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: Intuitive Explanation

Here’s one way to see why:

  • Let’s fix the outcomes of B’s \(n\) flips.
  • A’s first \(n\) flips can match that exactly—and the extra coin then acts as a tiebreaker.

When A and B tie on the first \(n\) flips:

  • A wins half the time (when the extra coin is heads).
  • So overall, A has exactly a 50% chance of ending up with more heads.

This matches formal derivations using summations and generating functions as well.

Final Answer

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

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

Reference