| Scissors 16:03 - Dec 15 with 2882 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 07:44 - Dec 16 with 473 views | Nthsuffolkblue |
| Scissors on 07:33 - Dec 16 by 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! |
This has already been covered in the thread. It is worrying that you are teaching maths if you cannot follow the logic. I hope it isn't at any particularly high level! Here is the explanation.
There are 4 blades to originally choose from. (25% chance for each blade.) Regardless of which blade is chosen, there are now 3 blades left. Now start with a left blade. In 2/3 of the cases you will get a pair. 1/3 you won't. The same is true with a right blade. This gives a total percentage of 67% is a pair.
Alternatively, if you take it as 2 individual blades there are 12 combinations (let's call each blade L1, L2, R1 and R2):
L1 L2
L1 R1
L1 R2
R1 R2
R1 L1
R1 L2
L2 L1
L2 R1
L2 R2
R2 L1
R2 L2
R2 R1
Of these 12 combinations, 8 are a working pair (2/3). |  |
| |
| Scissors on 08:00 - Dec 16 with 446 views | iamatractorboy |
| Scissors on 07:33 - Dec 16 by 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! |
I made the same mistake as you initially: you are assuming we have to go on and select the third and fourth halves. We don't, so some of the possible combinations don't need to be taken account of when working out the probabilities. If we get MF or FM, we don't need to keep going. |  | | |
| Scissors on 08:12 - Dec 16 with 434 views | DJR |
For those who think 2/3 is the answer, it is interesting to note that using the same logic the answer becomes 1/3 in the situation where there are 2 red buttons and 2 blue buttons, and you want to calculate the chance of picking two buttons of the same colour.
Such different odds for what seems essentially the same exercise doesn't seem right to me. As in the case of the scissors there are only 4 outcomes, so it seems to me that 50% is the right answer. And I am comforted in this view by the view of a Maths teacher. I might add that I did Pure Maths with Stats at A-level nearly 50 years ago but I must admit I have forgotten all I learnt.
For those who like this sort of thing, the Pyrgic Puzzles in the Guardian on Saturday are worth a go but they tend to be much more fiendish. [Post edited 16 Dec 8:24]
|  | | |
| Scissors on 08:17 - Dec 16 with 430 views | Nthsuffolkblue |
| Scissors on 08:12 - Dec 16 by DJR |
For those who think 2/3 is the answer, it is interesting to note that using the same logic the answer becomes 1/3 in the situation where there are 2 red buttons and 2 blue buttons, and you want to calculate the chance of picking two buttons of the same colour.
Such different odds for what seems essentially the same exercise doesn't seem right to me. As in the case of the scissors there are only 4 outcomes, so it seems to me that 50% is the right answer. And I am comforted in this view by the view of a Maths teacher. I might add that I did Pure Maths with Stats at A-level nearly 50 years ago but I must admit I have forgotten all I learnt.
For those who like this sort of thing, the Pyrgic Puzzles in the Guardian on Saturday are worth a go but they tend to be much more fiendish. [Post edited 16 Dec 8:24]
|
Your comfort is misplaced. The explanation is above (and earlier in the thread too). It is the same as the red and blue buttons. Why do you think it is different?
I did A level Maths with statistics slightly more recently than you, then an engineering degree and worked in science teaching to A level that involves plenty of maths. More importantly, though, I can understand and explain the logic.
Do you agree that there are 12 possible combinations? If not, why not? Do you agree that 8 of those 12 are a working pair? If not, why not? That gives 67% probability. Please give your working for 50%. [Post edited 16 Dec 8:53]
| |
| |
| Scissors on 08:37 - Dec 16 with 405 views | DJR |
| Scissors on 07:44 - Dec 16 by Nthsuffolkblue |
This has already been covered in the thread. It is worrying that you are teaching maths if you cannot follow the logic. I hope it isn't at any particularly high level! Here is the explanation.
There are 4 blades to originally choose from. (25% chance for each blade.) Regardless of which blade is chosen, there are now 3 blades left. Now start with a left blade. In 2/3 of the cases you will get a pair. 1/3 you won't. The same is true with a right blade. This gives a total percentage of 67% is a pair.
