# Cumulative Odds for Dynamic Probability

I want to know how to calculate the cumulative odds of an event being successful as more attempts are made. One caveat is that the odds each attempt are not static and increase as more attempts are made.

For example:

1st attempt is 1/4069 odds.

2nd attempt is 1/1365 odds.

3rd attempt is 1/819 odds.

4th attempt is 1/585 odds.

5th attempt is 1/455 odds.

How would I calculate the cumulative odds that the event should have been successful with increasing attempts?

My understanding is that it should be as follows:

1st attempt - (1/4069) = ~0.024576%

2nd attempt - (4068/4069)*(1/1365) = ~0.073242%

3rd attempt - (4068/4069)*(1364/1365)*(1/819) = ~0.121981%

Am I correct in my approach?

## 1 Answer

Suppose you want to compute the cumulative probability that you have a sucess in the first two tries. Then

\[P[n\leq 2]=P(1)+(1-P(1))P(2).\]

Note that $(1-P(1))P(2)$ is the probability that ther is a failure in the first attepmt and success in the second.

Similarly

\[P[n\leq 3]=P(1)+(1-P(1))P(2)+(1-p(1))(1-p(2))p(3)\]

and so on.

- 1 Answer
- 144 views
- Pro Bono

### Related Questions

- Statistical analysis on a 2 dimensional data set
- reading the output of a r regression analysis
- How do we describe an intuitive arithmetic mean that gives the following? (I can't type more than 200 letters)
- Applied Probability
- Sample size calculation - single mean method
- Bivariate Normality questions
- Probability and Statistics problem
- foundations in probability

What does "2nd attempt is 1/1365 odds" mean? Does it mean having success in the first two attempts or having success in the second attempt?

It means the chance of success on the 2nd attempt is now 1/1365 instead of 1/4069 as in the first attempt.