Probability • Symmetry
Alice tosses one more coin than Bob
Alice tosses \(n+1\) fair coins. Bob tosses \(n\) fair coins. What is \(P(\text{Alice gets strictly more heads})\)?
Explore different values of \(n\)
Bob's number of coins
5
Alice tosses
6
\(n+1\)
Bob tosses
5
\(n\)
Click the button to simulate one round.
The trick
Separate Alice's extra coin from her first \(n\) coins.
Alice's first \(n\) coins
\(X\)
\(X\sim\mathrm{Binomial}(n,\tfrac12)\)
vs.
Bob's \(n\) coins
\(Y\)
\(Y\sim\mathrm{Binomial}(n,\tfrac12)\)
Alice also has one extra fair coin \(E\), independent of \(X,Y\).
Since \(X\) and \(Y\) have exactly the same distribution, the events \(X>Y\) and \(Y>X\) have the same probability.
Case 1 • probability \(1/2\)
Extra coin is tails
Alice's total is just \(X\), so Alice wins exactly when \[ X>Y. \]
Let \[ p=P(X=Y). \] By symmetry,
\[
P(X>Y)=\frac{1-p}{2}.
\]
Case 2 • probability \(1/2\)
Extra coin is heads
Alice's total is \(X+1\), so she wins when \[ X+1>Y, \] which is equivalent to \[ X\ge Y. \]
\[
P(X\ge Y)=\frac{1+p}{2}.
\]
Average the two equally likely cases
\[
\begin{aligned}
P(\text{Alice wins})
&=\frac12\cdot\frac{1-p}{2}
+\frac12\cdot\frac{1+p}{2}\\[4pt]
&=\frac12.
\end{aligned}
\]
50%
The tie probability \(p\) cancels out completely.
Interview intuition:
Match Bob's \(n\) coins with \(n\) of Alice's coins. Those two head counts are symmetric.
Alice's one extra fair coin converts exactly half of the tie situations into strict wins,
making her overall winning probability exactly \(1/2\).