# % of % - Please Help Me Prove My Friend Wrong

Here is the situation: My friend and I are at an impasse. I believe I'm correct, but he's so damn stubborn he won't believe me. Also, I'm not the most articulate at explaining things. Hopefully some of you guys can help me explain this to him in a way he'll understand. Here is the problem:

A DVD is either at his parent's house or his own. The probability that it's at his house is 30%. If the DVD is at his own house, there is a 90% chance it's on the porch, and a 10% chance it's in the living room. What is the % chance the DVD is on the porch?

My friend says you take 90% of 30% which is 27 and that is the % chance it's on the porch. Is this correct? I don't believe so.

I believe that regardless of where the DVD is, the chance of it being anywhere in his house is still 30% overall. Location inside his house won't change those odd because the porch and the living room are both part of the house. If there is a 90% chance it's on the porch, it doesn't change the overall odds of it being in that location.

Now, if you rephrase the question and ask, "The DVD Is either at my parents, my porch, or my living room. What is the % chance it's on my porch?", the answer is 33%. If there are three places it could be, then there is 33.333% chance it's on the porch. Even if it's a 90% chance it's at his house, if there are only three places it can be, it remains the same.

I think the correct way of answering the question is: There is a 30% chance the DVD is at my house. If it is at my house, there is a 90% chance it's on my porch. They are two separate odds and you can't take a percentage of the overall odds since the locations are inside the house.

Is this correct or am I wrong? And regardless, please give me your explanation.

• I'm afraid your friend is correct. One of the flaws in your argument is "The DVD Is either at my parents, my porch, or my living room. What is the % chance it's on my porch?", the answer is 33%". By that argument, I am either human or I am not. Therefore there is 50% chance I'm not human... sounds dubious. – Mathmo123 Nov 26 '14 at 18:51
• @Mathmo123 Or there's a 50 % chance I win the lottery this week (either I win or I don't) – Hagen von Eitzen Nov 26 '14 at 18:54
• You came to the right place with this question! We all leap at the chance of telling somebody that they're wrong :-) – TonyK Nov 26 '14 at 19:06
• Nothing like a good pie chart to clear things up. (jsfiddle). – Digital Chris Nov 26 '14 at 19:33
• Y'know, you might have been able to figure this out on your own if you were not "so damn stubborn" and listened to your friend long enough for him to explain this in a way that you'd understand. You've obviously approached this from the start in a way that excludes the possibility that you are wrong. BUT then you contradicted that by coming here and asking about it, so you've improved ... even if you really just wanted us to help prove you right. ;) – Lightness Races in Orbit Nov 27 '14 at 16:19

## 13 Answers

Your friend is correct and I'll give you an experiment you can try:

Take two boxes labeled "my house" and "my friend's house" and in the first box put two bags labeled "porch" and "living room". You will also need a marble and a die. Take one die and roll it (we're going to approximate 70% ~ 2/3 here). If it's a 1,2,3 or 4, put your hand in the "my house" box but don't pick a bag yet. If it's a 5 or 6 put it anywhere in the "friend's house" box. If your hand is in the "my house" box then you need to roll again to figure out which bag to put it in. If it's a 1,2,3,4 or 5 (let's say 90% ~ 5/6) put it in the "porch" box, and if it's a 6 put it in the "living room" box. Repeat this exercise until you have a feel for how often it lands in "my porch." Record the trials and see what the odds are. You should be convinced now.

Your reasoning is incorrect here:

If there is a 90% chance it's on the porch, it doesn't change the overall odds of it being in that location.

No, you changed things. There is not a 90% chance it's on the porch. If - if! - it is in your house, then (and only then!) there is a 90% chance it's on the porch. The magic of the 27% comes from the fact that mathematically - and experimentally, as I hope the above box/bag/marble exercise shows - we know how the "in your house" 30% and the "on the porch" 90% interact. Namely, they interact multiplicatively.

How about this? There is a 30% chance you'll go to New York and a 100% chance you'll go to the Empire State Building if you go to New York (because why else would you go to New York? kidding...). Does this mean there's a 100% chance you'll go to the Empire State Building? Well, since you have to go to NY to go to the ESB, that would mean there's a 100% you'll go to New York - and now we're being contradictory! So this interpretation makes no sense and is never what we mean mathematically or in plain English.

"The DVD Is either at my parents, my porch, or my living room. What is the % chance it's on my porch?", the answer is 33%.

