Stock Market Investor: Probability of Retirement as a Winner

  • Thread starter Thread starter Lisawmi
  • Start date Start date
Lisawmi
Messages
1
Reaction score
0
I am struggling with this problem, and overall have found probability to be a very difficult subject. I was hoping someone could explain to me how to work this problem.

A stock market investor owns shares in a stock whose present value is 25. She has decided that she must sell her stock if it either goes down to 10 or up to 40. If each change of price is either up 1 point with probability .55 or down 1 point with probability .45, and the successive changes are independent, what is the probability the investor retired a winner?

Thanks
 
Physics news on Phys.org
A few hints:

1. Binomial tree model.

2. You want either a net number (n) of 15 successes(= stock up) or 15 failures (=stock down).

3. Work it out instead with 1 success or failure (should be easy), then with 2 successes or 2 failures, or n=2. Once you figure out n=2, n=15 should be pretty easy.

4. For n>=2, there are only 3 possible types of outcomes:
a. Retired winner in the stock market
b. Retired loser in the stock market
c. An infinite loop where the number of successes is equal to the number of failures.

I think we can ignore case (c) because the weighted probability of such an event is small.
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...

Similar threads

Replies
8
Views
2K
Replies
46
Views
5K
Replies
45
Views
5K
Replies
1
Views
3K
Replies
3
Views
2K
Replies
5
Views
2K
Replies
8
Views
2K
Replies
30
Views
4K
Back
Top