How to calculate "average" probability?
April 13, 2006 5:57 AM Subscribe
If there's a 20 percent chance of rain on Monday, a 20 percent chance on Tuesday, and a 30 percent chance on Wednesday, what's the overall chance that it'll rain during this three-day period? How do you calculate this?
It's not really average probability you're after here.
It depends upon whether or not the rain even is independent, but... it rains if it rains on any of the days (monday or tuesday or wednesday or any two or all three - that is M, T, W, M&T, T&W, M&W, M&T&W). So instead of and (in which case you'd mulitply the probabilities) you've got an or, which means you can simply add them.
So the probability is 0.2+0.2+0.3=0.7 so 70%
The bonus question you didn't ask - what's the probability it rains all three days? 0.2x0.2x0.3=0.012
However, rain's not independent. At least, it isn't here in England.
posted by handee at 6:09 AM on April 13, 2006
It depends upon whether or not the rain even is independent, but... it rains if it rains on any of the days (monday or tuesday or wednesday or any two or all three - that is M, T, W, M&T, T&W, M&W, M&T&W). So instead of and (in which case you'd mulitply the probabilities) you've got an or, which means you can simply add them.
So the probability is 0.2+0.2+0.3=0.7 so 70%
The bonus question you didn't ask - what's the probability it rains all three days? 0.2x0.2x0.3=0.012
However, rain's not independent. At least, it isn't here in England.
posted by handee at 6:09 AM on April 13, 2006
handee: That logic doesn't make any sense. If there's an 80% chance of rain on Mondy, Tuesday and Wednesday, then by your logic the chance it would rain at all is 240%. This is not correct.
posted by mge at 6:12 AM on April 13, 2006
posted by mge at 6:12 AM on April 13, 2006
Seconding Dasein, that is. MrMulan appears to have averaged the three probabilities, which falls under the category of "wild guess" or "please don't do that in the green."
(handee: Suppose there's a 20% chance of rain each day for the next two weeks. Use your technique to compute the probability of at least one day with rain. Little problem?)
posted by Wolfdog at 6:12 AM on April 13, 2006
(handee: Suppose there's a 20% chance of rain each day for the next two weeks. Use your technique to compute the probability of at least one day with rain. Little problem?)
posted by Wolfdog at 6:12 AM on April 13, 2006
If the chances are independant (they aren't, but let's say they are), then the chance of it raining at all is the chance it rains, given that it is a Monday, times the chance it is a Monday (which is a third), plus the same for Tuesday and Wednesday, so:
(0.2 + 0.2 + 0.3)*1/3 = roughly 23%
On preview, I thought handee's method was the correct way at first, but then what if the chance for rain was 70% on Wednesday? Then you'd add them and get a 110% of rain, which can't be right.
posted by Orange Goblin at 6:15 AM on April 13, 2006
(0.2 + 0.2 + 0.3)*1/3 = roughly 23%
On preview, I thought handee's method was the correct way at first, but then what if the chance for rain was 70% on Wednesday? Then you'd add them and get a 110% of rain, which can't be right.
posted by Orange Goblin at 6:15 AM on April 13, 2006
d'oh!. Sorry. Logical errors 'r' us. I'm blaming my hangover, and fetching my coat.
posted by handee at 6:16 AM on April 13, 2006
posted by handee at 6:16 AM on April 13, 2006
But I don't think futility closet was asking about calculating the possibility of it raining on this day or that one, or a combination. From the question it looks like he wants to know about the probability of it raining at all during those three days, given the percentages for each day. (20 + 20 + 30) / 3 = 23.3%
But the last math course I ever took was during my junior year in high school.
posted by emelenjr at 6:17 AM on April 13, 2006
But the last math course I ever took was during my junior year in high school.
posted by emelenjr at 6:17 AM on April 13, 2006
Dasein is correct.
handee: Your calculation isn't right. If, say, there is a 60% chance of rain on Monday and Tuesday, you can't simply add them and say there's a 120% chance of rain on those two days. However, the answer to your bonus question is correct.
I've always wondered about this variation: The weather office predicts a 60% chance of rain but their predictions are only right 20% of the time. What's the chance of rain?
posted by chrisch at 6:21 AM on April 13, 2006
handee: Your calculation isn't right. If, say, there is a 60% chance of rain on Monday and Tuesday, you can't simply add them and say there's a 120% chance of rain on those two days. However, the answer to your bonus question is correct.
I've always wondered about this variation: The weather office predicts a 60% chance of rain but their predictions are only right 20% of the time. What's the chance of rain?
posted by chrisch at 6:21 AM on April 13, 2006
he wants to know about the probability of it raining at all during those three days
Yes.
(20 + 20 + 30) / 3 = 23.3%
True equation, but not the answer to his question. This is worth thinking about, and understandable even if you haven't seen any math for a long time. Imagine again a simplified problem where there's a 20% chance of rain each day. And OP asks the question "what's the probability we get at least some rain over the next n days".
Now, if you try to solve this problem by averaging the probabilities for each day, you get 20% no matter how many days he's asking about. That should be a red flag that something's wrong with your "solution." To make it a little more dramatic: suppose a marksman has a 20% chance of hitting you each time he fires. Do you believe your chances of survival are exactly the same whether he gets to take 1 shot, 10 shots, or 100 shots?
The weather question is exactly the same.
posted by Wolfdog at 6:24 AM on April 13, 2006
Yes.
(20 + 20 + 30) / 3 = 23.3%
True equation, but not the answer to his question. This is worth thinking about, and understandable even if you haven't seen any math for a long time. Imagine again a simplified problem where there's a 20% chance of rain each day. And OP asks the question "what's the probability we get at least some rain over the next n days".
