Coming up with an optimal strategy.

Question

I've been given the following problem in the context of probability:

"The player is playing a game where they have to pick one of five wheels to spin. Each wheel has a different chance to win and a different prize if the wheel lands on a positive segment (See Wheel Details table). If the player spins a wheel and lands on a negative segment the game ends and they get to keep all prizes they have won up to that point. If the player spins a wheel and lands on a positive segment, the player wins the prize for that wheel and gets to spin another wheel.The player can only spin each wheel once. The player has all the information about the wheels when they select which to spin. In which order should the player spin the wheels to achieve the optimal strategy?"

So far, what've i've done is calculate the probability of winning on each wheel. This seems like the most obvious solution, however, given that one can win more money on wheel 4 than all others put together, it doesn't seem "optimal". My next idea was to calculate the expected value of each wheel, but they all give 10, so it's inconclusive. Finally, i calculated the standard deviations of each (35.35533906,7.071067812,3.889087297,176.7766953,14.14213562 respectively), however, this gave the exact same strategy as when considering the probabilities.

I'm not looking for someone to tell me the optimal strategy, far from it. I'm more interested in ways one could explore different strategies given the information available, so any hints on how to approa

Answer 1

For each permutation $(p_1,p_2,p_3,p_4, p_5)$ of $\{1,2,3,4,5\}$, the player can decide on the strategy of:

I will spin $p_1$ first, then, if I am still in the game, I will spin $p_2$, then $p_3$, then $p_4$, then $p_5$.

For each such strategy there are $5$ possible outcomes:

1. Wheel $p_1$ stops on negative segment.
2. Wheel $p_1$ stops on a positive segment, wheel $p_2$ stops on a negative segment.
3. Wheels $p_1$ and $p_2$ stop on a positive segment, wheel $p_3$ stops on a negative segment.
4. Wheels $p_1$, $p_2$ and $p_3$ stop on a positive segment, wheel $p_4$ stops on a negative segment.
5. Wheels $p_1$, $p_2, p_3$ and $p_4$ stop on a positive segment, wheel $p_5$ stops on a negative segment.
6. All wheels stop on a positive segment.

You can calculate the probability and winnings of each of the $6$ events, and from that, calculate the expected winnings of the strategy.

Repeating (hopefully with a computer) the thing with all $5!$ strategies can yield you the best one.

Answer 2

You should spin the wheels in descending order of probability of landing on a positive segment.

Suppose you have chosen a strategy -- number your wheels so that in your chosen turn order the first wheel is 1, the second 2, and so on.

Write $W$ for your payoff from the whole game, write $W_i'$ for your payoff from wheel $i$ if you get to spin it, and $W_i$ for the payoff from the wheel (which might be 0 if you didn't get to spin it at all), and write $p_i$ for the probability that you land positively on wheel $i$. Then $$ \mathbb E(W) = \mathbb E\left(\sum_{i=1}^n W_i\right) = \sum_{i=1}^n \mathbb E(W_i) = \sum_{i=1}^n \prod_{k = 1}^{i-1} p_i \mathbb E(W_i') = 10 \sum_{i=1}^n \prod_{k = 1}^{i-1} p_i. $$ Now it is clear that to maximize this sum of products, the element $p_i$ that you want to occur in every term is the highest, the element $p_j$ that you want to occur in every term but one is the second highest, and so on.

In your specific case you should therefore play $3, 2, 5, 1, 4$.