Alternatively, if you take it as 2 individual blades there are 12 combinations (let's call each blade L1, L2, R1 and R2):
L1 L2
L1 R1
L1 R2
R1 R2
R1 L1
R1 L2
L2 L1
L2 R1
L2 R2
R2 L1
R2 L2
R2 R1
Of these 12 combinations, 8 are a working pair (2/3). |
Wrong reply. [Post edited 16 Dec 8:40]
| | | |
| Scissors on 08:45 - Dec 16 with 390 views | DanTheMan |
| 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. |
https://www.mathwarehouse.com/monty-hall-simulation-online/
For people who want to see the Monty Hall problem in action. It's not intuitive in the slightest at first thought. |  |
| |
| Scissors on 08:53 - Dec 16 with 378 views | ITFCson |
| Scissors on 19:41 - Dec 15 by flykickingbybgunn |
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. |
Yes, because if you pick up M and then going to pick 1 more to make a pair FF is not possible. So ridding 1 possibility leaves you with 2 out of 3 (66%)
If you pick up 2 initially all outcomes are available including FF which is 2 out of 4 outcomes being a pair (50%)
It messes with the mind but its not a maths question really. In the first instance you know the 1st M or F, whereas in the second instant you do not. | | | |
Login to get fewer ads
| Scissors on 08:58 - Dec 16 with 370 views | Nthsuffolkblue |
| Scissors on 08:53 - Dec 16 by ITFCson |
Yes, because if you pick up M and then going to pick 1 more to make a pair FF is not possible. So ridding 1 possibility leaves you with 2 out of 3 (66%)
If you pick up 2 initially all outcomes are available including FF which is 2 out of 4 outcomes being a pair (50%)
It messes with the mind but its not a maths question really. In the first instance you know the 1st M or F, whereas in the second instant you do not. |
Try reading the thread. In both instances the probability is the same - 67%. The logic is explained several times within the thread (although several use false logic to attempt to explain the wrong answer as well). In logic it is important to carefully consider all the available options. Sometimes logic is counter-intuitive. However, if you end up with an answer that appears illogical (the premise in the OP that by randomly choosing an item initially you change the overall odds is a good example), it is likely you haven't considered all the possibilities.
EDIT: To simplify where you have made the error. You seem to think there are 4 combinations when picking 2 items together. There are actually 6 combinations only 2 of which are the same blades meaning the percentage chance remains 67%.
Another way to think of it is this. Try randomly colouring the blades yellow (Y), blue (B), green (G) and red (R). Now you can choose any of the following combinations: YB, YG, YR, BG, BR, GR. Now substitute in any combination of which colours are left and right and you will find there are 4 working pairs and 2 non-working ones. [Post edited 16 Dec 9:07]
| |
| |
| Scissors on 09:04 - Dec 16 with 364 views | ITFCson |
| Scissors on 08:58 - Dec 16 by Nthsuffolkblue |
Try reading the thread. In both instances the probability is the same - 67%. The logic is explained several times within the thread (although several use false logic to attempt to explain the wrong answer as well). In logic it is important to carefully consider all the available options. Sometimes logic is counter-intuitive. However, if you end up with an answer that appears illogical (the premise in the OP that by randomly choosing an item initially you change the overall odds is a good example), it is likely you haven't considered all the possibilities.
EDIT: To simplify where you have made the error. You seem to think there are 4 combinations when picking 2 items together. There are actually 6 combinations only 2 of which are the same blades meaning the percentage chance remains 67%.
Another way to think of it is this. Try randomly colouring the blades yellow (Y), blue (B), green (G) and red (R). Now you can choose any of the following combinations: YB, YG, YR, BG, BR, GR. Now substitute in any combination of which colours are left and right and you will find there are 4 working pairs and 2 non-working ones. [Post edited 16 Dec 9:07]
|
Apologies - I read the 50% and was focused on arriving on how that was achieved. You are indeed correct, it is 67% and now I feel a fool  | | | |
| Scissors on 09:08 - Dec 16 with 356 views | Smoresy |
| Scissors on 07:44 - Dec 16 by Nthsuffolkblue |
This has already been covered in the thread. It is worrying that you are teaching maths if you cannot follow the logic. I hope it isn't at any particularly high level! Here is the explanation.
