March 27th, 2012 at 2:18:38 PM
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...before you met one born on a Wednesday?

March 27th, 2012 at 2:40:40 PM
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Sounds like a question made for the Wizard but I'll take a shot. If we assume there's an even distribution of births over days of the week then you will have a 53% chance of meeting someone born on a Wednesday if you've met 5 people: 1 - (6/7)^5

A report from the CDC shows that births are significantly less common on Saturday and Sunday, most likely due to scheduled caesareans and induced labor. The birth rate they list for 1997 shows 11,723 births on Wednesdays out of 74,407 total births tracked or 15.75%. This would mean, at least for people born in 1997, that after meeting 5 people the chance that at least one was born on a Wednesday is 57%. The chance is 49.6% for 4 people. I'm sure I'm making several horrible assumptions there but I feel safe saying your answer is slightly over 4.

A report from the CDC shows that births are significantly less common on Saturday and Sunday, most likely due to scheduled caesareans and induced labor. The birth rate they list for 1997 shows 11,723 births on Wednesdays out of 74,407 total births tracked or 15.75%. This would mean, at least for people born in 1997, that after meeting 5 people the chance that at least one was born on a Wednesday is 57%. The chance is 49.6% for 4 people. I'm sure I'm making several horrible assumptions there but I feel safe saying your answer is slightly over 4.

March 27th, 2012 at 4:00:01 PM
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Not many...what doctor wants to work on a Friday or Monday, shiiit...

Gambling calls to me...like this ~> http://www.youtube.com/watch?v=4Nap37mNSmQ

March 27th, 2012 at 4:56:51 PM
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__6__

It's not whether you win or lose; it's whether or not you had a good bet.

March 27th, 2012 at 5:39:03 PM
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Care to show your work Wizard?

Vote for Nobody 2016!

March 27th, 2012 at 5:43:33 PM
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the old song says "wednesday's child is full of woe"...

i hope i don't meet one.

i hope i don't meet one.

In a bet, there is a fool and a thief.
- Proverb.

March 27th, 2012 at 5:48:42 PM
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I found this problem in an old book called Problematical Recreations - 10th in a series that I picked up at a swap meet for a buck this past weekend.

The answer given is 7. The explanation given in the book is this:

Not 3.5 since this a case of sampling with replacement.

Each person has a 1/7 probability of having been born on a Wednesday. In a sense then, each person is "1/7 of an expected Wednesday child." Since it requires 7 such to add up to a Wednesday child, the answer is 7 people.

I initially thought, for a moment, the answer was 3.5. But based upon some of the not-so-obvious answers to some of the other questions in the book, I decided 3.5 was probably not correct.

This afternoon I wrote a computer program to simulate this problem (10,000,000 trials) and I decided the answer was indeed 7. (A total of 70,015,454 people were "met" in all, giving the average number of people met for each trial to be 7.0015)

Once I refreshed memory on what the differences were between "sampling with replacement" and "sampling without replacement" I realized the answer could not be 3.5 or 4 or even 5 point something. I tweaked my program to let me know how many people the "unluckiest" person had to meet before he met someone born on a Wednesday. The answer for one 10,000,000 trial was 100 (!) and for a later trial it was 108!

In a case of sampling WITHOUT replacement, if I understand the differences, the correct answer would be 4. (1+2+3+4+5+6+7) / 7 = 4

The answer given is 7. The explanation given in the book is this:

Not 3.5 since this a case of sampling with replacement.

Each person has a 1/7 probability of having been born on a Wednesday. In a sense then, each person is "1/7 of an expected Wednesday child." Since it requires 7 such to add up to a Wednesday child, the answer is 7 people.

I initially thought, for a moment, the answer was 3.5. But based upon some of the not-so-obvious answers to some of the other questions in the book, I decided 3.5 was probably not correct.

This afternoon I wrote a computer program to simulate this problem (10,000,000 trials) and I decided the answer was indeed 7. (A total of 70,015,454 people were "met" in all, giving the average number of people met for each trial to be 7.0015)

Once I refreshed memory on what the differences were between "sampling with replacement" and "sampling without replacement" I realized the answer could not be 3.5 or 4 or even 5 point something. I tweaked my program to let me know how many people the "unluckiest" person had to meet before he met someone born on a Wednesday. The answer for one 10,000,000 trial was 100 (!) and for a later trial it was 108!

In a case of sampling WITHOUT replacement, if I understand the differences, the correct answer would be 4. (1+2+3+4+5+6+7) / 7 = 4

March 27th, 2012 at 6:09:23 PM
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I'm with the Wizard on this one. Ed, maybe you included the Wednesday guy that you met. There should only be 6 people that you meet BEFORE the Wednesday guy. Otherwise, you met 7 other people for every Wednesday guy, so his frequency is only 1/8.

I heart Crystal Math.

March 27th, 2012 at 6:10:48 PM
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Quote:EdCollins...The answer given is 7...

7 is the answer to this slightly different question: How many people would you expect to meet in order to meet one born on a Wednesday?

March 27th, 2012 at 6:16:29 PM
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Yes, I think this could be back to a matter of language, which often seems to cause us difficulty. My limited analysis suggests that the answer is 7 if you interpret the question to mean you should include the person born on Wednesday in your count. If you are only to include those you meet before you meet that person, I think the answer is 6, which agrees with the Wizard's answer.

My formula (for not counting the person born on Wednesday) is:

I suspect there is a way to simplify that, but my brain is too atrophied to do it. I just used Excel to figure the first 60 terms and got EV=5.9926. I suspect that if I included all ∞ terms, it would come out to 6. If you leave off the -1, you include the person born on Wednesday and get a total of 7.

My formula (for not counting the person born on Wednesday) is:

I suspect there is a way to simplify that, but my brain is too atrophied to do it. I just used Excel to figure the first 60 terms and got EV=5.9926. I suspect that if I included all ∞ terms, it would come out to 6. If you leave off the -1, you include the person born on Wednesday and get a total of 7.