Problem 33

Parrondo's Perplexing Paradox (1996)

Problem. Consider the following games G_{1}, G_{2}, and G_{3}, in each of which $1 is won if a head is obtained, otherwise $1 is lost (assuming the player starts with $0).

G_{1}: A biased coin that has probability of heads .495 is tossed.

G_{2}: If the net gain is a multiple of three, coin A is tossed. The latter has probability of heads .095. If the net gain is not a multiple of three, coin B is tossed. The latter has probability of heads .745.

G_{3}: G_{1} and G_{2} are played in any random order.

Prove that although G_{1} and G_{2} each result in a net expected loss, G_{3} results in a net expected gain.

Solution. For game G_{1} the net expected gain is $1(.495) − $1(.505) = −$.01.

For game G_{2}, let denote the net gain of the player after n tosses. Then, given all the prior values of the net gain, the value of depends only on , that is, is a Markov chain.^{1} Moreover, ^{2} is also a Markov chain with states {0, 1, 2} and transition probability matrix

After an infinitely large ...