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Alternately a fair coin is tossed and a fair die is thrown, beginning with the coin. What is the probability that the coin will register a "head" before the die registers a "5" or "6"?

The answer given in my textbook is $\frac{3}{4}$. I can't come up with it.

My attemp is as follows. Since both the coin and the die are fair each new toss and thrown is independent from the previous ones, so we shall think just about the single round of the game. What is the probability that the game will stop with the die "wins" over the coin? It is $\frac{1}{2}\cdot\frac{2}{6} = \frac{1}{6}$ since the coin must return a "tail" from two possible outcomes and the die "5" or "6" from six possible outcomes. What is the probability that the coin will "win" over a die? It is $\frac{1}{2}$, since the coin is tossed first and it sufficies just thrown a winning "head" that happens with $\frac{1}{2}$ probability. What is the probability that the game will turn into the next round? It happens if the coin registers a "tail" and die comes up with one of the values in a range from $1$ to $4$, so the probability of this event is $\frac{1}{2}\cdot\frac{4}{6}=\frac{1}{3}$. But as you can see $\frac{1}{2}+\frac{1}{3}+\frac{1}{6} \neq 1$. What am I missing and how to obtain original answer?

UPDATE: Surely $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, it was a stupid mistake in my quick verbal counting, but I still can't understand how $\frac{3}{4}$ and not $\frac{1}{2}$ may be derived as an answer. May there be a mistake in given answer?

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    $\begingroup$ Umm, $\frac12 + \frac13 + \frac16 = 1$... $\endgroup$ – Blaza Aug 27 '17 at 17:28
  • $\begingroup$ @Blaza, thanks, I uncounted here, but still can't understand why $\frac{3}{4}$ and not $\frac{1}{2}$ is an answer. $\endgroup$ – Hasek Aug 27 '17 at 17:34
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It's the coin versus the die . . .

Let $p\;$be the probability that the coin wins.$\;$Then $$p=\frac{1}{2}+\left(\frac{1}{2}\right)\left(\frac{4}{6}\right)p$$ Explanation:

If the coin comes up heads on the first flip (probability $1/2$), the coin wins right away.

If not (probability $1/2$), the coin wins if and only if the die comes up less than $5$ (probability $4/6$), and the coin wins from that point on (probability $p$).

Solving the equation for$\;p\;$yields$\;p=3/4$.

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