Consider a 10 mile length of road with mile markers exactly every mile including 0 and 10. Starting at the 0 marker, you drive towards 10 and randomly stop at some point with a position "p" defined as (0 marker)<= p <=(10 marker).

Part 1 (teaser): - What is the expected value of the distance from "p" to the nearest mile marker?

Part 2 (more difficult): - Three mile markers, randomly selected, have been removed by vandals, but 0 and 10 remain. Same question as part 1.

Part 3: - Consider part 2, but markers 0 and 10 are also potentially part of the vandalism.

Does the answer change from part 2, and if so what is it?

(You have no problem knowing where to start your trip, even if marker 0 happens to be missing)

If three mileposts are removed, the problem is indeed harder.

a) If the three are consecutive, then there is one gap of 4 miles, and six 1 mile gaps.

If stopping in the 4 mile gap, the expected distance is 4*(1/4) = 1 mile.

If stopping in a 1 mile gap, the expected distance is 1/4 mile.

There is a 40% chance of stopping in the 4 mile gap.

So the expected distance to a signpost is .4*1 + .6*(1/4) = .4 + .15 = .55

b) If two of the three are consecutive, then there is one gap of 3 miles, one gap of 2 miles, and five 1 mile gaps.

If stopping in the 3 mile gap, the expected distance is 3*(1/4) = 3/4 mile.

If stopping in a 2 mile gap, the expected distance is 2/4 mile.

If stopping in a 1 mile gap, the expected distance is 1/4 mile.

There is a 30% chance of stopping in the 3 mile gap and 20% chance of stopping in the 2 mile gap.

So the expected distance to a signpost is .3*(3/4) + .2*(2/4) + .5*(1/4) = (9+4+5)/40 = .45

b) If none of the three are consecutive, then there are three gaps of 2 miles, and four 1 mile gaps.

If stopping in a 2 mile gap, the expected distance is 2/4 mile.

If stopping in a 1 mile gap, the expected distance is 1/4 mile.

There is a 60% chance of stopping in a 2 mile gap and 40% chance of stopping in a 1 mile gap.

So the expected distance to a signpost is .6*(2/4) + .4*(1/4) = (12+4)/40 = .40

d) There are C(9,3) = 84 ways of choosing 3 of the 9 signposts to remove.

7 of those ways select 3 consecutive (starting with 1,2,...,7).

How many ways to remove 2 but not 3 consecutive?

You can remove 1,2 and one of 4-9, for a total of 6 ways.

Similarly, remove 8,9 and one of 1-7, for a total of 6 ways.

There are 6 other ways to remove two consecutive, and for each of those 5 which can be removed that are not adjacent, for 30 ways.

Ways to remove two adjacent = 6 + 6 + 30 = 42

Ways to remove 3 non-adjacent = 84 - 7 - 42 = 35.

e) Final calculation: (7/84)*.55 + (42/84)*.45 + (35/84)*.40 =

(1/12)*(11/20) + (6/12)*(9/20) + (5/12)*(8/20) =

(11 + 54 + 40)/240 = 105/240 = 7/16 = .4375