Ranking throws
Jason throws two darts at a dartboard, aiming for the center. The second dart lands farther from the center than the first. He throws a third dart. What's the probability the third is also farther from the center than the first?
Enumerating all rankings
Label the quality of each throw A (best), B (middle), C (worst). There are 6 possible orderings:
| Ranking | 1st | 2nd | 3rd | 2nd worse than 1st? | 3rd worse than 1st? |
|---|---|---|---|---|---|
| 1 | A | B | C | yes | yes |
| 2 | A | C | B | yes | yes |
| 3 | B | A | C | no | — |
| 4 | B | C | A | yes | no |
| 5 | C | A | B | no | — |
| 6 | C | B | A | no | — |
All 6 rankings are equally likely. The condition "2nd is worse than 1st" eliminates rankings 3, 5, and 6. Of the remaining three (rankings 1, 2, and 4), the 3rd dart is worse than the 1st in two of them.
Jason throws 3 darts. The 2nd is farther from center than the 1st. What's the probability the 3rd is also farther than the 1st?
A faster argument
The enumeration approach works for 3 darts but gets unwieldy for more. There's a better way to think about it.
The event "3rd dart is the best of all three" is independent of the relative ordering of the first two. Whether the 1st beat the 2nd or not, the 3rd still has a 1/3 chance of being the overall best. This independence is what makes the generalization work.
The general case
Jason throws darts. Each subsequent dart lands farther from the center than the first. He throws one more. What's the probability it's also farther than the first?
The event that the -th throw is the best among all throws is independent of the relative ordering of the first throws. Each throw has probability of being the overall best, regardless of any conditioning on the order of the others.
For (the original problem): . For : . As grows, you become more confident that the first dart is the best, but there's always a chance the next one beats it.