**(1) How many Plutonians would you need to gather before there is a 50:50 chance that two of them have the same birthday?**

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### #1

Posted 13 February 2009 - 10:31 PM

**(2) Planet Gplex takes a googolplex (10^(10^100)) of its days to orbit its sun. Approximately how many Gplexians would you need to gather before there is a 50:50 chance that two of them have the same birthday? Prove that your estimate is a good one.**(yeah, I'm trying to make direct computation impossible, and yes I have an answer/proof)

And now, a somewhat related original puzzle...

**(3) You are given a stack of cards numbered 1 to n, and their order is randomized. What is the probability that two consecutively numbered cards (lets say, a and a+1) are found consecutively (a immediately before a+1) somewhere in the stack?**

**Edited by EventHorizon, 13 February 2009 - 10:34 PM.**

### #2

Posted 13 February 2009 - 10:41 PM

### #3

Posted 13 February 2009 - 10:46 PM

2)10^(10^100)/2?

3)does n stand for a variable?

### #4

Posted 13 February 2009 - 11:04 PM

1) look it up. (though I did include the information needed in the spoiler)1) how many days is in their year?

2)10^(10^100)/2?

3)does n stand for a variable?

2) nope....

3) Yes, n is a variable.

### #5

Posted 13 February 2009 - 11:26 PM

I'll start with extensions of Prof. Templeton's Birthday and SSN puzzles...

(1) How many Plutonians would you need to gather before there is a 50:50 chance that two of them have the same birthday?Spoiler for Pluto facts needed

**(2) Planet Gplex takes a googolplex (10^(10^100)) of its days to orbit its sun. Approximately how many Gplexians would you need to gather before there is a 50:50 chance that two of them have the same birthday? Prove that your estimate is a good one.**(yeah, I'm trying to make direct computation impossible, and yes I have an answer/proof)

And now, a somewhat related original puzzle...

**(3) You are given a stack of cards numbered 1 to n, and their order is randomized. What is the probability that two consecutively numbered cards (lets say, a and a+1) are found consecutively (a immediately before a+1) somewhere in the stack?**

### #6

Posted 13 February 2009 - 11:59 PM

nope. For n=3, it is 1/2, but your equation gives 1/3.Spoiler for answer to #3 (I think)

### #7

Posted 14 February 2009 - 12:21 AM

### #8

Posted 14 February 2009 - 12:23 AM

### #9

Posted 14 February 2009 - 12:26 AM

### #10

Posted 14 February 2009 - 12:38 AM

Avalanche5x: You answer fails for n=3. the probability is 1/2, and you have 1/3.