# A probability question (outcomes)

John has $\ 6$ coins. He performs a test which involves tossing all coins simultaneously. He repeats this experiment until all coins shows up as heads. What is the probability that it takes him at least $\ 12$ tosses to achieve this feat?

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Until all coin shows up heads at least one or simultaneaously? –  Jean-Sébastien Dec 6 '12 at 16:26
Hint: it requires at least 12 repetitions of the experiment if and only if each of the first eleven experiments resulted in at least one of the coins showing a tail. –  David Mitra Dec 6 '12 at 16:30
Hint: Think of the geometric distribution. –  Patrick Dec 6 '12 at 17:52

The probability to fail at any toss is $p=1-\frac1{2^6}$ (why?). The probability of failing $11$ times in a row is $p^{11}$.

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The probability of having all $6$ coins come up heads simultaneously on any given throw is $(1/2)^6 = 1/64$

The Geometric Distribution is used to determine the number of Bernoulli trials needed to get $1$ success. What we are seeking here is the probability that a series of Bernoulli trials with probability $1/64$ takes longer than $12$ trials to achieve a success.

The cumulative distribution function for the Geometric distribution is $1 - (1-P)^k$, where $P$ is the probability of success and $k$ is the number of trials. The table below is computed by substituting $1/64$ for $P$ and the value in the first column for $k$.

$k$   $P(n\le k)$
$1$   $0.0156$
$2$   $0.031$
$3$   $0.046$
$4$   $0.061$
$5$   $0.075$
$6$   $0.090$
$7$   $0.104$
$8$   $0.118$
$9$   $0.132$
$10$   $0.145$
$11$   $0.159$

The bottom line tells us that there is approximately a $15.9\%$ chance that we will have a success by the $12$th trial. Therefore there is approximately and $84.1\%$ chance that it will take at least $12$ tries before getting all $6$ coins to show heads.

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