Puzzle | The Alien Extinction Riddle

An alien visits Earth one day. Each alien accomplishes one of the following four actions every day, with equal likelihood: 

• Kill himself 
• Do nothing 
• Divide into two aliens (while killing itself) 
• Divided itself into three aliens (while killing itself) 

What is the probability that the alien species eventually die out entirely?

 The Alien Extinction Riddle

Solution:

Suppose that the probability of aliens eventually dying out is x. Then for n aliens, the probability of eventually dying out is xn because we consider every alien as a separate colony. Now, if we compare aliens before and after the first day, we get:

• x = (1 /4) * 1 + (1 /4) * x + (1 /4) * x² + (1 /4) * x³
• x³ + x² − 3x + 1 = 0
• (x − 1)(x 2 + 2x − 1) = 0

We get,  x = 1, −1 − √ 2, or − 1 + √ 2

We claim that x cannot be 1, which would mean that all aliens eventually die out. The number of aliens in the colony is, on average, multiplied by 0 + 1 + 2 + 3 + 4 = 1.5 every minute, which means in general the aliens do not die out. (A more rigorous line of reasoning is included below.) Because x is not negative, the only valid solution is x = √ 2 − 1.

To show that x cannot be 1, we show that it is at most √2 − 1.

• Let xn be the probability that a colony of one bacteria will die out after at most n minutes. Then, we get the relation:
• xn+ 1 = 1/4 (1 + xn + x²n + x²n)
• We claim that xn ≤ √ 2 − 1 for all n, which we will prove using induction.
• It is clear that x1 = 1 /4 ≤ √ 2 − 1. Now, assume xk ≤ √ 2 − 1 for some k. We have:
• xk+1 ≤ 1/4 (1 + xk + x²k + x³k) ≤ 1/4 ( 1 + (√ 2 − 1) + (√ 2 − 1)² + (√ 2 − 1)³ ) = √ 2 − 1
• Which completes the proof that xn ≤ √ 2 − 1 for all n. Now, we note that as n becomes large, xn approaches x.

Using formal notation, x = lim (n →∞) xn ≤ √ 2 − 1, so x cannot be 1.