There are 4 blades to originally choose from. (25% chance for each blade.) Regardless of which blade is chosen, there are now 3 blades left. Now start with a left blade. In 2/3 of the cases you will get a pair. 1/3 you won't. The same is true with a right blade. This gives a total percentage of 67% is a pair.
Alternatively, if you take it as 2 individual blades there are 12 combinations (let's call each blade L1, L2, R1 and R2):
L1 L2
L1 R1
L1 R2
R1 R2
R1 L1
R1 L2
L2 L1
L2 R1
L2 R2
R2 L1
R2 L2
R2 R1
Of these 12 combinations, 8 are a working pair (2/3). |
Excellent summary. In technical language I believe we have 6 combinations, 12 permutations, providing my stats modules served me well (the latter being order specific). These two formulas weren't introduced at GCSE level, in my day at least, so hopefully not something on ADS's syllabus. | | | |
| Scissors on 09:15 - Dec 16 with 334 views | Nthsuffolkblue |
| Scissors on 09:04 - Dec 16 by ITFCson |
Apologies - I read the 50% and was focused on arriving on how that was achieved. You are indeed correct, it is 67% and now I feel a fool |
No problem. Easily done and why stats is potentially confusing. Part of it is having what we think should be the right answer and then looking to prove it rather than carefully working it out. If it is any consolation, I learnt the explanation of the 67% in the Monty Hall problem (and the fact it had that name) from lowhouse. Previously I had thought it had changed from 33% to 50% rather than the 67% it is. | |
| |
| Scissors on 09:17 - Dec 16 with 334 views | iamatractorboy |
Having misunderstood the problem myself earlier on, then realising where I went wrong, on reflection I think where it gets people confused might be the fact it is a finite (and very small) population rather than a large/infinite one. Selecting the first 'scissor' reduces the remaining population. If we were sampling from a much larger population then the probability would be 50%. (Think I've got that right anyway!!) | | | |
| Scissors on 09:20 - Dec 16 with 332 views | Nthsuffolkblue |
| Scissors on 09:17 - Dec 16 by iamatractorboy |
Having misunderstood the problem myself earlier on, then realising where I went wrong, on reflection I think where it gets people confused might be the fact it is a finite (and very small) population rather than a large/infinite one. Selecting the first 'scissor' reduces the remaining population. If we were sampling from a much larger population then the probability would be 50%. (Think I've got that right anyway!!) |
Nope. I don't think you have. 50% is the wrong assumption. At no point is it the correct answer. Selecting the first scissor makes no difference to the probabilities. | |
| |
| Scissors on 09:29 - Dec 16 with 314 views | NthQldITFC |
| Scissors on 07:33 - Dec 16 by 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! |
I love the doubt that this sort of apparently simple and obvious question brings into my mind, and despite being convinced of the 66. ̅6 (recurring) answer was correct, I've built a spreadsheet simulator over breakfast which does the following:
1 - Choose random number 1 in range 0 to 1
1a - assign M1,M2,F1,F2 to a fixed quarter of that range for the first half scissor chosen
2 - Choose random number 2, candidate 1, in range 0 to 1
2a - assign M1,M2,F1,F2 to a fixed quarter of that range for the second half scissor chosen
3 - Repeat 2 for a further 7 (total 8) candidates
4 - If 2a clashes with 1a, reject candidate 1 and try again with candidate 2 etc. (i.e. if the same precise part has been chosen, which cannot happen (=clash), reject until one of the actually possible remaining THREE parts is chosen)
5 - Once a non-clashing part is chosen for scissor half 2, look to see if the first character 'M' or 'F' of each is different to scissor part 1, if so, success (=pair")
Running sets of 100 tests, 20 times results in totals in the range 61 to 73 for me.
