How do you calculate the probability of an event occurring at least once over three attempts if the likelihood of success increases on each attempt?

Answer 1

Not knowing exactly what the 'increase' is, it is hard to answer this. Here is what happens if the event probability is the same each time (at the bottom is one example where probability increases):

Let p = probability that the event will occur on a trial (attempt).

Let q = probability against the event occurring on a trial = 1 - p. [100% - probability for]

Probability that the event will not occur on n trials = qn, so the probability that the event will occur at least once on n trials is 1 - qn = 1 - (1-p)n.

Example: Say you roll a 6-sided die twice and want the probability that you'll get a 1 at least one of the throws.

p = 1/6, n = 2 : Probability = 1 - (1 - 1/6)² = 30.56%

Let's look at another way to figure this. Rolling a die twice, it's easy enough to make a 2-dimensional table of the results. There are 36 possibilities, and 11 of them have a 1 occurring. So the probability is 11/36 = 30.56%

If you were to do 3 attempts, you would have a 3-dimensional table with 216 possibilities, and 91 of them have a 1 occurring at least once.

91/216 = 42.13%, which is the same using the formula: 1 - (1 - 1/6)³ = 42.13%

So for the problem where the probability increases each time. As an example, say you're playing BINGO with 75 balls. Say you're playing a blackout game and 50 balls have already been drawn, and you've filled your block except for a single square (say it's number 1). So there are 25 balls remaining. It will be similar to above, except you cannot just square or cube the probabilities because they change.

You still want to calculate the probabilities against, then multiply them together then subtract from 100%. Since this is an example, you're playing BINGO and you look around at several neighbors and see that everybody around you has three spaces to fill, so you want to hit #1 in the next three balls.

Assign the probabilities for picking #1 on each trial as p1, p2, p3.

So p1 = 1/25. If you don't draw #1 the first time, then there are only 24 balls left, so p2 = 1/24, and p3 = 1/23. We need to multiply the q's (probabilities against) then subtract that product from 100%.

So the probability of drawing #1 on any of the next 3 trials, starting with 25 balls is: 1 - ((1 - 1/25)(1 - 1/24)(1 - 1/23))= 1 - ((24/25)*(23/24)*(22/23)) = 12%. Notice the way I wrote it, the fractions cancel. Here's the interesting thing:

The probability of getting #1 on the first try is 1/25 = 4%. The first or second try is 1 - ((24/25)*(23/24)) = 2/25 = 8%. Each trial will become an increased number over 25: 3/25 = 12%, 4/25 = 16%, etc. I'm not sure how you would generalize the formula though.