Probability ?: Only 23 people needed in same room for 50% chance of same Bday (month/day)
With 23 individuals, there are
(23 × 22)/2 = 253 pairs to consider, far more than half the number of days in a year.
If that's the case, why 23?
365 days in a year.
20 individuals:
(20x19)/2 = 190, which is still more than half the number of days in a year.
Why not 20 people?
Quote: 100xOddshttps://en.wikipedia.org/wiki/Birthday_problem
With 23 individuals, there are
(23 × 22)/2 = 253 pairs to consider, far more than half the number of days in a year.
If that's the case, why 23?
365 days in a year.
20 individuals:
(20x19)/2 = 190, which is still more than half the number of days in a year.
Why not 20 people?
link to original post
(364/365)^X = 50%
Solve for X
Pr(# of matching birthdays >= 2) = 1 - (365/365 * 364/365 * 363/365 * ),
where the number of terms in the parentheses equals the number of people in the room.
With 20 people, the probability is only 41.14%, and with 23, its 50.73%.
By the time you get to 83 people, the probability is essentially 100% (actually 99.9959%, which rounds to 100.00%).
Quote: camaplBy the time you get to 83 people, the probability is essentially 100% (actually 99.9959%, which rounds to 100.00%).
link to original post
With 83 random people in a room, it's almost guaranteed that 2 people have the same bday?!
83/365day = 22.7%
22.7% get's you effectively 100%?!? Wished this applied to lotto tix.
and Apparently i failed probability/statistics class.
Quote: 100xOddsQuote: camaplBy the time you get to 83 people, the probability is essentially 100% (actually 99.9959%, which rounds to 100.00%).
link to original post
With 83 random people in a room, it's almost guaranteed that 2 people have the same bday?!
83/365day = 22.7%
22.7% get's you effectively 100%?!? Wished this applied to lotto tix.
and Apparently i failed probability/statistics class.
link to original post
That 22.7% is the odds that one of the 83 people shares one particular birthday, like *your* birthday. You need to consider that the shared birthday could be any day.
Quote: 100xOddsQuote: camaplBy the time you get to 83 people, the probability is essentially 100% (actually 99.9959%, which rounds to 100.00%).
link to original post
With 83 random people in a room, it's almost guaranteed that 2 people have the same bday?!
83/365day = 22.7%
22.7% get's you effectively 100%?!? Wished this applied to lotto tix.
and Apparently i failed probability/statistics class.
link to original post
I gave the equation to you.
(364/365)^X = 50%. Solve for X
You find that the answer is very close to (23 x 22)/2. Which is why 23 people is the breakeven(ish) number of people.
Quote: camaplNot exactly Another way to state the math in OPs link:
Pr(# of matching birthdays >= 2) = 1 - (365/365 * 364/365 * 363/365 * ),
where the number of terms in the parentheses equals the number of people in the room.
With 20 people, the probability is only 41.14%, and with 23, its 50.73%.
By the time you get to 83 people, the probability is essentially 100% (actually 99.9959%, which rounds to 100.00%).
link to original post
No my formula is exactly.
