Probability Hill
The circles in the diagram represent points on Probability Ski Hill. Each arrow represents a ski run
that allows a skier to travel from one point to another on the hill. When leaving a particular point,
a skier is equally likely to choose any of the available ski runs from that point.
If a skier starts at point A, what is the probability that she will reach point D?
Connection to the Real World
When you google the keywords “fever cough chill”, how does the search engine choose which related pages to display first?
Google’s PageRankT M algorithm ranks the pages in a way similar to the ski hill problem. Suppose that each circle in the diagram represents
a web page with keywords “fever cough chill”. Each arrow in the diagram represents a link on one page leading to another; for example, web
page A has a link leading to web page D. Once a reader arrives at a page, he may follow a link to a different page or stay on the current page.
The PageRank algorithm ranks each relevant page according to the probability that the user will stop clicking and stay on that page. This
method favours pages that may be reached via many other relevant pages.
For more Real-World Problems Being Solved by Mathematics, visit http://www.cemc.uwaterloo.ca/resources/real-world.html.
Solution: Let P (X, Y ) represent the probability that skier who is already at point X will take
the ski run from point X to point Y .
When leaving a particular point, a skier is equally likely to choose any of the available ski runs from
that point.
• From A there are two possible choices, so P (A, D) = P (A, B) = 12 .
• From B there are three possible choices, so P (B, C) = P (B, D) = P (B, E) = 31 .
• From C there are two possible choices, so P (C, D) = P (C, E) = 12 .
There are three possible routes from point A to point D:
1. Directly from A to D,
2. A to B to D,
3. A to B to C to D.
We will consider the probability of the skier taking each of these routes from A to D.
1. The probability of going directly from A to D is P (A, D) = 21 .
2. The probability of going from A to B is 21 .
Once at B, the probability of continuing to D is 13 .
Thus the probability of going from A to B to D is
P (A, B) × P (B, D) = 12 × 31 = 61 .
3. The probability of going from A to B is P (A, B) = 21 .
Once at B, the probability of continuing to C is P (B, C) = 13 .
Once at C, the probability of continuing to D is P (C, D) = 21 .
Thus the probability of going from A to B to C to D is
1
.
P (A, B) × P (B, C) × P (C, D) = 21 × 13 × 12 = 12
The probability of going from A to D along any of these routes is
1
2
+ 16 +
1
12
=
6+2+1
12
=
For more Real-World Problems Being Solved by Mathematics, visit http://www.cemc.uwaterloo.ca/resources/real-world.html.
9
12
= 34 .