Solving the Probability Problem of Flipping Coins with Two Players

Coin flipping is a common example in probability theory, used to demonstrate the concept of independent events. When two players, A and B, flip a fair coin independently n times, k times by A and n-k times by B, what is the probability that they flip the same number of heads? This problem involves understanding the binomial distribution, combinatorics, and the theory of independent events. In this blog post, we will provide a step-by-step solution to this problem and offer some tips on how to approach similar probability problems.

Let H be the event that a coin flip results in heads, and let T be the event that a coin flip results in tails. Since A flips the coin k times, the probability that A flips x heads is given by the binomial distribution formula:

P(A flips x heads) = (k choose x) * (1/2)^k * (1/2)^(k-x) = (k choose x) * (1/2)^k

where (k choose x) denotes the number of ways to choose x elements from a set of k elements, which can be calculated as follows:

(k choose x) = k! / (x! * (k-x)!)

Similarly, the probability that B flips y heads, where y = n-k-x, is:

P(B flips y heads) = ((n-k) choose y) * (1/2)^(n-k) * (1/2)^y = ((n-k) choose y) * (1/2)^(n-k)

The probability that A and B flip the same number of heads, denoted by P(A=B), can be calculated by summing the probabilities of all possible outcomes:

P(A=B) = ∑(k choose x) * (1/2)^k * ((n-k) choose y) * (1/2)^(n-k)

where the summation is taken over all values of x from 0 to k. Note that y is uniquely determined by x and is equal to n-k-x.

Now, suppose we want to calculate the probability that there are a total of k heads, regardless of who flips them. This can be done by considering all possible ways in which k heads can be distributed between A and B. Specifically, the probability that there are x heads flipped by A and k-x heads flipped by B is given by:

P(A flips x heads and B flips k-x heads) = (k choose x) * (1/2)^k * ((n-k) choose k-x) * (1/2)^(n-k)

Again, the summation over all values of x from 0 to k gives the probability that there are a total of k heads:

P(total of k heads) = ∑(k choose x) * (1/2)^k * ((n-k) choose k-x) * (1/2)^(n-k)

It can be shown that P(A=B) and P(total of k heads) are equal by equating the two expressions and simplifying using combinatorial identities. Therefore, the probability that A and B flip the same number of heads is equal to the probability that there are a total of k heads.