Take thrice thy money, bid me tear the bond

A rather fun riddle for which the pen&paper approach proved easier than coding in R, for once. It goes as follows: starting with one Euro, one sequentially predicts a sequence of four random bits, betting an arbitrary fraction of one’s money at each round. When winning, the bet is doubled, otherwise, it is lost. Under the information that the same bit cannot occur thrice in a row, what is the optimal minimax gain?

Three simplifications: (i) each bet is a fraction ε of the current fortune of the player, which appears as a product of (1±ε) the previous bets (ii) when the outcome is 0 or 1, this fraction ε can thus be chosen in (-1,1), (iii) while the optimal choice is ε=1 when the outcome is known, i.e., when both previous are identical. The final fortune of the player is thus of the form

(1±ε)(1±ε’)(1±ε”)(1±ε”’)

if the outcome is alternating (e.g., 0101 or 0100), while it is of the form

2(1±ε)(1±ε’)(1±ε”)

if there are two identical successive bits in the first three results (e.g., 1101 or 0110). When choosing each of the fractions ε, the minimum final gain must be maximised. This implies that ε=0 for the bet on the final bit  when the outcome is uncertain (and ε=1 otherwise). In case of an alternating début, like 01, the minimal gain is

min{(1±ε)(1±ε’)(1+ε”),2(1±ε)(1±ε’)(1-ε”)}

which is maximised by ε”=1/3, taking the objective value 4(1±ε)(1±ε’)/3. Leading to the gain after the first bit being

min{4(1±ε)(1+ε’)/3,2(1±ε)(1-ε’)}

which is maximised by ε’=1/5, for the objective value 8(1±ε)/5. By symmetry, the optimal choice is ε=0. Which ends up with a minimax gain of 3/5. [The quote is from Shakespeare, in the Merchant of Venice.]

One Response to “Take thrice thy money, bid me tear the bond”

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