Matt's Hi Ho! Cherry-O Analysis

"Hi Ho! Cherry-O" is a game in which you try to collect 10 cherries from a tree before your opponents. Both "game" and "try" are used loosely, since players get no choices. You just keep spinning and see what happens.

Actually, parents get one choice, which is who goes first. (Not according to the rules, which say that the youngest player goes first, but practically speaking.) How much does this matter?

The games is very easy to implement a Monte Carlo for. Here's one. It simply keeps randomly picking one of the spinner outcomes, which either give you 1, 2, 3, or 4 cherries, take away 2 (because of a dog), take away 2 (because of a bird), or take them all away.

Players #1 wins #2 wins #3 wins #4 wins Mean plies
1 100% 15.8
2 51.9% 48.1% 18.7
3 35.8% 33.3% 30.9% 21.4
4 27.7% 25.8% 24.1% 22.4% 24.2

where a "ply" is one turn taken by one player.

Warning: You need a good random number generator to do this analysis. I first used Linux's random(), but found that, for a 1-player game, it gave a very significant spurious peak in probability for finishing in 32 turns. Substituting ROOT's implementation of the Mersenne Twister algorithm instead gave much more believable results, with overall somewhat lower average game lengths. I'd never seen PRNG quality actually matter before!

The good news is that the total length of the game doesn't go up too much with more players. The bad news is that in a 4-player game, the 4th player is at a quite substantial disadvantage. The first player has a 24% higher chance of winning than the fourth. This effect gets stronger as you add players, so it's a good thing there are only four cherry trees in the box. As the number of players goes to infinity, the probability that one of them wins on the third round goes to one, with the Nth player having a 10/343(1−10/343)N−1 chance of winning.

As in other children's games of chance, the possibility of a player getting reset to zero means that it is possible for games to take a very long time. Once you have progressed several turns beyond the first at which it is possible to win — the third, by getting {4, 4, 4}, {4, 4, 3}, {4, 4, 2}, or {4, 3, 3} with a probability of 10/343 = 2.9% — the game quickly loses its memory of how it got into its current state. In the moment, this means the average number of remaining turns is always about the same. Viewed overall, it means the probability distribution is an exponential decay. For a 1-player game:

      6.90% |
      6.60% |     _
      6.30% |      ~
      6.00% |    -  -
      5.70% |        _
      5.40% |
      5.10% |         ~
      4.80% |          -
      4.50% |           _
Prob  4.20% |
 of   3.90% |            ~
 End  3.60% |             ~_
      3.30% |               _
      3.00% |   _            _
      2.70% |                 -
      2.40% |                  ~_
      2.10% |                    -_
      1.80% |                      -_
      1.50% |                        -_
      1.20% |                          ~-_
      0.90% |                             ~--_
      0.60% |                                 ~~--___
      0.30% |                                        ~~~----______
      0.00% |---                                                  ~~~~~~~~~~~~~~--------
             0        10        20        30        40        50        60        70

Once you get a bit past the peak, it's an exponential with a decay constant of 12.4 turns, i.e. a half-life of 8.6 turns.

But Wait

Upon close inspection of the spinner, the seven regions are not the same size. The 1-cherry wedge is noticeably smaller. Nor is the spinner centered quite where it looks like it should be based on the division lines. In my set, that makes 1-cherry even less likely. The probabilities look to be approximately:

Outcome Probability
1 cherry 0.91/7
2 cherries 0.97/7
3 cherries 1.04/7
4 cherries 1.03/7
Lose 2 (both) 2.09/7
Lose all 0.96/7

No doubt the way the spinner is positioned with respect to the backer varies by set, so yours might be somewhat different. Anyway, how does this change things? The only obvious consequence is that it is slightly more likely to get the 3-turn win.

It turns out that the mean number of plies edges down slightly by about 0.2 for each number of players. However, the probability of winning for each player doesn't budge at the 0.1% level, even for the maximally unbalanced 4 player game. No doubt the engineers who set the tolerances had this in mind *eye-roll*.

Other People Who Have Done This

Here's one.

Other Variants

I found someone else on the Internet who had analyzed this game, but some other version where one of the bad spins takes away one cherry instead of two. This makes the game about 1 ply shorter. And some people have a version where both mild bad spins take away 1 instead of 2. Unsurprisingly, that reduces the length by about another ply.