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Ask The Wizard #149

Wizard, I recently came across an intriguing concept known as Juan Parrondo’s paradox, and I thought you might find it fascinating too. It illustrates how alternating between two games that are both losing can actually result in a favorable outcome. I find it quite a clever twist for those studying game theory. Your website is great!

Richard

While I may not find Parrondo’s paradox particularly captivating, I have had several inquiries about it, so I’ll share my perspective. The core idea is that by switching back and forth between two specific losing games, a player can actually acquire an edge.

For instance, in Game 1, the chances of winning $1 are 49%, while the risk of losing $1 stands at 51%. In Game 2, when the player’s bankroll is divisible by 3, they have a mere 9% chance of winning $1, and a daunting 91% chance of a loss. Conversely, if the bankroll isn’t divisible by 3, the player has a much higher 74% chance of winning $1 and a 26% chance of losing.

It's evident that Game 1 produces an expected outcome of 49% multiplied by 1 plus 51% multiplied by -1, resulting in a -2% expected value.

In Game 2, you cannot simply calculate a straightforward average of the two scenarios because winning causes the game to quickly deviate from a bankroll remainder of 1, and it often shifts between remainders of 0 and 2. Essentially, this leads to a disproportionate likelihood of engaging with the game that has only a 9% winning chance. Thus, if you only play Game 2, the expected value lands at -1.74%.

By interchanging two rounds of Game 1 with two rounds of Game 2, we disrupt the alternating nature of Game 2. As a result, players end up playing the game with a 75% winning chance more frequently, while the game with a 9% winning chance becomes less prevalent. There are countless combinations to mix both games. Employing a strategy that alternates two games of each type gives an expected value of 0.48%.

It’s crucial to highlight that this notion has no practical application in a casino setting. Casino games do not alter their rules based on the modulo of a player’s bankroll. Nevertheless, I anticipate that it’s only a matter of time before someone markets a betting system based on Parrondo’s concept, by switching between roulette and craps, which will inevitably be just as ineffective as all other betting strategies.

I understand 'wait time' for an event as the inverse of that event’s probability. I'm curious to determine the wait time needed to roll consecutive 2s with a single die. Through simulation, I find that it takes an average of 42 rolls. How can I tie this back to the probability of rolling consecutive 2s?

Lee from Andover

Indeed, when looking at individual events, if the probability is p, the average wait time equals 1/p. However, this becomes more complex with consecutive events. Let x represent the condition where the last roll wasn't a two, which is also the initial state. Let y represent the condition where the last roll was a two. After the first roll, there’s a 5/6 chance we remain in state x, and a 1/6 chance of moving to state y. Let Ex(x) denote the expected number of rolls from state x, and Ex(y) 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

Thus, for two consecutive twos, the expected wait time remains at 42 rolls.

I’m facing a similar issue, only my query revolves around the expected number of flips to achieve two heads, in my case. math problems , see problem 128.

My boyfriend was in email contact with an ex, but he decided to cut ties after she revealed that her intentions weren't exactly honorable. However, her girlfriend, who identifies as bisexual, then sent a message to my boyfriend, chastising him. He claims this leaves him with no choice but to keep corresponding, even after attempting to end things before. I was initially okay with the situation, despite her sending holiday cards and occasional texts. However, one day her girlfriend bizarrely left a voice message acting as if she were me, asking him out. He insists that his replies are ordinary, unromantic, and that he harbors no feelings for her. Yet, just last night at a club we frequent, he became anxious, urging me to leave as he felt the couple was there—he didn’t want them to catch sight of him. This makes me suspect he’s hiding something, but he claims it’s simply because he finds her girlfriend unpredictable. What’s really happening?

Dawn from Sarasota

I believe that it’s nearly impossible for ex-partners to remain friends after a breakup. The most one can hope for is a distant relationship where they exchange holiday greetings. Any level of closeness implies at least one person is harboring hopes of reconciliation. While you didn’t ask for my opinion on breakups, I’d advise cutting ties cleanly to move forward with your life. I can’t say for sure what’s occurring here, but where there’s smoke, there’s fire. Although you lack concrete evidence to make accusations, it’s wise to remain vigilant.