This is definitely wrong. I'm pretty sure you either have ebola or you don't have ebola. Up to you whether you need to call 911 and get yourself quarantined right away because there's a 50% chance you have ebola. Or perhaps the doctors gave your sick relative 6 months to live, but your relative might live a year or two years or three years or four years or five years, which means there's at least an 86% chance the doctor is wrong.

Now put 10 red m&ms in a bag and 1 blue one. Grab one without looking. Since there's two possibilities, there's a 50% chance it's a blue one, right? So I'll bet you a dollar that it's red and you bet me a dollar that it's blue, and we'll see who's paying for lunch later.

• No! You're supposed to bet before you explain why you'll win! – ike Nov 26 '14 at 20:13
• A one dollar lunch? Where do you even get these things @AAA? – yuritsuki Nov 26 '14 at 23:00
• @venidividicivicini yeah... writing that bothered me but I didn't really want to bet the OP \$20. I should have bet \$10. – djechlin Nov 26 '14 at 23:16
• Just repeat the M&M test and you'll quickly have enough to buy a big lunch! – Stephan Nov 27 '14 at 5:39
• -1 as this doesn't really help proving friend wrong :) – Kuba Nov 28 '14 at 11:50

The DVD is

• at his parents' house with a probability of $70\,\%$ ($=100\,\%-30\,\%$)
• in the porch of his house with $27\,\%$ ($=90\,\%$ of $30\,\%$)
• in the living room of his house with $3\,\%$ ($=10\,\%$ of $30\,\%$)

Check: $70\,\% + 27\,\%+3\,\%=100\,\%$.

• (and these possibilities are all mutually exclusive) – djechlin Nov 26 '14 at 19:19
• This wouldn't convince anyone who originally agreed with the OP's arguments. No explanation. – ike Nov 26 '14 at 20:13
• @ike I think summing to 100% is an argument in itself – keyser Nov 26 '14 at 22:17
• @keyser 1/3 for each also adds to 100%. – ike Nov 26 '14 at 23:11
• @keyser It made sense to you, but not to the OP. Saying that it is obvious does not help them understand. Just because the numbers add to 100 does not make it the right answer; three 1/3 s also add to 100. It looks like you don't really understand what's going through OP's head, which makes it difficult for you to explain anything to them. – ike Nov 27 '14 at 0:58

Here are two ways to think about probability, which I often find helpful.

# The Frequentist Interpretation

In the frequentist interpretation of probability, you have a large number of situations set up the same way, and the probability of something being true tells you the fraction of those situations in which the thing is true.

Suppose there are a million parallel universes, each with its own version of you, your friend, your houses, and the DVD. Then the statements your friend makes have the following consequences:

The probability that it's at his house is 30%.

This means that in 30% of the universes, that's 300,000 of them, the DVD is at your friend's house.

If the DVD is at his own house, there is a 90% chance it's on the porch, and a 10% chance it's in the living room.

Let me take this piece by piece because it's the most important part:

• If the DVD is at his own house,

This means that you have to only consider the universes where the DVD is at your friend's house. There are 300,000 of these. You have to forget about the rest of the universes for now.

• there is a 90% chance it's on the porch

In 90% of the universes, the DVD is on the porch. But we're pretending there are only 300,000 universes. So in 90% of those, or 270,000 universes, the DVD is on the porch.

• and a 10% chance it's in the living room.

Again, we're pretending there are 300,000 universes in all. In 10% of those, or 30,000, the DVD is in the living room.

Okay, those are all the statements, so time to stop pretending and go back to considering all million universes. We have the following totals:

• 30,000 universes where the DVD is in the living room at your friend's house
• 270,000 universes where the DVD is on the porch at your friend's house
• 700,000 universes where the DVD is at his parents' house

To find the probability of the DVD being on the porch, you take the number of universes where the DVD is on the porch and divide it by the total number of universes.

$$P(\text{DVD on porch}) = \frac{\text{# of universes where it's on the porch}}{\text{total # of universes}} = \frac{270\,000}{1\,000\,000} = 27%$$

And similarly for the other cases:

\begin{align} P(\text{DVD in living room}) \\ &= \frac{\text{# of universes where it's in the living room}}{\text{total # of universes}} \\ &= \frac{30\,000}{1\,000\,000} \\ &= 3\% \end{align} and \begin{align} P(\text{DVD at parents' house}) &= \frac{\text{# of universes where it's at his parents' house}}{\text{total # of universes}} \\ &= \frac{700\,000}{1\,000\,000} \\ &= 70\% \end{align}

You can also combine cases: \begin{align} P(\text{DVD at friend's house}) &= \frac{\text{# of universes where it's at your friend's house}}{\text{total # of universes}} \\ &= \frac{30\,000 + 270\,000}{1\,000\,000} \\ &= 30\% \end{align} which was an assumption from the start, so of course that has to be the result - but it's good to see that the math works out.