Now, if you try to solve this problem by averaging the probabilities for each day, you get 20% no matter how many days he's asking about. That should be a red flag that something's wrong with your "solution." To make it a little more dramatic: suppose a marksman has a 20% chance of hitting you each time he fires. Do you believe your chances of survival are exactly the same whether he gets to take 1 shot, 10 shots, or 100 shots?
The weather question is exactly the same.
posted by Wolfdog at 6:24 AM on April 13, 2006
The problem, as Orange Goblin pointed out, is that these numbers aren't practical measurements of anything, so its tough to say. Most likely, the weatherman saw a storm system coming our way, but wasn't sure when it would hit. So he said "Well, at current speeds it will hit on Tuesday, but it could speed up and hit on Wednesday or slow down and hit on Monday. Then again, there's about a one in three chance it changes direction and misses us altogether. So let's say 20% on Monday, 20% on Tuesday, and 30% on Wednesday."
Since these probabilities are really just guesses that the weatherman made, by reverse engineering his train of thought, I'd say its probably 70% it rains at some point over those three days.
posted by ChasFile at 6:26 AM on April 13, 2006
Since these probabilities are really just guesses that the weatherman made, by reverse engineering his train of thought, I'd say its probably 70% it rains at some point over those three days.
posted by ChasFile at 6:26 AM on April 13, 2006
For chrisch's problem, the probability is
(0.2)(0.6)+(0.8)(0.4)=0.44.
In the cases when they're right (0.2) there's an 0.6 chance; in the cases when they're wrong (0.8) there's an 0.4 chance; those cases are disjoint, so we add them.
posted by Wolfdog at 6:30 AM on April 13, 2006
(0.2)(0.6)+(0.8)(0.4)=0.44.
In the cases when they're right (0.2) there's an 0.6 chance; in the cases when they're wrong (0.8) there's an 0.4 chance; those cases are disjoint, so we add them.
posted by Wolfdog at 6:30 AM on April 13, 2006
ChasFile's solution: ("It's only probability, so we can just guess!") is a very popular technique.
posted by Wolfdog at 6:30 AM on April 13, 2006
posted by Wolfdog at 6:30 AM on April 13, 2006
To elaborate on Dasein's correct answer, the question you're asking is "What is the chance it will rain in this 3 day period" is the opposite of "What is the chance it won't rain during this period."
It's easy to compute the chance that it won't rain: 80% day 1 * 80% day 2, * 70% day 3 = 44.8% overall.
So now, to answer your question, you simply get the opposite of the result above: 100% - 44.8% = 55.2%.
posted by knave at 6:33 AM on April 13, 2006
It's easy to compute the chance that it won't rain: 80% day 1 * 80% day 2, * 70% day 3 = 44.8% overall.
So now, to answer your question, you simply get the opposite of the result above: 100% - 44.8% = 55.2%.
posted by knave at 6:33 AM on April 13, 2006
OK, and a little more elaboration: if you want a solution that's not based on this "figure out the opposite probability" strategy, you can think of it this way:
There's an 0.2 chance that you get rain on Monday.
There's an 0.8*0.2 chance of no rain on Monday, but rain on Tue.
And an 0.8*0.8*0.3 chance of no rain on Mon/Tue but rain on Wed.
Those are all mutually exclusive outcomes, and cover all the ways you could get some rain. So the total is
0.2+(0.8)(0.2)+(0.8)(0.8)(0.3) = .552, confirmed.
posted by Wolfdog at 6:42 AM on April 13, 2006
There's an 0.2 chance that you get rain on Monday.
There's an 0.8*0.2 chance of no rain on Monday, but rain on Tue.
And an 0.8*0.8*0.3 chance of no rain on Mon/Tue but rain on Wed.
Those are all mutually exclusive outcomes, and cover all the ways you could get some rain. So the total is
0.2+(0.8)(0.2)+(0.8)(0.8)(0.3) = .552, confirmed.
posted by Wolfdog at 6:42 AM on April 13, 2006
The reason you calculate it Dasein's way is pretty simple. The question is, 'what is the chance it will rain on at least one of the three days'. Any day with rain fulfills this requirement.
So if it rains on Day 1, Day 2, OR Day 3, the requirement is fulfilled.
If it rains on Day 1, there's no reason to calculate the later days. There's a 20% chance that first day, so it's obvious that chances on later days must make that percentage higher. It's not just the average of the three probabilities, because rain on ANY of the days counts. As Wolfdog is saying, if a marksman is shooting at you three times with a 20% chance each time, his chance to hit you is higher than just shooting at you once.
So you can iterate it:
Day 1, 20% chance of hitting, 80% chance of missing.
Day 2 (only if Day 1 didn't rain): 20% chance of raining, 80% chance of missing.... that means we have a cumulative chance of .64 of missing. (0.8*0.8).
Day 3 only happens if we missed the first two days. It has a 70% chance of missing... so it's .70 * .64... a 0.448 chance of not raining all three days. So there is a 55.2% chance that it WILL rain.
Those of you waving your hands in the air and violating the terms of the problem by inventing how weathermen aren't accurate.... bah. You shouldn't be answering this kind of question.
posted by Malor at 6:49 AM on April 13, 2006
So if it rains on Day 1, Day 2, OR Day 3, the requirement is fulfilled.
If it rains on Day 1, there's no reason to calculate the later days. There's a 20% chance that first day, so it's obvious that chances on later days must make that percentage higher. It's not just the average of the three probabilities, because rain on ANY of the days counts. As Wolfdog is saying, if a marksman is shooting at you three times with a 20% chance each time, his chance to hit you is higher than just shooting at you once.
So you can iterate it:
Day 1, 20% chance of hitting, 80% chance of missing.
