I'm terrible at math!
August 14, 2006 9:50 PM Subscribe
If i have a 20% probability of something happening each time its checked, and make two checks of said event triggering, what is the overall chance of it happening?
Yeah, i haven't taken a math class in about a decade, and even then, the last bunch were gloriously bad "core plus" math.
Yeah, i haven't taken a math class in about a decade, and even then, the last bunch were gloriously bad "core plus" math.
Quick answer, %36
It's not simple addition, otherwise if you rolled six dice, there would be %100 odds that one of them would be a six.
posted by RobotHero at 9:58 PM on August 14, 2006
It's not simple addition, otherwise if you rolled six dice, there would be %100 odds that one of them would be a six.
posted by RobotHero at 9:58 PM on August 14, 2006
Right. I'm an idiot. Sorry.
posted by JakeWalker at 10:01 PM on August 14, 2006
posted by JakeWalker at 10:01 PM on August 14, 2006
A = the event triggers the first time. p(A) = 0.2
B = the event triggers the second time. p(B) = 0.2
You want p(A or B). The obvious way is to do what JakeWalker did, and say "p(A or B) = p(A) + p(B) = .4". What's wrong with that? You've over counted once (that is, you've double counted the p(A and B)). This is inclusion-exclusion principal, which says "p(A or B) = p(A) + p(B) - p(A and B) = 0.2 + 0.2 - 0.04 = 0.36".
Or, you could directly calculate it by checking all the possibilities, but you have to be careful. In this case, "p(A or B) = p(A)p(not B) + p(not A)p(B) + p(A)p(B) = 0.16 + 0.16 + 0.04 = 0.36".
You you can do it the easy way, and realize that p(A or B) is equal to "1 - p(neither A nor B)". That is "1 - 0.8*0.8 = 0.36".
posted by sbutler at 10:12 PM on August 14, 2006 [3 favorites]
B = the event triggers the second time. p(B) = 0.2
You want p(A or B). The obvious way is to do what JakeWalker did, and say "p(A or B) = p(A) + p(B) = .4". What's wrong with that? You've over counted once (that is, you've double counted the p(A and B)). This is inclusion-exclusion principal, which says "p(A or B) = p(A) + p(B) - p(A and B) = 0.2 + 0.2 - 0.04 = 0.36".
Or, you could directly calculate it by checking all the possibilities, but you have to be careful. In this case, "p(A or B) = p(A)p(not B) + p(not A)p(B) + p(A)p(B) = 0.16 + 0.16 + 0.04 = 0.36".
You you can do it the easy way, and realize that p(A or B) is equal to "1 - p(neither A nor B)". That is "1 - 0.8*0.8 = 0.36".
posted by sbutler at 10:12 PM on August 14, 2006 [3 favorites]
I agree with 36%.
If you'd like to do this yourself, the plain English explanation is this:
Everytime you do something, the odds of it occurring at least once is a running total of the happenings before plus the remaining total times the chance of it happening now.
So the first time would be...
Running total: 0%
Remaining total: 100%
Chance: 20%
So... 0 + (100 * .2) = .2 or 20% (obvious)
Second time would be...
Running total: 20%
Remaining total: 80%
Chance: 20% (never changes)
So... 20 + (80 * .2) = .36 or 36%
Third time would be
Running total: 36%
Remaining total: 64%
Chance: 20%
So... 36 + (64 * .2) = 48.8 or roughly 49%
And so on and so forth.
posted by unixrat at 10:20 PM on August 14, 2006
If you'd like to do this yourself, the plain English explanation is this:
Everytime you do something, the odds of it occurring at least once is a running total of the happenings before plus the remaining total times the chance of it happening now.
So the first time would be...
Running total: 0%
Remaining total: 100%
Chance: 20%
So... 0 + (100 * .2) = .2 or 20% (obvious)
Second time would be...
