Poll of the Day > Easy probability question. A guy has two white socks and two pink socks.

(+) Yeah! (+)

I dunno about the correct answer, but I would be happy to hear your reasoning.

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Gonna go with 1/3 :x
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I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.
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Tails 64 posted...

I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.

Agree. It doesn't matter which he draws first, the chance of drawing a matching sock is 1/3.
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I think 1/2
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the 4 outcomes are

ww
wp
pw
pp

Successful outcomes are

ww and pp

there for 0.5
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Tails 64 posted...

I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.

he said he picks TWO socks, not one sock and then another sock.

when pulling 2 out of a group of 2 pairs, you have 3 possibilities

two white
two pink
one white one pink (same as one pink one white. since we are pulling both at the same time, it is the same pull)
white white

the answer is 2/3
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The correct answer is 1/2.

PP = Two pink socks.
PW = One pink sock and one white sock.
WW = Two white socks.
WP = One white sock and one pink sock.

There are two matching pairs that he can pull out: Two pink socks and two white socks. There are four different combinations.

Two over four is 1/2, which is zero-point-five; therefore, the correct answer is 1/2.

You're welcome.
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Out of his sock drawer? 1.
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How can he have two pink socks if he only has one asshole

answer is 1/3.

there are 6 possible combinations. only 2 of them involve matching colors. 2/6 = 1/3
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Amuseum posted...

answer is 1/3.

there are 6 possible combinations. only 2 of them involve matching colors. 2/6 = 1/3

No there isn't six combinations.

It's PP, WP, PW, WW

No other way to spin it.
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Amuseum posted...

answer is 1/3.

there are 6 possible combinations. only 2 of them involve matching colors. 2/6 = 1/3

Would you care to type out those 6 combinations?
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Lokarin posted...

Amuseum posted...

answer is 1/3.

there are 6 possible combinations. only 2 of them involve matching colors. 2/6 = 1/3

Would you care to type out those 6 combinations?

P1P2, P1W1, P1W2, P2W1, P2W2, W1W2
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while there are two white and two pink they still might not be a matching pair. more info is needed.
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ANation posted...

The correct answer is 1/2.

PP = Two pink socks.
PW = One pink sock and one white sock.
WW = Two white socks.
WP = One white sock and one pink sock.

There are two matching pairs that he can pull out: Two pink socks and two white socks. There are four different combinations.

Two over four is 1/2, which is zero-point-five; therefore, the correct answer is 1/2.

You're welcome.

Also my conclusion and reasoning.
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dancer62 posted...

Tails 64 posted...

I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.

Agree. It doesn't matter which he draws first, the chance of drawing a matching sock is 1/3.

Also agreed. Doesn't matter if he draws them simultaneously or not, for the purposes of the problem, you can treat one draw as the "first."

Whichever color he draws first, there are now three socks left. One is the matching color, two are not. Therefore, the probability he will draw a match is 1/3.

streamofthesky posted...

dancer62 posted...

Tails 64 posted...

I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.

Agree. It doesn't matter which he draws first, the chance of drawing a matching sock is 1/3.

Also agreed. Doesn't matter if he draws them simultaneously or not, for the purposes of the problem, you can treat one draw as the "first."

Whichever color he draws first, there are now three socks left. One is the matching color, two are not. Therefore, the probability he will draw a match is 1/3.

But you are forgetting to multiply by the odds of picking the first sock.
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streamofthesky posted...

dancer62 posted...

Tails 64 posted...

I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.

Agree. It doesn't matter which he draws first, the chance of drawing a matching sock is 1/3.

Also agreed. Doesn't matter if he draws them simultaneously or not, for the purposes of the problem, you can treat one draw as the "first."

Whichever color he draws first, there are now three socks left. One is the matching color, two are not. Therefore, the probability he will draw a match is 1/3.

That doesn't make any sense. There are 4 possible outcomes, not 3 possible outcomes. The clue isn't about picking one and then the odds of picking one that matches, it's picking TWO at random and whether those two match.
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Lokarin posted...

streamofthesky posted...

dancer62 posted...

Tails 64 posted...

I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.

Agree. It doesn't matter which he draws first, the chance of drawing a matching sock is 1/3.

Also agreed. Doesn't matter if he draws them simultaneously or not, for the purposes of the problem, you can treat one draw as the "first."

Whichever color he draws first, there are now three socks left. One is the matching color, two are not. Therefore, the probability he will draw a match is 1/3.

But you are forgetting to multiply by the odds of picking the first sock.

If you pick pink, you have 1/3 of getting pink again.
If you pick white, you have 1/3 of getting white again.
It's 1/3 regardless.

