Question

Suppose we toss a fair coin. What is the probability that we obtain an equal number of heads and tails for the first time after 2n tosses?

Answer 1

The probability of obtaining an equal number of heads and tails for the first time after 2n tosses is
`1/√πn`.

Step-by-step explanation:

Consider the scenario where we want to achieve an equal number of heads (H) and tails (T) after 2n coin tosses. Let P(n) be the probability of reaching this outcome. To determine P(n), we need to count the number of valid sequences that result in an equal number of heads and tails.

The valid sequences can be represented by the Catalan numbers, which are given by
`C(n) = (2n)! / [(n + 1)! * n!]`. Each valid sequence corresponds to a combination of H and T in a specific order. However, not all sequences are valid because they must maintain an equal number of H and T at each step.

To find P(n), divide the number of valid sequences by the total number of possible sequences, which is
`2^(2n)`. Therefore,
`P(n) = C(n) / 2^(2n).`

Simplifying this expression, we get
`P(n) = (2n choose n) / 4^n`. The probability of obtaining an equal number of heads and tails for the first time after 2n tosses is given by
`P(n) = 1/√πn`. This result is derived using the Central Limit Theorem and Stirling's approximation. The square root of n emerges as a significant factor in the denominator, highlighting the probabilistic nature of achieving balance in a fair coin toss experiment.