Recently, I wrote an article on why the Monty Hall problem has perplexed so many brilliant minds where I showed that it was a corner case between 1 open and 1 closed door, while the intuitive but wrong answer is close to the probability curve of 1 open door.

Now a coin toss puzzle has appeared on Twitter that has gone viral as it goes against our common intuition of probability and random sequences (such as a series of coin tosses). The puzzle goes as follows.

## The Problem

Flip a fair coin 100 times—it gives a sequence of heads (H) and tails (T). For each HH in the sequence of flips, Alice gets a point; for each HT, Bob does, e.g., for the sequence THHHT Alice gets 2 points and Bob gets 1 point. Who is most likely to win?

- Option A: Alice
- Option B: Bob
- Option C: Equally likely

I invite you to pause here for a moment and come up with your own solution.

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- Since coin tosses are the very foundation of our understanding of probability theory that can be best understood by equally likely outcomes, most people tend to think that Option C is the correct answer.
- Since Alice gains two points in 3 heads appearing together (HHH) and any extension thereof, as opposed to Bob who does not have this luxury, the second most chosen answer was Option A.
- In reality, the correct answer is Option B. Bob wins most of the times.

Let us find out the reason from both intuitive and mathematical viewpoints.

## Intuitive Reasoning

For any sequence starting with HH, the next coin flip has one of the following two outcomes.

- HH.H: Alice wins with the score 2-0 (since in 100 tosses, Alice wins 99-0, 98-1, and so on, this margin of victory can easily be confused with the number of branches in which Alice wins).
- HH.T: It is a tie with the score 1-1 (HH for Alice and HT for Bob).

So we have 1 win for Alice and 1 draw between the two.

On the other hand, for any sequence starting with HT, the next coin flip has one of the following two outcomes.

- HT.H: Bob wins with the score 0-1.
- HT.T: Bob wins with the score 0-1.

So we have 2 wins for Bob in such a case, as opposed to Alice’s case above. Clearly, Bob wins in most branches of all possible sequences.

## Proof by Mathematical Induction

Since two coin tosses scenario produces equally likely outcomes, we start with $k=3$. The possible scenarios and their outcomes are drawn in the figure below.

From the above figure, Bob wins in 3 branches as compared to Alice who wins in 2. Therefore, Bob is the winner for $k=3$. Let us draw a table for flip $k+1$.

Flip $k$ | Flip $k+1$ | Outcome |
---|---|---|

H | H | Draw or Bob wins |

H | T | Bob wins |

T | H | Bob wins |

T | T | Bob wins |

Keep in mind that this table below does not represent sequences of possible outcomes. Instead, it simply describes the outcome for the next flip at $k+1$ as is customary in mathematical induction.

Now from the first row, we can see that Alice wins 1 point when an HH arrives. Hence, it is either a draw (if Bob has won until flip $k$ by a margin of 1) or Bob still wins (if he has won by a margin greater than 1 by flip $k$). The second row just consolidates his position due to HT while the third and fourth row do not change anything (i.e., Bob still wins).