But fine, let's look at it your way.
You have a 1/2 chance of the first pick being white. And a 1/3 chance after that of the 2nd pick being white. 1/2 * 1/3 = 1/6 chance.
You have a 1/2 chance of the first pick being pink. And a 1/3 chance after that of the 2nd pick being pink. 1/2 * 1/3 = 1/6 chance.

1/6 + 1/6 = still 1/3 probability of getting matching colors.

Zeus posted...

streamofthesky posted...

dancer62 posted...

Tails 64 posted...

I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.

Agree. It doesn't matter which he draws first, the chance of drawing a matching sock is 1/3.

Also agreed. Doesn't matter if he draws them simultaneously or not, for the purposes of the problem, you can treat one draw as the "first."

Whichever color he draws first, there are now three socks left. One is the matching color, two are not. Therefore, the probability he will draw a match is 1/3.

That doesn't make any sense. There are 4 possible outcomes, not 3 possible outcomes. The clue isn't about picking one and then the odds of picking one that matches, it's picking TWO at random and whether those two match.

The person is removing a sock after he picks each. He is not replacing them (at least as worded in the opening topic, otherwise he's not ending up w/ a pair of socks in hand anyway).
If he takes out a white sock, he cannot select that white sock again on his 2nd pick from what was originally four. His 2nd pick is among a pool of three because one possible option has been removed.

e = mc^2

a^2+b^2 = c^2

Therefor e = m(a^2+b^2)
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streamofthesky posted...

Zeus posted...

streamofthesky posted...

dancer62 posted...

Tails 64 posted...

I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.

Agree. It doesn't matter which he draws first, the chance of drawing a matching sock is 1/3.

Also agreed. Doesn't matter if he draws them simultaneously or not, for the purposes of the problem, you can treat one draw as the "first."

Whichever color he draws first, there are now three socks left. One is the matching color, two are not. Therefore, the probability he will draw a match is 1/3.

That doesn't make any sense. There are 4 possible outcomes, not 3 possible outcomes. The clue isn't about picking one and then the odds of picking one that matches, it's picking TWO at random and whether those two match.

The person is removing a sock after he picks each. He is not replacing them (at least as worded in the opening topic, otherwise he's not ending up w/ a pair of socks in hand anyway).
If he takes out a white sock, he cannot select that white sock again on his 2nd pick from what was originally four. His 2nd pick is among a pool of three because one possible option has been removed.

He's being handed two simultaneously. There are four possible outcomes. Ergo 2/4 or 1/2
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Maybe it'll be easier to visualize if we use shoes instead of socks. Let's say you have two pairs of sneakers, same brand, same make, but one pink pair and one white pair. Let's say, without looking, you randomly pick out two shoes. What are the odds you have a matching pair?

1 in 3.

There are six possibilities:

You pick the left pink shoe and the right pink shoe.
You pick the left pink shoe and the right white shoe.
You pick the left pink shoe and the left white shoe.
You pick the right pink shoe and the right white shoe.
You pick the right pink shoe and the left white shoe.
You pick the right white shoe and the left white shoe.

2 out of 6 will give you a matching pair of shoes. 2 out of 6 will give you a mismatched pair of shoes. And 2 out of 6 will have you trying to squeeze into a shoe not meant for that foot.
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But socks aren't footed, a left or a right can go on either
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The correct answer is to just grab three socks.

Zeus posted...

He's being handed two simultaneously. There are four possible outcomes. Ergo 2/4 or 1/2

Ok, first of all, the TC never said simultaneous, that's an assumption you and some others are inserting:

He picks out two socks at random. Probability of pulling a matching pair?

Second, even if it were simultaneous, as long as you're picking 2 different socks and not replacing the sock(s) you take out, you can treat the events as having a 1st and 2nd pick to make the calculations easier anyway.
Finally, even if you want to look at it your way, you're not representing the situation correctly. You reach in at the same time w/ your left and right hands, and grab two different socks. Your subset of choices for each hand is:
Left hand: the 4 socks in the drawer - the sock in your right hand = 3 socks to choose from
Right hand: the 4 socks in the drawer - the sock in your left hand = 3 socks to choose from

And I don't feel like doing that math from there, it's needlessly complicated and you're probably going to ignore what I type anyway. Instead, I'll re-post this response which shows how things work well enough for anyone to understand:
streamofthesky posted...

Tails 64 posted...

But you are forgetting to multiply by the odds of picking the first sock.

If you pick pink, you have 1/3 of getting pink again.
If you pick white, you have 1/3 of getting white again.
It's 1/3 regardless.

But fine, let's look at it your way.
You have a 1/2 chance of the first pick being white. And a 1/3 chance after that of the 2nd pick being white. 1/2 * 1/3 = 1/6 chance.
You have a 1/2 chance of the first pick being pink. And a 1/3 chance after that of the 2nd pick being pink. 1/2 * 1/3 = 1/6 chance.

