Thread: Tickets in pockets.

A company test 95% of its jeans and put a ticket in a certain pocket (left or right) if it has been tested. It always put's it in the same pocket but you don't know which one.

If you buy a pair of jeans and find there is no ticket in one pocket, what is the probability there is one in the other?

The ticket is only put in the left or the right pocket, but are you able to check any of the pockets, including rears and the mini/change-pocket that all jeans seem to have? That would make it more interesting.

If it's just the fronts, wouldn't it be 47.5%?

I'm probably missing something silly though.

95%?

95%.

All wrong so far, 95% seems reasonable when you do the maths, but you know it's wrong, as it would be 95% if you knew which pocket the ticket went in!!
I guess 95% initially.

Originally Posted by Feifer

The ticket is only put in the left or the right pocket, but are you able to check any of the pockets, including rears and the mini/change-pocket that all jeans seem to have? That would make it more interesting.

If it's just the fronts, wouldn't it be 47.5%?

I'm probably missing something silly though.

That's if it is put in a pocket a random.

I also say 95%.

All wrong so far, 95% seems reasonable when you do the maths, but you know it's wrong, as it would be 95% if you knew which pocket the ticket went in!!

I don't understand what this means.

Edit:

Upon further consideration, I'll change my answer to 90.48%.

Odds that I look in the correct pocket and they weren't tested: 2.5%
Odds that I look in the wrong pocket and they weren't tested: 2.5%
Odds that I look in the correct pocket and they were tested: 47.5%
Odds that I look in the wrong pocket and they were tested: 47.5%

I can rule out the third option (because that one would have resulted in me finding the tag). Looking at the remaining choices, the odds of the ticket being in the other pocket are 47.5%/(2.5%+2.5%) = 9.5 times greater than the odds of it not being in the other pocket. Remembering that the odds have to add up to 100% and doing some algebra, you can solve for the odds of the ticket being in the other pocket.

For everyone else who said 95% (as I initially did): Suppose the jeans have a billion pockets, and you look in all but the last one without finding the ticket - would you still say the odds are 95% that the ticket will be in that last pocket?

Originally Posted by Scifor Refugee

I also say 95%.

All wrong so far, 95% seems reasonable when you do the maths, but you know it's wrong, as it would be 95% if you knew which pocket the ticket went in!!

I don't understand what this means.

If would be a 95% if you knew the ticket went in that pocket, however you don't know that, you are just assuming it is the other pocket, however you might have a pair that was not tested so surely the chance of finding one when you switch is less than 95%?

Originally Posted by Scifor Refugee

I also say 95%.

All wrong so far, 95% seems reasonable when you do the maths, but you know it's wrong, as it would be 95% if you knew which pocket the ticket went in!!

I don't understand what this means.

prob you get a trousers with a ticket: 95/100
trousers without a ticket: 5/100

Prob that you get no ticket at first try = 0.5*95/100 + 5/100
Prob that you get no ticket at second try = 5/100

Originally Posted by evariste.galois

Originally Posted by Scifor Refugee

I also say 95%.

All wrong so far, 95% seems reasonable when you do the maths, but you know it's wrong, as it would be 95% if you knew which pocket the ticket went in!!

I don't understand what this means.

prob you get a trousers with a ticket: 95/100
trousers without a ticket: 5/100

Prob that you get no ticket at first try = 0.5*95/100 + 5/100
Prob that you get no ticket at second try = 5/100

I don't thik that's right.
The chanace you don't get a ticket first time is 100% as you are told that in the question, the 5/100 is wronog too.

Originally Posted by esbo

If would be a 95% if you knew the ticket went in that pocket, however you don't know that, you are just assuming it is the other pocket, however you might have a pair that was not tested so surely the chance of finding one when you switch is less than 95%?

I get when you mean now (see my edit to my last post). But is there a reason to say that the ticket always goes in the same (unknown) pocket rather than saying the ticket is put in a random pocket?

Originally Posted by Scifor Refugee

Originally Posted by esbo

If would be a 95% if you knew the ticket went in that pocket, however you don't know that, you are just assuming it is the other pocket, however you might have a pair that was not tested so surely the chance of finding one when you switch is less than 95%?

I get when you mean now (see my edit to my last post). But is there a reason to say that the ticket always goes in the same (unknown) pocket rather than saying the ticket is put in a random pocket?

it's not put in a random pocket.

Originally Posted by esbo

it's not put in a random pocket.

I know that. I was saying that I don't see why it would change anything if it was in a random pocket vs. a specific pocket that you don't know. Either way, from your perspective the ticket is in a random pocket.

Originally Posted by Scifor Refugee

Originally Posted by esbo

it's not put in a random pocket.

I know that. I was saying that I don't see why it would change anything if it was in a random pocket vs. a specific pocket that you don't know. Either way, from your perspective the ticket is in a random pocket.

WEll not really you know it always goes in the same pocket.

Originally Posted by esbo

WEll not really you know it always goes in the same pocket.

But since I don't know which pocket, it is functionally the same as if the tag was placed in a random pocket. Anyway, I'll stick with my 90.48% answer.

Originally Posted by Scifor Refugee

Originally Posted by esbo

WEll not really you know it always goes in the same pocket.

But since I don't know which pocket, it is functionally the same as if the tag was placed in a random pocket. Anyway, I'll stick with my 90.48% answer.