So I'm happy the answer is 66. ̅6 (recurring)
(btw. I downvoted your post and then immediately upvoted it again in shame because I realised that downvote was really about me and not your post, sorry) [Post edited 16 Dec 12:02]
|  |
| |
| Scissors on 09:30 - Dec 16 with 314 views | iamatractorboy |
| Scissors on 09:20 - Dec 16 by Nthsuffolkblue |
Nope. I don't think you have. 50% is the wrong assumption. At no point is it the correct answer. Selecting the first scissor makes no difference to the probabilities. |
Might have worded my comment clumsily but I was thinking of a situation analogous to tossing a coin, which would be an infinitely sized population of equal numbers of M and F scissor halves. In the coin situation, after a heads on the first toss, there would be equal chance of a heads or tails on the next toss. Or the reverse, after a tails. | | | |
| Scissors on 09:36 - Dec 16 with 295 views | lowhouseblue |
| Scissors on 09:29 - Dec 16 by NthQldITFC |
I love the doubt that this sort of apparently simple and obvious question brings into my mind, and despite being convinced of the 66. ̅6 (recurring) answer was correct, I've built a spreadsheet simulator over breakfast which does the following:
1 - Choose random number 1 in range 0 to 1
1a - assign M1,M2,F1,F2 to a fixed quarter of that range for the first half scissor chosen
2 - Choose random number 2, candidate 1, in range 0 to 1
2a - assign M1,M2,F1,F2 to a fixed quarter of that range for the second half scissor chosen
3 - Repeat 2 for a further 7 (total 8) candidates
4 - If 2a clashes with 1a, reject candidate 1 and try again with candidate 2 etc. (i.e. if the same precise part has been chosen, which cannot happen (=clash), reject until one of the actually possible remaining THREE parts is chosen)
5 - Once a non-clashing part is chosen for scissor half 2, look to see if the first character 'M' or 'F' of each is different to scissor part 1, if so, success (=pair")
Running sets of 100 tests, 20 times results in totals in the range 61 to 73 for me.
So I'm happy the answer is 66. ̅6 (recurring)
(btw. I downvoted your post and then immediately upvoted it again in shame because I realised that downvote was really about me and not your post, sorry) [Post edited 16 Dec 12:02]
|
the easier and traditional approach is just to draw a tree diagram. |  |
| 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 |
| |
| Scissors on 09:39 - Dec 16 with 290 views | NthQldITFC |
| Scissors on 09:30 - Dec 16 by iamatractorboy |
Might have worded my comment clumsily but I was thinking of a situation analogous to tossing a coin, which would be an infinitely sized population of equal numbers of M and F scissor halves. In the coin situation, after a heads on the first toss, there would be equal chance of a heads or tails on the next toss. Or the reverse, after a tails. |
I think that's the problem with some of the answers in here - people are giving a valid answer to a slightly different question! | |
| |
| Scissors on 09:40 - Dec 16 with 289 views | NthQldITFC |
| Scissors on 09:36 - Dec 16 by lowhouseblue |
the easier and traditional approach is just to draw a tree diagram. |
Fine, yes, but there are several ways to skin a cat.
(and indeed to confirm that said cat has no skin) [Post edited 16 Dec 10:02]
| |
| |
| Scissors on 09:43 - Dec 16 with 284 views | Nthsuffolkblue |
| Scissors on 09:30 - Dec 16 by iamatractorboy |
Might have worded my comment clumsily but I was thinking of a situation analogous to tossing a coin, which would be an infinitely sized population of equal numbers of M and F scissor halves. In the coin situation, after a heads on the first toss, there would be equal chance of a heads or tails on the next toss. Or the reverse, after a tails. |
That's a different problem but would give the probability you state. The probability remains 50% regardless. You can do strange things with probabilities if you aren't careful, though.
If you had a truly infinite number of opposite halves of the scissors, the probability would always be 50% I think.
You pick a L (50% chance) you then have 50% chance of a right still (gives a total of 25% chance).
You pick a R (50% chance) you then have a 50% chance of a left (gives a total of 25% chance).
Those two combined gives a 50% chance of a pair. All other possibilities are not a pair and must sum to 100%.
