How is this probability reasoning wrong?

[samh]

How is this probability reasoning wrong??

Homework Statement

There are light socks and dark socks. Francis reasons as follows about socks.

"If I pull out 3 socks, at least 2 will be alike. Suppose that they are dark. The third one will also be dark with probability 1/2. Similarly, if the pair is light, the probability that the remaining one will be light is also 1/2, so the probability of all socks being the same color must be 1/2."

What is wrong with this argument? What is the actual probability of pulling out three socks of the same color.

Homework Equations

The professor didn't say it explicitly, but I'm pretty sure that sock choices are taken to be independent, so P(A,B,C) = P(A)P(B)P(C).

The Attempt at a Solution

The last part is easy...the correct probability is P(three socks of the same color) = P(all light) + P(all dark) = 1/2*1/2*1/2 + 1/2*1/2*1/2 = 1/4.

But I CANNOT figure out where Francis's reasoning is wrong. I've been thinking about it for so long... He seems to be completely right, I mean, when you grab 3 socks you ALWAYS end up with 2 matching socks, and the probability of the other sock matching those two is 1/2...right? So from that reasoning the probability of all socks matching is 1/2.

I've thought about it and thought about it and thought about it and I can't figure it out! It's driving me crazy! If you can find the flaw I would really appreciate it if you could please explain it or give me some good hints. Thanks to anyone who responds.

 

[Dr Zoidburg]

What would be the probability of the 3rd sock being light if you only have 1 pair of each colour - and the first two socks you pulled out were light?

 

[samh]

If there was only one pair of each (2 light, 2 dark) and the first two were the two lights then the is no third light sock so the probability would be 0..? Hmm I don't really see what you're getting at, sorry...

 

[kenewbie]

Argh, someone spell it out. I see why it's wrong but I fail to verbalize it.

"The third one will also be dark with probability 1/2". You have already used the third sock in establishing that 2 out of 3 need to be the same color. Which means you cannot use it as a single case probability in the same set.

Thats the best way I can put it :/

k

 

[HallsofIvy]

"If I pull out 3 socks, at least 2 will be alike. Suppose that they are dark. The third one will also be dark with probability 1/2.

Okay and, assuming there are a huge number of socks so that each color is equally likely on each pull (and Samh's example of 4 socks can't hold) the probability of getting two dark socks is(1/2)(1/2)= 1/4 to begin with. The probabilty of getting 3 dark socks is 1/8.

Similarly, if the pair is light, the probability that the remaining one will be light is also 1/2, so the probability of all socks being the same color must be 1/2."

And, again, the probability of all light socks is (1/4)(1/2)= 1/8, just as you calculated.

What this argument shows is that the probability of all socks being the same color is 1/2 given that the first two socks were the same color.

 

[Jane1]

The probability of light or dark changes after every sock pull. The only time the probability of an individual sock color is 1/2, is when the pile happens to contain 50% light, 50% dark. But once you pull one sock from the 50/50 pile, the probability of the next sock being a particular color adjusts.

The probability of pulling the first dark sock is (# dark socks) / (# total socks). The probability of pulling the 2nd dark sock is (# dark socks - 1) / (# total socks -1).

Say there are 6 dark and 4 lights socks.

The probability of pulling 3 dark socks is:

6/10 * 5/9 * 4/8 = 120/720 = 16.67%.

Each time, you just count the new number of darks that are left (6, then 5, then 4) for the numerators. The total also decreases 10, then 9, then 8 in the denominators.

The probability of pulling 3 light socks is:

4/10 * 3/9 * 2/8 = 24/720 = 3.3%.

The probability of pulling out 3 socks of the same color is (120+24)/720 = 144/720 = 20%

The mistake Francis makes is if 2 socks are dark, he can't assume that the third one being dark will have a probability of 1/2.

The probability of the 3rd sock also being dark depends on what the pile held after the first 2 socks were pulled. In my example of 6 dark and 4 light, it just happens to be 1/2 for the darks, but it is 1/4 for lights.

Picture a pile with 1 million dark socks and just 4 light socks. Now pull 3 socks. Suppose 2 are dark (surprised?). Is it obvious the chance of the 3rd sock being dark is not 1/2? (it's .9999).

 

[Jane1]

Okay and, assuming there are a huge number of socks so that each color is equally likely on each pull (and Samh's example of 4 socks can't hold) the probability of getting two dark socks is(1/2)(1/2)= 1/4 to begin with. The probabilty of getting 3 dark socks is 1/8.

And, again, the probability of all light socks is (1/4)(1/2)= 1/8, just as you calculated.

What this argument shows is that the probability of all socks being the same color is 1/2 given that the first two socks were the same color.

We can't confust wanting to find the P(all 3 same color) vs P(3rd is same color) given the first 2 are same color (what your bolded statement alludes to).

The original question steers you into confusing these. It first mentions the probability of the remaining sock being the same color.

Then at the end, it asks the actual probability of pulling 3 the same color.

For infinite socks, as in your example (but not the original example), the probability the 3rd sock is the same color, given the first two are the same color, is 1/2.

