[superprismatic]

I am a member of "People Whose Birthday Is NOT On February 29^th Society",

a not very exclusive club. A total of 128 people had satisfactorily proven that they were

eligible to be members. I, being the society's secretary, was charged with mailing out the

announcement for a recent meeting. The first 100 announcements had the correct time for

the meeting (6 O'clock) but I had inadvertently put the time of 7 O'clock on the announcements

sent to the remaining 28 members. At 6 O'clock on meeting day, 100 members showed up. We

decided to wait for the remaining 28. Even though each of the members had proven that

he was not born on February 29^th, no records were kept of his actual birth day.

So, to kill time, we all annnounced our birthdays and were astonished to find that, of the

100 people in attendence, we had only 80 distict birth days! My question to you is this:

When the other 28 arrive and announce their birthdays, how many distinct birthdays will

the entire group of 128 be expected to have? Assume that members were randomly chosen from

the population of all persons whose birthday is not February 29^th and that the

distribution of birthdays is otherwise uniform for this population.

[Guest]

92

[Guest]

It seems if 80 of the 100 had distinct birthdays, that translates to 80%. Therefor when the 128 gather together 80% of them would be 102 (102.4 but I rounded down). Perhaps my logic is askewed or I'm not seeing the problem correctly ... guess I'll find out in a while.

[Guest]

It seems if 80 of the 100 had distinct birthdays, that translates to 80%. Therefor when the 128 gather together 80% of them would be 102 (102.4 but I rounded down). Perhaps my logic is askewed or I'm not seeing the problem correctly ... guess I'll find out in a while.

358 people were to show up late, that'd be 366 total unique birthdays?? There's a Feb 29th traitor among us!

Edited January 28, 2010 by ljb

[Guest]

Nice observation ljb (although I think you meant 458). A great discussion of how difficult it is for the paltry human mind to comprehend statistics and the concept of randomness can be found in "The Drunkard's Walk" by Mlodinow.

[Guest]

I think it's 101 (rounding from 101.0732..)

When the 101st person shows up, he either matches one of the 80 distinct birthdays (with probability 80/365), or he doesn't. So, after the 101st person, the expected number of distinct birthday, called e1, is:

80*(80/365)+81*(265/365)

e2 will then be e1*(e1/365)+(e1+1)*((365-ei)/265))

talking that out to e28 gives the result above

[superprismatic]

I think it's 101 (rounding from 101.0732..)

When the 101st person shows up, he either matches one of the 80 distinct birthdays (with probability 80/365), or he doesn't. So, after the 101st person, the expected number of distinct birthday, called e1, is:

80*(80/365)+81*(265/365)

e2 will then be e1*(e1/365)+(e1+1)*((365-ei)/265))

talking that out to e28 gives the result above

NoMoney is right on! Nice thinking! You didn't fall for the 100 red herring.

[Guest]

Since 20% of the group of 100 didn't have distinct birtthdays, there were 20 in the group that did not.

Since 20% of 28 is 5.6, you could expect to find between 5 & 6 of the remaining 28 not to have distinct birthdays. The number of distinct birthdays in total would be either 102 or 103.

BTW, those with indistinct birthdays were born on March 1 between leap years