| Scissors 16:03 - Dec 15 with 3975 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 17:06 - Dec 15 with 835 views | NthQldITFC |
| Scissors on 17:04 - Dec 15 by Cafe_Newman |
Can someone else downvote this please on my behalf. |
...oh sorry, you meant the other post  |  |
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| Scissors on 17:09 - Dec 15 with 830 views | J2BLUE |
| Scissors on 17:02 - Dec 15 by bartyg |
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 2025 17:04]
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I went on there AFTER I made that post as I wanted confirmation I was right. Going on there and asking before posting would have been cheating. |  |
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| Scissors on 17:10 - Dec 15 with 826 views | _CliveBaker_ |
The answer is 66.7%, because you're not specific whether you choose F or M first.
1st choice is 100% and 2nd is then 66.7%, hence 66.7%. |  | | |
| Scissors on 17:15 - Dec 15 with 801 views | iamatractorboy |
| 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 2025 16:51]
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You've forgotten M2M1 and F2F1.
Honestly everyone is overcomplicating this. The odds are 50/50. There are only 4 combinations, MM, FF, MF and FM. |  | | |
| Scissors on 17:25 - Dec 15 with 770 views | bartyg |
| Scissors on 17:15 - Dec 15 by iamatractorboy |
You've forgotten M2M1 and F2F1.
Honestly everyone is overcomplicating this. The odds are 50/50. There are only 4 combinations, MM, FF, MF and FM. |
They aren't forgetting, because the quantities are commutative. If you want to separate those out, do the other 4. The odds stay the same. |  |
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| Scissors on 17:34 - Dec 15 with 744 views | Kievthegreat |
| Scissors on 17:15 - Dec 15 by iamatractorboy |
You've forgotten M2M1 and F2F1.
Honestly everyone is overcomplicating this. The odds are 50/50. There are only 4 combinations, MM, FF, MF and FM. |
I haven't forgotten them, but I've treated the order as irrelevant i.e M1-M2 is the same as M2-M1.
If you want to include M2-M1 and F2-F1, then you also have to do the same for the others:
M1-F1
M1-F2
M2-F1
M2-F2
also gets
F1-M1
F1-M2
F2-M1
F2-M2
Then we have 12 combinations, 8 of which make a working pair of scissors. 2/3rds again. |  | | |
| Scissors on 17:38 - Dec 15 with 733 views | flykickingbybgunn |
Just to find a clear answer I did this experiment 50 times
32 of them made a pair first time.
I think I will take that as with most on here as 2/3.
Many thanks for all the contributions. | | | |
| Scissors on 17:47 - Dec 15 with 715 views | Kievthegreat |
| Scissors on 17:38 - Dec 15 by flykickingbybgunn |
Just to find a clear answer I did this experiment 50 times
32 of them made a pair first time.
I think I will take that as with most on here as 2/3.
Many thanks for all the contributions. |
Some good empirical data to back up the theory. Now do it 500 times to be sure! | | | |
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| Scissors on 17:50 - Dec 15 with 713 views | vapour_trail |
You say much better half, but if she could just be arsed to use one pair at a time, she’d have saved us all a lot of time, bother and declining levels of self esteem as the supposed hive mind of twtd grappled away with this fairly simple endeavour.
You’re the better half, flykickingbybgunn. |  |
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| Scissors on 17:58 - Dec 15 with 706 views | iamatractorboy |
| Scissors on 17:15 - Dec 15 by iamatractorboy |
You've forgotten M2M1 and F2F1.
Honestly everyone is overcomplicating this. The odds are 50/50. There are only 4 combinations, MM, FF, MF and FM. |
I now realise the error of my ways...
I was assuming we go on to pick the remaining halves, after the first two are chosen. This of course isn't true because there's no need, if you get MF or FM after the first two. So yeh, it is 2/3 (it's 8/12 in terms of actual possible choices after two picks) | | | |
| Scissors on 18:03 - Dec 15 with 692 views | flykickingbybgunn |
| Scissors on 17:47 - Dec 15 by Kievthegreat |
Some good empirical data to back up the theory. Now do it 500 times to be sure! |
Err ....... | | | |
| Scissors on 19:38 - Dec 15 with 639 views | ITFCson |
The 66% is based on there being 3 outcomes because you already determined the 1st M or F.
The 50% is based on 4 outcomes as you have not yet picked up any of the M or F. | | | |
| Scissors on 19:41 - Dec 15 with 637 views | flykickingbybgunn |
| Scissors on 19:38 - Dec 15 by ITFCson |
The 66% is based on there being 3 outcomes because you already determined the 1st M or F.
The 50% is based on 4 outcomes as you have not yet picked up any of the M or F. |
So what you are saying is if I pick up one and look at it I have a 66% chance of a pair.
But if I pick up two together it is 50%.
I'm no further forward. | | | |
| Scissors on 19:51 - Dec 15 with 620 views | Kievthegreat |
| Scissors on 19:38 - Dec 15 by ITFCson |
The 66% is based on there being 3 outcomes because you already determined the 1st M or F.