Day 2 (only if Day 1 didn't rain): 20% chance of raining, 80% chance of missing.... that means we have a cumulative chance of .64 of missing. (0.8*0.8).
Day 3 only happens if we missed the first two days. It has a 70% chance of missing... so it's .70 * .64... a 0.448 chance of not raining all three days. So there is a 55.2% chance that it WILL rain.
Those of you waving your hands in the air and violating the terms of the problem by inventing how weathermen aren't accurate.... bah. You shouldn't be answering this kind of question.
posted by Malor at 6:49 AM on April 13, 2006
It depends on how much the wedding costs, and the balance of luck for the couple, combined with the weather forecast. The math is beyond the scope of this life.
posted by Goofyy at 7:36 AM on April 13, 2006
posted by Goofyy at 7:36 AM on April 13, 2006
50% chance of rain ... either it will, or it won't. The weathermen really don't know what they're talking about ... they're just guessing for the most part.
posted by indigo4963 at 7:36 AM on April 13, 2006
posted by indigo4963 at 7:36 AM on April 13, 2006
50% chance of rain ... either it will, or it won't.
That's like saying there's a 50% chance that you're pregnant, because either you are or you aren't.
Not good odds.
posted by Jeanne at 7:41 AM on April 13, 2006
That's like saying there's a 50% chance that you're pregnant, because either you are or you aren't.
Not good odds.
posted by Jeanne at 7:41 AM on April 13, 2006
ChasFile's answer gets at an important point -- the comment that "bah, you shouldn't be answering this kind of question" is absolutely inappropriate. All the computations that have been offered as answers to this question rely on a very specific model, namely one where the probability of rain on any given day is independent of what happens on any other day.
I don't know how the weathermen actually produce their daily probabilities. The most obvious method would be to run say 1000 simulations, and then look at how many of them have rain on day N and call that the probability of rain on day N. In that case the correct answer to this question is absolutely not the one that you would compute by assuming that the days are independent.
(A quick way of seeing that there is a problem with the independent-events reasoning is to imagine that the weatherman gives you a forecast of the probability of rain every hour -- as they indeed sometimes do -- or even every MINUTE. Now suppose you want to know the probability that it rains sometime during the day; are you going to assume that all hours or minutes are independent? Looking at the hourly forecasts the weathermen indeed produce, it seems obvious that this computation would lead to an overestimate -- you can even compare it to the per-day probability that the weathermen also produce and see explicitly that it is an overestimate!)
posted by em at 7:49 AM on April 13, 2006 [1 favorite]
I don't know how the weathermen actually produce their daily probabilities. The most obvious method would be to run say 1000 simulations, and then look at how many of them have rain on day N and call that the probability of rain on day N. In that case the correct answer to this question is absolutely not the one that you would compute by assuming that the days are independent.
(A quick way of seeing that there is a problem with the independent-events reasoning is to imagine that the weatherman gives you a forecast of the probability of rain every hour -- as they indeed sometimes do -- or even every MINUTE. Now suppose you want to know the probability that it rains sometime during the day; are you going to assume that all hours or minutes are independent? Looking at the hourly forecasts the weathermen indeed produce, it seems obvious that this computation would lead to an overestimate -- you can even compare it to the per-day probability that the weathermen also produce and see explicitly that it is an overestimate!)
posted by em at 7:49 AM on April 13, 2006 [1 favorite]
em, this is obviously a question about how to calculate a set of probabilities... the actual scenario doesn't matter. Wolfdog's alternate scenario of a bad guy shooting at you is just as good.
Don't be distracted by the fluff. The poster is asking about the math. If you notice, the question doesn't mention a weatherman. That got added.
posted by Malor at 8:25 AM on April 13, 2006
Don't be distracted by the fluff. The poster is asking about the math. If you notice, the question doesn't mention a weatherman. That got added.
posted by Malor at 8:25 AM on April 13, 2006
Metafilter: You're oh so very wrong. (No, you're oh so very wrong!)
I appreciate the refresher course in calculating probability.
posted by emelenjr at 8:27 AM on April 13, 2006
I appreciate the refresher course in calculating probability.
posted by emelenjr at 8:27 AM on April 13, 2006
em sort of has a point. The 55% answer only holds true if futility closet's original statements are true independant of each other. This isn't the case with weather forecasts.
As such, if this is an actual weather forecasting problem, the correct answer is..... I guess... to hire a meteorologist to look at some models for you.
posted by I Love Tacos at 8:28 AM on April 13, 2006
As such, if this is an actual weather forecasting problem, the correct answer is..... I guess... to hire a meteorologist to look at some models for you.
posted by I Love Tacos at 8:28 AM on April 13, 2006
I have no idea what you mean by telling me not to be distracted by the fluff, or that this is a math question rather than a meteorological question. The question as stated is about how to calculate the chance of rain during a 3-day period given the chances of rain during each individual day.
Without the bizarre, unphysical assumption that "rain on day N" is independent of "rain on day N+1", the answers being offered are dead 100% wrong, and the correct answer is that there is not enough information to determine the chance of rain over the whole 3-day period.
This is a practical point: as far as I can see, there is not even any reasonable approximation in which the days are independent. If you actually follow the procedure proposed upthread, you will get the wrong answer.
You're right that I shouldn't have involved the weatherman; I forgot that the question was phrased in terms of the known probabilities rather than the probabilities as announced by the weatherman.
posted by em at 8:34 AM on April 13, 2006
Without the bizarre, unphysical assumption that "rain on day N" is independent of "rain on day N+1", the answers being offered are dead 100% wrong, and the correct answer is that there is not enough information to determine the chance of rain over the whole 3-day period.
This is a practical point: as far as I can see, there is not even any reasonable approximation in which the days are independent. If you actually follow the procedure proposed upthread, you will get the wrong answer.
