This puzzle is a classic twist on probability:

Two gamblers, A and B, flip coins:

  • Gambler A flips \(n+1\) fair coins.
  • Gambler B flips \(n\) fair coins.

What is the probability that A ends up with strictly more heads than B?

Step 1: Understand the Setup

Let \( H_A \) and \( H_B \) be the number of heads obtained by A and B, respectively.

We’re interested in computing:

\[ P(H_A > H_B) \]

Given that all coin flips are independent and fair (i.e., probability of heads = 0.5), we want to find this probability exactly.

Step 2: Symmetry Insight

Let’s define a fair framework.

Think of B flipping all their coins first. Then A flips their \(n\) coins plus one extra.

There is a deep symmetry hidden in this setup.

Step 3: Core Result

A beautiful and perhaps surprising fact:

The probability that A gets strictly more heads than B is exactly 0.5.

This is true for all values of \(n\).

Step 4: The Symmetry Argument

Consider the first \(n\) coins flipped by both A and B. Let \(H_A(n)\) be A’s heads in these first \(n\) flips, and \(H_B\) be B’s heads. There are three mutually exclusive scenarios:

  1. \(H_A(n) > H_B\): A already has strictly more heads. A has won, regardless of what happens on the extra \((n+1)\)-th flip.
  2. \(H_A(n) < H_B\): A has fewer heads. Even if A flips a head on the extra flip, A can at best tie. A has lost.
  3. \(H_A(n) = H_B\): They are tied. A’s extra flip now acts as the tiebreaker!

Because A and B flip the exact same number of coins in this first phase, scenarios 1 and 2 are perfectly symmetric. This means \(P(H_A(n) > H_B) = P(H_A(n) < H_B)\).

Let \(p_{tie}\) be the probability they tie. Then the probability A wins outright in the first phase is exactly half of the non-tie probability: \(\frac{1 - p_{tie}}{2}\).

If they tie, A wins if their final extra coin lands heads (which happens half the time): \(\frac{p_{tie}}{2}\).

Adding A’s winning probabilities together yields the exact answer: \[ P(\text{A wins}) = \frac{1 - p_{tie}}{2} + \frac{p_{tie}}{2} = \frac{1}{2} - \frac{p_{tie}}{2} + \frac{p_{tie}}{2} = \frac{1}{2} \]

Final Answer

The probability that Gambler A ends up with strictly more heads than Gambler B is:

\[ \boxed{\frac{1}{2}} \]

Reference