One extra coin

Two gamblers sit across from each other. Player A has 6 fair coins. Player B has 5. They both flip all their coins at once. A wins if she gets strictly more heads than B.

How likely is that?

What do you think?

Player A has n+1 = 6 coins, Player B has n = 5. What's the probability A gets more heads?

Try it. Watch the rate converge:

Coin Toss Game

n = 5A: 6 coins, B: 5 coins

Drag the slider to change $n$ . Whether B has 1 coin or 12, A's win rate stubbornly sits at around 50%.

The symmetry trick

Computing $P(\text{A gets more heads})$ directly requires summing over all possible head counts for both players. With $n+1$ and $n$ coins, that's a double sum of binomial coefficients — messy.

There's a faster approach. Set aside A's last coin. Now both players have exactly $n$ coins each. Compare their head counts. Three things can happen:

The Three Events

After both players flip their first $n$ coins:

$E_1$ : A has more heads than B

$E_2$ : A has the same number of heads as B

$E_3$ : A has fewer heads than B

By symmetry, $P(E_1) = P(E_3)$ . Both players have $n$ fair coins, so "A beats B" and "B beats A" are equally likely. Call this probability $x$ , and let $y = P(E_2)$ .

Since the three events cover everything: $2x + y = 1$ .

What does the extra coin do?

P(E_1) = P(E_3) = x, \quad P(E_2) = y, \quad 2x + y = 1

Symmetry gives us that A-beats-B and B-beats-A are equally likely with n coins each.

Step 1 of 5

The proof never computes $x$ or $y$ . It doesn't need to. The structure of the problem forces the answer to be 1/2 regardless of those values. Symmetry arguments let you skip the computation entirely.

Why this matters for interviews

This problem appears in nearly every quant finance interview prep book. Interviewers use it to test whether candidates reach for brute-force computation or pause to find structural shortcuts.

The lesson generalizes. Whenever a problem has a natural symmetry, exploiting it can collapse pages of algebra into a few lines.

What do you think?

What if A has n+2 coins and B has n coins? Is the probability still 50%?

A has 101 fair coins, B has 100. P(A gets strictly more heads)? (decimal like 0.5)

1/3

The takeaway

When you see asymmetry in a problem (one player has more coins, one team is bigger), ask: "What if I remove the asymmetry?" Stripping down to the symmetric core often reveals a clean decomposition into cases. The algebra that looked scary cancels itself out. This technique shows up in independence arguments, conditional probability, and throughout competition mathematics.