But the probability that all 3 are same color is 1/8 + 1/8 = 1/4.

 

[Dr Zoidburg]

I think the problem here is that Francis is getting confused between dependent vs independent probability. (sorry there's prob a better way to say that but that's the best I can come up with!)
By this I mean that he (assuming Francis is a he) is correct in saying the probability of the next sock being pulled out is light coloured is 50% (assuming an infinite number of socks). However, the probability of it being the same colour as the previous two socks is not 50%, as others have already shown. The first case is independent, whereas the in the 2nd it's dependent upon what's already happened (and the proportion of like-coloured socks remaining in the drawer).
Easiest way to see this is to draw a probablity tree diagram.

 

[samh]

Thanks for the replies everyone, but I'm still kind of confused as usual... Forget my first post and tell me what you think of this:

There are infinite socks. Choose 3 socks. 2 of them match, and have color X, where X=light or X=dark. The probability of the other sock being color X is 50%. Therefore, the probability of 3 matching is 50%.

Which sentence in that reasoning is wrong?

 

[kenewbie]

The probability of the other sock being color X is 50%.

This is wrong. The probability of the 3rd sock being of the correct color is 50% only if the first two are of the same color, which happens to be a 50% chance as well. 50% of 50% is 25%, which is 1/4.

When you pick 3 socks at once to make sure you get 2 of the same color, you can no longer say that the last sock has a 50% chance of matching the 2 of the same color, because that last sock contributed to the fact that you got 2 of the same color in the first place!

k

 

[MeJennifer]

The probability of getting three socks with the same color is not 1/8 but less, unless you can select from an infinite number of socks.

 

[D H]

There are infinite socks. Choose 3 socks. 2 of them match, and have color X, where X=light or X=dark. The probability of the other sock being color X is 50%. Therefore, the probability of 3 matching is 50%.

Which sentence in that reasoning is wrong?

You did two things wrong. The first is assuming that dark socks and light socks are evenly distributed: i.e., drawing one sock yields 50-50 odds of light versus dark. That is a reasonable Bayesian assumption, but be aware that it is an assumption.

Secondly, and more importantly, you are implicitly assigning equal probabilities to events that are not equiprobable. A different question on which many people make the same error: A woman has two children, at least one of them is male. What is the probability both are male? Many will say 50%. The correct answer is 1/3.

Suppose you draw three socks, blindfolded. I'll take a matching pair and tell you the color of that pair. The probability that your remaining sock is the same color is not 50%. You drew one of the following four sets: XXX, XXY, XYX, YXX. These are the equiprobable events assuming an even distribution of light and dark socks. How many have all three socks the same color?

 

[HallsofIvy]

We can't confust wanting to find the P(all 3 same color) vs P(3rd is same color) given the first 2 are same color (what your bolded statement alludes to).

The original question steers you into confusing these. It first mentions the probability of the remaining sock being the same color.

Yes, and that was my point.

Then at the end, it asks the actual probability of pulling 3 the same color.

For infinite socks, as in your example (but not the original example), the probability the 3rd sock is the same color, given the first two are the same color, is 1/2.

But the probability that all 3 are same color is 1/8 + 1/8 = 1/4.

Yes, exactly what I said. (Actually I said that the probability of all light socks was 1/8, the probability of all black socks was 1/8: it follows that the probability of "all 3 socks of the same color" is 1/8+ 1/8= 1/4.

But the original question was what was wrong with the reasoning:

If I pull out 3 socks, at least 2 will be alike. Suppose that they are dark. The third one will also be dark with probability 1/2. Similarly, if the pair is light, the probability that the remaining one will be light is also 1/2, so the probability of all socks being the same color must be 1/2.

And my answer was that the error was exactly what you were (I think) accusing me of: confusing "probability all socks are the same color" with "probability all socks are the same color given that the first two socks are of the same color".

And the reason I assumed "infinite socks" was that no number of socks was given but the original post said

The professor didn't say it explicitly, but I'm pretty sure that sock choices are taken to be independent, so P(A,B,C) = P(A)P(B)P(C).

 

[Jane1]

And my answer was that the error was exactly what you were (I think) accusing me of: confusing "probability all socks are the same color" with "probability all socks are the same color given that the first two socks are of the same color".

Actually I was agreeing with you and highlighting what you were saying. B/c you bolded the "given", I wanted to reinforce to others how that makes it a different probablility.

And the reason I assumed "infinite socks" was that no number of socks was given but the original post said

The professor didn't say it explicitly, but I'm pretty sure that sock choices are taken to be independent, so P(A,B,C) = P(A)P(B)P(C).

I go with the problem as written, not a student's assumption. It's clear to me that they would have said "There is an infinite number of dark socks and light socks" if that is what they meant.

Instead, the problem just said "There are light socks and dark socks". This means it's a drawer of mixed socks.

 

[D H]

I go with the problem as written, not a student's assumption. It's clear to me that they would have said "There is an infinite number of dark socks and light socks" if that is what they meant.

That is exactly what the poster said in post #9.

The problem is that the OP has mistakenly applied the principle of indifference to events that are not equiprobable.