# Gambling in a rigged casino

A well-known crooked casino is offering the following bet with a pair of (identically looking and indistinguishable) dice that both show the numbers $1,2,3,4,5,6$ on their faces.

• The first die is fair and perfectly standard: throwing it yields each of the outcomes $1,2,3,4,5,6$ with equal probability $1/6$.
• The second die is loaded: throwing it yields each of the five outcomes $1,2,3,4,5$ with equal probability $1/5$ (while $6$ is impossible to throw).

We stress that there is no way whatsoever for an honest gambler to tell the two dice from each other. Now, in the bet that we are considering, the gambler is allowed to throw these dice as often as he likes, subject to the following casino rules:

• The gambler has to pay $1$ Euro for every single throw of a single die.
• Whenever he switches from throwing one die to throwing the other die, he must pay $5$ Euro.
For his very first throw, he may arbitrarily choose either of the dice (without paying anything).
• As soon as he throws a $6$, the game is over. The gambler wins $20$ Euro.
• At any moment in time, the gambler may decide to quit the game and leave the table at no further cost.

Question: What is the best strategy for this game and what is the expected profit?

• I assume each game begins with a randomly chosen die? Feb 6 '15 at 13:15
• @frodoskywalker: Yes, the gambler is free to choose the die with which he wants to start. Feb 6 '15 at 13:39
• a best strategy: come with a friend, each take a dice. On average you will spend 12 eur in total and win 20, so 4 each :) Feb 6 '15 at 14:29
• Rather than crooked, this casino seems very generous. Legitimate casinos always defines the rules of the games to have a positive house edge even with optimal strategy, this game gave a negative house edge even after the crooked rule. Seems like they need to hire a better mathematician. Feb 7 '15 at 5:13
• The interesting question is, if it is ever better to switch dice for 5 Euro instead of resetting the game for free. Resetting gives you the right die with 50%, if you pay 5 Euro to switch, you will have to be a lot more confident! Feb 7 '15 at 14:02

What follows is a strategy for earning an expected profit of about $3.63$ Euro, and an argument that this strategy is optimal.

First we'll pick some positive integer $k$. We will switch dice after every roll number that is an odd multiple of $k$, and continue to roll until we get a 6. So we will roll the first die $k$ times, then the second die $2k$ times, then the first die $2k$ times, and so on.

With probability $1$ we will eventually roll a 6 and win the 20 Euro. To compute the expected cost, we consider two cases, depending on which die we pick up first.

First die is fair. We expect to roll the fair die $6$ times. We expect to roll the loaded die $$\sum_{n\geq 0}\left(\frac{5}{6}\right)^{2nk}2k = \frac{2k\cdot(5/6)^k}{1-(5/6)^{2k}}$$ times. The expected number of switches is $$\left(\frac{5}{6}\right)^k+2\cdot\sum_{n\geq 1}\left(\frac{5}{6}\right)^{(2n+1) k} = \frac{2(5/6)^k}{1-(5/6)^{2k}}.$$ So if we start with the fair die, our expected profit is $$20-6-\frac{2k\cdot(5/6)^k}{1-(5/6)^{2k}}-5\cdot\frac{2(5/6)^k}{1-(5/6)^{2k}}.$$

First die is loaded. We again expect to roll the fair die $6$ times. We expect to roll the loaded die $$k+\sum_{n\geq 1}\left(\frac{5}{6}\right)^{2nk} 2k=\frac{k\cdot(1+(5/6)^{2k})}{1-(5/6)^{2k}}$$ times. The expected number of switches is $$1+2\cdot\sum_{n\geq 1}\left(\frac{5}{6}\right)^{2nk}=\frac{1+(5/6)^{2k}}{1-(5/6)^{2k}}.$$ So if we start with the loaded die, our expected profit is $$20-6-\frac{k\cdot(1+(5/6)^{2k})}{1-(5/6)^{2k}}-5\cdot \frac{1+(5/6)^{2k}}{1-(5/6)^{2k}}.$$

Expected profit Our overall expected profit is the mean of the expected profit if the first die is fair and the expected profit if the first die is loaded. The expression simplifies to $$14-\frac{1}{2}\frac{(5+k)\left(1+(5/6)^{k}\right)^2}{1-(5/6)^{2k}}.$$ Numerically, I found that profit is maximized taking $k=9$. The expected profit using this strategy is $$\frac{4218321}{1160653}\approx 3.63\,\,\,\text{Euro}.$$

Is this strategy optimal? Here's a sketch of a proof that this strategy is optimal.

Recall that the odds a given die is loaded is the ratio $$\frac{\text{probability that die is fair}}{\text{probability that die is loaded}}.$$ The product of the odds that each die is fair is $1$.

