| Scissors 16:03 - Dec 15 with 825 views | flykickingbybgunn |
Anybody not into maths read no further.
My much better half has two pairs of identical kitchen scissors. They come apart into male and female for cleaning.
She often uses both pairs and they sit on the draining board awaiting drying.
My problem is this.
What is the probability that I pick up a M and a F so that I can make a pair straight off ?
As I see it there are two conflicting probabilities.
If I pick up one, at random, it does not matter if it is M or F because what is left gives me a probability of 2-1 (66%) of picking up one to make the set.
On the other hand if I pick up two together then the results are MM, MF, FM or FF.
In other words 2-4 (50%)
They cant both be right.
P!ease help me. Answers on a post card to .....
Puzzled @ TWTD |  | | |  |
| Scissors on 16:13 - Dec 15 with 619 views | Zx1988 |
I think both are right, but can't really put into words why.
It feels like a utensil-based version of the Monty Hall problem. |  |
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| Scissors on 16:14 - Dec 15 with 611 views | Cafe_Newman |
Your first scenario is incomplete.
Yes, there is a 2/3 chance of picking the other sex, but there's also a 1/3 chance of picking the same sex.
So you need to take the average of these two events occurring:
(2/3 + 1/3) / 2 = 50% |  | | |
| Scissors on 16:15 - Dec 15 with 604 views | Illinoisblue |
Isn’t the probability of making the set 1 in 3? |  |
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| Scissors on 16:16 - Dec 15 with 603 views | blue_curacao |
50% probability that you can make a pair from the first two you pick, whether you pick one and then another or two at once.
In your first scenario, 66% is the probability of the second one matching the first. Not the probability of you having a pair. | | | |
| Scissors on 16:22 - Dec 15 with 542 views | Cafe_Newman |
| Scissors on 16:14 - Dec 15 by Cafe_Newman |
Your first scenario is incomplete.
Yes, there is a 2/3 chance of picking the other sex, but there's also a 1/3 chance of picking the same sex.
So you need to take the average of these two events occurring:
(2/3 + 1/3) / 2 = 50% |
I should add that I've oversimplified that a bit as there's a greater chance of selecting M if you pick up F first.
And vice versa.
So the average of the two possible outcomes for each M or F first pick up is 50%. | | | |
| Scissors on 16:22 - Dec 15 with 541 views | _CliveBaker_ |
| Scissors on 16:16 - Dec 15 by blue_curacao |
50% probability that you can make a pair from the first two you pick, whether you pick one and then another or two at once.
In your first scenario, 66% is the probability of the second one matching the first. Not the probability of you having a pair. |
Why is it the probability of the 2nd one matching the first?
If he's picked up M then he's got FFM left, surely that gives a 2/3 chance of picking up a F to make the set? |  | | |
| Scissors on 16:23 - Dec 15 with 522 views | bartyg |
The first is the correct probabilities. In the second the outcomes do not have the same probability as each one you take cannot be the other.
The six outcomes in full at 1/6th possibility:
Ma Mb
Fa Fb
Ma Fa
Ma Fb
Mb Fa
Mb Fb |  | | |
| Scissors on 16:24 - Dec 15 with 521 views | flykickingbybgunn |
| Scissors on 16:16 - Dec 15 by blue_curacao |
50% probability that you can make a pair from the first two you pick, whether you pick one and then another or two at once.
In your first scenario, 66% is the probability of the second one matching the first. Not the probability of you having a pair. |
No.
If I pick say a F. Then there remain FMM to pick from therefore the chance is 66% of making a pair.
The probability for M is exactly the same.
My head is boiling. | | | |
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| Scissors on 16:26 - Dec 15 with 509 views | blue_curacao |
| Scissors on 16:22 - Dec 15 by _CliveBaker_ |
Why is it the probability of the 2nd one matching the first?
If he's picked up M then he's got FFM left, surely that gives a 2/3 chance of picking up a F to make the set? |
"If he's picked up M then he's got FFM left, surely that gives a 2/3 chance of picking up a F to make the set?" Yes, so it's a 66% probability that the 2nd will match the first.
The probability of making a pair from any two picked (whether picked at once or one after another) is 50% i.e. MM or FF rather than MF or FM | | | |
| Scissors on 16:28 - Dec 15 with 491 views | _CliveBaker_ |
| Scissors on 16:26 - Dec 15 by blue_curacao |
"If he's picked up M then he's got FFM left, surely that gives a 2/3 chance of picking up a F to make the set?" Yes, so it's a 66% probability that the 2nd will match the first.
The probability of making a pair from any two picked (whether picked at once or one after another) is 50% i.e. MM or FF rather than MF or FM |
Why? If he's picked M and he has FFM left, surely its a 33.3% it matches, and 66.6% chance of a pair.
EDIT: We might be getting confused on our definitions of matching and pairs. I meant match being the same, pair being M/F [Post edited 15 Dec 16:29]
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| Scissors on 16:30 - Dec 15 with 480 views | bartyg |
| Scissors on 16:14 - Dec 15 by Cafe_Newman |
Your first scenario is incomplete.
Yes, there is a 2/3 chance of picking the other sex, but there's also a 1/3 chance of picking the same sex.
So you need to take the average of these two events occurring:
(2/3 + 1/3) / 2 = 50% |
This might be the funniest logic I've ever seen.
So if I roll a die, the chance of getting a 6 is 1/6, but there's a 5/6 chance of rolling not a 6, so I should average the probability to 50%. | | | |
| Scissors on 16:36 - Dec 15 with 427 views | NthQldITFC |
| Scissors on 16:14 - Dec 15 by Cafe_Newman |
Your first scenario is incomplete.
