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?