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The Pace of the Race (Posted on 2007-06-19) Difficulty: 4 of 5
In a simplified version of a child’s board game, players "race" around the board going from Start to Finish, moving their tokens based on the roll of a six sided die. The board has 101 spaces where the Start space represents "0" and the Finish space represents "100". An exact roll is not required to cross the finish line. To ensure that one child does not have an advantage, all the children roll the die separately, but move simultaneously.

Part I. With one child playing, how many turns (on average) will it take to have a winner?
With 2, 3, or 4 children playing, how many turns will it take (on average) to have a winner?

Part II. With 4 children playing, when one of the children crosses the finish line, how far along should the child in 2nd place, 3rd place, and 4th place be? (Again, looking for the statistical average position.)

Part III. What would be the solution to the previous questions if an exact roll is required to cross the Finish line? (If the roll is too high, the player loses a turn and does not move.)

No Solution Yet Submitted by Leming    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Part 1 for 1 player | Comment 3 of 4 |

These are pretty close to Charlie's simulation values.  I based everything on a normal distribution, since the central limit theorem says I should get a good approximation.  Everything here is therefor approxiamate but probably fine to 4 or 5 decimal places.  I used my TI-83 Plus graphing calculator because that's what I know how to quickly calculate normal probabilities with. 

It should be noted that the mean sum of X rolls of a die is 3.5X and the standard deviation is sqrt(X*35/12)

The calculator's format for normal probabilities is normalcdf(LB,UB,Mean,St.Dev)

I made a list L1 of the numbers 17 to 57 (17 being the minimum rolls to cross the finish, and 57 because beyond it the probabilities get negligably small.)  Let X be the number of rolls, the approximate probability of winning in X is normalcdf(100,1000,3.5X,sqrt(X*35/12)) - normalcdf(100,1000,3.5(X-1),sqrt((X-1)*35/12)) put these into L2

The number of roll with the highest probability is 29 p(29)=.152355
The mean number of rolls = 29.190476
St.Dev. = 2.637598

This seems reasonable, and it is close to the simulation.

  Posted by Jer on 2007-06-22 19:41:05
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