You're right that I shouldn't have involved the weatherman; I forgot that the question was phrased in terms of the known probabilities rather than the probabilities as announced by the weatherman.
posted by em at 8:34 AM on April 13, 2006
Tacos, the problem is clearly stated; there are 20%, 20%, and 30% chances. Period. That's the problem we were given; it doesn't matter how those numbers were reached.
Inventing the weatherman and complexifying the problem is fluff. We don't get to argue with the problem, we're just supposed to solve it and show how it's done.
The poster is infallible. All hail the poster! :)
posted by Malor at 8:35 AM on April 13, 2006
Inventing the weatherman and complexifying the problem is fluff. We don't get to argue with the problem, we're just supposed to solve it and show how it's done.
The poster is infallible. All hail the poster! :)
posted by Malor at 8:35 AM on April 13, 2006
Dasein: "This is a math question, not a meteorological question"
Malor: "The poster is asking about the math."
And just how the hell do you know? They never say that this is just an example. Maybe they really wanted to know how to work this out for weather specifically.
posted by reklaw at 8:39 AM on April 13, 2006
Malor: "The poster is asking about the math."
And just how the hell do you know? They never say that this is just an example. Maybe they really wanted to know how to work this out for weather specifically.
posted by reklaw at 8:39 AM on April 13, 2006
What they mean, reklaw, is that the same skeleton of math underlies many questions like this, for example, a coin toss. If you changed the percentages to fifty percent each day, it's equivalent to calculating the odds of getting one heads in three coin tosses. It helps to simplify. If you understand the math, you can apply it to other situations.
posted by weapons-grade pandemonium at 9:17 AM on April 13, 2006
posted by weapons-grade pandemonium at 9:17 AM on April 13, 2006
em« writes 'the correct answer is that there is not enough information to determine the chance of rain over the whole 3-day period.'
Actually, probability theory is all about incomplete information, that's why you compute probabilities rather than certainties. As stated, with probabilities given for each day, the problem is solvable. If you dispute whether or not you can state a probablity of rain for a day or not, you're arguing with the problem, which is not helping to answer it.
posted by signal at 9:39 AM on April 13, 2006
Actually, probability theory is all about incomplete information, that's why you compute probabilities rather than certainties. As stated, with probabilities given for each day, the problem is solvable. If you dispute whether or not you can state a probablity of rain for a day or not, you're arguing with the problem, which is not helping to answer it.
posted by signal at 9:39 AM on April 13, 2006
Paging Futility Closet!
I think we really need the OP to let us know if they wanted to derive the overall chance of rain, or if it was just an example.
Arguing over what he meant is a bit pointless.
posted by I Love Tacos at 9:39 AM on April 13, 2006
I think we really need the OP to let us know if they wanted to derive the overall chance of rain, or if it was just an example.
Arguing over what he meant is a bit pointless.
posted by I Love Tacos at 9:39 AM on April 13, 2006
I now think that 55% is the correct answer.
I got distracted for a bit, but I'm back to that. If the question involved rain on specific combinations of days, I'd go back to agreeing with em.
posted by I Love Tacos at 9:47 AM on April 13, 2006
I got distracted for a bit, but I'm back to that. If the question involved rain on specific combinations of days, I'd go back to agreeing with em.
posted by I Love Tacos at 9:47 AM on April 13, 2006
The general (ugly, brute force) approach is to work out all possible outcomes and add up the chances of each that apply:
(R = rain, C = clear)
RRR = .2 * .2 * .3
RRC = .2 * .2 * .7
RCR = .2 * .8 * .3
CRR = .8 * .2 * .3
CCR = .8 * .8 * .3
CRC = .8 * .2 * .7
RCC = .2 * .8 * .7
CCC = .8 * .8 * .7
Everything else is just shortcuts!
For example, notice that the sum of all these is 1 = 100%. Since you want to know "everything but CCC" you can do 100% - 80% * 80% * 70%. That's Dasein's answer.
But say the question were "what's the chance that it won't rain 2 days in a row?" you can do 100% - (RRC + CCC).
posted by fleacircus at 9:52 AM on April 13, 2006
(R = rain, C = clear)
RRR = .2 * .2 * .3
RRC = .2 * .2 * .7
RCR = .2 * .8 * .3
CRR = .8 * .2 * .3
CCR = .8 * .8 * .3
CRC = .8 * .2 * .7
RCC = .2 * .8 * .7
CCC = .8 * .8 * .7
Everything else is just shortcuts!
For example, notice that the sum of all these is 1 = 100%. Since you want to know "everything but CCC" you can do 100% - 80% * 80% * 70%. That's Dasein's answer.
But say the question were "what's the chance that it won't rain 2 days in a row?" you can do 100% - (RRC + CCC).
posted by fleacircus at 9:52 AM on April 13, 2006
answer 1) - in michigan, the probability is 100%
answer 2) - the straight dope
answer 3, similar to answer 1, from same link) - "In Washington, Oregon, and British Columbia, we have a different interpretation of precipitation forecasts. If the weatherman says there's a 20 percent chance of rain, that means it will rain 20 percent of the day. If there's a 50 percent chance, it will rain 50 percent of the time, etc., up to 100 percent, which means, of course, a typical January day. This interpretation seems to be quite accurate."
posted by pyramid termite at 9:52 AM on April 13, 2006
answer 2) - the straight dope
answer 3, similar to answer 1, from same link) - "In Washington, Oregon, and British Columbia, we have a different interpretation of precipitation forecasts. If the weatherman says there's a 20 percent chance of rain, that means it will rain 20 percent of the day. If there's a 50 percent chance, it will rain 50 percent of the time, etc., up to 100 percent, which means, of course, a typical January day. This interpretation seems to be quite accurate."
posted by pyramid termite at 9:52 AM on April 13, 2006
signal, no, there isn't enough information in the question to answer it. As stated, the problem is not solvable, unless you make assumptions about the unstated information--namely, how these three probabilities depend upon one another.