If you tried to say

If there are three places it could be, then there is 33.333% chance it's on the porch.

then you would be claiming that in 333,333 of the universes, the DVD is on the porch. That clearly conflicts with what we calculated, that the DVD is on the porch in only 270,000 universes! So if the earlier statements about probability (30% that it's at his house, etc.) are correct, this latest statement cannot also be correct. You can't assume that the probability of something being true is $1/N$ just because there are $N$ possibilities! Not unless you know, somehow, that all the possibilities are equally likely. (In fact, I didn't say this before, but when I invented the million universes, I assumed that each universe is equally likely. That's important.)

# The Bayesian Interpretation

In the Bayesian interpretation, probability is a reflection of how much you know or don't know about some system. It seems quite similar to the frequentist interpretation, at first, but it works a little differently when you start talking about conditional probabilities ("if X then the probability of Y is Z").

To explain this, let me go back to the universes. We started with a million of them. Then you said

The probability that it's at his house is 30%

which means that 300,000 universes have the DVD at your friend's house. OK, that much is the same as the frequentist interpretation.

Then move along to the next statement, and again I'll take it piece by piece:

• If the DVD is at his own house,

OK, now we're saying you have determined that the DVD is at your friend's house. This is where the Bayesian intepretation differs from the frequentist interpretation: for the rest of this statement, we'll say you know that the DVD is at your friend's house. So you can literally throw out the 700,000 universes where that is not the case. They don't exist anymore.

Like I said, it's a pretty subtle difference.

One consequence of this, by the way, is that the probability of the DVD being at your friend's house is now 100%. Or to be more precise, when you found out that the DVD was at your friend's house, you updated the probability of the DVD being at your friend's house from 30% (that's 300,000/1,000,000) to 100% (that's 300,000/300,000).

• there is a 90% chance it's on the porch

In 90% of the universes, the DVD is on the porch. There are 300,000 universes, so in 90% of those, or 270,000 universes, the DVD is on the porch.

• and a 10% chance it's in the living room.

In 10% of 300,000 universes, or 30,000 of them, the DVD is in the living room.

If you now back up to the point where you didn't know the DVD was at your friend's house, you'll see that the probabilities wind up being the same as in the frequentist interpretation. (That's true in general. These interpretations are just different ways to think about probability, but they produce the same results.)

There are mathematical procedures for "reversing" an assumption that you made, but I won't get into that level of detail. The point is just that Bayesian probability is a reflection of your knowledge of the system, and that you update probabilities as you learn more about it.

• While I think this answer does an effort to explain how OP should correct his reasoning, I think throwing the freqentist-Bayesian opposition at it is totally beside the point, and therefore contraproductive and unhelpful. – Marc van Leeuwen Nov 30 '14 at 8:38
• @MarcvanLeeuwen I included the Bayesian interpretation because I thought it would be useful for the OP to see why the probability of the same event can be a different number depending on context. In particular, why the probability of the DVD being on the porch can be quoted as 90% or as 27%, and how both numbers are correct in their appropriate contexts. It seemed that the OP's confusion stemmed in part from not understanding this; they assumed 90% simply means 90% overall probability, period. – David Z Nov 30 '14 at 8:47
• That being said, perhaps including the names and links to Wikipedia articles may be more confusing than necessary. – David Z Nov 30 '14 at 8:47
• OK, I see. But then the confusion you address is the one between conditional probability and absolute probability, which to me seems to be relevant independently of the interpretation. – Marc van Leeuwen Nov 30 '14 at 9:58
• @MarcvanLeeuwen this is true. I thought it might be useful to provide two different ways of thinking about it, since I use each of them in certain circumstances. I would note that I posted my answer when there were already several perfectly good answers, and I see it (mine) as a followup or complement to those, for readers who are interested in a different perspective. Perhaps I should edit that in. – David Z Nov 30 '14 at 10:47

Your friend is right, and his explanation is right. Ask yourself this, using the reasning you put forth in your explanation:

My hat is either in my closet, on my head, or I left it on Mars. What is the probability that it is on Mars?

