Probability of winning HiHo Cherry-O

HiHo Cherry-O is a kid's board game. There is an option for "team" play that, mathematically has the following rules -- I am trying to determine what the odds of winning the team game are:

Spin the wheel for 1 of 7 results (equal odds of all outcomes):

A) +1 points
B) +2 points
C) +3 points
D) +4 points
E) -1 points
F) -2 points
G) "Puzzle" (will explain)

The goal is to get 40 points before you get 9 "Puzzles." Additionally, you can't go negative. So if you start with a -2 points you stay at 0. If you get 40 points before you get 9 "Puzzles" you win. If you get 9 "Puzzles" before 40 points, you lose.

What are the odds of winning?

Quote: masondc

HiHo Cherry-O is a kid's board game. There is an option for "team" play that, mathematically has the following rules -- I am trying to determine what the odds of winning the team game are:

Spin the wheel for 1 of 7 results (equal odds of all outcomes):

A) +1 points
B) +2 points
C) +3 points
D) +4 points
E) -1 points
F) -2 points
G) "Puzzle" (will explain)

The goal is to get 40 points before you get 9 "Puzzles." Additionally, you can't go negative. So if you start with a -2 points you stay at 0. If you get 40 points before you get 9 "Puzzles" you win. If you get 9 "Puzzles" before 40 points, you lose.

What are the odds of winning?

Using interation on an Excel sheet with columns labeled 0 puzzles to 9 puzzles and rows labeled 0 points to 40 points, I get 61.6056% as the probability of winning.

(I edited this after finding one error.)

I took a whack at this and got 82.97%.

I used a similar strategy to what ChesterDog explains: I wrote out a 41-by-10 matrix on the left designated as "inputs", and then a 41-by-10 matrix on the right designated as "outputs", and then two summary values "Win" and "lose".

My left-hand matrix runs from B2 (0 points, 0 puzzles) to K42 (which is actually impossible: 40 points and 9 puzzles). Row i is (i-2) points; column B is 0 puzzles, column C is 1 puzzle, and so on.

My right-hand matrix runs from M2 (0 points, 0 puzzles) to V42 (40 points, 9 puzzles) similarly.

There are formulas in the right-hand matrix.

For example, M2 = 1/7*(2*B2 + 2*B3 + B4). In other words, you get to 0 points and 0 puzzles from 0 points and 0 puzzles (2/7 of the time), 1 point and 0 puzzles (2/7 of the time), or 2 points and 0 puzzles (1/7 of the time).

The first four rows get progessively more terms, since you can get to 1 point from 0, 2, 3. You can get to 2 points from 0, 1, 3, 4, and so on. The second through eighth columns have an extra term because of adding a puzzle.

A typical cell in the middle of the table looks like this: R16 = 1/7 * (G12+G13+G14+G15+G17+G18+F16). In other words, you have a 1/7 chance of getting to 5 puzzles and 14 points from any of the states (5 puzzles and 10 points, 5 puzzles and 11 points, 5 puzzles and 12 points, 5 puzzles and 13 points, 5 puzzles and 15 points, 5 puzzles and 16 points, and 4 puzzles and 14 points).

The last three rows get fewer terms. The very last row, you have to be careful to add up the extra chances of going over 40. So, for example, M42 = 1/7*(B38 + 2*B39 + 3*B40 + 4*B41). In other words, getting to 40 points and 0 puzzles is a 1/7 chance from 36 points and 0 puzzles, and a 2/7 chance from 37 points and 0 puzzles, and a 3/7 chance from 38 points and 0 puzzles, and a 4/7 chance from 39 points and 0 puzzles.

The last column is simpler, as well. Each entry in the last column is just 1/7 times the corresponding eight-puzzle cell. For example, V2 = 1/7*J2 (in other words, 0 points and 9 puzzles is only reachable from 0 points and 8 puzzles). And so on.

With me so far?

The sum values I stuck into cells AA41 (Win) and AA42 (Lose).

AA41 = AA41 + SUM(M42:U42). The chance of winning up through this turn is the chance of winning up through last turn, plus all the states where we just hit 40 points and 0 through 8 puzzles.

AA42 = AA42 + SUM(V2:V41). The chance of losing this turn is the chance of losing up through last turn, plus all the states where we just hit 9 puzzles.

Now I made sure that AA41, AA42, and all the entries in the left-hand matrix are zero. I set Excel for manual computation and one iteration.

Now I put "1" in B2 (0 points, 0 puzzles) and I record an excel macro:

F9
select the right hand matrix (M2..V42) and copy
Select B2 and hit "paste special" "values"

and I assigned that to control-q. Then I mashed control-q and let it auto-repeat until the values in AA41 and AA42 stopped moving. The value in AA41 (Win) settled to 0.829745285, and the value in AA42 (Lose) settled to 0.170254715. That adds up to 1 (to rounding error) which eliminates certain kinds of errors I might have made in the spreadsheet.

I went through all this detail here since we got different answers and I'm reasonably sure of mine. I'm hoping someone will either reproduce my result or try it and show me what I screwed up.

This would have been easier in MATLAB -- I've done a similar analysis to show that the original Hi Ho Cherrio game is something like 52 or 53% to the first player, you can go look it up on google usenet news -- but I don't have handy access to MATLAB these days.

--JMike

Quote: JMike

I took a whack at this and got 82.97%....

I rechecked my work and found an error. My result (82.9745%) agrees with yours, now.