A nearby casino is discontinuing their Caribbean Stud game but must first pay out the entire jackpot according to MGC regulations. The game has a $5 ante coupled with a $1 Progressive side bet. They’ll be paying out Flush at 150, Full House at 300, 4 of a kind at 1500, and a Straight Flush at the entire jackpot (155,000), effective from 12/1. My calculations suggest a staggering 270% player edge in the side bet, primarily tied to the straight flush. Analyzing just the lower three payouts, the player edge stands at 8.7%. Does this provide sufficient leverage to negate the house advantage of roughly 5.25% on the primary bet? How should I merge the two edges? Clearly, there's a favorable outcome if I believe I can score a straight flush, but is it worth playing if I assume I have no shot at a straight flush? Thank you for your help.

James from St. Louis

Interestingly, I’ve heard of a casino in Vegas that’s taking similar actions regarding their Caribbean Stud game. Here’s a formula to compute the expected return when a straight flush pays the complete 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 instance, the calculation looks like this: (((5108*150+3744*300+624*1500+40*155000)/2598960)-5*0.052243-1)/(5+1) = 36.858%. This means the player’s advantage is 36.858% of the total ante plus the $1 side bet, resulting in an expected profit of $2.21 per hand.

Imagine a hotel boasting 10,000,000 rooms equipped with 10,000,000 electronic keys. Due to an error, every key is randomly programmed, resulting in a 1 in 10,000,000 chance of being correct. The hotel is at full capacity. What are the odds that at least one guest possesses a functioning key?

Danny from London, U.K.

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

There seems to be an ongoing discussion within our office regarding the etiquette of revealing cards in Texas Hold’em. Can a player disclose their hole cards to the table if they decide to fold while other players are still actively betting? Are there any official rules concerning this situation?

Rick from Ottawa, Canada

Such behavior would be deemed poor poker etiquette. If you attempted this in Vegas, you might receive a warning on the first instance, but a second violation could result in your removal from the table.

I engage in three-card, Caribbean stud, and four-card poker at machine-shuffled tables. I am continuously surprised by the frequent appearance of a playable three-card hand in a four-card setting, and similarly, the frequency of a viable four-card hand in Caribbean stud. This raises the question of whether these shuffling machines might be programmed to benefit the house. Are these machines genuinely random, or are they manipulated in favor of the house, and if so, isn’t that illegal?

Cherrice from North Carolina

I firmly believe that the manufacturers of shuffling machines strive to make them as fair and random as possible. If a machine is deliberately tampered with, it would likely breach Nevada regulations. It’s relatively common to observe strong x-card hands when dealt x+1 cards. For instance, the probability of obtaining three of a kind with three cards is just 0.235%, whereas it rises to 0.922% when four cards are involved, making it almost four times more probable.

I have developed a close friendship with a coworker over the past year, and I can’t get her out of my mind. I’m a bit apprehensive about moving forward because a) I fear rejection and the potential loss of our wonderful friendship b) dating someone at work can lead to complications if it doesn’t work out. All in all, what are the chances that I’m falling for her? And what are the odds that she may share those feelings?

Jason from Vancouver, Canada

Firstly, I want to express that I am supportive of dating coworkers! I do not adhere to rules that prohibit interoffice relationships. Meeting new people can be challenging enough without limiting your interactions with familiar faces. Nevertheless, I would suggest caution if both parties fall under the same chain of command. It doesn't sound like this is the case, so I wouldn’t let that hinder your pursuits.

The fact that you have established a friendship is a great indicator. I know this may sound immature, but could you enlist the help of someone you trust to gauge her feelings about you? Alternatively, you could arrange a casual dinner after work with her and some colleagues. In a more relaxed setting, possibly with a few drinks, you could subtly explore her sentiments without outright professing your emotions. It’s crucial to raise the temperature of the situation without risking your friendship by declaring your feelings outright.

In response to your question, the likelihood that you are developing feelings for her is, without a doubt, nearing 100%. As for her feelings, it’s more ambiguous—probably around 10%. Nevertheless, don’t let this deter you; she might just need more time to understand her own emotions. Wishing you the best of luck! I would love to hear how things unfold.