I'm guessing you can see that just because there are 3 possibilities, that does not mean we should assign a 33.33% probability to each.

• As you have not supplied probabilities of each outcome it might be reasonable to assume that each is equally likely until further evidence is provided :-). – Bob Jarvis Nov 29 '14 at 18:03
• @BobJarvis No, it would not be reasonable to assume that. – Miles Rout Nov 30 '14 at 0:30
• No, all of those have a 100% chance of being true, because he is in his closet on Mars, wearing his hat. – rodolphito Nov 30 '14 at 7:37

Probabilities like this can be represented as a tree:

100% --- 70% Parent's House - 70%
|
|- 30% His House ------ 27% (30%*90%) Porch
|-  3% (30%*10%) Living Room


Notice all the leaves of the tree add up to 100%.

The probability of it being on the porch is conditional on it being in his house. Conditional probabilities are calculated based on the condition of the parent probability being true.

They are two separate odds and you can't take a percentage of the overall odds since the locations are inside the house.

They are not separate. There are four places described: parent's house, his house, porch, living room. Consider these two statements, one is obviously wrong:

• The DVD can be at his house, and also be on the porch. (This makes sense because the porch is at his house.)
• The DVD can be at the parent's house, and also be on the porch. (Makes no sense, because the porch is not at his parent's house)

If they were truly separate as you rationalize, then either statement could be true. However, only one of the above is true because there the chance of it being on the porch is conditional on it being in the house, and therefore is conditional on the probability of it being in the house. So the probability of it being on the porch is a portion of the probability of it being in his house.

If we added another probability, such as "There is a 40% chance of the DVD being scratched." then that would be independent of where it was. Thus it would be considered an independent probability and how you calculated the probability of combinations of where it was and whether or not it was also scratched would be quite different.

If you Google conditional probability, you will find alot of examples.

• Minor nitpick (because I don't have the rep to make trivial edits): the paragraph just below the bullet points says "However, onyl one" rather than only one. – a CVn Nov 27 '14 at 13:36

I believe that regardless of where the DVD is, the chance of it being anywhere in his house is still 30% overall.

And it is! And 90% of the time that it is, it is on the porch, and 10% of the time it's in the living room.

That makes, in total, 90% of 30% is 27% chance that it's on the porch, and, a 10% of 30% is 3% chance that it's in the living room.

Since 27% + 3% = 30%, that means that the chance that it is anywhere in his house is indeed still 30% overall. That part of your intuition was fine.

It's just that that 30% can be subdivided in portions based on the probability of each place in the house it could be, if it is in the house.

See, it's true that there are only three possibilities: at your house, the porch or the living room, but it is not equally likely that one may occur. Consider this: if we have to compute the probability that the today is Sunday, you might say 1/2, either it is a Sunday or it's not. But being both is not equally likely. The number of possibilities that it is not a Sunday is 6, and that it is a Sunday is 1. So the probability that today is a Sunday is 1/7.

Similarly the above goes in your case as well. There are two possibilities: either the DVD is in your house, or it is in your friends house. If it is in your friends house, it's either in the living room or the porch. The probability that it is in your friends house is $30%$, so the sub cases that exist here will have their sum as $30%$. The sub cases are $90%$ and $10%$ are percentages of the percentage of the actual value, which is $30%$. So it's 27 and 3 respectively.

You wanted confirmation of your view... well tough luck: You are confused and your friend is right.

I think this is as much a language as a math problem, the "%" ist confusing you, so do away with it for a moment.

To do this you need to do some/a lot running around (at least virtually/in your head):

Assume your friend and you are going to a pub every weekend, afterwards you watch a DVD together. You do this 100 times. 70 times, you need to go to his parents to get the DVD. 30 times you need to go to you friend's place to get it.

Now to get back to this "%"-thing... "percent" or (remember: language problem) perhaps think about it as "per cent", which means "per a hundred" or "from a hundred".

So 30 times from 100 times is 30%, right? In 10% of the times you go to his place, the DVD is in the living room, and 10% from 30 is 3 times. Three times from a hundred, per a hundred, per cent, or 3%.

So the probabilites for your given problem are:

• 70% DVD is at friend's parent's
• 27% DVD is on friend's porch
• 3% DVD is in friend's living room

To answer your general question, it's valid to calculate percent of percent to reach the overall probability.

There are already really nice answers to your question that explain why you are wrong.

I am going to try and show you where your thinking goes wrong.

