Alice and Bob's Coin Toss Challenge

Solution & Explanation

To solve the problem of determining the probability that Alice ends up with more heads than Bob when Alice tosses $n+1$ coins and Bob tosses $n$ coins, we can break down the problem into simpler steps.

Step 1: Understanding the Problem

Alice and Bob are tossing coins, with Alice tossing one more coin than Bob. We need to find the probability that Alice has more heads than Bob at the end of their respective tosses.

Step 2: Defining the Events

• Let $X$ be the number of heads Alice obtains.
• Let $Y$ be the number of heads Bob obtains.
• We are interested in finding $P(X > Y)$.

Step 3: Analyzing the Coin Tosses

• Both $X$ and $Y$ are binomially distributed random variables with a probability $p = 0.5$ for obtaining heads on any single coin toss.
• $X \sim \text{Binomial}(n+1, 0.5)$
• $Y \sim \text{Binomial}(n, 0.5)$

Step 4: Using Symmetry

• If Alice and Bob were to toss the same number of coins, the probability of Alice having more heads than Bob would be the same as Bob having more heads than Alice due to symmetry. Thus, $P(A_n > B_n) = P(A_n < B_n)$.
• However, Alice tosses one extra coin, which influences the outcome.

Step 5: Calculating the Probability

• The extra coin toss by Alice can either result in a head or a tail, each with a probability of 0.5.

• If Alice and Bob have the same number of heads after $n$ tosses, Alice can only have more heads than Bob if she gets a head on the $n+1$th toss, which has a probability of 0.5.

• Therefore, the probability that Alice has more heads than Bob is:

  $P(A(n+1) > B(n)) = P(A(n) > B(n)) + 0.5 \times P(A(n) = B(n))$

Step 6: Solving for the Probability

• By symmetry, $P(A_n > B_n) = P(A_n < B_n) = x$ and $P(A_n = B_n) = 1 - 2x$.

• Substituting these into the equation:

  $P(A(n+1) > B(n)) = x + 0.5 \times (1 - 2x)$

• Simplifying, we get:

  $P(A(n+1) > B(n)) = 0.5$

Conclusion

The probability that Alice ends up with more heads than Bob when she tosses $n+1$ coins and Bob tosses $n$ coins is $0.5$. This result is due to the symmetry of the problem and the fact that the extra coin toss by Alice has an equal chance of tipping the balance in her favor or having no effect if they were previously tied.