# Does the casino have an edge on a commission-free roulette wheel?

I have an interesting discussion with someone on YouTube.

Assume a commission-free roulette wheel (without the number 0) so you have a true 50/50 chance on black or red. Also, assume that all players will use the martingale betting strategy. Every player starts with $$\1000$$ and the maximum bet is also $$\1000$$ per game. Also assume that all players will play as long as they can until they are bankrupt.

According to the other person, the casino has an edge due to betting limits. Because every player will eventually go bankrupt and, thus, stop playing, the casino has taken all the players money ($$\1000$$) and then the next player comes in with a fresh $$\1000$$ and the process starts again. So due to betting and bankroll limits, the casino has always an edge, even in this commission-free roulette.

Although this example seems very logical, I don't agree on this. If you look to it on a mathematical perspective, every 50/50 game is independent and has an expected return of $$0\%$$ (neither the casino nor the player has an edge). It does not matter which betting strategy or limit there is. If you play a million games, all with different betting amounts (whatever strategy) all these games will add up to the same $$0\%$$ expected result.

According to me, if you consider the mathematics, which seems to be bullet-proof, it is useless to think of any strategy that will give the house an edge. But it is also very unnatural to not agree to the fact that the casino has the edge when every player stops on bankruptcy.

In my defense, I state that some players will never get bankrupt and those compensate for all bankrupt player. But is this so?

Which leads to the final question: does the casino has an edge in this example?

• if yes, how does this align with the expected result of $$0\%$$ of each game?

• if not, how can you easily explain the paradox that the casino does not make money when every player stops on a loss?

• The issue is that you're forcing players to make bad decisions which influences the outcome. The martingale strategy doesn't work unless you have infinite money otherwise it's guaranteed to bankrupt you. If you choose a strategy that will always bankrupt you this is good for the casino. Commented Jan 26, 2022 at 9:56
• @CyclotomicField In a comission-free game, there are no bad decisions (also no good decisions of course). The expected outcome of every game is always zero. Commented Jan 26, 2022 at 10:24
• @CyclotomicField how is it possible to choose a strategy that benefits the casino? That doesn't fit the math..... I could then just inverse-mirror player A's strategy as player B on the same table and take all the casino's profit. Commented Jan 26, 2022 at 10:24
• In a fair game, there are no bad decisions. In the real game, you could force your loss (for example by betting 1 dollar on each number), but no (reasonable) strategy is a "bad decision" a priori. Explain someone cracking the lottery Jackpor that it was a bad decision to participate. Commented Jan 26, 2022 at 10:44
• The argument that each player eventually goes bankrupt with probability $1$ is correct if (a) the casino has infinite money and (b) there is no limit on the number of games (or the size of bets), which are both dubious assumptions. But if you are allowing these infinities then you could equally assume an infinite number of players, and at any point in time some of them would still be playing. Commented Jan 26, 2022 at 10:47

First, lets ignore the martingale and say instead that you always play for \$1, until you are out of money. Does the casino have an edge over you? Answer: This is not actually well defined. 'having an edge' means that the expected outcome (i.e. averaged over all possible games) is non-zero. But the problem is that it is possible for a game to never end, because the player just keeps winning more than loosing. What is the result of such a game? It is not defined, and therefore the average over all results is not defined. If you have an upper limit of games, for example "after a million games, the player stops, even if not bankrupt yet", then the answer is, that there is exactly no house-edge. The chance of a player not being bankrupt after a million games is tiny, but the winnings of that player are huge. In the average, it cancels exactly and the expected outcome is zero. If you increase the maximum number of games, the number of winning players gets smaller and smaller, but their winnings get larger and larger. It always cancels exactly. Secondly: The Martingale is irrelevant. As long as there is a maximum bet (or the player only has a finite amount of money to bet), Martingale certainly does not change anything. Lets say the minimum bet is \$1 and maximum is \$1000, then you can do at most 9 steps of martingale. If you win at least one of the 9 steps, you win \$1 overall. If you loose all 9 steps, you loose \\$511 overall (because $$1+2+4+..+256=511$$). The loss has a chance of exactly one in $$2^9$$, or $$1/512$$. So the expected result is \begin{align} \1\cdot\frac{511}{512} - 511\cdot\frac{1}{512}=\0. \end{align} This means that by playing a martingale, the expected result did not change. The only thing you did is replacing a commision-free 50:50 game with a commision-fre 1:511 game. You move these numbers around as much as you like with complicated betting strategies, but the overall expected result will always be zero.
In one sense, a casino running such a game would usually have an edge over any individual player, but the edge has nothing to do with betting limits. It arises because the casino's capital is typically a large multiple of any individual player's. Although the probability of any individual player's winning all of the casino's capital is very small, however, it is not zero, and in the long run the casino will go broke with probability $$1$$.
The problem of determining the probability that the casino, playing against a single player, will win all of his or her money is known as the fair gambler's ruin problem. If the casino's current capital is $$\ C\$$, and the player's current capital is $$\ P\$$ ($$\ =\1,000\$$ in your scenario) then the probability of the casino's winning all of the player's money is $$\ \frac{C}{C+P}\$$, largely independent of any strategy adopted by the player. The only thing the player can do to affect this probability is to bet more than $$\ C-P\$$ at some stage, in which case the casino's chances of winning go up to $$\ \frac{C}{4P-2C}\$$ (provided the player makes no further mistakes).
After $$\ i\$$ players, all with the same initial capital $$\ P\$$, have gone broke, the casino's capital will have risen to $$\ C+iP\$$ (ignoring any running costs), so the probability it will win all the $$\ (i+1)^\text{th}\$$ player's money is $$\ \frac{C+iP}{C+(i+1)P}=1-\frac{1}{C+(i+1)P}\$$. The probability of its winning all the money of $$\ n\$$ players in succession is therefore $$\prod_{i=0}^{n-1}\left(1-\frac{1}{C+(i+1)P}\right)\ .$$ Since the infinite sum $$\ \sum_\limits{i=0}^\infty\frac{1}{C+(i+1)P}\$$ diverges to $$\ \infty\$$, a well-known criterion for the convergence or divergence of infinite products tells us that $$\ \prod_\limits{i=0}^\infty\left(1-\frac{1}{C+(i+1)P}\right)\$$ diverges to zero. Thus, the probability of the casino's surviving indefinitely is zero.