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Adam, Bertrand, and Carissa toss a coin in sequence until one person wins by tossing the first head.

If the coin is fair, find the probability that Adam wins.

Can somebody tell me if I'm on the right track?

Adam can win on the first round if he rolls a head. Or he can win on the second round if all three of them roll tails and then he rolls a head. Or he can win on the third round if they roll tails six times and then he rolls a head.

P(Adam wins) $= .5 + .5^4 + .5^7 + \dots + .5^{3n-2}$ where $n=$ number of rounds.

Thoughts? Should I be getting a numeric answer here...?

pocketlizard
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  • This is a geometric series and so has a numeric answer. http://mathworld.wolfram.com/GeometricSeries.html – E W H Lee Sep 16 '14 at 17:34

4 Answers4

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Another approach to get the same answer: If the probability that Adam wins is $p$, you can say $p=\frac 12 + \frac 18 p$, where the $\frac 12$ comes from winning immediately, and the $\frac 18p$ comes from the $\frac 18$ chance they all throw tails and we are back in the same condition as the start, so Adam must have $p$ chance of winning from here. This gives the same answer as Adriano.

Ross Millikan
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Good reasoning; you're almost there! Since Adam could win on round 1 OR round 2 OR any round really, we want to sum up an infinite series and take the limit as $n \to \infty$: $$ 0.5 + 0.5^4 + 0.5^7 + \cdots = \sum_{n=1}^\infty 0.5^{3n-2} = \sum_{n=1}^\infty 0.5^10.5^{3n-3} = 0.5\sum_{n=1}^\infty ((0.5)^3)^{n-1} = \frac{0.5}{1 - 0.5^3} $$

Adriano
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A variation on Ross Millikan’s answer: If the probability that Adam wins is $p$, then the probability that Bertrand wins is $\frac 12 p$, because, if Adam throws a tails, then Bertrand is, essentially, starting a new game.  Likewise, the probability that Carissa wins is $\frac 14 p$.  But somebody has to win, so $$p + \frac 12 p + \frac 14 p = 1$$

Scott - Слава Україні
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4

You are absolutely right. The probability forms an infinite series:

$$0.5 + 0.5^4 + 0.5^7 + 0.5^{10}+ 0.5^{13}+\cdots$$

This series is a Geometric Series. The first term, $a$, is $0.5$ and the common ratio, $r$, is $0.5^3$.

Since $|r|<1$, the infinite sum converges to a sensible answer. Any geometric series $$a + ar + ar^2 + ar^3 + ar^4 + \cdots$$ with $|r|<1$ has the infinite sum $$S_{\infty} = \frac{a}{1-r}$$

In your case $a=0.5=\frac{1}{2}$ and $r=0.5^3 = \frac{1}{8}$. Putting these into the formula:

$$S_{\infty} = \frac{1/2}{1-1/8} = \frac{1/2}{7/8} = \frac{4}{7}$$

Fly by Night
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