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Say I roll two twenty-sided dice and always take the higher number rolled, and dispose of the lower number. What are the percent chances of rolling each number on the die? I tried the equations in this question: What is the average of rolling two dice and only taking the value of the higher dice roll? But for some reason I couldn't get the correct answer.

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Tobias Fizzlewig
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  • The die are numbered ${1,\dots,20}$? So there are $400$ equally likely pairs. A bit hard for direct enumeration, but still. The only ways to get $k$ as the maximum are to toss ${k,j}$ or ${j,k}$ with $j<k$, or to throw ${k,k}$. Not that hard to count! – lulu Dec 10 '15 at 21:13
  • How many times are you planning to roll the dice? For example, if you roll them infinitely many times, then the probability is $100%$. – barak manos Dec 10 '15 at 21:14
  • @barakmanos I want to know the percentages for getting each number in one roll. – Tobias Fizzlewig Dec 10 '15 at 21:16
  • There are $20$ numbers. How can you possibly get all of them in $1$ roll of $2$ dice????? – barak manos Dec 10 '15 at 21:21
  • @barakmanos I could have been clearer there. I meant I want to know the percent chances for any given number in 1-20 in one roll. – Tobias Fizzlewig Dec 10 '15 at 21:26

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There is $1$ way to get a $1$, namely $(1,1)$

There are $3$ ways to get a $2$, namely $(1,2),(2,2),(2,1)$

There are $5$ ways to get a $3$, namely $(1,3),(2,3),(3,3),(3,2),(3,1)$

In general there are $2k-1$ ways to get a $k$.

The basic percentage unit is $0.25\%$, so the percentage chance to throw a $k$ is $(\frac{2k-1}{4})\%$, e.g. for $k=5$ the percentage is $1.25\%$.

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