Jtdylktuy

Here's an example but it can be different too, like about people and their children, etc.

http://www.cut-the-knot.org/Probability/AmoebaSurvival.shtml

A population starts with a single amoeba. For this one and for the generations thereafter, there is a probability of 3/4 that an individual amoeba will split to create two amoebas, and a 1/4 probability that it will die out without producing offspring. What is the probability that the family tree of the original amoeba will go on forever?

The solution is stated but I can not make sense of it. I once found a site that explained it in very great detail, using a family tree, but unfortunately do not recall the address. My problem is I keep getting confused between the small $p$ and large $P$, the formula used makes no sense to me: "The probability that at least one of them will have an infinite family tree is $1 - (1 - P)^2$, because $(1 - P)^2$ is the probability that both of them will perish undivided. Therefore, $P = p(1 - (1 - P)^2)$."

My level of math is first year college (though that was long ago). So when I looked up similar problems, I got real complicated explanations, like with branching process and Markov chain and all that. I appreciate it if someone, perhaps with help of a picture or something, can explain this to me baby steps?

p.s. I hope the formula shows up okay, I don't know how to write math forumlae in text like fashion.

Little $p$ is the probability that a given amoeba produces two descendants, and big $P$ is the probability that a given amoeba produces a line that never dies out. Since the latter implies the former, we can assume that $P leq p$.

The analysis examines the situation recursively: In order for a given amoeba's line not to die out, it must (a) produce two offspring, and (b) at least one of those offspring must produce a line that never dies out. In this way, we create an expression for $P$ in terms of itself.

The probability (a) is trivial; it's stated in the problem description as $3/4$. Once the two offspring are produced—let's call them A and B—the probability (b) that at least one of them produces a line that never dies out is the probability that A does and B doesn't (probability $P(1-P)$), plus the probability that B does and A doesn't (probability $(1-P)P$), plus the probability that both A and B do (probability $P^2$). That sums up to $2P-P^2$. (Alternatively, we could express that as one minus the probability that neither A nor B produces a line that never dies out: $1-(1-P)^2 = 2P-P^2$.)

So now we have an expression for $P$ in terms of itself:

$$
P = frac{3}{4} left(2P-P^2right)
$$

We can rewrite that as

$$
3P^2-2P = 0
$$

which gives us two solutions, $P = 0$ and $P = 2/3$. Since the average number of amoebas in each generation goes up by a factor of $2 times 3/4 = 3/2$, the amoeba line must have some positive chance of survival, and we choose the larger solution, $P = 2/3$.

My problem is I keep getting confused between the small $p$ and large $P$,

Then to avoid that, let $p$ be the probability that an amoeba divides into daughters and $Q$ the probability that the amoeba will spawn an eternal family tree.

Now, $;1-Q;$ is the probability that an amoeba does not spawn an eternal tree.   That is that eventually there will be no descendants of this amoeba.   An application of the rule of complements.

Then, $;(1-Q)^2;$ is the probability that two amoeba will both not spawn eternal trees.   An application of the product rule for independent events.

So, $;1-(1-Q)^2;$ is the probability that of two amoeba at least one will spawn an eternal tree.   Another application of the rule of complements.

Finally, $;p,(1-(1-Q)^2);$ is the probability that one amoeba will divide and that at least one of its daughters will spawn an eternal family tree.   This is the probability that that amoeba itself spawns an eternal family tree.

$$Q = p;(1-(1-Q)^2)$$

Expanding and rearranging to solve for $Q$ gives the solutions:$$left[Q = 2-frac 1 p wedge pneq 0right] ; vee ; [Q=0]$$

Now since we need $0leq Qleq 1$, (to be a valid probability) then to have a non-zero solution requires $frac 1 2lt p leq 1$.   This is the case for our given probability.

As $p=3/4$ we thus have $Q = 2/3$ as the non-zero solution.