A very long line...
Assume there are 7.6 billion people in the world. They all line up in a very long line and are given the numbers 1 to 7.6B:
Then two are randomly picked out and it is determined that the first one has a number higher than the second one.
What is the approximate probability that the first one has the highest value of all in the world?
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4 solutions
Small typo on Bayes theorem. The denominator should be P(B) instead of P(A).
Originally, the first person had a 7 . 6 B 1 chance of having the largest number in the world. However, the new information eliminates exactly half of the permutations, without eliminating any of those involving him/her having the highest number. Therefore, he/she has twice the chance they had originally, or 7 . 6 B 2 .
Note, this is not only approximate, but in fact an exact answer! :^)
And in general, if n individuals so numbered are picked randomly one after the other from N people, the the probability that the first person has the highest overall value, given that they have the highest value among the group of n people, is N n .
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Good point... And I wonder what the probability is that they are the m th highest... 🤔
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With just two people chosen, I think that the probability that they are the m th highest is N ( N − 1 ) 2 ( N − m ) . So for m = 1 we get N 2 as found previously, and with, say, m = 2 N , (supposing even N ), the probability would be N − 1 1 .
Nice.
And oddly enough, I seem to recognize Ella, Bella and Frabduzella. :D
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Hahaha... Yup! Guess I was two lazy to draw up a new image! :0)
All the others solutions to this problem are pretty logical, this might seem a little stupid :-
Since the second person picked is already numbered lower than the first, his probability of being the highest in the world is vanished.
And since the total probability should be 1 , and the other people remain unaffected, we could say the second person gave his initial probability to the first, hence the answer.
Yes. I know what you mean Parth... This is similar to the popular argument often given for the Monty Hall problem! :)
Bayes' Theorem is probably the "correct" way to determine the answer, but this is the reason "why" the solution is what it is, intuitively. The question is essentially "If two numbers are chosen uniformly and randomly from the numbers between 1 and n, what is the probability that one of them is n?", the answer to which is of course 2/n. Replacing "one of them" with "the higher of the two" doesn't change the problem, because if one is n, it has to be the higher one.
A: number of first person, B: number of second person
There are 7.6B! different sorting alternatives. In ( 7 . 6 B − 1 ) ! of them A is the highest number. The information A>B eliminates 2 7 . 6 B ! alternatives where B>A
Obiviously none of these eliminated alternatives contains the previously mentioned ( 7 . 6 B − 1 ) ! alternatives where A is the highest number. So the new probablity is 7 . 6 B ! − 2 7 . 6 B ! ( 7 . 6 B − 1 ) ! = 7 . 6 B ! 2
According to Bayes' Theorem: P(A|B) = P(B) P(A) ⋅ P(B|A)
Let A be the event of the first one being the highest, and let B be the event of the first one being the higher of the 2.
P(A) is obviously 7 . 6 B 1 , since there is only 1 possibility out of the 7.6 billion.
P(B|A) is 1, because if we are given that the first one is the highest value, then it is obviously guaranteed to be higher than the second choice.
P(B) is a bit trickier. Imagine choosing the first person. On average, half of the population will be below him. Therefore, the probability is 2 1 .
Therefore,
2 1 7 . 6 B 1 ⋅ 1 = 7 . 6 B 2 β ⌈ ∣ ⌉