It is funcitionally different as it affects the likly hood of finding a ticket in the other pocket.

Originally Posted by Scifor Refugee

I also say 95%.

All wrong so far, 95% seems reasonable when you do the maths, but you know it's wrong, as it would be 95% if you knew which pocket the ticket went in!!

I don't understand what this means.

Edit:

Upon further consideration, I'll change my answer to 90.48%.

Odds that I look in the correct pocket and they weren't tested: 2.5%
Odds that I look in the wrong pocket and they weren't tested: 2.5%
Odds that I look in the correct pocket and they were tested: 47.5%
Odds that I look in the wrong pocket and they were tested: 47.5%

I can rule out the third option (because that one would have resulted in me finding the tag). Looking at the remaining choices, the odds of the ticket being in the other pocket are 47.5%/(2.5%+2.5%) = 9.5 times greater than the odds of it not being in the other pocket. Remembering that the odds have to add up to 100% and doing some algebra, you can solve for the odds of the ticket being in the other pocket.

For everyone else who said 95% (as I initially did): Suppose the jeans have a billion pockets, and you look in all but the last one without finding the ticket - would you still say the odds are 95% that the ticket will be in that last pocket?

yes. There is a chance that a given pair of jeans has a ticket. The element of the unknown is why I will still say that not finding a ticket doesn't impact the chances, though, I feel that If I'm wrong then you must be right.

Originally Posted by Arcane_Mathematician

Originally Posted by Scifor Refugee

For everyone else who said 95% (as I initially did): Suppose the jeans have a billion pockets, and you look in all but the last one without finding the ticket - would you still say the odds are 95% that the ticket will be in that last pocket?

yes. There is a chance that a given pair of jeans has a ticket. The element of the unknown is why I will still say that not finding a ticket doesn't impact the chances, though, I feel that If I'm wrong then you must be right.

If there are a 1000 pockets and you have only one left to check, it means that either the pants didn't have a ticket (0.05 chance of occurring) or that the pants do have a ticket and you managed to randomly check the 999 out of 1000 pockets that didn't have it (.95 * 1/1000 chance of occurring). So if you get to the last pocket in a par of jeans with 1000 pockets, it's more likely that this pair never had a ticket than that the ticket is in the last pocket that you have yet to check. Using an example with a huge number of jeans makes it more intuitive, but the same math applies if there are only 2 pockets.

Unless someone can explain why I'm wrong (it certainly wouldn't be the first time).

Originally Posted by Scifor Refugee

Originally Posted by Arcane_Mathematician

Originally Posted by Scifor Refugee

For everyone else who said 95% (as I initially did): Suppose the jeans have a billion pockets, and you look in all but the last one without finding the ticket - would you still say the odds are 95% that the ticket will be in that last pocket?

yes. There is a chance that a given pair of jeans has a ticket. The element of the unknown is why I will still say that not finding a ticket doesn't impact the chances, though, I feel that If I'm wrong then you must be right.

If there are a 1000 pockets and you have only one left to check, it means that either the pants didn't have a ticket (0.05 chance of occurring) or that the pants do have a ticket and you managed to randomly check the 999 out of 1000 pockets that didn't have it (.95 * 1/1000 chance of occurring). So if you get to the last pocket in a par of jeans with 1000 pockets, it's more likely that this pair never had a ticket than that the ticket is in the last pocket that you have yet to check. Using an example with a huge number of jeans makes it more intuitive, but the same math applies if there are only 2 pockets.

Unless someone can explain why I'm wrong (it certainly wouldn't be the first time).

Lets take a sample of 100.
Imagine the pockets are see through, and an X is a ticket and a 0 is no ticket
You would see something like this:-


X 0
X 0
X 0
X 0
X 0
/* there are 95 lines of X 0 , I have skipped a pile for brevity */
X 0
X 0
X 0
X 0
X 0
X 0
X 0
X 0
X 0
X 0
X 0
X 0
X 0
X 0
0 0 //these last 5 are the untested ones with no ticket in any pocket
0 0 // they don't have to be the last 5 they could be anywhere
0 0 // it just clearer if we group them at the end.
0 0
0 0


It is probably clear what it happening now.
Note there are 105 zeros in total, and you picked one of them first time.
Now what happens when you look across at the other pocket?
For the first 95 you will get an X, however for the last 10, (105 in total) you will get another 0.

Thus the chance of getting an X in the other pocket is 95/105 = 90.476%

It is so much simpler when you look at it this way, assuming of course my answer is correct. Anyone disagree?

Originally Posted by Scifor Refugee

Originally Posted by esbo

WEll not really you know it always goes in the same pocket.

But since I don't know which pocket, it is functionally the same as if the tag was placed in a random pocket. Anyway, I'll stick with my 90.48% answer.

You are correct.

This is a problem in condidtional probability. It doesn't matter that the ticket always goes in the same pocket, since you don't know which one it is.

So it could be in pocket A or pocket B or somewhere else -- C. You know that the probability of C is .05 and the probability of AUB is .95. So P(A)=P(B)=.475

Now suppose that you picked pocket B and found nothing. You are then asked the probability of A given ~B (not B) or P( A |~ B) = P( A and ~B)/P(~B) =P(A)/P(~B)=.475/.525 which is about 90.476%.