| Maths boffins 18:53 - Jun 22 with 6296 views | BrixtonBlue |
I just put my hand into a bag of skittles and pulled out 4 sweets ...all green. What are the chances of that?!? |  |
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| Maths boffins on 18:57 - Jun 22 with 5016 views | BudapestByBlimp |
You'll need to count all the different colours of Skittles in that bag and get back to us. |  | | |
| Maths boffins on 18:58 - Jun 22 with 5011 views | Blue_Order |
Any self-respecting Town fan discards all yellow and green skittles before they begin eating them.
Traitor. |  | | |
| Maths boffins on 18:59 - Jun 22 with 5003 views | BrixtonBlue |
| Maths boffins on 18:58 - Jun 22 by Blue_Order |
Any self-respecting Town fan discards all yellow and green skittles before they begin eating them.
Traitor. |
I don't eat the yellow and green ones together though. | |
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| Maths boffins on 19:22 - Jun 22 with 4951 views | StokieBlue |
| Maths boffins on 18:59 - Jun 22 by BrixtonBlue |
I don't eat the yellow and green ones together though. |
I think you need to accept defeat.
He's done you like a kipper.
SB |  |
| Avatar - IC410 - Tadpoles Nebula |
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| Maths boffins on 19:23 - Jun 22 with 4942 views | BrixtonBlue |
| Maths boffins on 19:22 - Jun 22 by StokieBlue |
I think you need to accept defeat.
He's done you like a kipper.
SB |
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| Maths boffins on 20:13 - Jun 22 with 4880 views | Fixed_It |
Am I the only sad one who refuses to eat a green one after a yellow one (and vice versa)? Same with M&Ms.... |  |
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| Maths boffins on 22:02 - Jun 22 with 4735 views | NorthstaNder |
50/50
either you'll pull out 4 green sweets. or you won't. |  | | |
| Maths boffins on 10:27 - Jun 23 with 4491 views | mos |
Presuming that there are 5 different colours of skittles in your bag, and there are equal amounts of them, then...
Red, Green, Purple, Orange and Yellow are the flavours present in your bag, with the chance of pulling any of those colours at 1/5 or 0.2 chance, then,
0.2 x 0.2 x 0.2 x 0.2 or 0.2^4 = 0.0016
So you had a 0.16% chance or 1 in 625 chance of picking 4 greens in a row.
Incredible. |  |
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| Maths boffins on 10:34 - Jun 23 with 4472 views | BrixtonBlue |
| Maths boffins on 10:27 - Jun 23 by mos |
Presuming that there are 5 different colours of skittles in your bag, and there are equal amounts of them, then...
Red, Green, Purple, Orange and Yellow are the flavours present in your bag, with the chance of pulling any of those colours at 1/5 or 0.2 chance, then,
0.2 x 0.2 x 0.2 x 0.2 or 0.2^4 = 0.0016
So you had a 0.16% chance or 1 in 625 chance of picking 4 greens in a row.
Incredible. |
Brilliant, thanks for that! | |
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| Maths boffins on 10:36 - Jun 23 with 4463 views | mos |
| Maths boffins on 10:34 - Jun 23 by BrixtonBlue |
Brilliant, thanks for that! |
No problem  | |
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| Maths boffins on 10:39 - Jun 23 with 4448 views | StokieBlue |
| Maths boffins on 10:27 - Jun 23 by mos |
Presuming that there are 5 different colours of skittles in your bag, and there are equal amounts of them, then...
Red, Green, Purple, Orange and Yellow are the flavours present in your bag, with the chance of pulling any of those colours at 1/5 or 0.2 chance, then,
0.2 x 0.2 x 0.2 x 0.2 or 0.2^4 = 0.0016
So you had a 0.16% chance or 1 in 625 chance of picking 4 greens in a row.
Incredible. |
Not quite right as the chance of pulling out a green one is no longer 0.2 after the first one is taken. Your formula is correct when the events are independent such as tossing a coin or if the skittle was placed back into the packet after it was picked.
If we assume there are 4 of each of the 5 types of Skittle (this is important) then:
0.2 * 0.158 * 0.111 * 0.059 = 0.00021 = 1/4845
Sorry to be a pedantic bore. Really you need to know the exact number of each in the bag as it affects the calculation.
SB | |
| Avatar - IC410 - Tadpoles Nebula |
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| Maths boffins on 10:44 - Jun 23 with 4426 views | mos |
| Maths boffins on 10:39 - Jun 23 by StokieBlue |
Not quite right as the chance of pulling out a green one is no longer 0.2 after the first one is taken. Your formula is correct when the events are independent such as tossing a coin or if the skittle was placed back into the packet after it was picked.
