Probability of More Heads Than Tails With Odd Flips
You flip a fair coin $2n + 1$ times (an odd number). What is the probability of getting more heads than tails?
Hints
- With an odd number of flips, can the number of heads ever equal the number of tails?
- If ties are impossible, how do "more heads" and "more tails" partition the sample space?
- Use the symmetry $P(H = k) = P(H = 2n+1-k)$ for a fair coin to relate the two events.
Worked Solution
How to Think About It: With an odd number of flips, you can never tie -- the number of heads and tails must differ by at least 1. So the only two outcomes are "more heads" and "more tails." Since the coin is fair, by symmetry these two events have the same probability. Two equally likely events that partition the entire sample space each have probability $1/2$. That is the whole argument -- no computation needed.
Quick Estimate: Take $n = 1$ (3 flips). Outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. More heads (at least 2): HHH, HHT, HTH, THH -- that is 4 out of 8 = $1/2$. Checks out.
Approach: Symmetry argument on the binomial distribution.
Formal Solution:
Let $H$ denote the number of heads in $2n+1$ flips. We have $H \sim \text{Binomial}(2n+1, 1/2)$.
"More heads than tails" means $H \geq n+1$ (since there are $2n+1$ flips total, more heads requires at least $n+1$).
"More tails than heads" means $H \leq n$.
Since $2n+1$ is odd, $H$ cannot equal $(2n+1)/2$, so ties are impossible. Therefore:
$$P(H \geq n+1) + P(H \leq n) = 1$$
By symmetry of the fair coin, $P(H = k) = \binom{2n+1}{k}(1/2)^{2n+1} = \binom{2n+1}{2n+1-k}(1/2)^{2n+1} = P(H = 2n+1-k)$.
This maps $H \geq n+1$ bijectively to $H \leq n$, so:
$$P(H \geq n+1) = P(H \leq n)$$
Combining with the partition:
$$P(H \geq n+1) = \frac{1}{2}$$
Answer: The probability of getting more heads than tails in $2n+1$ fair coin flips is exactly $\dfrac{1}{2}$.
Intuition
This problem is really about recognizing when symmetry gives you the answer for free. The fair coin makes heads and tails interchangeable -- swapping every H with T in any sequence gives an equally likely sequence, but flips the outcome from "more heads" to "more tails" and vice versa. Since ties are impossible with an odd number of flips, these two events are complementary and equally likely, so each has probability exactly $1/2$.
The deeper lesson for interviews: always check for symmetry before computing. If you start writing out binomial sums, you are working too hard. This principle extends broadly -- many probability puzzles with fair coins, symmetric distributions, or exchangeable random variables have elegant symmetry-based solutions. The interviewer wants to see if you spot the shortcut.