This means the original possibilities are 50% as they wouldn't change. This is due to the infinite number of blades available so there is no impact of the handedness of one being chosen.
I will await being proven wrong on that by a better logician! | |
| |
| Scissors on 09:49 - Dec 16 with 274 views | ADStephenson |
| Scissors on 07:44 - Dec 16 by Nthsuffolkblue |
This has already been covered in the thread. It is worrying that you are teaching maths if you cannot follow the logic. I hope it isn't at any particularly high level! Here is the explanation.
There are 4 blades to originally choose from. (25% chance for each blade.) Regardless of which blade is chosen, there are now 3 blades left. Now start with a left blade. In 2/3 of the cases you will get a pair. 1/3 you won't. The same is true with a right blade. This gives a total percentage of 67% is a pair.
Alternatively, if you take it as 2 individual blades there are 12 combinations (let's call each blade L1, L2, R1 and R2):
L1 L2
L1 R1
L1 R2
R1 R2
R1 L1
R1 L2
L2 L1
L2 R1
L2 R2
R2 L1
R2 L2
R2 R1
Of these 12 combinations, 8 are a working pair (2/3). |
I didn't read most of the thread before replying and based my answer on the chance of getting two heads or two tails, which is, of course, a different problem. I teach A-Level maths, so the "it is worrying" comment was a bit rough (although I'm not here to get into an argument on the internet, which is why I rarely post anything - lesson learnt!).
Now I don't have my 4 year old daughter incessantly talking to me about anything that pops into her head and my 6 year old son dropping a pint of milk on the floor a minute before we have to leave the house I've managed to draw a tree diagram with the actual probabilities in it (during a maths lesson, nonetheless) and agree that we get two branches with a probability of a third, making two thirds overall... |  |
| |
| Scissors on 09:58 - Dec 16 with 253 views | iamatractorboy |
| Scissors on 09:43 - Dec 16 by Nthsuffolkblue |
That's a different problem but would give the probability you state. The probability remains 50% regardless. You can do strange things with probabilities if you aren't careful, though.
If you had a truly infinite number of opposite halves of the scissors, the probability would always be 50% I think.
You pick a L (50% chance) you then have 50% chance of a right still (gives a total of 25% chance).
You pick a R (50% chance) you then have a 50% chance of a left (gives a total of 25% chance).
Those two combined gives a 50% chance of a pair. All other possibilities are not a pair and must sum to 100%.
This means the original possibilities are 50% as they wouldn't change. This is due to the infinite number of blades available so there is no impact of the handedness of one being chosen.
I will await being proven wrong on that by a better logician! |
Yeah the infinite population of 50/50 M/F is the same as just replacing the picks. Like picking a back or white ball out of a bag, noting what you get and replacing it, then doing that again. Same with a die (albeit more outcomes possible) or a coin. | | | |
| Scissors on 10:08 - Dec 16 with 235 views | Nthsuffolkblue |
| Scissors on 09:49 - Dec 16 by ADStephenson |
I didn't read most of the thread before replying and based my answer on the chance of getting two heads or two tails, which is, of course, a different problem. I teach A-Level maths, so the "it is worrying" comment was a bit rough (although I'm not here to get into an argument on the internet, which is why I rarely post anything - lesson learnt!).
Now I don't have my 4 year old daughter incessantly talking to me about anything that pops into her head and my 6 year old son dropping a pint of milk on the floor a minute before we have to leave the house I've managed to draw a tree diagram with the actual probabilities in it (during a maths lesson, nonetheless) and agree that we get two branches with a probability of a third, making two thirds overall... |
I did think the comment may have been rather harsh and I apologise for it. I was concerned that you were doubling down on false logic but I accept you recognise you got in a muddle. If we can admit our mistakes as teachers that is a good thing.
I apologise for the bluntness of my earlier comment. | |
| |
| Scissors on 10:12 - Dec 16 with 226 views | Kievthegreat |
| Scissors on 09:43 - Dec 16 by Nthsuffolkblue |
That's a different problem but would give the probability you state. The probability remains 50% regardless. You can do strange things with probabilities if you aren't careful, though.
