Coin Toss Problem

Created on November 07, 2025

2025   ·   probability   counting   symmetry symmetry     ·   quant-finance-prep quant-finance-prep  

A note for the probability problem on Coin Toss Problem.

Problem Statement

Two gamblers are playing a coin toss game. Gambler $A$ has $(n+1)$ fair coins; $B$ has $n$ fair coins. What is the probability that $A$ will have more heads than $B$ if both flip all their coins?

Personal Approach

In this section, I’ll describe my own approach to solving the problem. The idea is to apply basic counting methods from probability theory.

Simplified problem

To start, I simplified the problem by setting $n = 2$. In this case, gambler $A$ has 3 fair coins, while gambler $B$ has 2.

Let $h_A$, $h_B$ denote the number of heads obtained by $A$ and $B$ respectively. The goal is to find all possible pairs ($h_A$, $h_B$) that satisfy the following conditions:

  • $h_A > h_B$
  • $0 \leq h_A \leq n+1$
  • $0 \leq h_B \leq n$

With the simplified setup ($n=2$), I listed all possible combinations of $h_A$ and $h_B$ in a table, where each column corresponds to each $h_A$ value and each row corresponds to each $h_B$ value:

  0 1 2 3
0   $A$ win $A$ win $A$ win
1     $A$ win $A$ win
2       $A$ win


To interpret the table, take the cell where the $column = 1$ and the $row = 0$. This cell corresponds to ($h_A = 1$, $h_B = 0$), meaning gambler $A$ gets one head while gambler $B$ gets none. In this case, $A$ wins, so I labeled this cell as A win.

To find the probability that $A$ wins, we simply count all the cells labeled A win and divide by the total number of cells.

In this simplified case:

  • The 1st row has 3 cells labeled A win.
  • The 2nd row has 2 such cells.
  • The 3rd row has 1 such cell.

In total, there are $3+2+1=6$ winning cells for $A$. Since the table has $4 \times 3 = 12$ cells in total, the probability that $A$ has more heads than $B$ is:

\[\mathbb{P}(\text{$A$ wins}) = \frac{6}{12} = 0.5 = 50\%\]

Generalised problem

With this intuition, we can now extend the reasoning to a general case with any arbitrary $n$.

From the pattern observed:

  • The 1st row contains $n$ A win cells.
  • The 2nd row contains $n-1$ A win cells.
  • …
  • The $n^{th}$ row contains just $1$ A win cell.

Adding them together gives:

\[n+(n-1)+(n-1)+...+2+1 = \frac{n\times(n+1)}{2}\]

The total number of cells in the table is $n\times(n+1)$.

Therefore, the probability that gambler $A$ gets more heads than $B$ is:

\[\mathbb{P}(\text{$A$ wins}) = \frac{\frac{n\times(n+1)}{2}}{n\times(n+1)} = \frac{1}{2}\]

In conclusion, regardless of the number of coins, gambler $A$ always has a $50\%$ chance of getting more heads than gambler $B$.