stats help ?

Yes, it is a binomial distribution of 5 trials where p = 0.5 and we want to know the probability of 3 successes (ie r = 3), which we can write as p(3). So you could look that up in a table, or do it from first principles as below, which may be actually easier.

Since the probability of winning is the same as the probability of losing, then any ordered sequence of 5 outcomes is equally likely. Thus there are 2^5 = 32 possible ordered outcomes, all equally likely. And there are 5C3 ways to have a 3 wins in 5 games.

Thus,

p(3) = 5C3 / 32

p(3) = (5 x 4 x 3) / (3! x 32)

p(3) = 0.3125

0.3125

I used the binomial probability distribution function.

P(X=3)= 5C3 * 0.5^3 * 0.5^2= 0.3125