# Odds of Coming Out Ahead in Roulette

I have an interesting roulette problem that I initially thought was easy but now I'm second guessing my self. The problem is as follows:

A friend of yours thinks that he has devised a purely mathematical way of beating the standard European roulette wheel, which has 37 pockets. His plan is to come to the roulette table with 100 units with which to bet. On each spin, he places a one-unit, single-number bet. (Recall that winning such a bet returns 36 units to the winner.) He will make exactly 100 bets, no matter how much he wins or loses. He claims that this strategy results in a greater than 50% chance that he will come out ahead at the end of 100 spins, so using this strategy repeatedly will make him a winner.

a) Does he have a greater than 50% chance of coming away with more money than he started with?

b) If the answer to the previous question is yes, then does that make his strategy a winning one, meaning that it will lead to a long-term increase in bankroll?

So for part A I think he does have a greater than 50% chance of coming away ahead. He needs to win at least 3 times in order to walk away with more than he started with. And the probability of winning at least three times is 50.939% (I did 1 - the probability of winning 0, once, or twice with p=(1/37).)

So based on that it seems like he does have a greater than 50% chance of walking away ahead...

This goes against my intuition, especially for part b. If everyone could do this strategy and come out ahead then they would. But surely it can't be this simple... I don't see how this would lead to a long term increase in bank roll. Any thoughts?

• Keep in mind: it's easy to design games in which you win more than half the time yet your expected gain is negative. Say you are tossing a fair die and you win $\$1$if you throw a$1,2,3,4,5$and you lose$\$10$ if you throw a $6$. In that game, you win five games out of six but it is still a losing game. That's what is going on with your freind's strategy. – lulu Jun 20 '18 at 14:09
• I really like this question and I would like to add that these pseudo arguments aren't just found in gambling settings the world over. You will find, if you pay close attention, that politicians for example are very skilled at using these and other pseudo arguments to convince their audience as well. – Stefan Mesken Jun 20 '18 at 14:45
• There are betting patterns such that a gambler has a better than 50% chance to walk away with more money than he started. However the expected value of each wager remains negative. The expected distribution of outcomes has a large probability of a small gain and a small probability of a very large loss. – Doug M Jun 20 '18 at 15:16

Hint: Odds of winning aren't relevant to part b). What is relevant is the expected value. Calculate that and you will see whether it's positive (which would mean you've discovered a winning strategy) or negative.

edit: Let me add a little story to this answer which I think might be entertaining.

As a young adult I played poker rather successfully and thus spent many, many nights in casinos. It always struck me how much the casino, in a subtle but distinct fashion, encourages the suspicions and false believes of its customers. Roulette tables record the previous ~20(?) outcomes and advertise them to every bystander in big, illuminated letters (to encourage the belief that past outcomes somehow affect the likelyhood of future outcomes), slot machines have correct but irrelevant calculations printed on them (suggesting that there may be a winning strategy), you are constantly bombarded by special deals and jackpots both visually and audibly, ...

The later decision to become a mathematician, I think, has been fueled at least in part by the desire to truly understand these misleading claims.

In any way: The situation you describe here must be any casino-owner's wet dream. It's not the casino that convinced the customers of his perceived advantage -- it's the customer himself who came up with a fake recipe for his success.

• Exactly. This is what every Martingale system relies on - a high probability of winning a bit, a low but non-zero probability of losing very much more. – Jaap Scherphuis Jun 20 '18 at 14:16
• Hmm. that is helpful. So the expected loss per play is $. 027 and therefore after 100 plays your expected loss is$2.70. I'm still just struggling with why this is...if your odds are better than the house it seems like you should come out ahead! Maybe i just need to think about it a bit more. Feel like I am not understanding something very obvious... – Eliza Watts Sells Jun 20 '18 at 14:25
• @ElizaWattsSells If you are struggling to see why you should not come out ahead, take the situation to an extreme. For example, if we played a game where a diced is rolled and on a 1 you pay me £100,000 but otherwise I pay you £1. You are most likely to come out ahead, would you play? Why not?. It works a lot like this but less extreme. You are most likely to come out on top in one roll but after a few rolls I am most likely to come out with (a lot) more money – DaveP Jun 20 '18 at 14:34
• @ElizaWattsSells Richard Feynman: "The first principle is that you must not fool yourself – and you are the easiest person to fool." That quote, I think, is very important to keep in mind for any scientist. You have this calculation that doesn't prove you have a winning strategy but you feel it should say exactly that. Your job is to find out whether the suggested meaning is the actual meaning of the calculation. In this case it isn't and the expected value enables you to prove just that. – Stefan Mesken Jun 20 '18 at 14:34
• @StefanMesken Thanks - that is a great quote. Will put that one in my back pocket as I continue my studies! – Eliza Watts Sells Jun 20 '18 at 14:40