Problem 202 Solution

Question

Every generaion of red amoeba has a 5% chance of spawning 5 more red amoeba, a 10% chance of spawning 3 blue amoeba, a 1% chance of doing both, and a 84% chance of not spawning.

Every generaion of blue amoeba has a 3% chance of spawning 5 more blue amoeba, a 2% chance of spawning 10 red amoeba, a 1% chance of doing both, and a 94% chance of not spawning.

Your petri dish starts with one red amoeba. What is the expected number of red and blue amoeba observed before all amoeba die off?

Solution

Let r = expected number of red amoeba observed per initial red amoeba.
Let b = expected number of red amoeba observed per initial blue amoeba.

Next, from the first piece of information, "Every generaion of red amoeba has a 5% chance of spawning 5 more red amoeba, a 10% chance of spawning 3 blue amoeba, a 1% chance of doing both, and a 84% chance of not spawning," we can put in equation form:

r = 1 + 0.05×5×r + 0.1×3×b + 0.01×(5r+3b)
r = 1 + 0.25r + .3b + 0.05r + 0.03b
r = 1 + 0.3r + 0.33b

Next, from the second piece of information, "Every generaion of blue amoeba has a 3% chance of spawning 5 more blue amoeba, a 2% chance of spawning 10 red amoeba, a 1% chance of doing both, and a 94% chance of not spawning," we can put in equation form:

b = 0.03×5b + 0.02×10r + 0.01×(5b+10r)
b = 0.15b + 0.2r + .05b + 0.1r
b = 0.2b + 0.3r

Using basic algebra, we can solve for r and b , given the two equations, and get:

r = 800/461 =~ 1.7354
b = 300/461 =~ 0.6508

So, one initial red amoeba will spawn into 1.7354 on average. We also know that one blue amoea will spawn into 0.6508 red amoeba on average. However, that doesn't tell us how many blue amoeba a red amoeba will spawn into. To answer that question, let's redefine the variables as follows:

Let r = expected number of blue amoeba observed per initial red amoeba.
Let b = expected number of blue amoeba observed per initial blue amoeba.

Putting the initial conditions into equations:

r = 0.3r + 0.33b
b = 1 + 0.2b + 0.3r

Using algebra to solve the equations we get:

r = 330/461 =~ 0.7158 b = 700/461 =~ 1.5184

So, one red amoeba will spawn into 0.7158 blue amoeba on average.

Simulation

I did a random simulation of 74,016,000,000 initial states of one red amoeba. Following were the total counts observed:

Red amoeba: 128,445,045,020
Blue Amoeba: 52,983,662,357

That is an averoage of 1.735368637 red and 0.715840661 blue.

Michael Shackleford, ASA — Apr. 1, 2009