What Is the Winning Probability for Player One in an N-Sided Die Game?

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    2017
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SUMMARY

The probability of Player One winning in an N-sided die game, where players alternate rolling the die and must roll higher than the previous roll, is calculated using combinatorial probability. Player One has a distinct advantage by rolling first, as they set the initial threshold for Player Two. The winning probability can be expressed mathematically as P(Player One wins) = 1 - (1/N)^(N-1), where N represents the number of sides on the die. This formula accounts for the scenarios where Player Two fails to roll higher than Player One's initial roll.

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anemone
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Here is this week's POTW:

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Two people take turns rolling an N-sided die. The object of the game is to roll a higher number than the previous roll. Failure to roll a higher number results in losing the game. Assuming player one rolls first, find the probability that player one wins the game.

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No one answered last week's problem.(Sadface)

You can find the suggested solution as follows:

The probability that player one loses is

$P_L=0$ for $N=1$

$P_L=\dfrac{1}{4}$ for $N=2$

$P_L=\dfrac{8}{27}$ for $N=3$

$P_L=\dfrac{81}{256}$ for $N=4$

We can see the pattern where $P_L=\left(1-\dfrac{1}{n}\right)^n$ for $N=n$.

Therefore, the probability that player one wins $=P_W=1-P_L=1-\left(1-\dfrac{1}{n}\right)^n$.
 

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