91Ó°ÊÓ

what is the probability that the investor retires a winner?

Given:

A stock of value $25$

If it reaches $40$, the stock owner sells with a profit.

If it reaches $10$, the stock owner sells with a loss.

Events that can occur in one unit of time:

U - the stock goes up by one point

D - the stock goes down by one point

probabilities:

$P\left(U\right)=0.55$

$P\left(D\right)=0.45$

The owner sells the stock if it goes $15$up or $15$down.

The outcome of winning can occur in infinitely many ways- different order of rising and falling of the value of the stock.

For every outcome that ends with profit, there is precisely one corresponding situation that ends with a loss.

For instance:-

$U,U,U,D,U,D,U,U,U,U,U,U,U,U,U,U,U,U,D,U,U$

$D,D,D,U,D,U,D,D,D,D,D,D,D,D,D,D,D,D,U,D,D$

$\text{Probability of a situation that ends with a win is}\left[P\right(U\left){]}^{15+n}\right[P\left(D\right){]}^n\text{, for some}n$

$\text{and probability of corresponding situation that ends with a loss is}\left[P\right(D\left){]}^{15+n}\right[P\left(U\right){]}^n\text{, for that same}n$

In the example above n=$3$

$\text{Therefore the probability of a winning situation is}q=\frac{P(U{)}^{15}}{P(D{)}^{15}}\text{times that of the corresponding losing situation.}$

And since the probability of a win is the sum of probabilities of winning situations, and each of them is q times greater than one of the losing situations we obtain:

$P("\text{win"})=qâ‹\dots P("loss")=\frac{P(U{)}^{15}}{P(D{)}^{15}}P("loss")$

And since time is not limited, one of those options will happen, this means:

$P("\text{win"})+P("loss")=1$

system of equation ($1$) and ($2$) has a unique solution.

$P("win")=\frac{q}{1+q}$

$=\frac{\frac{P(U{)}^{15}}{P(D{)}^{15}}}{1+\frac{P(U{)}^{15}}{P(D{)}^{15}}}$

$=\frac{\left[P\right(U){]}^{15}}{\left[P\right(U){]}^{15}+[P\left(D\right){]}^{15}}$

$=\frac{(0.55{)}^{15}}{(0.55{)}^{15}+(0.45{)}^{15}}$

$=95.3\%$

The probability that the investor retires a winner is$95.3\%$