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)
1: The distance function is a sawtooth. Between each pair of markers the distance goes from zero to 1/2 and back to zero, each half-mile segment being linear. The expected distance is 1/4 mile.
2: There are three cases: the three mile markers are all adjacent, two are adjacent but one is separate, no two adjacent markers have been removed.
Case 1: There is a 4/10 probability the stopping is in the 4-mile stretch between successive markers and the expected value there is 1 mile, and there's a 6/10 probability the stop is in the normal area where the expected value is 1/4. Weighted, the overall expected value is 0.4 * 1 + 0.6 * 1/4 = 2/5 + 6/40 = 11/20.
Case 2: There's a 3/10 probability that the expected value is 3/4; a 2/10 probability that the expected value is 1/2; and a 1/2 probability that the expected value is the normal 1/4. Combined, we get the weighted expected value as being 9/20.
Case 3: There's a 6/10 probability the stop will be in a 2-mile stretch, with expected value of 1/2 and a 4/10 probability of being in a 1-mile stretch with 1/4 expected value; making an overall expected value of 4/10.
Now these cases themselves must be weighted by the probabilities each would be produced by the vandals. Overall there are C(9,3)=84 possible sets of markers that could have been taken.
Case 1 (three in a row) could start at any of markers 1 - 7, so the probability of case 1 is 7/84 = 1/12.
In case 2, the two in a row could start anywhere from marker 1 through marker 8. In the two extreme cases there are 6 lone markers that could have been taken, but if the first marker of the adjacent pair were 2 through 7, there would be only 5 places for the lone marker taken. So that's 5*6 + 6*2 = 42, so the probability of case 2 is 42/84 = 1/2.
That leaves the probability of case 3 as 1 - 1/2 - 1/12 = 5/12.
The overall expected value is therefore:
1/12 * 11/20 + 1/2 * 9/20 + 5/12 * 4/10 = 7/16.
3: Now there are C(11,3) = 165 possible sets of markers taken, of which we've considered 84, leaving 81 possibilities where one or both of the end markers have been taken.
There are 5 cases:
0-2 or 8-10 have been taken
0,1 and any of 3-9 or 9 and 10 and any of 1-7 have been taken
0,1 and 10 or 9,10 and 0 have been taken
0 and any adjacent pair starting at 2-8 or 10 and any adjacnt pair starting at 1-7
0 and 10 and any individual marker from 2 - 8
We can call these cases 4 through 8:
To start:
Case 4: Markers 0-2 or 8-10 have been taken. There's 3/10 probability the stop has taken place in the affected area so the conditional expected value is 3/10 * 3/2 + 7/10 * 1/4 = 5/8.
But, the remaining cases have been left to the reader, as well as combining them with the cases from part 2.
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Posted by Charlie
on 2017-08-30 12:28:14 |
parts 1 & 2 only; outline of 3
