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!
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!
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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.
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.