If you assume that the probabilities are independent, then you can solve the problem. However, that is not explicitly stated in the question; and in fact, as other posters are pointing out, weather probabilities like this are in fact not independent.
On a different note, the reason you can't just add the probabilities together (assuming they are independent) is that you are not excluding the middle. The probability of either of two independent events A and B occurring is the sum of their individual probabilities, minus the probability of both happening. You subtract that part since you've already counted it once when you added them together; you don't want to count it twice.
(Imagine a Venn diagram with two overlapping sections. If you want the total area of the diagram, you'll add the area of each section, and then subtract the area of the middle section. Otherwise you'll have added it in twice.)
Or in mathematical notation:
P(A or B) = P(A) + P(B) - P(A and B)
This is equivalent to:
P(A or B) = P(A) + P(B) - P(A) P(B)
P(A or B) = P(A) + [1 - P(A)] P(B)
P(A or B) = P(A) + P(not A) P(B)
P(A or B) = P(A) + P(not A and B)
Or, for three events:
P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(A and C) + P(A and B and C)
Work out similar math and you'll see this is equivalent to:
P(A or B or C) = P(A) + P(not A and B) + P(not A and not B and C)
That last equation is the same one that Malor gave.
posted by Khalad at 10:05 AM on April 13, 2006
If you assume that the probabilities are independent, then you can solve the problem. However, that is not explicitly stated in the question; and in fact, as other posters are pointing out, weather probabilities like this are in fact not independent.
On a different note, the reason you can't just add the probabilities together (assuming they are independent) is that you are not excluding the middle. The probability of either of two independent events A and B occurring is the sum of their individual probabilities, minus the probability of both happening. You subtract that part since you've already counted it once when you added them together; you don't want to count it twice.
(Imagine a Venn diagram with two overlapping sections. If you want the total area of the diagram, you'll add the area of each section, and then subtract the area of the middle section. Otherwise you'll have added it in twice.)
Or in mathematical notation:
P(A or B) = P(A) + P(B) - P(A and B)
This is equivalent to:
P(A or B) = P(A) + P(B) - P(A) P(B)
P(A or B) = P(A) + [1 - P(A)] P(B)
P(A or B) = P(A) + P(not A) P(B)
P(A or B) = P(A) + P(not A and B)
Or, for three events:
P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(A and C) + P(A and B and C)
Work out similar math and you'll see this is equivalent to:
P(A or B or C) = P(A) + P(not A and B) + P(not A and not B and C)
That last equation is the same one that Malor gave.
posted by Khalad at 10:05 AM on April 13, 2006
Actually, probability theory is all about incomplete information, that's why you compute probabilities rather than certainties.
What you say is true, but nevertheless, the information provided in this problem is too incomplete to allow one to compute the desired answer.
The general (ugly, brute force) approach is to work out all possible outcomes and add up the chances of each that apply:
This would be the correct approach assuming that rain on the three different days are independent events.
What they mean, reklaw, is that the same skeleton of math underlies many questions like this, for example, a coin toss.
They are both questions of probability theory, but in the case of the coin toss one has the extra information that successive tosses are independent of one another. That extra information is absent here.
I am not saying that probability theory doesn't apply to the weather! Let me make my claim more precisely. Suppose we posit that there is some distribution of possible weather phenomena, and the actual phenomena will be chosen randomly from this distribution -- the usual starting point for problems in probability. Suppose for a second that we knew the distribution. Then we could compute various different things from this distribution. In particular, we could compute the probabilities P_1, P_2, P_3 of rain on day 1, day 2, or day 3 respectively. One could also compute the probability P that it will rain on any of the three days. All of these would be well defined numbers. But unless you make additional assumptions about the nature of the distribution, these numbers would not satisfy any simple relation; in particular, it would be impossible to calculate P from P_1, P_2, P_3.
The scenario described by ChasFile, where a storm is coming with probability .70 and it may arrive on one of three different days, is an example of a possible distribution: there P_1 = .20, P_2 = .20 and P_3 = .30 as stipulated in the problem, but P = .70.
Another possible distribution is one in which the weather on every day is independent from the weather on every other day, and where the probabilities on each individual day are as stipulated in the problem. In that case, and only in that case, the computation of P given by various posters upthread is correct.
Without knowing the distribution, there is no way to calculate P, even given P_1, P_2 and P_3.
posted by em at 10:36 AM on April 13, 2006
What you say is true, but nevertheless, the information provided in this problem is too incomplete to allow one to compute the desired answer.
The general (ugly, brute force) approach is to work out all possible outcomes and add up the chances of each that apply:
This would be the correct approach assuming that rain on the three different days are independent events.
What they mean, reklaw, is that the same skeleton of math underlies many questions like this, for example, a coin toss.
They are both questions of probability theory, but in the case of the coin toss one has the extra information that successive tosses are independent of one another. That extra information is absent here.
I am not saying that probability theory doesn't apply to the weather! Let me make my claim more precisely. Suppose we posit that there is some distribution of possible weather phenomena, and the actual phenomena will be chosen randomly from this distribution -- the usual starting point for problems in probability. Suppose for a second that we knew the distribution. Then we could compute various different things from this distribution. In particular, we could compute the probabilities P_1, P_2, P_3 of rain on day 1, day 2, or day 3 respectively. One could also compute the probability P that it will rain on any of the three days. All of these would be well defined numbers. But unless you make additional assumptions about the nature of the distribution, these numbers would not satisfy any simple relation; in particular, it would be impossible to calculate P from P_1, P_2, P_3.
