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:

*p*_{win}= (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 *P _{i}(t)* of

**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:

P_{0}= 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:

* p*_{win}= (1-P_{0})
(3/4-e) + P_{0} (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-P_{0})
(5/8-e) + P_{0} (3/10 - e) = 727/1418-486795/502681 e = 0.5127 - 0.9684 e

The paradox occurs for e small enough to have:

* p*_{win}<0.5
and * p'*_{win}>0.5.