Two Gamblers, Unequal Numbers of Coins

Gambler A flips n+1 fair coins. Gambler B flips n fair coins. What is the probability that A gets strictly more heads than B?

Problem Setup

  • A flips n+1 independent fair coins.
  • B flips n independent fair coins.
  • We want: P(A has more heads than B).

Key Idea

Let HA be A's number of heads and HB be B's number of heads. Then A wins exactly when:
HA > HB
Since B flips n coins, its number of tails is:
TB = n - HB
So
HA > HB  ⟺  HA + TB > n
Now interpret HA as “successes” among A's n+1 coins, and TB as “successes” among B's n coins. Altogether, that gives 2n+1 fair independent trials.
Therefore
X = HA + TB ~ Binomial(2n+1, 1/2)
and A wins exactly when
X > n
Because 2n+1 is odd, the binomial distribution is perfectly symmetric, and there is no middle tie case. So:
P(X > n) = 1/2

Final Answer

1/2
The probability that A gets strictly more heads than B is always 1/2.
A flips
5
B flips
4
Combined trials
9
Quantity Value
A wins if HA > HB
Equivalent condition HA + TB > 4
Random variable X ~ Binomial(9, 1/2)
Desired probability P(X > 4) = 1/2

This result does not depend on n. For every positive integer n, the answer stays 1/2.

Binomial Symmetry Visualization

The bars show the distribution of X = HA + TB, where X ~ Binomial(2n+1, 1/2). Red bars correspond to X ≤ n (A does not win), and green bars correspond to X > n (A wins).

Distribution for X ~ Binomial(9, 1/2)
A does not win: X ≤ n A wins: X > n

One-Line Proof

Let X = HA + TB. Then X ~ Binomial(2n+1, 1/2), and A wins iff X > n. Since 2n+1 is odd, symmetry gives P(X > n) = 1/2.