If you had a truly infinite number of opposite halves of the scissors, the probability would always be 50% I think.
You pick a L (50% chance) you then have 50% chance of a right still (gives a total of 25% chance).
You pick a R (50% chance) you then have a 50% chance of a left (gives a total of 25% chance).
Those two combined gives a 50% chance of a pair. All other possibilities are not a pair and must sum to 100%.
This means the original possibilities are 50% as they wouldn't change. This is due to the infinite number of blades available so there is no impact of the handedness of one being chosen.
I will await being proven wrong on that by a better logician! |
Used Co-pilot and Python to simulate randomly picking scissors 100k times for 2 pairs all the way up to 30 pairs. got my computer to run it 2.9 million times based on the different number of pairs, so should be pretty reflective. Probability clearly tends to 50% as number of pairs increases.
Probability of picking a valid M–F pair:
2 pairs: 0.6668
3 pairs: 0.5971
4 pairs: 0.5724
5 pairs: 0.5535
6 pairs: 0.5452
7 pairs: 0.5390
8 pairs: 0.5351
9 pairs: 0.5275
10 pairs: 0.5264
11 pairs: 0.5225
12 pairs: 0.5229
13 pairs: 0.5199
14 pairs: 0.5181
15 pairs: 0.5181
16 pairs: 0.5145
17 pairs: 0.5146
18 pairs: 0.5145
19 pairs: 0.5118
20 pairs: 0.5106
21 pairs: 0.5108
22 pairs: 0.5129
23 pairs: 0.5121
24 pairs: 0.5102
25 pairs: 0.5085
26 pairs: 0.5118
27 pairs: 0.5123
28 pairs: 0.5078
29 pairs: 0.5069
30 pairs: 0.5081
If we do it mathematically:
With 2 pairs, it's a 66.7% chance mathematically (2 bad and 4 good)
With 3 pairs, it's 60% (6 bad pairs and 9 good)
With 4 pairs, it's 57.1% (12 bad pairs and 16 good)
With 10 pairs, it's (90 bad pairs and 100 good)
To calculate good outcomes, it is just n^2. To calculate bad, it is =n(n-1). In both equations it is where n is the number of pairs. |  | | |
| Scissors on 10:14 - Dec 16 with 221 views | Nthsuffolkblue |
| Scissors on 10:12 - Dec 16 by Kievthegreat |
Used Co-pilot and Python to simulate randomly picking scissors 100k times for 2 pairs all the way up to 30 pairs. got my computer to run it 2.9 million times based on the different number of pairs, so should be pretty reflective. Probability clearly tends to 50% as number of pairs increases.
Probability of picking a valid M–F pair:
2 pairs: 0.6668
3 pairs: 0.5971
4 pairs: 0.5724
5 pairs: 0.5535
6 pairs: 0.5452
7 pairs: 0.5390
8 pairs: 0.5351
9 pairs: 0.5275
10 pairs: 0.5264
11 pairs: 0.5225
12 pairs: 0.5229
13 pairs: 0.5199
14 pairs: 0.5181
15 pairs: 0.5181
16 pairs: 0.5145
17 pairs: 0.5146
18 pairs: 0.5145
19 pairs: 0.5118
20 pairs: 0.5106
21 pairs: 0.5108
22 pairs: 0.5129
23 pairs: 0.5121
24 pairs: 0.5102
25 pairs: 0.5085
26 pairs: 0.5118
27 pairs: 0.5123
28 pairs: 0.5078
29 pairs: 0.5069
30 pairs: 0.5081
If we do it mathematically:
With 2 pairs, it's a 66.7% chance mathematically (2 bad and 4 good)
With 3 pairs, it's 60% (6 bad pairs and 9 good)
With 4 pairs, it's 57.1% (12 bad pairs and 16 good)
With 10 pairs, it's (90 bad pairs and 100 good)
To calculate good outcomes, it is just n^2. To calculate bad, it is =n(n-1). In both equations it is where n is the number of pairs. |
Excellent work. | |
| |
| |
By continuing to use the site, you agree to our use of cookies and to abide by our Terms and Conditions.
