# 6) Solution - Non-Transitive Bet

This is more of an explanation than a solution of the Walter Penny non-transitive bet. Let us remember the mnemonic trick for identifying the winning strategy: take Victor’s triplet ABC. Remove C. Then reverse B and place it before A. We now have B(rev)-A-B. The following explanation is adapted from the page by Yutaka Nishiyama and Steve Humble (Click Here).

We need to work out 2 probabilities:

- The probability of Victor’s triplet coming up with the very first cards 1, 2 and 3 and
- The probability of your triplet coming up just before Victors.

We have 8 combinations of B’s and R’s in 3 places. But due to symmetry, the logic of working out BBB is the same as that of working out RRR, and that of working out RRB is the same as that of working out BBR, etc.

1) Victor chooses RRR (or BBB)

Suppose that RRR does not come up with the first 3 cards dealt, say it happens with the 12th card, i.e., cards 12, 13 and 14 are RRR. Since Victor is looking for the first appearance of RRR, card 11 must have been a B otherwise, RRR would have been seen at cards 11, 12 and 13. So, if you choose BRR, you will always win (unless Victor gets an RRR on the first dealt card). To calculate the probability of your winning in the long run for this choice, we know that the probability of getting an RRR with the first 3 cards is 1/8. As per the above logic, BRR will always appear before RRR except with the first 3 cards. Therefore, the probability of BRR winning over RRR is certainty reduced by the probability of RRR appearing the first time, i.e., 1 - 1/8 = 7/8 as per our table. (The same logic applies to BBB).

2) Victor chooses RRB (or BBR)

Suppose Victor chooses RRB. According to our table, you should choose BRR. In order for you to win, a B must appear before RRB . What is the probability of getting an RRB or an RRRB or an RRRRB, etc. which allows you (with BRR) to win?

Using some probability math, the probability of getting an RRB in the first 3 cards is 1/8. The probability of getting an RRRB in the first 4 cards is 1/16, an RRRRB in the first 5 cards is 1/32. Summing these in a series is the sum of a geometric series. Let us assume the series is infinite: A + AR + AR^2 + AR^3 . . . . The sum is given by the formula = A / (1 - R) where A is the first term (1/8 in our case) and R is the multiplicand (1/2 in our case). The answer is 1/4. It follows that the probability of BRR winning is 1 - 1/4 = 3/4 (as per our table).

3) Victor chooses BRB (or RBR)

If Victor chooses BRB, you should choose BBR. The probability of getting a BRB in the first 3 cards is again 1/8. If BRB appears somewhere down the line, the probability of getting a B before that (so you can win) is 1/2. So, the probability your choice will come up first is 1 - 1/2 = 1/2.

4) Victor chooses RBB (or BRR)

If Victor chooses RBB, you should choose RRB. The probability of getting RBB in the first 3 cards is again 1/8. The probability of getting an R before RBB is 1/2. In the same manner as the previous case, the probability of your choice coming up first is 1 - 1/2 = 1/2.

A more mathematical solution

If you wish follow a solution which is based on probabilistic formulas, this YouTube clip does it all (Click Here and Here).

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