Problem Statement

You and your friend are playing a game with a fair coin, tossing it and writing down the outcomes. You win if HTH appears before HHT, else your friend wins. What is the probability that your friend wins?

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Solution

Step 1: Understanding the Game

The game continues until either HTH or HHT appears first. To analyze this, we consider the probability of winning from the instance the first H appears.

Let \(p\) be the probability that I win from this point onward. We now determine \(p\) using conditional probability.

Step 2: Breaking Down Possible Sequences

After the first H, the next flip can be:

  1. H → Sequence: HH_
  2. T → Sequence: HT_

We analyze both cases separately.

Case 1: The sequence becomes HH_

  • The next flip can be H (forming HHH_) or T (forming HHT).
  • If we get HHT, my friend immediately wins.
  • If we get HHH, the sequence will continue indefinitely, but since the only way to eventually break out is forming HHT, I am guaranteed to lose.
  • Thus, if HH occurs, I cannot win.
    \(P(\text{Winning} \mid HH) = 0\)

Case 2: The sequence becomes HT_

  • If the next flip is H, then HTH forms, and I immediately win.
  • If the next flip is T, then the sequence extends to HTT_ and continues.
  • Now, at some point in the future, an H must eventually appear (since an infinite T sequence has probability 0).
  • When this H appears, the game resets to the same original scenario, meaning my probability of winning from here is still \(p\).

Thus,

\[P(\text{Winning} \mid HTT) = p.\]

and

\[P(\text{Winning} \mid HTH) = 1.\]

Now, consider that I said earlier “consider the case from the first H”. Since the game is an inifinte sequence, a series of TTTTT… is not possible, hence the first H is guarateed to appear. Therefore, I can argue that this probability \(p\) is the same as me winning in any case from the start of the game, and not just from the first H.

Step 3: Setting Up the Probability Equation

Obviously,

\[P(HTT) = P(HTH) = \frac{1}{4}\]

and

\[P(HH) = \frac{1}{2}\]

Using the law of total probability, we can write

\[p = \frac{1}{4} (1) + \frac{1}{4} (p) + \frac{1}{2} (0)\] \[p = \frac{1}{4} + \frac{1}{4} p\] \[\frac{3}{4} p = \frac{1}{4}\] \[p = \frac{1}{3}\]

Thus, the probability that I win is \(\frac{1}{3}\), and the probability that my friend wins is:

\[1 - \frac{1}{3} = \frac{2}{3}.\]

Thus the probability my friend wins is \(\mathbf{\frac{2}{3}}\)

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