## A detailed discussion

Let us call X(t), the capital at time t.  A game is winning (loosing) if  the average <X(t)> is a monotonously   increasing (decreasing) funtion of t and it is fair if <X(t)> is constant.

It is easy to prove that game A is a loosing game if  e positive:

<X(t+1)>-<X(t)> = <X(t+1)-X(t)> = (1/2 - e) - (1/2 + e) = -2e

 Game A COIN 1 Prob. of winning Prob. of losing 1/2 - e 1/2 + e

The analysis of game B is not that easy. Remember the rules:

 Game B Is X(t) a multiple of 3? NO YES Coin 2 Prob. of winning Prob. of losing Coin 3 Prob. of winning Prob. of losing 3/4 - e 1/4 + e 1/10 - e 9/10 + e

a naive (and wrong) argument is as follows: coin 3 is used 1/3 of the time, whereas coin 2 is used 2/3 of the time. Then, the probability of winning is:

pwin= (2/3) (3/4-e) + (1/3) (1/10 - e) = 17/30 - e

which is slightly bigger than 1/2 for e = 0. Then, game B, according to this argument, is not a loosing game.

The argument is not correct because coin 3 is not used one third of the time, but a little bit more often. If X(t) is a multiple of three, say 9, then we use coin 3 and the most likely result is loosing. Then, the most likely value of X(t+1) is 8. We have now to use coin 2, and the most likely result is now winning. Therefore, oscillations between 9 and 8, or between 3n and 3n-1 have a high probability. As a consequence, the frequency of using coin 3 is between one third and one half.

The calculation of this frequency is a little technical. Let us define the variable:

Y(t) = X(t) mod 3

which can take only three values:  0,1, and 2. Y(t) is a Markov chain with three states and a transition matrix given by:

 P= 0 1/4 + e 3/4 - e 9/10 + e 0 1/4 + e 1/10 - e 3/4 - e 0

The probability distribution P(t), i.e., the vector whose components are the probabilities Pi(t) of Y(t)=i with i=0,1,2, obeys the evolution equation:

P(t+1) = P P(t)

The matrix P has three eigenvalues: one is equal to 1 and the modulus of the other two is smaller than 1. The eigenvector corresponding to eigenvaue 1 is the stationary probability distribution, i.e., its components are the probability of Y(t)=0,1, and 2 for t very large.

We are interested on the probability of Y(t)=0, which reads, for e small:

P0= 5/13 - (440/2197) e   = 0.3846 - 0.2003  e

which is the probability of using the bad coin 3. The probability of winning is:

pwin= (1-P0) (3/4-e) + P0 (1/10 - e) = 1/2 - 147/169 e = 0.5 - 0.87 e

which is smaller than 0.5 for e positive.

Now, when games A and B are combined, we have:

P'0= 245/709 - 48880/502681  e   = 0.3456 - 0.9724  e

and

p'win= (1-P0) (5/8-e) + P0 (3/10 - e) = 727/1418-486795/502681 e = 0.5127 - 0.9684 e

The paradox occurs for e small enough to have:

pwin<0.5 and   p'win>0.5.