Probability (1 Viewer)

[seanieg89]

 

[Carrotsticks]

Them sigmoids.

 

[Shadowdude]

i c wut u did thar

 

[Sindivyn]

I have a feeling this is completely wrong and illegitimate, but anyway.
Set a date in a year for one birthday.
Probability of everyone NOT sharing that birthday = (364/365)^22
Now, this event repeats 22 more times (setting each person).
= ((364/365)^22)23
= 0.22495.....
Therefore conjugate (that is, sharing birthdays) is 1-0.2495
=0.75047... which is greater than the chance of none of the people sharing birthdays.

Most likely completely wrong but meh

Currently working on an alternate method (realized a mistake)
Alternate (seems more legit)
Set 1 date in the year for one person
P(no shared dates) = (364!/342! divided by (365)^22) Having trouble inputting that into the calculator, so not sure if it's <0.5.

Edit: online scientific calculator showed it to be 0.4927.... Therefore the probability of shared dates = 1-0.4927 = 0.5073.
That is, probability of a shared date > probability of no shared dates.
Is this right?

 

[RealiseNothing]

Such a classic problem.

 

[seanieg89]

I have a feeling this is completely wrong and illegitimate, but anyway.
Set a date in a year for one birthday.
Probability of everyone NOT sharing that birthday = (364/365)^22
Now, this event repeats 22 more times (setting each person).
= ((364/365)^22)23
= 0.22495.....
Therefore conjugate (that is, sharing birthdays) is 1-0.2495
=0.75047... which is greater than the chance of none of the people sharing birthdays.

Most likely completely wrong but meh

Currently working on an alternate method (realized a mistake)
Alternate (seems more legit)
Set 1 date in the year for one person
P(no shared dates) = (364!/342! divided by (365)^22) Having trouble inputting that into the calculator, so not sure if it's <0.5.

Edit: online scientific calculator showed it to be 0.4927.... Therefore the probability of shared dates = 1-0.4927 = 0.5073.
That is, probability of a shared date > probability of no shared dates.
Is this right?

Yep your second method is correct. Incidentally, this is a fantastic bet to make at a party. People who are non-mathematical seem to find it counter-intuitive and when you have say 30 people in the room, the probability jumps to >70%.

 

[Shadowdude]

Yep your second method is correct. Incidentally, this is a fantastic bet to make at a party. People who are non-mathematical seem to find it counter-intuitive and when you have say 30 people in the room, the probability jumps to >70%.

make dat cash money

 

[Sindivyn]

Yep your second method is correct. Incidentally, this is a fantastic bet to make at a party. People who are non-mathematical seem to find it counter-intuitive and when you have say 30 people in the room, the probability jumps to >70%.

Yeah, I subbed in the wrong value at first and got something like 0.9 chance. I would hate to get this question in a test though, kept on getting math errors on my calculator because of the permutations .