The 50% is based on 4 outcomes as you have not yet picked up any of the M or F. |
It's not 50%. There are 6 outcomes (12 if you care about order) when picking up 2 at once. 4 (or 8 if you care about order) will make a pair of scissors. You've missed that there are 4 ways of making a pair of scissors because you can mix and match the sets. | | | |
| Scissors on 20:01 - Dec 15 with 597 views | Nthsuffolkblue |
| Scissors on 16:23 - Dec 15 by 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 |
And so 4/6 are a pair and hence 67% probability either way.
Which leads me on to the following.
If you have 3 boxes with only one containing an item of value, you choose one and then someone shows you that one of the other two is empty. Should you now change your mind or stick with your original choice? What are the odds in each case? |  |
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| Scissors on 20:05 - Dec 15 with 587 views | Smoresy |
4C2 = 6 on my calculator.
One is MM (grrr). One is FF (grrr). Four are neither combination (yay).
66% Chris final answer. | | | |
| Scissors on 20:08 - Dec 15 with 564 views | lowhouseblue |
| Scissors on 20:01 - Dec 15 by Nthsuffolkblue |
And so 4/6 are a pair and hence 67% probability either way.
Which leads me on to the following.
If you have 3 boxes with only one containing an item of value, you choose one and then someone shows you that one of the other two is empty. Should you now change your mind or stick with your original choice? What are the odds in each case? |
you should change. |  |
| And so as the loose-bowelled pigeon of time swoops low over the unsuspecting tourist of destiny, and the flatulent skunk of fate wanders into the air-conditioning system of eternity, I notice it's the end of the show |
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| Scissors on 20:08 - Dec 15 with 576 views | bluelagos |
Was told of a family friend who was a regular lottery ticket buyer and who justified her £20 weekly purchase as the chances of winning were 50%, cos "you either win, or you don't"
True story. |  |
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| Scissors on 20:11 - Dec 15 with 566 views | Nthsuffolkblue |
| Scissors on 20:08 - Dec 15 by lowhouseblue |
you should change. |
Feel free to explain why! | |
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| Scissors on 20:12 - Dec 15 with 560 views | Nthsuffolkblue |
| Scissors on 20:08 - Dec 15 by bluelagos |
Was told of a family friend who was a regular lottery ticket buyer and who justified her £20 weekly purchase as the chances of winning were 50%, cos "you either win, or you don't"
True story. |
She must have reckoned on being the unluckiest person ever! Surely, they said it tongue-in-cheek. | |
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| Scissors on 20:14 - Dec 15 with 545 views | lowhouseblue |
because there's a 1 in 3 chance of giving away the box with something in it, and a 2 in 3 chance of swapping nothing for the box with something in it.
2 in 3 of course being the chance that you initially choose an empty box. | |
| And so as the loose-bowelled pigeon of time swoops low over the unsuspecting tourist of destiny, and the flatulent skunk of fate wanders into the air-conditioning system of eternity, I notice it's the end of the show |
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| Scissors on 20:18 - Dec 15 with 546 views | Smoresy |
| Scissors on 20:08 - Dec 15 by bluelagos |
Was told of a family friend who was a regular lottery ticket buyer and who justified her £20 weekly purchase as the chances of winning were 50%, cos "you either win, or you don't"
True story. |
I make that unfunny jest myself on the rare occasions I buy a ticket, gets a sympathetic laugh sometimes. | | | |
| Scissors on 20:24 - Dec 15 with 533 views | Nthsuffolkblue |
| Scissors on 20:14 - Dec 15 by lowhouseblue |
because there's a 1 in 3 chance of giving away the box with something in it, and a 2 in 3 chance of swapping nothing for the box with something in it.
2 in 3 of course being the chance that you initially choose an empty box. |
I think that is correct.
The explanation given to me was to consider the same scenario with 100 boxes and after choosing one, 98 empty ones are removed.
By your explanation you have gone from 1 in 100 to 99 in 100. My thought was that you had gone from 1 in 3 to 1 in 2 but your explanation makes far more sense with the 100 box scenario. | |
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| Scissors on 20:39 - Dec 15 with 507 views | bartyg |
| Scissors on 20:01 - Dec 15 by Nthsuffolkblue |
And so 4/6 are a pair and hence 67% probability either way.
Which leads me on to the following.
If you have 3 boxes with only one containing an item of value, you choose one and then someone shows you that one of the other two is empty. Should you now change your mind or stick with your original choice? What are the odds in each case? |
In the monty hall you always swap. | |
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| Scissors on 07:33 - Dec 16 with 403 views | ADStephenson |
The probability is definitely 50%. There are only 4 options overall: MM, MF, FM, FF. Of those four options, two result in a pair. The first pick doesn't make a difference overall as the final results are limited.
Source: I'm a maths teacher and have been teaching this to confused teenagers for 15 years! |  |
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