Yes, there is a 2/3 chance of picking the other sex, but there's also a 1/3 chance of picking the same sex.
So you need to take the average of these two events occurring:
(2/3 + 1/3) / 2 = 50% |
I thought this was right at first, but now I'm doubting myself. |  |
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| Scissors on 16:39 - Dec 15 with 407 views | blue_curacao |
| Scissors on 16:28 - Dec 15 by _CliveBaker_ |
Why? If he's picked M and he has FFM left, surely its a 33.3% it matches, and 66.6% chance of a pair.
EDIT: We might be getting confused on our definitions of matching and pairs. I meant match being the same, pair being M/F [Post edited 15 Dec 16:29]
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Now my head hurts. I think I've changed my mind! I now think that the answer is 66% in either scenario :)
If we label the first pair of scissors M1 & F1 and the second pair M2 & F2. Then the possible outcomes in either scenario are:
M1M2
M1F1
M1F2
M2M1
M2F1
M2F2
F1F2
F1M1
F1M2
F2F1
F2M1
F2M2
So 8 out of 12 chances of getting a pair (M/F) [Post edited 15 Dec 16:40]
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| Scissors on 16:50 - Dec 15 with 360 views | NthQldITFC |
I'm going with 66% now!
The 'error' in your two cases in the the OP is:
'On the other hand if I pick up two together then the results are MM, MF, FM or FF.
In other words 2-4 (50%)'
which should read:
On the other hand if I pick up two together then the results are
M1 & M2 - fail
M1 & F1 - pass
M1 & F2 - pass
M2 & F1 - pass
M2 & F2 - pass
F1 & F2 - fail
In other words 4/6 (66%) | |
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| Scissors on 16:51 - Dec 15 with 359 views | Kievthegreat |
The odds are always 2/3rds.
"On the other hand if I pick up two together then the results are MM, MF, FM or FF.
In other words 2-4 (50%)" - This line is wrong because MF and FM implies only 2 combinations of Female and Male work. However if you label the M and F, 1 and 2, you'll find there are 4 working combinations:
M1-F1
M1-F2
M2-F1
M2-F2
Add those to M1-M2 and F1-F2, you get 4 out of 6 combinations are valid, or 2/3rds again. [Post edited 15 Dec 16:51]
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| Scissors on 16:55 - Dec 15 with 300 views | J2BLUE |
| Scissors on 16:51 - Dec 15 by Kievthegreat |
The odds are always 2/3rds.
"On the other hand if I pick up two together then the results are MM, MF, FM or FF.
In other words 2-4 (50%)" - This line is wrong because MF and FM implies only 2 combinations of Female and Male work. However if you label the M and F, 1 and 2, you'll find there are 4 working combinations:
M1-F1
M1-F2
M2-F1
M2-F2
Add those to M1-M2 and F1-F2, you get 4 out of 6 combinations are valid, or 2/3rds again. [Post edited 15 Dec 16:51]
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I might be a bellend [Post edited 15 Dec 16:58]
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| Scissors on 16:58 - Dec 15 with 284 views | Cafe_Newman |
| Scissors on 16:39 - Dec 15 by blue_curacao |
Now my head hurts. I think I've changed my mind! I now think that the answer is 66% in either scenario :)
If we label the first pair of scissors M1 & F1 and the second pair M2 & F2. Then the possible outcomes in either scenario are:
M1M2
M1F1
M1F2
M2M1
M2F1
M2F2
F1F2
F1M1
F1M2
F2F1
F2M1
F2M2
So 8 out of 12 chances of getting a pair (M/F) [Post edited 15 Dec 16:40]
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That's exactly what I meant to say, but didn't. | | | |
| Scissors on 17:02 - Dec 15 with 257 views | Cafe_Newman |
| Scissors on 17:00 - Dec 15 by NthQldITFC |
66% chance  |
Yep, irrespective of whether he's male or female.
The same goes for me. | | | |
| Scissors on 17:02 - Dec 15 with 257 views | bartyg |
| Scissors on 16:51 - Dec 15 by J2BLUE |
Went on chat gpt to prove I was right and it said I was wrong.
I am still arguing with it  [Post edited 15 Dec 17:00]
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It's irrelevant if you check what it is- you're removing it from the pool of options.
Try and visualise it- if you're picking up one item in your left hand, your right hand cannot take the same item. It is picking from the remaining pool, of which you have just removed one of the variables, making the other variable more likely.
EDIT: To combat your edit, this is yet more proof that using LLMs reduces your critical thinking capabilities [Post edited 15 Dec 17:04]
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| Scissors on 17:03 - Dec 15 with 248 views | J2BLUE |
| Scissors on 17:00 - Dec 15 by NthQldITFC |
66% chance |
Surprisingly low! | |
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| Scissors on 17:04 - Dec 15 with 237 views | Cafe_Newman |
| Scissors on 16:14 - Dec 15 by Cafe_Newman |
Your first scenario is incomplete.
Yes, there is a 2/3 chance of picking the other sex, but there's also a 1/3 chance of picking the same sex.
So you need to take the average of these two events occurring:
(2/3 + 1/3) / 2 = 50% |
Can someone else downvote this please on my behalf. | | | |
| Scissors on 17:06 - Dec 15 with 212 views | NthQldITFC |
| Scissors on 17:02 - Dec 15 by Cafe_Newman |
Yep, irrespective of whether he's male or female.
The same goes for me. |
...and me | |
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