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Birthday Probability


Best Answer superprismatic, 25 February 2014 - 09:47 PM

Spoiler for a simulation of a billion trials

Go to the full post


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7 replies to this topic

#1 BMAD

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Posted 25 February 2014 - 05:43 PM

At a movie theater, the manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket.  You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout a 365 day year, what position in line gives you the best chance of being the first duplicate birthday?


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#2 bonanova

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Posted 25 February 2014 - 06:59 PM

Spoiler for First thoughts


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#3 superprismatic

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Posted 25 February 2014 - 09:47 PM   Best Answer

Spoiler for a simulation of a billion trials


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#4 Rainman

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Posted 25 February 2014 - 11:51 PM

Spoiler for doing the math


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#5 superprismatic

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Posted 26 February 2014 - 02:03 AM

Spoiler for doing the math

Nice analysis!  I wish I had thought of that.


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#6 bonanova

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Posted 26 February 2014 - 11:32 AM

 

Spoiler for doing the math

Nice analysis!  I wish I had thought of that.

 

 

 

Spoiler for Agree, nice analysis

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#7 Rainman

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Posted 26 February 2014 - 08:32 PM

 

 

Spoiler for doing the math

Nice analysis!  I wish I had thought of that.

 

 

 

Spoiler for Agree, nice analysis

 

 

The method I used is very similar to differentiation. Differentiation also compares f(x) to some nearby value f(x+h), in order to tell whether the function is growing or declining. However, this particular function is only defined for integers, so it's not differentiable unless we first extend it in some way.


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#8 bonanova

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Posted 27 February 2014 - 10:48 AM

Right ... it's "discrete" differentiation.

And that permits a (nice) direct calculation as opposed to simulation.

 

I simulated the average first-duplicate position and got the kind-of well-known 23.

I am modifying the program now to show the median value.

The distribution is skewed, so the mean is slightly larger.

 

And, I guess, the median must be the point of greatest slope of the cumulative probability.


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