Nate got a series of classic kids games for Christmas: Hi-Ho-Cherry-Oh, Candyland, Memory and Cooties. With the exception of Memory, these are all games of complete and total chance. For those that haven’t played, three of them basically go like this:
- Invoke a random event.
- Execute required action.
- Has victory condition been met?
- If so, yippee.
- If not, repeat from step 1.
For Hi-Ho-Cherry-Oh, the randomizer is a spinner with 7 positions. Four positions let you take from 1 to 4 cherries off your tree, two positions penalize you 2 cherries and the last causes you to loose all of your cherries. Once you have collected 10 cherries, you win.
So being the geek that I am, I found myself wondering the average number of cherries gained per spin. At first this is simple (1+2+3+4-2-2)/6=1. Unfortunately, there is that nasty bucket which is a variable penalty. Rather than create the entire statistical equation, I simply created a Monte Carlo simulation in Excel. After about 100 ‘games’ it was clear that the average payout is between 1 and 1.1 cherries per turn.((This is determined as the 10 cherries needed to win divided by the number of spins required to get there. Even though the last spin might put you over 10, you can’t get that many off the tree.))
The shortest game is three spins (4,4,4;4,4,3;4,4,2;4,3,3). There are 10 winning series as the spins could come in any order and 343 possible outcomes. The odds are 2.9% of winning in the minimum amount of time. Of course the odds of a shorter game improve with more players as the game ends if any of the players succeeds, so roughly this increases to 11%. The maximum game is theoretically infinite, but about 40% of the time 10 spins or fewer are required (85% of the time for four players). By 20 spins, a solitaire game is over 75% of the time (with four players this would increase to 99.6%).
Ok, I now return you to your regularly scheduled intra-holiday period.
Maybe I’ll work on Cooties tomorrow. Candyland is just too annoying.