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Suppose I have $6$ fair coins, all facing heads ups. For each turn I will flip each of the $6$ coins. If a coin lands tails up, I set it aside. The next turn I take the remaining coins facing heads up and repeat the process. Now I want to know what is the expected number of turns $T$ until I have exactly $6$ tails. I know $$P(T=1)=\frac{1}{2^6}$$ But how do I find $P(T=x)$ with $x\gt 1$? For $T=2$ it seems I should count the different ways this might happen i.e. $1$ tail on turn $1$, $5$ tails on turn $2$, etc. There are $5$ different ways, so is the probability $5/64$? That does not seem right to me. Any hint is appreciated.

meiji163
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2 Answers2

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See this question: Expected days to finish a box of cookies. Practically cookies are your coins, and eating a cookie is equivalent to flipping a tail. And you have 6 coins, instead of 10.

Community
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Momo
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Let's follow your way (find $P(T=x)$ for every $x > 0$).

For $x=2$ you are right about the general method (find all the possible ways this can happen), but you are not getting the right result. The main flaw is that you are assuming that you flip only 6 coins (hence the 64 on your denominator), where in reality we are flipping anywhere between 6 to 12 coins for the two turns.

Here's what we want instead. We want to find the probability to have $k$ tails in the first turn i,e., $P(tails=k | \space flips = 6)$ and then multiply it with the probability to get all the remaining coins flipped as tails in the second turn, i.e. $P(tails=6-k | \space flips = 6-k)$. We add these for all possible $k$. Notice that $k$ should take values $0,1,\dots,5$, if $k=6$ this means we are done in the first turn.

$$P(T=2) = \sum_{k=0}^{5} P(tails=k | \space flips = 6) \cdot P(tails=6-k | \space flips = 6-k) = \sum_{k=0}^{5} {6 \choose k}\cdot \frac{1}{2^6}\cdot \frac{1}{2^{6-k}} = \sum_{k=0}^{5} {6 \choose k}\cdot \frac{1}{2^{12-k}} = \frac{665}{4096} \approx 0.162$$

What would $P(T=3)$ be? It gets more complicated following this method. Have a try yourself and I will post my findings afterwards.

Thanassis
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