Physically Incorrect

This is a solution to this puzzle.

The key insight here is, I think, the “lack of memory” the game displays. Suppose Alica plays, then Bob, then Cindy, then Alice (who doesn’t win) - now, do you care what Alice, Bob and Cindy got in the previous round when trying to determine, say, whether Bob will win in the next round? Nope. A lack of memory usually, albeit not always, implies recursion might be useful.

Here is a diagram that looks like it came out of biology textbook, but actually portrays are all possible events after 4 turns. Remember we’re trying to compute the probability of Bob winning the game. Blue marks desirable states in which Bob wins. Red marks undesirable states in which either Alice or Cindy win:

Alice goes first and can get either heads or tails. Then Bob tosses his coin, and might win. If he doesn’t, Cindy tosses her coin, then it’s back at Alice. If Alice doesn’t win, then we’re really back at square one, i.e. the first step in our diagram - Alice getting either heads or tails and not winning.

I’ve used P, P’ and P” to mark the conditional probabilities of Bob (aka B) winning given different scenarios.  All that’s left to do is to write the recursion equations for P, P’ and P”:

• First, note that Prob(Bob wins) = Prob(Bob wins and Alice got heads on her first toss) + Prob(Bob wins and Alice got tails on her first toss). Using Bayes’ theorem we can express this in terms of conditional probabilities: Prob(Bob wins) = Prob(Bob wins | Alice got heads on her first toss)*Prob(Alice got heads on her first toss) + Prob(Bob wins | Alice got tails on her first toss)*Prob(Alice got tails on her first toss) = P/2 + P/2 = P.
• Next, examine the scenario where Alice got heads on her first toss. Then: P = Prob(Bob wins and Bob got heads on his toss|Alice got heads on her toss)  + Prob(Bob wins and Bob got tails on his toss|Alice got heads on her toss) =  Prob(Bob wins | Bob got heads on his toss and Alice got heads on her toss before him)*Prob(Bob got tails) + Prob(Bob wins | Bob got tails on his toss and Alice got heads on her toss before him)*Prob(Bob got heads) =  ½(1 + P’).
• Similarly, P’ = ½(0 + P”).
• Finally, P” = ½(0 + P).

Combining all of these equations, one can solve for P and obtain P=4/7, which is the desired probability.