If we assume there are 4 of each of the 5 types of Skittle (this is important) then:
0.2 * 0.158 * 0.111 * 0.059 = 0.00021 = 1/4845
Sorry to be a pedantic bore. Really you need to know the exact number of each in the bag as it affects the calculation.
SB |
You’re right, I’m wrong.
But how did you get to multiplying by 0.158? | |
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| Maths boffins on 10:50 - Jun 23 with 4408 views | StokieBlue |
| Maths boffins on 10:44 - Jun 23 by mos |
You’re right, I’m wrong.
But how did you get to multiplying by 0.158? |
Taking your assumptions of 5 types of Skittles and the assumption there are 4 of each type in the bag:
(4 / 20) * (3 / 19) * (2 / 18) * (1 / 17)
It's early though so feel free to point out any errors, my brain isn't up to speed yet today.
SB | |
| Avatar - IC410 - Tadpoles Nebula |
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| Maths boffins on 10:52 - Jun 23 with 4391 views | DropCliffsNotBombs |
| Maths boffins on 10:44 - Jun 23 by mos |
You’re right, I’m wrong.
But how did you get to multiplying by 0.158? |
because let's say theres 20 in total (4 of each colour). the first chance is 4÷20 = 0.2.
the second pick are 3 greens left out of 19 (3÷19 = 0.158)) |  | | |
| Maths boffins on 10:56 - Jun 23 with 4383 views | BrixtonBlue |
| Maths boffins on 10:39 - Jun 23 by StokieBlue |
Not quite right as the chance of pulling out a green one is no longer 0.2 after the first one is taken. Your formula is correct when the events are independent such as tossing a coin or if the skittle was placed back into the packet after it was picked.
If we assume there are 4 of each of the 5 types of Skittle (this is important) then:
0.2 * 0.158 * 0.111 * 0.059 = 0.00021 = 1/4845
Sorry to be a pedantic bore. Really you need to know the exact number of each in the bag as it affects the calculation.
SB |
"Not quite right as the chance of pulling out a green one is no longer 0.2 after the first one is taken."
As per the OP, I pulled out 4 all at once. | |
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| Maths boffins on 10:57 - Jun 23 with 4380 views | mos |
Through research, I believe there are about 180 skittles in a bag, presuming that there are an equal amount to start off with then,
36/180 x 35/179 x 34/178 x 33/177 = 1413720/1015123320
= 0.00139265838
= 0.14%
Or 14 in 10000 chance.
Now I’m confused. | |
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| Maths boffins on 10:57 - Jun 23 with 4375 views | BrixtonBlue |
| Maths boffins on 10:50 - Jun 23 by StokieBlue |
Taking your assumptions of 5 types of Skittles and the assumption there are 4 of each type in the bag:
(4 / 20) * (3 / 19) * (2 / 18) * (1 / 17)
It's early though so feel free to point out any errors, my brain isn't up to speed yet today.
SB |
It's also a family size bag, 196g, so there are more than 4 of one colour in each.
Sorry, I should've stated this earlier! | |
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| Maths boffins on 10:57 - Jun 23 with 4373 views | mos |
| Maths boffins on 10:52 - Jun 23 by DropCliffsNotBombs |
because let's say theres 20 in total (4 of each colour). the first chance is 4÷20 = 0.2.
the second pick are 3 greens left out of 19 (3÷19 = 0.158)) |
I see. Understood. That was what I thought. | |
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| Maths boffins on 10:58 - Jun 23 with 4371 views | StokieBlue |
| Maths boffins on 10:56 - Jun 23 by BrixtonBlue |
"Not quite right as the chance of pulling out a green one is no longer 0.2 after the first one is taken."
As per the OP, I pulled out 4 all at once. |
You can't really pull all four at once, the probability still decreases with each one.
SB | |
| Avatar - IC410 - Tadpoles Nebula |
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| Maths boffins on 11:07 - Jun 23 with 4335 views | BanksterDebtSlave |
| Maths boffins on 10:58 - Jun 23 by StokieBlue |
You can't really pull all four at once, the probability still decreases with each one.
SB |
Wouldn't you have to work out all possible combinations of 4 colours taken (in one go) from the bag (x) and then it would be 1/x For the answer?
Edit....although once you have one yellow in your hand the chances of another being yellow (even in that moment) are slightly less than all other colours! (This may effect the answer......not sure.) [Post edited 23 Jun 2018 11:12]
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