The scenario described by ChasFile, where a storm is coming with probability .70 and it may arrive on one of three different days, is an example of a possible distribution: there P_1 = .20, P_2 = .20 and P_3 = .30 as stipulated in the problem, but P = .70.
Another possible distribution is one in which the weather on every day is independent from the weather on every other day, and where the probabilities on each individual day are as stipulated in the problem. In that case, and only in that case, the computation of P given by various posters upthread is correct.
Without knowing the distribution, there is no way to calculate P, even given P_1, P_2 and P_3.
posted by em at 10:36 AM on April 13, 2006
Em, I'm confident that everyone upthread who proposed answers based on an independence assumption knew they were doing it, and knew the importance of making that. I'm also confident that the original poster knows that claims like "on Tuesday there's a 20% chance of rain" have to be taken with an appropriate grain of salt. That's why the word if - which he uses to introduce his hypotheses - is so important.
A short comment that the hypotheses are questionable in the real world might be appropriate, but that's about all that's needed in that direction.
posted by Wolfdog at 10:49 AM on April 13, 2006
A short comment that the hypotheses are questionable in the real world might be appropriate, but that's about all that's needed in that direction.
posted by Wolfdog at 10:49 AM on April 13, 2006
Response by poster: I'm the original poster, and actually, I really did mean it as a weather question. I'm in Raleigh, which is in a severe drought, and we're trying to decide what the real prospects for rain are.
The data we have are e.g. these from weather.com; chance of precipitation is given as a percentage for each day, and we want to understand how to put them together.
I gotta say, this is way more entertaining then I thought it would be.
posted by futility closet at 11:13 AM on April 13, 2006
The data we have are e.g. these from weather.com; chance of precipitation is given as a percentage for each day, and we want to understand how to put them together.
I gotta say, this is way more entertaining then I thought it would be.
posted by futility closet at 11:13 AM on April 13, 2006
Em, I'm confident that everyone upthread who proposed answers based on an independence assumption knew they were doing it, and knew the importance of making that.
Then they shouldn't have attacked ChasFile, who provided an equally correct (perhaps more correct) answers based on a different assumption, by making comments like
Those of you waving your hands in the air and violating the terms of the problem by inventing how weathermen aren't accurate.... bah. You shouldn't be answering this kind of question.
The exquisite combination of condescension and wrongness in this reply left me breathless.
I'm also confident that the original poster knows that claims like "on Tuesday there's a 20% chance of rain" have to be taken with an appropriate grain of salt.
Just to be sure: the issue of independence or non-independence has nothing to do with the issue of whether the stated probabilities for a given day are accurate. Even if one knows the probabilities for every given day with complete accuracy, without additional information one still cannot determine the probability for a longer period.
posted by em at 11:36 AM on April 13, 2006
Then they shouldn't have attacked ChasFile, who provided an equally correct (perhaps more correct) answers based on a different assumption, by making comments like
Those of you waving your hands in the air and violating the terms of the problem by inventing how weathermen aren't accurate.... bah. You shouldn't be answering this kind of question.
The exquisite combination of condescension and wrongness in this reply left me breathless.
I'm also confident that the original poster knows that claims like "on Tuesday there's a 20% chance of rain" have to be taken with an appropriate grain of salt.
Just to be sure: the issue of independence or non-independence has nothing to do with the issue of whether the stated probabilities for a given day are accurate. Even if one knows the probabilities for every given day with complete accuracy, without additional information one still cannot determine the probability for a longer period.
posted by em at 11:36 AM on April 13, 2006
Since there's a surprising amount of disagreement, here's one more "vote" confirming that Dasein's answer really is correct.
posted by dfan at 1:05 PM on April 13, 2006
posted by dfan at 1:05 PM on April 13, 2006
Reading the responses to this question, I am reminded of just how poor the average person's grasp on probability is.
posted by iconjack at 1:06 PM on April 13, 2006
posted by iconjack at 1:06 PM on April 13, 2006
I'm the original poster, and actually, I really did mean it as a weather question. I'm in Raleigh, which is in a severe drought, and we're trying to decide what the real prospects for rain are.
I hope all of you who were so snotty about ChasFile and others who answered based on what turned out to be a correct assumption are now blushing. Try not to be so sure of yourselves from now on, huh?
Since we are talking about actual weather forecasts, there is no real way to answer the question; I'd basically average the probabilities and figure there's some chance of rain over that period, but not a hell of a lot. Mathematical calculations make no sense because 1) the chances on consecutive days are not independent and 2) the "probabilities" are actually descriptions of previous weather conditions; as the Straight Dope puts it, "When you hear there's a 10 percent chance of rain, that means that out of the last 100 times the weather conditions were just like they are now, it rained 10 times."
Also, as I keep reminding my wife, weathermen are usually wrong.
posted by languagehat at 1:28 PM on April 13, 2006
I hope all of you who were so snotty about ChasFile and others who answered based on what turned out to be a correct assumption are now blushing. Try not to be so sure of yourselves from now on, huh?
Since we are talking about actual weather forecasts, there is no real way to answer the question; I'd basically average the probabilities and figure there's some chance of rain over that period, but not a hell of a lot. Mathematical calculations make no sense because 1) the chances on consecutive days are not independent and 2) the "probabilities" are actually descriptions of previous weather conditions; as the Straight Dope puts it, "When you hear there's a 10 percent chance of rain, that means that out of the last 100 times the weather conditions were just like they are now, it rained 10 times."
