EdCollins
EdCollins
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March 27th, 2012 at 7:33:30 PM permalink
Ah! I think that's it. I agree with Crystal and Chester and Doc's post about interpreting the question properly. Excellent work.
PopCan
PopCan
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March 27th, 2012 at 8:37:42 PM permalink
Quote: PopCan

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



Ouch. Where did I go horribly wrong on this? I figured that you'd "win" if you met someone born on Wednesday so I tried to figure out the average number of consecutive losses.
Doc
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March 27th, 2012 at 9:11:00 PM permalink
Quote: PopCan

Ouch. Where did I go horribly wrong on this? I figured that you'd "win" if you met someone born on Wednesday so I tried to figure out the average number of consecutive losses.


I think what you calculated there is the probability that you will meet the first "Wednesday" person at some point before having met four people born on other days. That is a different issue than the "expected" number of people you will meet before meeting someone born on Wednesday. You need to calculate the expected value of the number of people you will meet.

The formula I presented calculates the Expected Value of the number of people met including the Wednesday person and subtracts 1.

EV(excluding) = -1 + EV(including) = -1 + 1*P(1) + 2*P(2) + 3*P(3) + ... + n*P(n) + ...
boymimbo
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March 27th, 2012 at 9:56:31 PM permalink
I don't think the answer is 6.

Consider a number from 1 to 7. If you knew that the sequence of numbers is 1,2,3,4,5,6,7, then the answer would be six. But the sequence of numbers can be any combination. Therefore, the answer is 3.

I'm thinking of a number between 1 and 7. What is the number of guesses it will take before you get it right? It won't be 7 guesses? It will be 4.

Oh yeah, the replacement thing! I'm wrong.

The answer is 6.
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ewjones080
ewjones080
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March 28th, 2012 at 3:26:34 AM permalink
So this reminds me of that birthday problem. I think that states: How many people do you need to have in a room for there to be a 95% chance at least two share a birthday, and its a surprisingly low answer, like 29 right?

This doesn't seem much different. If I asked "How many people do I need in a room to have better than a 95% chance two were born on the same day of the week?" that would be more like 5 right?

Now that I'm thinking through it, this problem is quite different than the sharing birth DATE problem.
Wizard
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Wizard
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March 28th, 2012 at 5:38:56 AM permalink
Here is my solution.

Let x be the answer. x has a memory-less property, in which no matter how many people you meet, you're never overdue to meet someone born on a Wednesday. So, when you meet the first person there is a 1/7 chance he was born on Wednesday, so you would have met 0 before him. There is a 6/7 chance he was not born on a Wednesday, so you met one person and you can still expect to meet x more. So the equation is:

x=(6/7)*(1+x) + (1/7)*0
x=(6/7)*(1+x)
7x=6*(1+x)
7x=6+6x
x=6

Note that I'm assuming a 1/7 chance of being born on a Wednesday. If the probability for being born on Wednesday is w, then the answer would be (1-w)/w.
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FinsRule
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March 28th, 2012 at 6:46:09 AM permalink
As was discussed on an earlier thread, births are not evenly distributed over days of the week. I think Tuesdays and Wednesdays were the heaviest days.

I know that's sort of not the point of the question, but I wanted to bring it up.
DJTeddyBear
DJTeddyBear
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March 28th, 2012 at 7:26:39 AM permalink
I "conditionally" agree with the Wiz.

If you take into account the CDC's report that there are more people born mid-week than on weekends, etc. then the answer is slightly less than 6. For the point of argument though, the correct answer is 6.



But let me see if I can cut thru the remaining clutter.

If you shuffled a deck of cards and started flipping cards, how many cards can you expect to flip before flipping the [specific card] ?

Did you say 26? Good! (Note, the answer could be 25, or 25.5 depending on how anal you are about the math and phrasing.)

What about if you shuffled the deck, flipped one card, and put it back and reshuffled? How many times would you expect to do that before flipping the desired card?

Did you say 52? Great. (or 51 if you're anal.)

That's the difference between "sampling without replacement" and "sampling with replacement"

In the case of birthdays, with 7 billion people on this planet, although we may be careful not to ask the same person twice, the pool size is large enough that the math works by "sampling with replacement".

With this type of question, the answer is always the number of possibilities. Or minus 1 depending on the wording.



Quote: ewjones080

So this reminds me of that birthday problem. I think that states: How many people do you need to have in a room for there to be a 95% chance at least two share a birthday, and its a surprisingly low answer, like 29 right?

The reason this number is so low is because you're comparing each new response, to every response alreay provided, so the number of combinations increases exponentially with each person added to the pool.


Here's another way to look at things.

If you started asking people their birthday (month and day), how many people would you expect to ask before you get a response that is the same as the previous person's response?

The answer is 365.25
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PopCan
PopCan
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March 28th, 2012 at 10:22:48 AM permalink
Quote: Doc

I think what you calculated there is the probability that you will meet the first "Wednesday" person at some point before having met four people born on other days. That is a different issue than the "expected" number of people you will meet before meeting someone born on Wednesday. You need to calculate the expected value of the number of people you will meet.

The formula I presented calculates the Expected Value of the number of people met including the Wednesday person and subtracts 1.

EV(excluding) = -1 + EV(including) = -1 + 1*P(1) + 2*P(2) + 3*P(3) + ... + n*P(n) + ...



Thanks. The moment you said "Expected Value" it made sense.
Nareed
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March 28th, 2012 at 12:12:47 PM permalink
Quote: Wizard

Note that I'm assuming a 1/7 chance of being born on a Wednesday. If the probability for being born on Wednesday is w, then the answer would be (1-w)/w.



I was going to bring that up. Is there any kind of data of the weekday distribution of births? Did you run across any in your actuarial work?
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