I play in a skeeball league. Yes, it's a real thing.
Each player is given 9 balls to roll every game. I'm hoping to maximize my score based on the probability that I'll get the ball in a certain scoring ring. For example, if I can get the ball into the 40 point ring 60% of the time, should I try for the 40 point ring for all 9 balls does it make more sense to try for the 100 point ring every time even though my make chance is only 20-25%? Or should it be a hybrid approach and try for 40 point ring for a certain number of balls and then go for the 100 scoring ring the remainder?
To make it a little more complicated, if I were to miss a 40 scoring ring, it doesn't result in 0, but most likely a 20 (see link to skeeball scoring ring below). And if I were to miss a 100 point ring, it often results in a 10. Here's what the scoring rings look like.
Based on my previous year, I believe if I aim at a particular ring my chances to getting the ball in that scoring rings are as follows -
Scoring Rings / if aiming at the ring my chances of getting it / ring likely to get if target missed
- 0 scoring ring / my odds 1% (very rare) / 0
- 10 scoring ring / my odds 99% chance of making / 0
- 20 scoring ring / my odds 80% chance / 10
- 30 scoring ring / my odds 70% chance / 20
- 40 scoring ring / my odds 60% / 20
- 50 scoring ring / my odds 40-50% / 10
- 100 scoring ring / my odds 20-25% / 10
Can anyone point me in the right direction to determine how I should approach my game of skeeball in order to maximize my final score?
NOTE: Final scores can only be in 10's i.e. 240, 350, 310, 400 (not 275, 387, etc)
antkam has the right idea, but I'll flesh it out a bit more for completeness' sake.
The expected value is computed as a sum of [value] * [probability of getting that value]. You can focus on a given scoring ring and compute the expected points scored for that ring. For example: if you aim for the 40 scoring ring, then the two outcomes that can happen are that you'll get 40 points (with probability $.6$) or get 20 points (with probability $.4$). Thus, the expected value of a throw that aims at the 40 ring is $$40 \cdot 0.6 + 20 \cdot 0.4 = 32.$$
Expected values are additive, meaning that if you aim for the 40 ring twice, your expected total score will be 64. A consequence of this is that as antkam said, all you really have to do is compute the expected value for each ring, find the best one, and then spend every ball aiming at that.
As a closing note, I'll point out that if you want to get a bit fancier with the probabilities, you can easily do that too. For instance, suppose you decide that aiming for the 40 gives you a 60% chance of hitting it, a 30% chance of falling to the 20, and a 10% chance of falling all the way to 10. In that case, your expected value would be $$40 \cdot 0.6 + 20 \cdot 0.3 + 10 \cdot 0.1 = 31.$$
You can adapt the expected value to as many scenarios as you like in this fashion.