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Players A and B take turns in answering trivia questions, starting with player A answering the first question. Each time A answers a question, she has probability $p_1$ of getting it right. Each time B plays, he has probability $p_2$ of getting it right.

Suppose that the first player to answer correctly wins the game (with no predetermined maximum number of questions that can be asked). Find the probability that A wins the game.

Can someone post step-by-step solution? This is a question from Joe Blitzstein's Stat 110 homework question

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  • $\begingroup$ $\frac{p_1}{p_1 + p_2}$ i guess $\endgroup$ – user76568 Jul 21 '17 at 12:31
  • $\begingroup$ @Dror Can you explain it step-by-step? $\endgroup$ – NeoButFemale Jul 21 '17 at 12:32
  • $\begingroup$ @Dror: that does not take account of A answering first. Try $\frac{d_1}{d_1+d_2-d_1d_2}$ $\endgroup$ – Henry Jul 21 '17 at 12:53
  • $\begingroup$ How'd you get that? $\endgroup$ – user76568 Jul 21 '17 at 13:05
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    $\begingroup$ @Dror There's precious little "nuance" in regard to which formula is correct. For example, when $p_1=p_2=\frac23,$ your formula says A has altogether a $\frac12$ probability to win. But $p_1=\frac23$ means A has a $\frac23$ probability to win on the very first guess, so altogether A's probability to win must be at least that great. $\endgroup$ – David K Jul 21 '17 at 16:44
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Solution #$1$:

Let $p$ be the probability that $A$ wins the game.

Then $$p = p_1 + (1-p_1)(1-p_2)p$$ Solve for $p$.

Explanation:

Either $A$ wins on the first question (probability $p_1$), or if not (probability $(1-p_1)$), then if $B$ doesn't win on his first question (probability $(1-p_2))$, player $A$ is effectively at the start of a brand new game, hence the probability of winning from that point is exactly $p$.

Solution #$2$:

$A$ wins on the first question with probability $p_1$.

$A$ wins on $A's$ second question with probability $(1-p_1)(1-p_2)p_1$.

$A$ wins on $A's$ third question with probability $(1-p_1)^2(1-p_2)^2p_1$.

. . .

Thus, the probability that $A$ wins is $$\sum_{n=0}^\infty(r^n)p_1$$ where $r=(1-p_1)(1-p_2)$.

It's the sum of a geometric series. Can you finish it?

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It can be solved considering just the first two rounds.

Odds in favor of $A = p_1 : (1-p_1)p_2$

Convert odds to probability, $Pr = \dfrac{p_1}{p_1+(1-p_1)p_2}$

$\underline{Added\; explanation}$

Probabilities for every subsequent two rounds will just get multiplied by some factor,
which means that the odds in favor (which is a ratio) remains unchanged

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