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Ask the Wizard #149

Wizard, I just read about Juan Parrondo’s paradox and thought it might interest you. It shows how two losing games can be played alternatively to make a winning game. Anyway, I thought it was an interesting "slight of hand" for game theorists. I like your site!


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.

My understanding of "wait time" for an event is the reciprocal of the probability of that event. I’m interested in calculating the wait time to roll consecutive 2s using one die. In a simulation I get 42 rolls on average. How do I make the connection with the probability of rolling consecutive 2s?

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.

My boyfriend maintained an email relationship with an ex-lover, but ended it when she admitted her intentions with him weren’t honorable. BUT, the ex’s girlfriend (she’s bisexual) then emailed my guy, berating him, so says he is "forced" to continue corresponding because he, "Tried to end it once." I was not uncomfortable her intentions, and this drama were disclosed, despite the ex sending him cards on holidays and ’just because,’ and texting messages occasionally. But one time her girlfriend left him a bizarre cell phone message pretending to be me asking him on a date. He still maintains his return emails are nondescript, mundane, and nonsexual and he has no feelings for her. Yet tonight, at a club we frequent, he became panic-y, dragging me away from our friends, saying the two were there and WE had to leave--he didn’t want them to see him. I say he’s hiding something. He says he’s not; just that her girlfriend is so unstable he doesn’t know what crazy thing/s she might do in public. What is going on here?

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.

A local casino is eliminating their Carribean Stud game, but by MGC rules they have to pay out the whole jackpot first. The table is $5 ante and $1 Progressive side bet. They are making the payouts Flush-150, Full House-300, 4 of a kind-1500, Straight Flush- Whole Jackpot (155,000), on 12/1. From my calculations my edge on the side bet is some insane 270% player edge, but almost all of that is in the straight flush. Just looking at the lower 3 payouts the player edge is 8.7%. Is this enough to overcome the house edge on the main bet of 5.25% or so? How do I combine the two edges? Obviously the bet is a winner if I think I have a chance to make a straight flush, but if I presume I have no chance to make a straight flush, is the game worth playing? Thanks for your time.

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.

Suppose a hotel has 10,000,000 rooms and electronic 10,000,000 keys. Due to a computer mistake each key is programmed with a random code, having a 1 in 10,000,000 chance of being correct. The hotel is sold out. What is the probability at least one customer has a working key?

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.

We seem to have a debate in our office about showing your cards in Texas Hold’em. Can a player show his hole cards to the table if he decides to fold, even though there are players still betting? Is there an actual rule?

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.

I play three card, Caribbean stud, and four card poker at machine shuffled tables. I am amazed the number of times a playable three-card hand frequently is dealt in a four card game, and a playable four-card hand is dealt in a Caribbean stud game. It makes me wonder if those shuffling machines aren’t pre-programmed to the house’s advantaged. Are these machines really random or are they programmed for the house, and if programmed, isn’t that illegal?

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.

I have been good friends with a coworker for a year now. I can’t stop thinking about her. I’m somewhat hesitant about taking the next step because a) I don’t want to get rejected and possibly lose our great friendship b) dating coworkers can get messy especially if it doesn’t work out. Anyways, what are the odds that I’m falling in love with her? What are the odds that she’s falling in love with me?

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.