Nonplussed! by Julian Havil Page B

solutions is itself a solution. This results in the completely general solution of (1) of

    for constants P and Q .
    We can now invoke the condition that p 0 = 1 to get P + Q = 1, which provides us with one equation in two unknowns and to find unique values of p and Q we need a second, independent equation. This is not quite so easy.
    If the opponent had a known capital, which would allow us to write the total initial capital between the two players as, say, N , we could also invoke the condition PN = 0, which would result in that second equation for p and Q and this would mean that the values could be established. As it is, we have no such condition, but we can make progress by using the following intuitive argument.
    It must be that (1 − p )/ p is equal to, greater than or less than 1, and we can consider the three cases separately.
    If ( 1 − p )/ p = 1 (that is,), then P r = P + Q = 1 and in the long term we are assured of losing all of the capital against an opponent of much greater wealth.
    If (1− p )/ p > 1, as r increases P r is bound to fall outside [0, 1] and so violate the laws of probability, that is, unless P = 0, which makes P r = 1 for all r .
    Now suppose that (1 − p )/ p < 1, then, as r increases,

    but as we approach an infinite resource our probability of losing must approach 0, which means that Q = 0 and so P = 1. In this case,

    and we conclude that

    Of course, the inequalities for p simplify toandand this accords with our intuitive idea that if the game is fairor biased against us, eventually we will probably lose all of our fortune. As we mentioned before, even a fair game is a losing game if we have no chance of exhausting the opponent’s fortune.

    Figure 11.2. Game A tree diagram.
    The Parrondo Setup
    This analysis is repeated three times to expose the Parrondo paradox: there will be a game A and another game B, both of which are losing. We will then combine them, to result in a multiple game C, which will prove to be winning.
    Game A
    This is simply the game we have already described. Think of it as tossing a biased coin and getting a winning head with a fixed probability p or a losing tail with probability 1 − p . The winning and losing conditions are then represented by the same diagram as before, which is repeated in shortened form in figure 11.2 . The solution is provided by equation (2) on page 118:

    This is more complicated, with the chance of winning on any play of it dependent on the size of the capital at the time. To be exact, if the capital happens to be a multiple of 3, we win withprobability p 1 ; otherwise, we win with probability p 2 , summed up in figure 11.3 .

    Figure 11.3. Game B tree diagram.
    Since any positive integer must be one of {3 r , 3 r + 1, 3 r + 2}, we must consider three recurrence relations, the first generated by the left branch of the diagram and the remainder by its right side.
    With P r defined as before we then have

    together with the condition p 0 = 1.
    Again, we want an explicit formula for P r in terms of r and we work towards this by first finding a formula for p 3 r .
    The algebra is devious and we start by writing equations (4) and (5) as

    and

    and so think of them as two equations in the two unknowns p 3 r +1 and P 3 r +2 .
    These have the solutions (after some standard but messy algebra)

    Rewriting P 3 r +2 with r − 1 replacing r then gives

    and substituting these expressions for p 3 r +1 and p 3 r −1 into equation (3) and simplifying gives

    which can be rewritten as

    or

    This may be messy but a careful look reveals that the sum of the coefficients is 1 and therefore that this has the form of equation (1) with

    and r replaced by 3 r .
    We then have that

    and so

    Exactly the same arguments as before, only this time with

    result in

    Repeat the same (rather tedious) working with p 3 r +1 and p 3 r +2 and we would get the equivalent expressions with 3 r replaced by 3 r + 1 and 3 r − 1, respectively, and we can compress the whole