# Ask the Wizard #149

Richard

Personally I don’t see what is so interesting about Parrondo’s paradox but you are not the first to ask me about it so I’ll give you my thoughts on it. The thrust of it is that if you alternate between two *particular* losing games the player can gain an advantage.

As an example, consider Game 1 in which the probability of winning $1 is 49% and losing $1 is 51%. In Game 2 if the player’s bankroll is evenly divisible by 3 he has a 9% chance of winning $1 and 91% of losing $1. In Game 2 if the player’s bankroll is not divisible by 3 he has a 74% chance of winning $1 and 26% of losing $1.

Game 1 clearly has an expected value of 49%*1 + 51%*-1 = -2%.

In Game 2 you can not simply take a weighed average of the two possibilities. This is because the game quickly gets off of a bankroll remainder of 1 with a win, and often alternates between remainders of 0 and 2. In other words the bankroll will disproportionately play the game with a 9% chance of winning. Overall playing Game 2 only the expected value is -1.74%.

However by alternating two games of Game 1 and two games of Game 2 we break the alternating pattern of Game 2. This results in playing the 75% chance game more and the 9% less. There are an endless number of ways to mix the two games. A 2 and 2 strategy of playing two rounds of Game 1 and two of Game 2, then repeating, results in an expected value of 0.48%.

I should emphasize this has zero practical value in the casino. No casino game changes the rules based on the modulo of the player’s bankroll. However I predict it is only a matter of time before some quack comes out with Parrondo betting system, alternating between roulette and craps, which of course will be just as worthless as every other betting system.

Lee from Andover

It is true that for single events if the probability is p then the average wait time is 1/p. However it gets more complicated with consecutive events. Let x be the state that the last roll was not a two. This is also the state at the beginning. Let y be the state that the last roll was a two. After the first roll there is a 5/6 chance we will still be in state x, and 1/6 chance we will be in state y. Let Ex(x) be the expected number of rolls from state x, and Ex(y) the expected number from state y. Then...

Ex(x) = 1 + (5/6)*ex(x) + (1/6)*ex(y), and

Ex(y) = 1 + (5/6)*ex(x)

Solving for these two equations...

Ex(x) = 1 + (5/6)*ex(x) + (1/6)*( 1 + (5/6)*Ex(x))

Ex(x) = 7/6 + (35/36)*Ex(x)

(1/36)*Ex(x) = 7/6

Ex(x) = 36*(7/6) = 42

So the average wait time for two consecutive twos is 42 rolls.

I have the same type of problem, only the expected flips to get two heads, in my site of math problems, see problem 128.

Dawn from Sarasota

I’m of the opinion that it is almost impossible for a couple to remain friends after breaking up. The best you can hope for is a Christmas card relationship. Anything more than that and at least one party is thinking about getting back together. Although you didn’t ask, my advice about breaking up is to do it cold turkey and get on with your life. I of course don’t know exactly what is going on here but where there’s smoke there’s fire. You don’t have enough evidence to make any accusations but continue to be suspicious.

James from St. Louis

Coincidentally I heard of a Vegas casino doing the same thing because they wanted to take out their Caribbean Stud game. Here is a general formula for calculating the expected return when a straight flush pays the full jackpot.(((5108*FL+3744*FH+624*FK+40*J)/2598960)-M*0.052243-1)/(M+1)where

FL = Flush win

FH = Full house win

FK = Four of a kind wi

n J = Jackpot amount

M = Minimum ante bet

In your case we have (((5108*150+3744*300+624*1500+40*155000)/2598960)-5*0.052243-1)/(5+1) = 36.858%. So the player advantage is 36.858% of the combined ante plus $1 side bet, or an expected profit of $2.21 per hand.

Danny from London, U.K.

The exact answer 1-(9,999,999/10,000,000)^{10,000,000} = 0.632121. This is also the same as (e-1)/e to seven decimal places.

Rick from Ottawa, Canada

This is in very poor poker etiquette. If you did that in Vegas you would probably be warned not to do it again the first time. A second time and you would probably be made to leave the table.

Cherrice from North Carolina

I strongly believe the makers of the shuffling machines at least attempt to make the shufflers as fair and random as possible. A deliberately gaffed machine I’m sure would violate Nevada law. It is fairly easy to see good x-card hands in x+1 cards. For example the probability of a three of a kind in three cards is 0.235%, and in four cards 0.922%, or almost four times higher.

Jason from Vancouver, Canada

First, let me express my opinion about dating coworkers. I’m all in favor of it! I also do not respect rules that forbid inter-office relationships. It is hard enough to meet people without restricting those faces you see every day. However I would make an exception if the two people were in the same chain of command. It doesn’t sound like this is the case here so I wouldn’t let point B get in your way.

The fact that you are friends is a very good sign. I know this sounds terribly junior high, but is there someone you trust you could send to feel her out for how she feels about you? If not perhaps you could organize an after work dinner sometimes with her and some other coworkers. Perhaps in a more casual setting, and preferably with a couple drinks in her, you can pry a little deeper. Do anything to increase the temperature without coming straight out and declaring your feelings. If you do that and she doesn’t share the same feelings it will ruin your friendship.

To answer your question the probability you are falling in love with her is clearly pushing 100%. It is hard to say the other way but probably only about 10%. However don’t let that discourage you, she just might need more time. Good luck to you. I’d be interested to know what happens.