Also, as I keep reminding my wife, weathermen are usually wrong.
posted by languagehat at 1:28 PM on April 13, 2006
What em and languagehat said.
posted by Count Ziggurat at 1:47 PM on April 13, 2006
posted by Count Ziggurat at 1:47 PM on April 13, 2006
Woah there Wolfdog, chrisch's problem doesn't work like that. If the weatherman is wrong, that doesn't mean a %40 chance of rain rather than %60. In fact chrisch's problem doesn't strike me as answerable.
posted by Shutter at 1:52 PM on April 13, 2006
posted by Shutter at 1:52 PM on April 13, 2006
Yes, I had some misgivings about the chrisch problem after posting that.
But I don't have any misgivings about answering the original poster's question as stated, and I'm not blushing. There was a set of hypotheses laid out, and in that context there is a correct answer to the problem. Subsequent comments about "that's not how weather works in the real world" fall in the "no shit" category, and remind one of the insufferable wise-ass at the back of the room ("Isn't that assuming we're in a vaccuum?" "That only works in characteristic zero, right?"). It's hard to see how "No one knows! The weathermen are always wrong! I'd just average and guess!" is more helpful than a guide to how a model of the problem is solved under a certain set of assumptions.
posted by Wolfdog at 2:04 PM on April 13, 2006
But I don't have any misgivings about answering the original poster's question as stated, and I'm not blushing. There was a set of hypotheses laid out, and in that context there is a correct answer to the problem. Subsequent comments about "that's not how weather works in the real world" fall in the "no shit" category, and remind one of the insufferable wise-ass at the back of the room ("Isn't that assuming we're in a vaccuum?" "That only works in characteristic zero, right?"). It's hard to see how "No one knows! The weathermen are always wrong! I'd just average and guess!" is more helpful than a guide to how a model of the problem is solved under a certain set of assumptions.
posted by Wolfdog at 2:04 PM on April 13, 2006
The original question is:
If there's a 20 percent chance of rain on Monday, a 20 percent chance on Tuesday, and a 30 percent chance on Wednesday, what's the overall chance that it'll rain during this three-day period? How do you calculate this?
Where do you see the hypothesis that these probabilities are independent? I am getting the sense that you believe the facts as described would be incoherent without the additional statement that the probabilities are independent, so that even though it was not explicitly stated, it must have been implicitly meant. This is not the case. As several people have stated (including explicit examples) it is quite possible for the probabilities to be as described without the events being independent. The actual world, with its actual weather forecasts, is precisely an example of that phenomenon.
And once again, this issue has nothing to do with the question of whether the weatherman correctly computes the probabilities or not.
posted by em at 2:23 PM on April 13, 2006
If there's a 20 percent chance of rain on Monday, a 20 percent chance on Tuesday, and a 30 percent chance on Wednesday, what's the overall chance that it'll rain during this three-day period? How do you calculate this?
Where do you see the hypothesis that these probabilities are independent? I am getting the sense that you believe the facts as described would be incoherent without the additional statement that the probabilities are independent, so that even though it was not explicitly stated, it must have been implicitly meant. This is not the case. As several people have stated (including explicit examples) it is quite possible for the probabilities to be as described without the events being independent. The actual world, with its actual weather forecasts, is precisely an example of that phenomenon.
And once again, this issue has nothing to do with the question of whether the weatherman correctly computes the probabilities or not.
posted by em at 2:23 PM on April 13, 2006
I once heard this explained. From what I remember, that 20% chance of rain is actually meant to indicate that there is a 100% chance of rain in roughly 20% of the forecaster's specified area. Which has nothing to do with solving any sort of math equation.
posted by ninjew at 2:46 PM on April 13, 2006
posted by ninjew at 2:46 PM on April 13, 2006
it is quite possible for the probabilities to be as described without the events being independent
I disagree with that, and I'll explain how I'm reading it, and that will have to be the end of my contributions on this.
I read the quoted sentence as being exactly equivalent to the follwing:
"Suppose the following three conditions:
1. The probability of rain on Monday is 20%.
2. The probability of rain on Tuesday is 20%.
3. The probability of rain on Wednesday is 30%.
Then what is the probability etc."
Each of the three numbered statements is an assertion about an unconditional probability. Or, to make my own judgement explicit, I think that the only reasonable interpretation of a statment of the form "The probability of E is p" is as an unconditional probability. At step two, if the poster wanted to introduce into the problem some conditional probability then that needs an explicit statement ("The probability of rain on Tuesday is 20% unless it has previously rained on Monday"). The default thing to do in a case like this is, as a reader, to take the one and only reading that makes the problem solvable; solve that; and point out the extra assumption that was made. Quite a few people did that.
Questions like this one are notoriously contentious because when we calculate probabilities for real world events we are always making assumptions and using a model. When we agree about the meaning of the hypotheses, there's only one correct answer for the problem. When we disagree about the hypotheses, then we come to different answers. That is fine, and far from earth-shattering, but it doesn't excuse answers of the form, "Your facts could be misinterpreted, so any old guess will do." One or two people did that, and got some criticism for it.
Any alternate reading of the question seems to interpret it as if it said "If ... [facts I know are uselessly incorrect as a model] .. then what is the probability etc" which really seems uncharitable to the questioner. If the question was not "how do you compute that" but rather "I know you can't compute that, but what are some entertaining guesses?", then sure, anything goes.
posted by Wolfdog at 2:55 PM on April 13, 2006 [1 favorite]
I disagree with that, and I'll explain how I'm reading it, and that will have to be the end of my contributions on this.
I read the quoted sentence as being exactly equivalent to the follwing:
"Suppose the following three conditions:
1. The probability of rain on Monday is 20%.
2. The probability of rain on Tuesday is 20%.
3. The probability of rain on Wednesday is 30%.
Then what is the probability etc."
