Maths question

tonyyeb

tonyyeb

tonyyeb

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If one in 9 people are left handed what would be the odds of two people meeting who were left handed? How is that worked out?

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It depends on how many meetings there are. If you have enough meetings (and there are at least two left handed people in your sample) then eventually there will be a meeting between two lefties.

It also depends on how many people are in the group. If the group of people is small then you would have to consider the change in the fraction of lefties when you selected the first person for a meeting, but if the group was large enough then you wouldn't have to consider it.

I think the binomial expansion is what you are after if you want to take multiple meetings into account in your calculations, but i haven't done statistics in a while so i may be wrong
 
It turns out that the chances of two people meeting that are left handed are around 1.2% or ~1 in 100. Still not found an explanation why that is the case mathematically.
 
Maybe it's

Chance of person 1 is left handed = 1 in 9
Chance of person 2 is left handed = 1 in 9

Chance of person 1 meeting person 2 = 1/9 x 1/9 = 1/81 = 0.0123 = 1.23%
 
The question is badly worded.

If someone is left handed then for every person they meet the chance of that person being left handed is 1/9. (This may not be strictly true as maybe lefties stick together, certain families are prone to more lefties etc.)

The chance of randomly selecting two people (from a large group of people) and both of them being left handed is 1/81
 
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If p(left handed) = 1/9, then the probability of two randomly selected people being left handed is 1/9 * 1/9 which is 1/81.

If you have some other definition of "meeting" in mind (small groups, etc.) then that will differ.
 
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KingOfTheDaves said:
Maybe it's

Chance of person 1 is left handed = 1 in 9
Chance of person 2 is left handed = 1 in 9

Chance of person 1 meeting person 2 = 1/9 x 1/9 = 1/81 = 0.0123 = 1.23%
Click to expand...

This is right - but only if the population can be assumed to be large.

For example if the population was only 18 people, two of whom were left handed, it would still be correct to say one in 9 people were left handed however the chance of them meeting would be 1/9 * 1/18 = 0.6%. We're picking 'without replacement'. The without replacement term is irrelevant when talking of large populations, but with small populations it becomes important.

The other way to think about it is that there are 153 possible combinations from 18 people, only 1 of which would be the two left handed people (0.6%)
 
Is this one of those trick questions where you actually have a left handed person in your basement who is never going to meet anyone again? The odds of them escaping is 1.23%
 
clv101 said:
This is right - but only if the population can be assumed to be large.

For example if the population was only 18 people, two of whom were left handed, it would still be correct to say one in 9 people were left handed however the chance of them meeting would be 1/9 * 1/18 = 0.6%. We're picking 'without replacement'. The without replacement term is irrelevant when talking of large populations, but with small populations it becomes important.

The other way to think about it is that there are 153 possible combinations from 18 people, only 1 of which would be the two left handed people (0.6%)
Click to expand...

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