

Hence $m^3-(m-1)^3$ is the number of ways to have all values $\le m$ but not $\le m-1$, in other words, one of the numbers is $m$. Then we divide by the total number of possibilities ($6^3=216$) to find the probability.