During the course of the game, we only get information of one kind, namely that the die we just rolled did not turn up 6 (if it turns up 6, the game ends). Bayes' Theorem says that whenever a die turns up something other than 6, the odds that it is fair get multiplied by $\frac{5}{6}$ (and so the odds the other die is fair get multiplied by $\frac{6}{5}$). This means that assuming no 6's have been rolled, the odds that a die is fair depend only on the different between the number of times that die has been rolled, and the number of times the other die has been rolled.

When we are deciding whether to roll our current die, switch dice, or walk away, we need only consider the odds our current die is fair. Any costs already paid (sunk costs) should not be taken into account, as future payments and earnings do not depend on them. This means that possible strategies fall into two possible kinds:

• If the odds the current die is fair are below a certain threshold, leave the game. Otherwise roll.
• If the odds the current die is fair are below a certain threshold, switch dice. Otherwise roll.

The strategy described above is of the second kind, where the threshold is $\left(\frac{5}{6}\right)^k$. We choose $k$ to optimize profit among strategies of this kind. The only other possibility is a strategy where we roll the current die $n$ times (for some fixed $n$), then leave the game if we haven't rolled a 6. It's not too hard to write down the expected profit here as a function of $n$, find the $n$ that maximizes profit (if I computed correctly it is $n=5$), and see that this doesn't do better.

• In hindsight this doesn't seem like a very good strategy Feb 6 '15 at 17:50
• Actually, I think it is the best. A gambling game with an expected return of 3.6x the buy-in? I'd play that. Feb 6 '15 at 19:14
• My intuition says that if you haven't won after 2*k rolls, either die is as likely to be the fair die, so switching at that point is a waste of five euros - the expected game length should be equivalent if you skip this switch, no? Feb 6 '15 at 22:09
• @waxwing In this strategy, you don't switch after $2k$ rolls. You roll the die you start with $k$ times, then switch. You need to roll the other die $2k$ times before you have rolled it $k$ times more than the first die. You switch after roll numbers that are odd multiples of $k$. Feb 6 '15 at 22:16
• I think I know why the casino is corrupt. It needs to rig all its other games in order to make up the 3.6 euros it typically loses whenever someone plays this game! Feb 7 '15 at 1:46

I'm not expecting that this is the best answer, just helping out the thoughts.

As a baseline, the strategy to sit down and roll once and walk away is a positive bet.

$EV = (\frac{1}{2})(-1) + (\frac{1}{2})(\frac{1}{6})(19) + (\frac{1}{2})(\frac{5}{6})(-1) = \frac{2}{3}$

There's a 50% chance to get the bad die and finish -1. There's a 50% chance of getting the good die and then a 1/6 chance of winning and coming out +19 or a 5/6 chance of coming out -1.

• totals up to 1/12 chance of winning the € 20 on just the first cast Feb 6 '15 at 14:12
• If I've understood the question correctly, there would be a $1/12$ chance of winning 19 Euros (net), not 20, since you still would have had to pay 1 Euro at the beginning in order to play. Feb 6 '15 at 18:14
• I actually just edited it, and then realized that was wrong and then edited it back. It should be right now. Feb 6 '15 at 18:23

Following from Jason's answer, if we ignore the possibility of switching dice, we can look at this as a simple game where paying €1 gives you a 1/12 chance of winning €20, and 11/12 chance of getting nothing. It is clear that this simplified version is a winning game. The chance of playing 20 times without winning is $\left(\dfrac{11}{12}\right)^{20}\approx0.175$, so you have about $82.5\%$ chance of winning (or breaking even). Furthermore, if you play this game indefinitely, until you win, your expected return is $$\sum_{n=0}^{\infty}\left(\frac{1}{12}\right)\left(\frac{11}{12}\right)^n \times(19-n)=8$$ So, on average, a player of this simplified game will win €8. Of course, this isn't a completely accurate picture, since well before you get to 20 failed rolls, it becomes much more likely that you have the rigged die, and will never win. So, the real question is, at what point does it become worthwhile to pay the €5 to switch?

My guess is that you should play up to 7 times, and if you haven't won by then, switch dice and and play up to 7 more times. I will try to verify (or disprove) this, and then update my answer.

Update:
If you choose the fair die initially, your expected return after 7 rolls is $$\sum_{n=0}^6\left(\frac16\right)\left(\frac56\right)^n(19-n)\approx12.05$$ However, you also have a $\left(\dfrac56\right)^7\approx28\%$ chance of losing, in which case you will switch and continue with the loaded die, losing a total of €19, which lowers the expected return to about €6.75

If you choose the loaded die first, you will end up in the same €12-return situation as above, except after spending €12 on 7 die rolls and one switch, giving an expected return of about €0.05. Well, it's better than nothing.

Given that these two situations are equally likely, this strategy gives an expected return of about $(6.75+0.05)/2=3.4$.

• Your first paragraph is a brilliant insight, with an obvious conclusion: it is a decent strategy to play many games of 1 round each. Each round has an expected winnings of € 0.67. I would +1 your answer for that, except that all the rest of your answer ends up in the weeds . . . Feb 7 '15 at 6:54