I have thought up a paradox, that may already exist, but I do not know what it's called. It's bothering me though, so any help regarding solving it or proving it impossible would be appreciated.

In this paradox, you have a gambler. The gambler is \$200 in debt, but the gambler has a wealthy friend who lets the gambler bet using money he does not have to play a game. In this game, the gambler bets \$1, and a random number generator generates a number from 1 to 100. If the number roll over, or exactly 55, the gambler wins \$2. If the number rolls under 55 the gambler loses that \$1 he bet.

From my understanding of statistics, over a certain expected period of time, the gambler should hit the unlikely scenario to get out of debt using this method. However using my computer simulations, it seems to take more time than I can allow my simulations to run.

Is it possible to guess an expected number of times the gambler would have to play the game in order to get out of debt using some mathematical model?

I am also concerned that the nature of random-number generators may make it impossible for the gambler to get out of debt, as random number generators could be heavily biased to avoid situations such as randomized decks to be fully sorted, or getting out of debt with negative debt and unlikely odds.

What I want to get out of this question is, how to explain why it's possible, how to calculate expected number of times the game has to be played to reach the goal, or why it's impossible, or some existing question I can try to study to better understand the problem.

• If a person flips a coin $n$ times, will he eventually flip $100$ heads in a row? The probability approaches $1^-$ as $n$ increases without bound. Is this similar to your gambler? – GFauxPas Apr 26 '15 at 20:53
• Seems similar, but does not satisfy me :\ – Dmitry Apr 26 '15 at 20:57
• galton.uchicago.edu/~lalley/Courses/312/RW.pdf – Memming Apr 26 '15 at 21:09
• I think that is exactly what I am looking for, thanks Memming! I will try to study it. – Dmitry Apr 26 '15 at 21:12
• No random number generator tries to avoid situations such as randomized decks to be fully sorted. With a good generator it may happen, but the probability is the same as in real life, i.e., practically zero. With a bad generator, it may or may not be possible, as it's incapable of producing certain sequences. But this is not by design. – maaartinus Apr 27 '15 at 3:09

This has little to do with random number generators.
As Tony K noted, this is a random walk where each step is $+1$ with probability $p < 1/2$ and $-1$ with probability $1-p$. Starting at $0$, the probability of ever reaching positive integer $m$ is $(p/(1-p))^m$. In your case, with $p = 0.46$ and $m = 200$, that probability is about $1.18 \times 10^{-14}$.

• I have never heard about random walks until TonyK brought it up. I am now excited to study them. Out of curiosity, what does that probability mean? 1/(1.18*10^(-14)) is the approximate expected number of games to play until you get out of debt? Or what does it refer to. Thanks again. – Dmitry Apr 26 '15 at 21:23
• @Dmitry: It means that there is only a one in $1.18\times10^{14}$ chance that you will ever get out of debt. Alternatively, if you have $1.18\times10^{14}$ gamblers all playing the game repeatedly forever, probably all but one of them will never ever get out of debt no matter how long they play. – Rahul Apr 26 '15 at 22:06
• @Dmitry Interesting consequence of that equation: If the probability of a successful roll is exactly 50% (or better), then the gambler will always get out of debt, no matter how much his starting-debt is. Assuming he has infinite time and starting cash, that is. – BlueRaja - Danny Pflughoeft Apr 26 '15 at 22:14
• @Rahul - you've made a minor mistake - if the probability is $1.18\times10^{-14}$, then it's a one in $8.47\times10^{13}$ chance. – Glen O Apr 27 '15 at 2:53

"From my understanding of statistics...": Your understanding is wrong. You are probably thinking of an unbiased random walk, where given any integer $n$, the probability of reaching $n$ at some point is 1. But this random walk is not unbiased, because the probability of losing is 0.54, not 0.5. So there is a non-zero probability (indeed, a very high probability) that the gambler never gets out of debt.

• Actually, I never heard of a random walk until you brought it up, and now I am excited to study it. However my question does not care about how unlikely the scenario is, rather, i am interested in a number of times the game would need to be played for the odds of coming out of debt would be over 0.5. I understand this number could be very large, but I am wondering in whether this number exists, or if the nature of the problem makes it impossible or too computationally expensive to compute. Thanks for bringing it up though! – Dmitry Apr 26 '15 at 21:29
• @Dmitry: The odds will never be over 0.5. That's the point. – TonyK Apr 26 '15 at 21:37

When you say he wins \$2 does that mean he gets back his \$1 plus another \$1? Or that he gets back his \$1 plus another \$2? If he has -\$200 and bets \$1 (now having -\$201) and loses, he'd have -\$201 of course. If he has -\$200 and bets \$1 (now having -\$201) and wins, would he then have -\$199 or -\$198?

If it's -\$199, then the probability of getting out of debt is very small as there is slightly more chance (p = 0.54) of the \$1 change away from zero (vs p = 0.46). As far as I know, calculating the expected number of events to reach an outcome that has a probability of < 0.5 does not make sense. Check out the Bernoulli distribution.

If it's -\\$198 that changes things. That's where my stats gets a bit rusty and I'm can't remember if there's such a thing called a 'weighted' Bernoulli distribution or if that has a different name, but it makes it much more likely that your poor addict might get out of debt.

I know this doesn't answer the question fully but perhaps someone else can pick up from here?