1/6 + 1/6 = still 1/3 probability of getting matching colors.

streamofthesky posted...

If you pick pink, you have 1/3 of getting pink again.
If you pick white, you have 1/3 of getting white again.
It's 1/3 regardless.

If you grab two at once, like TC heavily implied, the probability is 50%.

XX
XY
YX
YY

Yes XY and YX are functionally identical, but permutations are important for the math, as it's two possible occurrences of XY.

Sigh, just throw them all in the washer with some bleach and blam, you'll always have a matching set of socks from now on.


So much easier solution.

Much wow.

Total happy.
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wolfy42 posted...

Sigh, just throw them all in the washer with some bleach and blam, you'll always have a matching set of socks from now on.

Uh... what if they're different stitching?

Basically it comes down to if people think this is 4p2, or 2p2.
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1/2 being at there are only two matching pairs to begin with.
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https://www.random.org/

I roll it twice and if both numbers are even/odd it's a match. If I get the same number twice I re-roll. My "Hypothesis" is 1/3 given you can't pick the same sock twice.

1 4 not match
2 4 match
4 2 match
3 1 match
2 1 not match
4 4->1 not match (missed)
2 4 match
4 1 not match
2 3 not match
2 1not match
4 4->3 not match (missed)
2 2->4 match
4 4->2 match
2 2->2->4 match
2 1 not match
2 3 not match
3 4 not match
4 2 match
4 4->4->4->2 match
1 1->3 match
3 3->2 not match (missed)
2 3 not match

not match : match
12 : 9

Let's go again..

4 2 match
2 1 match
4 4->1 not match (missed)
2 4m
3 4n
4 4 2m
4 3 n
4 2 m
31 m
12 n
11113 m
24 m
31 m
2221 n
24 m
12 n
13 m
22 m
34 n

not match : match
12 : 7

total;
24 : 16

41 n1
41 n2
21 n3
42 m1
332 n4
221 n5
21 n6
223 n7
443 n8
32 n9
13 m2
43 n10

not match : match
10 : 2 (ouch)

total;
34 : 18

At a sample size of 52 pulls that's a telling 52% (those are coincidentally matching numbers I just so happened to stop there). Feel free to keep going. This probably has a blind spot where people don't understand the first sock is in a quantum position.

Questionmarktarius posted...

streamofthesky posted...

If you pick pink, you have 1/3 of getting pink again.
If you pick white, you have 1/3 of getting white again.
It's 1/3 regardless.

If you grab two at once, like TC heavily implied, the probability is 50%.

XX
XY
YX
YY

Yes XY and YX are functionally identical, but permutations are important for the math, as it's two possible occurrences of XY.

If you grab two at once you also have a chance of grabbing two parts of the same sock.

Why can't this man either match his socks before putting them away, or buy more than 2 pairs of socks?
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This program follows the same logic I was doing with the other rng, but 100,000,000 times.

Dim Match As Integer = 0
Dim NotMatch As Integer = 0
Dim Sock1 As Integer = 0
Dim Sock2 As Integer = 0

For I As Integer = 1 To 100000000
Sock1 = CInt(Int((4 * Rnd()) + 1))
Do
Sock2 = CInt(Int((4 * Rnd()) + 1))
Loop While (Sock1 = Sock2)


If ((Sock1 Mod 2) = (Sock2 Mod 2)) Then
Match += 1
Else
NotMatch += 1
End If

Next

lblMatch.Text = "Matched - " + Match.ToString()
lblNotMatch.Text = "Didn't Match - " + NotMatch.ToString()

It makes a number for sock one and sock 2. If sock 2 is the same sock as sock 1, it rerolls sock 2.



It suggests the chances of the sock matching are exactly 1/3.

I modify it to reroll both Sock1 and Sock2 if it picks the same sock twice..

Same result.

It's only one/third if we assume he only has the four socks he listed. The way it's worded could mean he has more socks than the ones that are listed thus it is impossible to answer.
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https://redd.it/66mezq
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I actually feel like I'm wrong and I'm waiting to be proven wrong because you'd think it'd be 50%.

Maybe the wording is confusing.

1/2 cause you pull them at the same time and obviously my interpretation of the question is the most important part of the answer
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Im going with 1/3 as well. Because regardless of he colors you will have a matching 1/3 of the time.

WW
PW
WP
PP

WW and PP are both a matching pair so they are basically the same variable. So the math is more like

Pair
W/P
P/W

I know komd of crazy but thats the way i see it.
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streamofthesky posted...

Zeus posted...

He's being handed two simultaneously. There are four possible outcomes. Ergo 2/4 or 1/2

Ok, first of all, the TC never said simultaneous, that's an assumption you and some others are inserting:

He picks out two socks at random. Probability of pulling a matching pair?

