# Need help with a probability problem

• sdclaw
In summary, the probability of the desired outcome (A) occurring is 75% over the next 24 trials, given that the desired outcome has occurred 25% of the time in the past.

#### sdclaw

Guys and girls, I need help calculating the probability of the following event happening (or feel free to just tell me the answer )

I know an event has two possible outcomes, let's call the outcome A and outcome B. I know that the desired outcome (let's call it A) has occurred 25% of the time over a sample size of 200 events.

What is the probability that A will occur 75% (or greater) of the time over the next 24 occurances of the event happening?

I have a basic understanding of probability but I'm having trouble wrapping my head around this.

Last edited:
Use the binomial distribution with p(A)=1/4. Although you don't know p(A) exactly, 200 trials gives you a good estimate.

mathman said:
Use the binomial distribution with p(A)=1/4. Although you don't know p(A) exactly, 200 trials gives you a good estimate.

How do we account for the fact that after each additional trial p(A) changes. In this instance let's say trials 201-210 all yield the desired result. That being the case p(A) will increase after each of these trials.

Let's say you're wondering about the next 2 trials. For the next trial you predict "Heads" with probability 0.25, "Tails" with prob. 0.75. Let's say you have Heads. So you'd need to update 0.25 up to 51/201. But, let's say you have Tails. Then you'd need to update 0.25 down to 50/201.

Since you don't know whether you'll get Heads or Tails, you decide to calculate the "average" probability of Heads that you expect to have after the next trial, using the probabilities that you have currently (0.25 for Heads and 0.75 for Tails). The average probability that you can expect to have after your next trial is (51/201) x 0.25 + (50/201) x 0.75 = 50.25/201 = 0.25 exactly.

The interpretation is that before you run the next trial, all you can say is that you expect the probability of Heads to stay constant at 0.25. The same logic applies to the second-next trial, and so on...