Not sure where to turn for this, but I’ve met some pretty smart people here so I thought I’d give it a shot:
For another game I play I’m trying to figure out something.
The best way I can explain it is to picture a roulette wheel with 9 slots, and I’m trying to get all slots filled with one ball in each.
The first ball goes in 100% of the time, but the second ball might land in the same spot as the first(a fail)
Best case scenario all 9 of my first 9 attempts all land in a different slot (game over, I win)
As a twist, every 10 fails gets me an “auto win” where I can place a ball in an empty slot without fail.
so worst case I fail the maximum number of times and I need 81 balls to win.
Any idea what the average number of balls needed is?
Cheers, and thanks if you even read this, let alone solved it for me =)
Is that something like a Pachinko machine?
Had to google pachinko, and I guess roulette is sorta like that.
The actual game I’m playing is a city building flash game, and you need to insert 9 “runes” into the slots of a ring on an ancient wonder to complete it.
Here is what’s easy to compute:
987654321 <- favorable case on each ball
999999999 <- total number of cases
That’s approximately 0.0937% chance to win after 9 plays.
Directly computing the average is a nightmare. However, if you disregard the “rule of 1 win after 10 fails”, you can linearize the expected value to simply get:
1 + 9/8 + 9/7 +9/6 +… +9/2 + 9/1
Expect to play 25 times on average.
Awesome, thank you Dandy.
Also, if you count 25 moves to win on average, that means you fail 16 times. So you get one freebie every time.
For obvious reasons, you should keep your freebie for the last rune. If that’s possible.