A Gambling Problem (statistics)

Question

I was watching a twitch stream the other day and the streamer was giving away his money based on a dice game. The game works as follows:

First, he rolls a 5 sided die to determine your initial starting money (from 1 to 5 dollars). From then on, you can either take the total money, or roll again. If you roll again, the value of the die will be added to your total money, unless you roll a 1 where you will lose it all.

He created a second game where you roll an 8 sided die with similar rules to the first game, except you lose all your money if you roll a 1 or an 8.

People in chat said that it was worse than the first game (since now you have a 75% chance to lose all your money instead of 80%), but there seems to be something intuitively wrong about it I can't explain.

I'm not a statistician, but if someone could point me towards some sort of way to figure out which game is better for the player it'd be much appreciated!

Answer 1

In the first case, if you have $x, your gain by playing is -x, 2, 3, 4, 5 with an average of (14-x) / 5 so you stop playing when x >= 14.

Second case your gain is -x, 2, 3, 4, 5, 6, 7, -x, average (27-2x) / 8, so again you stop playing with x >= 14.

You’d have to make a table with your expected outcome for each game to be sure.

Game 1:

x >= 14 -> x
x = 13 -> 0, 15, 16, 17, 18 average 13.2
x = 12 -> 0, 14, 15, 16, 17 average 12.4
x = 11 -> 0, 13.2, 14, 15, 16 average 11.64
and so on.

The probability of losing is higher in game 2, but if you win you win 4.5 on average instead of 3.5, and you start on average with 4.5 instead of 3, so you really need to create this table for the expected value.

An alternative was mentioned where your winnings are multiplied with the next number. Both variants have to stop when the banker runs out of money, say with a rule “when the calculated winning are larger than one billion, one billion is paid out and the game ends.

Or we could not optimise the number of dollars, but the benefit for the player. Getting one billion dollars or ten are basically the same benefit for me. Close enough that you wouldn’t risk losing the billion. Benefit grows linear with dollars only when the dollar value is within my normal means, say a year or a few years of my income.

Answer 2

As shown in the comments, the optimum strategy in both games is to continue rolling as long as your total winnings are $13$ dollars or less and stop when the winnings are $14$ dollars or more. With that strategy, the expected winnings in the first game are $5.81443$ and those in the second game are $5.836674$, so the second game is preferable, but not by much. We will analyze the first game; the analysis of the second game is sufficiently similar that you should be able to work it out.

It is evident from the optimal strategy and the rules that if $n$ is the number of rolls then $n \le 7$, and if $w$ is the total winnings then $w \le 18$. We will find it convenient for computational purposes to let $n=0$, so we have $0 \le n \le 7$ and $0 \le w \le 18$. Let's define $p(n,w)$ to be the probability that we have won $w$ dollars (cumulatively) on roll $n$. We will develop a recursive equation to compute $p(n,w)$. For an initial condition, $$p(0,w) = \begin{cases} 1 \qquad \text{if w = 0} \\ 0 \qquad \text{otherwise} \end{cases}$$ For $n > 0$, $p(n,w) = \sum_i (1/5) \;p(n-1,w-i)$ with appropriate limits on $i$. What are the constraints on $i$? For one, $2 \le i \le 5$ since the roll on turn $n-1$ is in the range from $1$ to $5$ but we only continue when the roll is greater than one. Clearly we had non-negative winnings on the previous roll, so $w-i \ge 0$. Finally, we only continue when the winnings so far are $13$ or less, so $w-i \le 13$. If we combine all these constraints, we find $$p(n,w) = \sum_{\max(2,w-13)}^{\min(5,w)}\frac{1}{5}\; p(n-1,w-i)$$ for $n>0$. Using this recursion, we can calculate $p(n,w)$ for all values of $1 \le n \le 7$ and $0 \le w \le 18$. Once we have done so, the expected total winnings are $$\sum_{w=14}^{18} \sum_{n=1}^7 w \;p(n,w)$$ which turns out to be $5.814438$.