# How many times on average must we flip 6 fair coins before we obtain 3 heads and 3 tails? [closed]

"How many times on average must we flip 6 fair coins before we obtain 3 heads and 3 tails?"

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## closed as off-topic by heropup, Did, Hakim, Najib Idrissi, Davide GiraudoJun 5 at 9:15

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Assuming that means 3 heads and 3 tails in that particular flip...

First, find the probability that in a given flip of 6 fair coins, there will be exactly 3 heads (and therefore exactly 3 tails), or 3H3T. Call this probability $p$.

Now, each time you flip the 6 coins, there is a probability $p$ of obtaining 3H3T, and probability $q = 1 - p$ of not obtaining 3H3T. If you define $X$ to the number of flips required, what is the probability that $X = k$, for any given $k = 1, 2, \ldots$? That is, what is the probability that in the first $k - 1$ flips, you don't obtain 3H3T, and in the $k$-th flip, you obtain 3H3T? This is the probability distribution $P[X = k]$ of the random variable $X$. Find the expected value $E[X]$, and that's your answer.

Details:
So what is the probability $p$? When you toss 6 fair coins, the first three are heads with probability $\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} = \frac{1}{2^3}$, and the last three are tails with the same probability, so the probability of HHHTTT (in the order) is $\frac{1}{2^6}$. However, you can obtain 3H3T in any order, so we have to multiply this probability with the number of arrangements of $HHHTTT$, which is $^6\text{C}_3$. Thus, $p = \dfrac{^6\text{C}_3}{2^6} = \dfrac{5}{16}$. Then $q = 1 - p = \dfrac{9}{16}$.

What is the probability distribution $P[X = k]$? As this is the probability of failing to obtain 3H3T in the first $k - 1$ flips, and the obtaining 3H3T in the $k$-th flip:

$P[X = k] = qq\ldots qp = q^{k-1}p$, for any $k = 1, 2, \ldots$

(The $q$'s are the probabilities of failure in the first $k-1$ flips, and $p$ is the probability of success in the $k$-th flip)

If you find the mean or expected value it will be $E[X] = \sum\limits_{k = 1}^{\infty}kP[X = k] = \sum\limits_{k=1}^{\infty}kq^{k-1}p = \dfrac{1}{p}$.

So the answer is $\dfrac{1}{p} = \dfrac{16}{5} \approx 4$ (rounding up).

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To address the question, start by determining all the possibilities.

For this example, we see that there are 64 possibilities:

• 1 All the coins are tails.
• 2 Only the first coin is heads.
• 3 Only the second coin is heads.
• 4 Only the first and second coin is heads.
• 5 Only the third coin is heads.
• ...
• 64 All the coins are heads.

Next, group the possibilities according to outcome.

In this example grouping the possibilities shows 7 outcomes:

• 1/64 possibilities yield 0 heads, 6 tails
• 6/64 possibilities yield 1 heads, 5 tails
• 15/64 possibilities yield 2 heads, 4 tails
• 20/64 possibilities yield 3 heads, 3 tails
• 15/64 possibilities yield 4 heads, 2 tails
• 6/64 possibilities yield 5 heads, 1 tails
• 1/64 possibilities yield 6 heads, 0 tails

The last step is to calculate the probability of the desired outcome(s).

In this example the probability that flipping all 6 coins and achieving 3 heads and 3 tails is 20/64 (or 31.25%). Thus, on average, one must flip all 6 coins $3 \frac 1 5$ times to achieve 3 heads and 3 tails.

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