### A Gambler’s Dream: Parrondo’s Paradox

Every gambler wishes there was a way to be guaranteed infinite money. Even better so if it goes against expectations — using *losing *strategies.

Enter Parrondos’ paradox, which says just that. Using the stochatisc process classes’ commonplace example of biased coin games, one can devise a strategy for alternating between 2 games, A and B, which guarantee wealth — despite a guaranteed bankruptcy playing A or B separately.

For game A, let there be a biased coin, P, where the probability of winning on P is *p*=0.5-*x*. Winning gains you a dollar, and losing loses you a dollar. After each game of A, examine your current bankroll — if it is a multiple of 3, play biased coin Q; else, play biased coin R. For coin Q, the probability of winning is *q*=0.75-*x*; for coin R, the probability of winning is *r*=0.1-*x*. The winnings are, once again, a dollar and losing subtracts a dollar.

In other words, you have a 50% chance of winning game A and either a 10% or 75% of winning game B.

When played alternately, the end result is a seeming paradox — that the limit of the money quickly ascends into infinity! (note that not ALL alterations of the game are winning — playing ABABABAB…. is a losing game, while ABBABBABBABB….is winning).

Why? Let’s look a simpler example (since I’m NOT about to type out transition matrices 3897234 times):

- In A, you lose 100% of the time and each time, you lose $3.
- In B, you win $8 if your bankroll is a multiple of 5 (0 counts as a multiple of 5), and lose $6 if it is not.

Let game A just be throwing your money into a blender. Just chuck $3 right in. Game B, if you have any number of $5 bills, you get $8. Because why not. Otherwise, they take $6 and poop on it, THEN throw it in the blender. Because fuck you, that’s why.

Let c(t) be the bankroll at game instance *t*, and consider c(0)=*m*. We will play the games alternatively ABABAB….

Suppose *m* will never reach a multiple of 5 for all t. This means that m and 5 are coprime. Suppose you play A then B — this gives you m-3-6=m-9, which must also be coprime with 5 (since we are assuming that m is never divisible by 5 at any time). So, suppose there are natural numbers a,b,c,d, for 0< b,d <5, such that m=5a+b and m-9=5c+d by the division algorithm. Generalizing this for any time *t:*

m-9(t)=5(c-2t)+d+2t

m-9t=5c-10t+d-2t

m-9t=**5c+d-12t.**

Now, since 0<d<5, we can explore all the possibilities as thus:

d=1; m-9(t)=5c+1-12t —> m=5c+1-12t+9t=5c+1-3t. Suppose this were coprime with 5 for all t. Then 1-3(t)=\{1,-2,-5,…\} and we can stop — t=2 says m=5c+1-3(2)=5c+1-6=5c-5=5(c-1), which is a multiple of 5. Contradiction.

d=2; m=5c+2-3t —> 2-3t=\{2, -1, -4, -7, -10,…\}, and again, we stop; t=4 —> 2-3(4)=-10, so m=5c-10, which is divisible by 5. Contradiction.

I’ll leave the rest up to the reader, since they’re all the same proof.

As this shows, no matter what your starting capital is, (well, you need $8 or more, since only $3 gets you the infinite sequence of 5 multiples needed), you will always gain infinite money!

Too bad the casinos know about this stuff.