Coin Toss Game 1

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1

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Problem Statement

A and B take turns tossing a fair coin, with A going first. The player who gets the first tails after a heads wins. What is the probability of A winning?

Solution

This is a classic conditional probability problem that requires careful analysis of the game’s structure.

Step 1: Set Up the Conditional Probability Framework

Since A goes first, let P(A) be the probability that player A wins. We can use the Law of Total Probability by conditioning on what happens in A’s first toss:

P(A) = \frac{1}{2} \left[ P(A \mid H) + P(A \mid T) \right]

where H means A gets heads and T means A gets tails.

Step 2: Analyze P(A \mid T)

If A gets tails on their first toss, then:

  • The game effectively resets

  • B becomes the new “A” (first player)

  • A becomes the new “B” (second player)

Therefore:

P(A \mid T) = P(B) = 1 - P(A)

Step 3: Analyze P(A \mid H)

If A gets heads on their first toss, then B can:

  • Throw a tail: B wins immediately, so P(A \mid H, B=T) = 0

  • Throw a head: In this case, B is now in the same position A was in initially (has H, opponent’s turn)

The key insight: When A gets H and B gets H, B is in the position A was in (has H, opponent’s turn). Therefore:

  • B’s probability of winning from this position is P(A \mid H)

  • A’s probability of winning from this position is 1 - P(A \mid H)

Step 4: Derive the Equation for P(A \mid H)

Using conditional probability on B’s first toss:

P(A \mid H) = \frac{1}{2} \cdot 0 + \frac{1}{2} \cdot [1 - P(A \mid H)]

The first term is 0 because if B throws tails, A loses immediately. The second term represents B throwing heads, which puts B in A’s original position.

Simplifying:

P(A \mid H) = \frac{1}{2}[1 - P(A \mid H)]
P(A \mid H) = \frac{1}{2} - \frac{1}{2}P(A \mid H)
\frac{3}{2}P(A \mid H) = \frac{1}{2}
P(A \mid H) = \frac{1}{3}

Step 5: Solve for P(A)

Substituting back into our original equation:

P(A) = \frac{1}{2} \left[ \frac{1}{3} + (1 - P(A)) \right]
P(A) = \frac{1}{2} \left[ \frac{1}{3} + 1 - P(A) \right]
P(A) = \frac{1}{2} \left[ \frac{4}{3} - P(A) \right]
P(A) = \frac{2}{3} - \frac{1}{2}P(A)
P(A) + \frac{1}{2}P(A) = \frac{2}{3}
\frac{3}{2}P(A) = \frac{2}{3}
P(A) = \frac{4}{9}

Final Answer

P(A) = \frac{4}{9}

Key Insights

  1. Conditional Probability Structure: Breaking down the problem by conditioning on the first outcome is crucial.

  2. Symmetry and Role Reversal: When a player gets tails first, the game resets with roles reversed, creating a recursive structure.

  3. Recursive Equations: The key insight is recognizing that P(A \mid T) = P(B) = 1 - P(A), which creates a self-referential equation.

  4. First-Step Analysis: Conditioning on what happens in the first step (or first few steps) is a powerful technique for solving sequential probability problems.

  5. Positional Symmetry: When both players have heads, the situation becomes symmetric with respect to who has the advantage, allowing us to relate probabilities through the recursive structure.

2

Suggestion and Correction:
P(a) = 4/ 9

Suggestion:
further, I think so its more better to put this problem in a later stage, when you add thread such as markov chain.
People might not get the exact chain of thought, when attempting this problem

3

Thanks for the heads up! Of course you can use Markov chains to solve the problem as well. However this will be a fairly simple chain that doesn’t really require a Markov formalization. We’ll probably add more complex problems in that topic.