From Algorithms to Z-Scores - matloff - University of California, Davis

2.10. EXAMPLE: A SIMPLE BOARD GAME 17
for ALL j in 1,...,5.
That may seem counterintuitive. Yet the example here is in essence the same as one found as an
exercise in many textbooks on probability:
A man has five keys. He knows one of them opens a given lock, but he doesn’t know
which. So he tries the keys one at a time until he finds the right one. Find P(N = j), j
= 1,...,5, where N is the number of keys he tries until he succeeds.
Here too the answer is 1/5 for all j. But this one makes intuitive sense: Each of the keys has chance
1/5 of being the right key, so each of the values 1,...,5 is equally likely for N.
This is then an example of the fact that sometimes we can gain insight into one problem by
considering a mathematically equivalent problem in a quite different setting.
2.10 Example: A Simple Board Game
Consider a board game, which for simplicity we’ll assume consists of two squares per side, on four
sides. A player’s token advances around the board. The squares are numbered 0-7, and play begins
at square 0.
A token advances according to the roll of a single die. If a player lands on square 3, he/she gets
a bonus turn. Let’s find the probability that a player has yet to make a complete circuit of the
board—i.e. has reached or passed 0—after the first turn (including the bonus, if any). Let R denote
his first roll, and let B be his bonus if there is one, with B being set to 0 if there is no bonus. Then
(using commas as a shorthand notation for and)
P (doesn’t reach or pass 0) = P (R + B ≤ 7) (2.30)
= P (R ≤ 6, R = 3 or R = 3, B ≤ 4) (2.31)
= P (R ≤ 6, R = 3) + P (R = 3, B ≤ 4) (2.32)
= P (R ≤ 6, R = 3) + P (R = 3) P (B ≤ 4) (2.33)
= 5 1 4
+ ·
6 6 6
(2.34)
= 17
18
(2.35)
Now, here’s a shorter way (there are always multiple ways to do a problem):