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• # Throwing a Hundred Sided Dice

• ### Puzzle Prime

October 16, 2018 at 11:35 pm

Two people are taking turns throwing a hundred sided die. The game continues until someone throws a number larger than the number thrown by the opponent on the previous turn. What is the probability that the person who throws first wins the game?

Example: 4 – 3 – 3 – 2 – 5 => First player wins
Example: 6 – 3 – 2 – 1 – 1 – 4 => Second player wins

• ### Josefina

Member
November 5, 2018 at 6:58 am

What type of assumption do we have to make over here?

• ### Puzzle Prime

November 5, 2018 at 1:58 pm

Assume that the die is fair, i.e. there is 1% chance that it lands on any of its sides. Also, the outcome of every throw is independent of the previous outcomes.

• ### Vivek

Member
October 21, 2020 at 3:52 am

(100/101)^100 approximates to 1/e

• ### Vivek

Member
October 21, 2020 at 1:09 pm

Is this right? used recurrence relation to get this.

• ### Puzzle Prime

October 21, 2020 at 3:18 pm

Yeah, it seems right:) I don’t remember the original solution, but if we set F(n) = P(win after side n), we get:

F(1) + F(2) + … + 101F(m) = 100

for m = 1, 2, … , 100. Then, it is easy to check that F(m) = (100/101)ᵐ, and therefore F(100) = (100/101)¹⁰⁰.

I feel there is some more elegant, combinatorial solution though, it is worth thinking about it:)

• ### Vivek

Member
October 22, 2020 at 1:57 am

It really is (n/(n+1))^n for n sided dice, approximating to 1/e as n increases. Confirmed it by simulating in R program. This is the nice way to see the e value with probabilistic simulation. I remember that the probability of derangement approaches 1/e. But This one is really super cool!

• ### Vivek

Member
October 22, 2020 at 2:42 am

Another interesting question with the interesting answer as well! What is the expected number of throws till somebody wins?

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