Suppose you are attending a Gambler’s anonymousmeeting with 49 other people. A f
ID: 2953701 • Letter: S
Question
Suppose you are attending a Gambler’s anonymousmeeting with 49 other people. A fellow attendee wants to bet you$1000 that no 2 people in the meeting were born on the same day ofthe year. Statistically speaking, would it be wise to accept thebet? I am so lost here.... Suppose you are attending a Gambler’s anonymousmeeting with 49 other people. A fellow attendee wants to bet you$1000 that no 2 people in the meeting were born on the same day ofthe year. Statistically speaking, would it be wise to accept thebet? I am so lost here....Explanation / Answer
Statistically it would be wise to bet if Probability ofwinning is > 0.5 . That is when P(winning) >P(losing)
so P(losing ) = P(no two attendees were born on the same dayof the year)
= (# of ways no two attendees were born on the same day/ Total# of ways they can be born)
let's assume all 366 days (including Feb 29th ) arepossible.
so # of ways no two attendees were born on the same day can be thought about like you are choosing 50 days for eachof them to be born.
So for first attendee you have 366 days to choose. For secondone you have 365 days because you can't choose the day that youhave chosen for first one. and for thirds (366-2) = 364 daysetc....
so # of ways no two attendees were born on the same day =366 * 365 * .....* (366-49)
= 366 ! / 316!
And total # of ways of 50 delegates being born = (# of ways ofchoosing one number from 1 to 366 each denoting one day)^ 366 ....this is because if you start choosing the days the attendees shouldbe born, you have 366 choices for all of them .
= 36650
therefore P(losing) = (366*365*364*.....317)/36650 = 2.99%
Therefore P(winning) = 1- P(losing) =97.01%
Therefore you should bet.
Hope this is helpful. Feel free to ask forany clarifications.