Running total: 20%
Remaining total: 80%
Chance: 20% (never changes)
So... 20 + (80 * .2) = .36 or 36%
Third time would be
Running total: 36%
Remaining total: 64%
Chance: 20%
So... 36 + (64 * .2) = 48.8 or roughly 49%
And so on and so forth.
posted by unixrat at 10:20 PM on August 14, 2006
Not to pick on JakeWalker, but a correction to his second paragraph is in order: the chance of the event happening once and only once is 0.32 (32%)-- sixteen percent for happening the first time, but not the second, and sixteen for happening the second time but not the first. Add the four percent (20% squared) chance of it happening both times and you're back up to the 36% (i.e. the "checking all the possibilities method shown by sbutler).
posted by alexei at 10:38 PM on August 14, 2006
posted by alexei at 10:38 PM on August 14, 2006
Response by poster: so 36%.
and looking at some of the explainations, no wonder i couldn't figure it out myself.
thanks everyone.
posted by teishu at 9:15 AM on August 15, 2006
and looking at some of the explainations, no wonder i couldn't figure it out myself.
thanks everyone.
posted by teishu at 9:15 AM on August 15, 2006
What's the event? Suppose I give you a closed box: 80 percent chance it's just an empty box, but 20 percent chance I scratched my initials on the inside of the lid.
Suppose that the event is that my initials are on the inside of the lid. Then each time you open the box, the probability that my initials are on the inside of the lid is 20%. (If you don't believe me, think about keeping your eyes closed the first two times you open the lid.) But given that the first time you opened the box, my initials were there, the probability that they're there the second and subsequent times is 100%.
For this event, the probability that the event happens either the first or the second time, is exactly the same as the probability it happens the first time, i.e., it is 20%, not 36%.
In probability-speak, this is because the two events are not independent. The 36% calculation is valid if the second time you check, the event "triggers" independently of the first time, which everyone (probably correctly) assumed that you meant. But this independence does not hold for every possible pair of events.
posted by louigi at 12:03 PM on August 15, 2006
Suppose that the event is that my initials are on the inside of the lid. Then each time you open the box, the probability that my initials are on the inside of the lid is 20%. (If you don't believe me, think about keeping your eyes closed the first two times you open the lid.) But given that the first time you opened the box, my initials were there, the probability that they're there the second and subsequent times is 100%.
For this event, the probability that the event happens either the first or the second time, is exactly the same as the probability it happens the first time, i.e., it is 20%, not 36%.
In probability-speak, this is because the two events are not independent. The 36% calculation is valid if the second time you check, the event "triggers" independently of the first time, which everyone (probably correctly) assumed that you meant. But this independence does not hold for every possible pair of events.
posted by louigi at 12:03 PM on August 15, 2006
Follow on question that may help the OP (but which secretly also interests me): what are the formulas for calculating the following, given an event with p = 0.2:
1) the event occurs at least once in n trials,
2) the event occurs exactly once in n trials,
3) the event occurs n times in n trials (OK, I know this one, it's p^n, right?).
4) As a bonus, where can an otherwise math-competent adult get info like this? I suppose a beginning stats/prop textbook is in order... any suggestions?
I think one source of confusion in learning probability is that a lot of sources start with degenerate cases, like a coin flip with p=0.5 where distinct concepts like p and !p get muddy, and/or go straight to a numeric solution without showing work along the way. *Sigh*... oh, for a sophomore-year math class!
posted by rkent at 12:04 PM on August 15, 2006
1) the event occurs at least once in n trials,
2) the event occurs exactly once in n trials,
3) the event occurs n times in n trials (OK, I know this one, it's p^n, right?).
4) As a bonus, where can an otherwise math-competent adult get info like this? I suppose a beginning stats/prop textbook is in order... any suggestions?
I think one source of confusion in learning probability is that a lot of sources start with degenerate cases, like a coin flip with p=0.5 where distinct concepts like p and !p get muddy, and/or go straight to a numeric solution without showing work along the way. *Sigh*... oh, for a sophomore-year math class!
posted by rkent at 12:04 PM on August 15, 2006
1) the event occurs at least once in n trials,
The opposite of this that the event never happens, i.e. doesn't happen the first time, and doesn't happen the second time, and ... = (1-p)n. So the probability it happens at least once: 1 - (1-p)n.
2) the event occurs exactly once in n trials,
This is "(E happens the first time and never any of the other times) or (E happens the second time and never any of the other times) or ... " Each of those terms is p(1-p)n-1, so the probability of the whole string is np(1-p)n-1.
Look up the binomial distribution on the web.
where C(n,i) is the number of ways you can choose i of n things, which is n!/(i! (n-i)!).
Check out any of the books by John Allen Paulos or Ivars Peterson for easy to follow popular treatments of math(s).
posted by phliar at 5:12 PM on August 15, 2006
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If you want to know the chance of just 1 thing happening, that is, that the event happens once and only once on two presses, that gets a bit more complicated. This would be .2 * .8 (the opposite of the event happening odds), or 16%.
posted by JakeWalker at 9:57 PM on August 14, 2006