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Throwing a Hundred Sided Dice
updated 1 year, 2 months ago
3 Members
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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 -
What type of assumption do we have to make over here?
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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.
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(100/101)^100 approximates to 1/e
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Is this right? used recurrence relation to get this.
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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:)
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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!
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Another interesting question with the interesting answer as well! What is the expected number of throws till somebody wins?
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