Each of the three numbered statements is an assertion about an unconditional probability. Or, to make my own judgement explicit, I think that the only reasonable interpretation of a statment of the form "The probability of E is p" is as an unconditional probability. At step two, if the poster wanted to introduce into the problem some conditional probability then that needs an explicit statement ("The probability of rain on Tuesday is 20% unless it has previously rained on Monday"). The default thing to do in a case like this is, as a reader, to take the one and only reading that makes the problem solvable; solve that; and point out the extra assumption that was made. Quite a few people did that.
Questions like this one are notoriously contentious because when we calculate probabilities for real world events we are always making assumptions and using a model. When we agree about the meaning of the hypotheses, there's only one correct answer for the problem. When we disagree about the hypotheses, then we come to different answers. That is fine, and far from earth-shattering, but it doesn't excuse answers of the form, "Your facts could be misinterpreted, so any old guess will do." One or two people did that, and got some criticism for it.
Any alternate reading of the question seems to interpret it as if it said "If ... [facts I know are uselessly incorrect as a model] .. then what is the probability etc" which really seems uncharitable to the questioner. If the question was not "how do you compute that" but rather "I know you can't compute that, but what are some entertaining guesses?", then sure, anything goes.
posted by Wolfdog at 2:55 PM on April 13, 2006 [1 favorite]
Each of the three numbered statements is an assertion about an unconditional probability . . . if the poster wanted to introduce into the problem some conditional probability then that needs an explicit statement.
Sure! Nobody is arguing with that. But even given these three statements about the unconditional probability, you can't conclude that the events are independent. Look, with any prior distribution, every event has some unconditional probability. It doesn't mean that every event is independent of every other event.
The default thing to do in a case like this is, as a reader, to take the one and only reading that makes the problem solvable
In order to solve the problem, you have to guess the underlying distribution. One possible guess is that the events are independent. Another guess is the guess ChasFile indicated. Both guesses lead to definite answers. I am hard pressed to come up with any property of the guess that the events are independent that makes it better than any other guess.
posted by em at 3:30 PM on April 13, 2006
Sure! Nobody is arguing with that. But even given these three statements about the unconditional probability, you can't conclude that the events are independent. Look, with any prior distribution, every event has some unconditional probability. It doesn't mean that every event is independent of every other event.
The default thing to do in a case like this is, as a reader, to take the one and only reading that makes the problem solvable
In order to solve the problem, you have to guess the underlying distribution. One possible guess is that the events are independent. Another guess is the guess ChasFile indicated. Both guesses lead to definite answers. I am hard pressed to come up with any property of the guess that the events are independent that makes it better than any other guess.
posted by em at 3:30 PM on April 13, 2006
Response by poster: Many thanks to everyone who contributed. Your discussion of how to reach the precise answer, with independent probabilities, gave me the intuitive understanding that I was looking for (even in a messy/approximate application like weather).
I'm always struck at what a smart and earnest crowd this is, intent on doing a proper job even for some bozo in North Carolina. Thanks.
posted by futility closet at 5:19 PM on April 13, 2006
I'm always struck at what a smart and earnest crowd this is, intent on doing a proper job even for some bozo in North Carolina. Thanks.
posted by futility closet at 5:19 PM on April 13, 2006
From what I remember, that 20% chance of rain is actually meant to indicate that there is a 100% chance of rain in roughly 20% of the forecaster's specified area.
Actually that is incorrect, but you are not alone. 3% of respondents gave that answer to this multiple choice question #3 in this interesting study.
The percentage at the end of each answer is the percentage of people with that response.
Question -- A precipitation probability forecast of 30% means:
a. Measurable precipitation will occur 30% of the forecast time period and not occur 70% of the time period. (3%)
b. At any particular point in the forecast are (for example, your house) there is a 30% chance that there will be measurable precipitation and a 70% chance that there will be no measurable precipitation during the forecast period. (39%)
c. There is a 30% chance that measurable precipitation will occur somewhere (i.e. in at least one place) within the forecast area during the forecast period and a 70% chance that it will not occur anywhere during the period. (56%)
d. If precipitation occurs during the forecast period, 30% of the total are will experience precipitation and 70% of the area will not have any measurable precipitation. (3%)
The correct answer is B. POP is the probability of rain at any particular random point in the forecast area. More than half of people gave the incorrect answer C.
If you think about it, B is actually the most useful information for most people because it tells you your chances of getting wet. C and D concern other people getting wet.
posted by JackFlash at 5:47 PM on April 13, 2006 [1 favorite]
Actually that is incorrect, but you are not alone. 3% of respondents gave that answer to this multiple choice question #3 in this interesting study.
The percentage at the end of each answer is the percentage of people with that response.
Question -- A precipitation probability forecast of 30% means:
a. Measurable precipitation will occur 30% of the forecast time period and not occur 70% of the time period. (3%)
b. At any particular point in the forecast are (for example, your house) there is a 30% chance that there will be measurable precipitation and a 70% chance that there will be no measurable precipitation during the forecast period. (39%)
c. There is a 30% chance that measurable precipitation will occur somewhere (i.e. in at least one place) within the forecast area during the forecast period and a 70% chance that it will not occur anywhere during the period. (56%)
d. If precipitation occurs during the forecast period, 30% of the total are will experience precipitation and 70% of the area will not have any measurable precipitation. (3%)
The correct answer is B. POP is the probability of rain at any particular random point in the forecast area. More than half of people gave the incorrect answer C.
If you think about it, B is actually the most useful information for most people because it tells you your chances of getting wet. C and D concern other people getting wet.
posted by JackFlash at 5:47 PM on April 13, 2006 [1 favorite]
This thread is closed to new comments.
posted by MrMulan at 6:09 AM on April 13, 2006