Probability of Infection

Date: 03/06/98 at 21:14:03
From: John Lictum
Subject: Probability

Six sociable hermits live on an otherwise deserted island. An 
infectious disease strikes the island. The disease has a 1-day 
infectious period, and after that, the person is immune.

Assume one of the hermits gets the disease (maybe from a piece of 
Skylab). He randomly visits one of the other hermits during his 
infectious period.

If the visited hermit has not had the disease, he gets it and is 
infectious the following day. The visited hermit then randomly visits 
another hermit. The disease is transmitted until an infectious hermit 
visits an immune hermit, and the disease dies out.

There is one hermit visit per day.

Assuming this pattern of behavior, how many hermits can be expected, 
on average, to get the disease?

Your help with this problem is greatly appreciated. I would not have a 
clue how to start! Any suggestions are greatly appreciated, thanks!

Date: 03/08/98 at 19:26:11
From: Doctor Anthony
Subject: Re: Probability

Suppose the hermits are A, B, C, D, E, F, and that A is the one hermit 
first infected. He visits another hermit who is not immune, say B, 
who is then infected.

A is now immune and B has probability 1/5 of visiting an immune 
hermit, and 4/5 of visiting a hermit who is not immune. This means 
there is probability 1/5 that only 2 get the disease.

This situation is best illustrated with a tree diagram.

The tree starts with A infecting B with probability 1, since no matter 
whom he visits that hermit is not immune. So assuming he visited B, we 
have the first branch at B having probability 1/5 that B visits A and 
the process ends, or probability 4/5 that he visits a vulnerable 
hermit. It is not easy to do diagrams in ASCII, but the tree is 
something like that shown below.
   
       Ends (2 get the disease)
     /
    1/5
   /
  /
B \        Ends (3 get the disease)
   \      /
    4/5  2/5
     \  /
      \/         Ends (4 get the disease)
       \        /
        \      /
         3/5  3/5
          \  /
           \/         Ends (5 get the disease)
            \        /
             \      /
              2/5  4/5
               \  /
                \/
                 \
                  \
                   1/5
                    \
                     \
                      Ends (6 get the disease)

The chance that 2 get the disease is 1/5.

The chance that 3 get the disease is 4/5 x 2/5 = 8/25.

The chance that 4 get the disease is 4/5 x 3/5 x 3/5 = 36/125.

The chance that 5 get the disease is 4/5 x 3/5 x 2/5 x 4/5 = 96/625.

The chance that 6 get the disease is 4/5 x 3/5 x 2/5 x 1/5 = 24/625.

If we add these probabilities, they give a total of 1, as they should.

The expected number getting the disease is then

        2 * 1/5 + 3 * 8/25 + 4 * 36/125 + 5 * 96/625 + 6 * 24/625

      = 3 and 319/625

      = 2194/625

      = 3.5104

So, on average, 3.5 of the hermits can be expected to be infected.

-Doctor Anthony, The Math Forum
 http://mathforum.org/dr.math/