You have a biased coin. On any given flip, there is a 2/3 chance it will land on heads and a 1/3 chance it will land on tails. Given four flips of this biased coin, what is the probability of finishing with a 50-50 split (two heads and two tails), regardless of order?

I tried to arrive at the answer more intuitively. Calculating the odds of getting all tails and all heads as:

P(4 tails) = (1/3)^4 = 0.012


P(4 heads) = (2/3)^4 = 0.197

The odds of arriving at an outcome other than 4 tails or 4 heads is 1 - (0.197 + 0.012), or 0.815.

So there is an 81.5% chance the result is either 3 Heads 1 Tails, 1 Heads 3 Tails, or 2 Heads 2 Tails. This is where I got stuck. How can you figure out the remaining probabilities from here?

  • 4
    $\begingroup$ It's good to present some reasoning over the problem even if you can't reach a solution. Otherwise you might get some downvotes over your question. $\endgroup$ – Nikola Mar 9 '18 at 17:09

The answer can be found out using the Binomial Distribution

$$P = {n \choose r}p^r(1-p)^{n-r}$$ where,

  • n is the total no. of trials

  • p is the probability of favourable case

  • r is the number of favourable cases

In this case, $n=4, r=2, p=\frac{2}{3}\ and\ q=\frac{1}{3}$ $$P = {4\choose 2}\left({\frac{2}{3}}\right)^2\left({\frac{1}{3}}\right)^2$$ $$P = 6*\frac{4}{9}*\frac{1}{9}$$ $$P = \frac{8}{27}$$


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