Unfair Heads and Tails

$1/2$1/2.

Let’s separate A’s coins into two groups: the first $n$n coins and the extra coin.

When Player A and B each flip $n$n coins, three outcomes are possible: player A gets more heads, B gets more heads, or they tie. By symmetry, player A and B have equal probability of getting more heads from their $n$n coins. Let’s call this probability $p$p.

Case 1: Player A already has more heads from the first $n$n coins. In which case, player A wins no matter what the $(n+1{)}^{\text{th}}$(n+1)th coin lands on. This case happens at $p$p chance, with player A always winning.

Case 2: Player B has more heads from the first $n$n coins. In which case, player B wins: even if player A’s last coin lands on a heads, at best they will have the same number of coins. But player A needs MORE coins to win! This case happens at $p$p chance, with player A always losing.

Case 3: Player A and B have the same number of coins from the first $n$n flips. Then player A has a 50% chance of flipping heads on the $(n+1{)}^{\text{th}}$(n+1)th coin to win, and they lose otherwise. This case happens at $1-2p$1−2p chance, and player A has a 50% chance of winning in this case.

The probability that A wins overall is:

• A wins from the first $n$n coins: $p$p
• Plus A ties in the first $n$n coins and the extra coin is heads: $(1-2p)/2$(1−2p)/2

This makes for a total of $p+\frac{(1-2p)}{2}=\frac{1}{2}$p+2(1−2p)​=21​.

Player A wins with probability $1/2$1/2, regardless of the value of $n$n.

Here’s another way to see it: since player A has exactly one more coin than player B, either player A has more heads than player B, or player A has more tails than player B — but not both. By symmetry (swapping “heads” and “tails”), these two events are equally likely. Since one of them must happen, each occurs with probability $1/2$1/2.