Lets say you have 2 stones. One is blue and the other is red.

Case 1: If I ask you to pick a stone, what is the probability that you will pick the red one?

50%. since you have two options and you can only pick one. This gives you 1/2 aka 50%.

Now lets say I hide them. One in a black box and the other in a white box.

Case 2: And now I ask you to pick a box. What is the probability that you will pick a box with the red stone?

Again 50%.

Here is a list of all the possibilities for Case 2:

1. White Box, Red stone 2. White Box, Blue stone 3. Black Box, Red Stone 4. Black Box, Blue Stone

And since there are:

2 out of 4 ways to pick a red stone, you have a 50% chance

Case 3: And what is the probability that you will find the red stone in the white box?

25%! The difference here is that first you will have to pick a box and then you have to deal with the probability of finding the red rock in that box.

You can list out all the possibilities for Case 3:

1. White Box, Red stone 2. White Box, Blue stone 3. Black Box, Red Stone 4. Black Box, Blue Stone

and you will find that only:

1 out of the 4 possibilities gives you the correct combination. Hence 25%! This is the same as 50% to pick a box and 50% that you get the red rock.

What you and your friend are arguing about is similar to Case 3 but you are mixing it up with Case 2 and hence your friend is correct.

Your friend is right. Let $A$ = "at his house" and $B$ = "on his porch". Then $A \cap B$ = "at his house AND on his porch". $P\left(X\right)$ means "the probability that $X$ is true"; $P\left(X|Y\right)$ means "the probability that if $Y$ is true, then $X$ is also true". $P\left(X|Y\right)$ is referred to as a conditional probability.

You have given that $P\left(A\right) = .30$ and $P\left(B|A\right) = .90$

The very definition of conditional probability is that $P\left(B|A\right) = \frac{P\left(B \cap A\right)}{P\left(A\right)}$

And since it cannot both be on his porch and not be at his house, $B \cap A = B$

Therefore, $P\left(B|A\right) = \frac{P\left(B\right)}{P\left(A\right)}$; $P\left(B\right) = P\left(B|A\right)P\left(A\right)=\left(.90\right)\left(.30\right)=.27$

Think of it this way: X = "on his porch"; Y = "at his house but not on his porch"; Z = "not at his house" (Therefore, "either X or Y" = "at his house")

XXXXXXXXXX
XXXXXXXXXX
XXXXXXXYYY
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ


If you choose a random letter, then your random choice is consistent with the information you provided:

• there is a 30% chance that you will get either X or Y

• There is a 90% chance that if you get X or Y, then it is X

Note that 27 of the 100 letters are X.

You've conflated unconditional probability with conditional probability.

You've said that

1. p(parents house) = 70%
2. p(friend's house) = 30%.

Here p stands for probability. So far so good. Then you gave the conditional probabilities:

1. p(porch | at friends house) = 90%
2. p(living room | at friends house) = 10%

Here the vertical bars are common notation indicating that the probability that it is on the porch or living room is conditioned on it being at your friend's house.

In general the joint probability, p(A,B), is given by p(A,B) = p(A|B)p(B). In this case A is "porch" and B is "at friend's house":

p(porch, at friend's house) = p(porch|at friends house) x p(friends house) = .9 x .3 = 27%

Here's the minimal answer, with no philosophy or extraneous material.

Define the following events:

• its at your own house $=O$

• its on the porch (of your own house) $=O_p$

We're trying to find $\mathbf{P}(O_p).$

The following information is given.

• (Implicitly: $O_p \subseteq O$)
• $\mathbf{P}(O) = 0.3$
• $\mathbf{P}(O_p \mid O) = 0.9$

By the definition of conditional probability: $$\mathbf{P}(O_p \mid O) = \frac{\mathbf{P}(O_p \cap O)}{\mathbf{P}(O)}.$$

But since $O_p \subseteq O$, hence:

$$\mathbf{P}(O_p \mid O) = \frac{\mathbf{P}(O_p)}{\mathbf{P}(O)}.$$

Rearranging: $$\mathbf{P}(O_p) = \mathbf{P}(O_p\mid O)\mathbf{P}(O) = 0.9 \times 0.3 = 0.27$$

So the probability that its on the porch of your own house is $0.27$ (your friend was right).

If you have already dictated the odds, you can't then change them and treat the house and places as isolated cases. Your friend is right.

• I don't see how this explains anything. – Null Nov 29 '14 at 1:45

## protected by user642796Nov 29 '14 at 21:08

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