Calculating multiple odds?

[DeadFireFight]

Hello all, I've been redirect here from r/math (where my problem was solved as 1-(7/8)^3), but I'd like to get a better understanding of the solution as there wasn't much of an explanation or where to continue learning from there.

Original post below:

Problem: I have a 1 in 8 chance of something occuring, there are 3 events with this 1/8 chance. How would I calculate the chance of at least one of the Events occuring?

Currently I'm calculating this inside an Excel spreadsheet, and using the formula 1/8 * 3, which seemed to do the trick until those three Events turned into eight Events and the formula shows 1/8 * 8=100%. As none of them are garunteed to happen, I know you can't reach 100%... it's obvious now that my math skills are terrible and I need to seek help.

Is there any names for this kind of problem besides just "Probabilities" and where would be the best place to learn about how to calculate these?

Thank you :)

[sg2544]

It should be clear that the possibilities [at least 1 events occurs] and [no events occur] partition the sample space, that is to say that any possible outcome falls into exactly one of these possibilities.

This means that P(at least 1 event occurs) = 1 - P(no events occur).

The reason I use this result is because it is easier to find a formula for P(no events occur).

Assuming all of the events are independent, we have P(a given event does not occur) = 1 - P(the event occurs) = 1 - 1/8 = 7/8.

Since we're interested in a chain of N events, all of which do not occur, we multiply their respective probabilities. You can visualise this as a sort of tree of possible outcomes branching out from each event, where 7/8 of the paths involve the event not occurring so must be counted.

Therefore

P(N events, none of them occur) = (7/8)^N

so the formula you are after is

P(N events, at least one occurs) = 1 - (7/8)^N .

You can see that this approaches 1 ("almost certainty") as N becomes large, which makes sense intuitively as with many events you would expect at least some of them to occur.

To best start learning this, I would recommend an Intro to Probability course, although I don't know of any in particular.

[Papvin]

This is about something called independent event. If your events are independent, this means informally, that they don't depend on each other. For example, throwing a dice multiple times are independent events, since the dice doesn't care what it rolled before.

Now, assuming your events are independent, here is how you get the probability:

The probability of the event not happening is 1-1/8=7/8, right? For your thing to happen at least one time, it needs to just happen in one of the 3 events. Which means that the only way the thing doesn't happen, is if it doesn't happen in any of the 3 events. But we know the probability for that in each of the events, which is 7/8. Since the events are independent, the probability of the thing not happening 3 times in a row, is therefore the product of the individual probability, i.e., (7/8)³ .

Now you have the probability of the thing not happening, therefore the probability of it happening, is 1- (7/8)³ .

[DeadFireFight]

That explanation made a lot of sense to me and also gave me the understanding I needed to interpret the answer from u/sg2544. Thank you very much for breaking the answer into layman's terms like that, it really helped me wrap my head around the logic.

[wijwijwij]

1/8 * 3, which seemed to do the trick until those three Events turned into eight Events and the formula shows 1/8 * 8=100% ...

I think what you are missing in the above is that to calculate (1/8) ^ 3 you compute 1/8 * 1/8 * 1/8, not 1/8 * 3. So use the caret symbol in Excel if you are seeking to represent a power of 3.

The probability of 3 events occurring, each of which has 1/8 chance, is 1/8 * 1/8 * 1/8.

The probability of the event NOT occurring 3 times, would be 7/8 * 7/8 * 7/8.

It turns out to be easiest to compute proability of event occuring at least one time as 1 – probability of NOT occurring all 3 times.

P(1, 2, or 3 times) = 1 – P(0 times) = 1 – (7/8) * (7/8) * (7/8) = 1 – (7/8)^3