Calculate the expected number of throws until a game stops when specific coin toss sequences occur | Step-by-Step Solution

🎲 Let's Explore Game Length Distribution!

What We're Solving:

We need to find the probability distribution for N, the number of throws until a game stops. The game can end after 2, 3, 4, or more throws, and we want to understand what sequences lead to each possible game length.

The Approach:

Think of this like analyzing all the different ways a story can end! We'll work systematically:

• First, figure out what conditions make the game stop
• Then, count all the ways to reach each possible game length
• Finally, calculate the probability for each scenario

The key insight is that we need to consider all possible sequences that lead to the game ending at exactly throw k, while making sure the game doesn't end before throw k.

Step-by-Step Solution:

Step 1: Understand the stopping condition First, we need to know what makes the game stop. Since this isn't fully specified in your problem, I'll assume a common scenario: the game stops when we get two consecutive identical outcomes (like HH or TT in coin flips).

Step 2: Analyze the minimum case (N = 2)

• Game ends after 2 throws: We need identical outcomes
• Possible sequences: HH, TT
• If each throw has probability 1/2: P(N = 2) = P(HH) + P(TT) = 1/4 + 1/4 = 1/2

Step 3: Work through N = 3

• For the game to end at throw 3, it must NOT end at throw 2, but MUST end at throw 3
• This means throws 1 and 2 are different, but throws 2 and 3 are the same
• Possible sequences: HTT, THH
• P(N = 3) = P(HTT) + P(THH) = 1/8 + 1/8 = 1/4

Step 4: Find the general pattern for P(N = k) For the game to end at throw k:

• The game must "survive" throws 1 through k-1 (not end early)
• The stopping condition must be met at throw k

Following this logic:

• P(N = k) = (1/2)^(k-1) × (1/2) = (1/2)^k for k ≥ 2

Step 5: Verify our distribution Check that probabilities sum to 1: ∑(k=2 to ∞) (1/2)^k = 1/4 + 1/8 + 1/16 + ... = 1/4 × (1/(1-1/2)) = 1/2 × 1 = 1/2

The Answer:

The distribution depends on the specific game rules, but the general form is: P(N = k) = (probability of surviving k-1 throws) × (probability of stopping at throw k)

For the example I used: P(N = k) = (1/2)^k for k ≥ 2

Memory Tip:

Remember the "survival story" approach: For any throw k, ask yourself "How can I survive until throw k-1, then stop at exactly throw k?" This systematic thinking helps you avoid missing cases or double-counting sequences!