Math probability problem
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Math probability problem

[From: ] [author: ] [Date: 11-08-25] [Hit: ]
..and you dont fall below 50% until you have 23 people.That low number is surprising to most people.......
If each person randomly gets a number between 1 and 10
then how many many people do you need together for there to be more than a 50 percent chance for two of them to have the same number.

Show how you did it please!

-
P(unique for 1 person) = 10/10
P(second person's number is different) = 10/10 * 9/10
P(third different) = 10/10 * 9/10 * 8/10 = 720/1000
P(4th different) = 720/1000 * 7/10 = 5040/10000
P(5th different) = 5040/10000 * 6/10 = 30240/100000

Since P(all different) fell below 50% after 5 people involved,
that is where it is > 50% that two will be the same.


A more common version of this problem involves shared birthdays.

In that case the numbers are
365/365 * 364/365 * 363/365 * ....
and you don't fall below 50% until you have 23 people.

That low number is surprising to most people.
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