Second, even if it were simultaneous, as long as you're picking 2 different socks and not replacing the sock(s) you take out, you can treat the events as having a 1st and 2nd pick to make the calculations easier anyway.
Finally, even if you want to look at it your way, you're not representing the situation correctly. You reach in at the same time w/ your left and right hands, and grab two different socks. Your subset of choices for each hand is:
Left hand: the 4 socks in the drawer - the sock in your right hand = 3 socks to choose from
Right hand: the 4 socks in the drawer - the sock in your left hand = 3 socks to choose from

And I don't feel like doing that math from there, it's needlessly complicated and you're probably going to ignore what I type anyway. Instead, I'll re-post this response which shows how things work well enough for anyone to understand:
streamofthesky posted...

Tails 64 posted...

But you are forgetting to multiply by the odds of picking the first sock.

If you pick pink, you have 1/3 of getting pink again.
If you pick white, you have 1/3 of getting white again.
It's 1/3 regardless.

But fine, let's look at it your way.
You have a 1/2 chance of the first pick being white. And a 1/3 chance after that of the 2nd pick being white. 1/2 * 1/3 = 1/6 chance.
You have a 1/2 chance of the first pick being pink. And a 1/3 chance after that of the 2nd pick being pink. 1/2 * 1/3 = 1/6 chance.

1/6 + 1/6 = still 1/3 probability of getting matching colors.

And again, I'll post the math: There are FOUR possible outcomes. It's either white/white, pink/pink, white/pink, or... well, white/pink again. To get it down to a 1/3 chance, you'd need to add additional unmatched socks.

The first draw is part of that same random sequence (and, for the 1/3 to work, he'd have to pick the same starting sock each time). You talk about me making assumptions, but you're arbitrarily breaking it up which directly contradicts the question. And to further drive home the fact that you're the one making assumptions, look at his potential answers and see that your 1/3 choice isn't even up there.
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Why would a guy have pink socks?
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Why is he picking the socks at random
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either he does or he doesnt. therefore 50%
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Violet_Blooded posted...

Why would a guy have pink socks?

red clothes with white clothes in the wash and they weren't colorfast
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The answer is 1/3.

The colour of the first sock drawn is not important; the second sock always has a 1/3 chance of matching it (since there will be two non-matching socks and one matching sock left in the drawer).

Note that drawing simultaneously does not affect this condition. If our socks are W1/W2 and P1/P2, the total combinations of socks that could be drawn are:

W1/W2
W1/P1
W1/P2
W2/P1
W2/P2
P1/P2

Two of the six combinations satisfy the requirement of matching, so the answer is still 1/3.

Zeus posted...

And again, I'll post the math: There are FOUR possible outcomes. It's either white/white, pink/pink, white/pink, or... well, white/pink again. To get it down to a 1/3 chance, you'd need to add additional unmatched socks.

Not so.

Well... I mean, you could say there's four possible outcomes (drawing white/white, white/pink, pink/white and pink/pink) and that's technically correct, but the problem is that not all of those draws are equally likely. The second and third of those combinations are twice as likely as the first and last.

Keep in mind, this isn't like "coin tosses", where the results are purely random. We are drawing from a sample and, thus, each draw affects the odds of subsequent draws. Basically, if I had a sock drawer with 100 white and 100 red socks, each white sock I draw increases the odds that subsequent draws will be red, because there are less white socks to draw. Contrast that where if I toss 100 coins, each heads result does NOT affect the odds of getting a subsequent tails.

In order for your answer to be correct, you would have to replace the sock you drew after you drew it. Basically you would draw a sock, note its colour, then return it to the drawer and draw again - your subsequent draw would have a 50% chance of being the noted colour and a 50% chance of being the opposite colour. But in the question as stated, where you draw a sock and keep it with you, the next draw will have a greater chance of being the opposite colour simply because there are now more of that colour to draw.
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Golden Road posted...

Maybe it'll be easier to visualize if we use shoes instead of socks. Let's say you have two pairs of sneakers, same brand, same make, but one pink pair and one white pair. Let's say, without looking, you randomly pick out two shoes. What are the odds you have a matching pair?

1 in 3.

There are six possibilities:

You pick the left pink shoe and the right pink shoe.
You pick the left pink shoe and the right white shoe.
You pick the left pink shoe and the left white shoe.
You pick the right pink shoe and the right white shoe.
You pick the right pink shoe and the left white shoe.
You pick the right white shoe and the left white shoe.

2 out of 6 will give you a matching pair of shoes. 2 out of 6 will give you a mismatched pair of shoes. And 2 out of 6 will have you trying to squeeze into a shoe not meant for that foot.

This is actually a really good way of phrasing it.
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