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A fair die is alternately thrown by two persons . The first one wins if one dot appears and the second one wins if 2-dots or 3-dots appear. The first one starts throwing the die. What is the probability that the first one ultimately wins ?

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3 Answers 3

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if the probability of player 1 winning is P, then you can note that if both players fail to win on their first roll, then player 1 has returned to an identical position with a probability of winning of P. The chances of each player failing on round one is ($\frac{5}{6})(\frac{4}{6}) = \frac{5}{9}$

and the chance of player 1 winning on roll 1 is $\frac{1}{6}$

P = prob(winning first roll) + P x prob(game goes to second round)

$P = \frac{1}{6} + \frac{5}{9} P$

$P = \frac{3}{8}$

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HINT

Denote by $W$ (or $L$) the event that the first one wins (or loses). $W$ consists of some possibilities:

  • in one rolls is $1/6$
  • in 3 rolls is $5/6$ for rolling not 1, $4/6$ for the second rolling not 2 nor 3, and $1/6$ for rolling 1 on the 3rd roll
  • all subsequent ones just add extra factors of $5/6$ and $4/6$ for another iteration.

Now clear pattern emerges: $$ \mathbb{P}[W] = \frac16 + \frac 56\frac46\frac16 + \left(\frac 56\frac46\right)^2\frac16 + \ldots $$ Can you sum the geometric series?

You should also compute $\mathbb{P}[L]$ that way and check $\mathbb{P}[W]+\mathbb{P}[L]=1$...

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Let $p$ denote the probability that first one wins and let $q$ denote the probability that the first one wins if one dot does not appear by his first throw.

Then $p=\frac16+\frac56q$ and $q=\frac46p$ (do you see why?) so that $p=\frac16+\frac56\frac46p$ leading to $p=\frac38$.

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