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20 replies to this topic

### #1 EventHorizon

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Posted 13 February 2009 - 10:31 PM

(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?

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

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### #2 Largaroth

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Posted 13 February 2009 - 10:41 PM

how many days is their year?
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### #3 Largaroth

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Posted 13 February 2009 - 10:46 PM

1) how many days is in their year?
2)10^(10^100)/2?
3)does n stand for a variable?
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### #4 EventHorizon

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Posted 13 February 2009 - 11:04 PM

1) how many days is in their year?
2)10^(10^100)/2?
3)does n stand for a variable?

1) look it up. (though I did include the information needed in the spoiler)
2) nope....
3) Yes, n is a variable.
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### #5 Professor H

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Posted 13 February 2009 - 11:26 PM

Spoiler for answer to #3 (I think)

(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?

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### #6 EventHorizon

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Posted 13 February 2009 - 11:59 PM

Spoiler for answer to #3 (I think)

nope. For n=3, it is 1/2, but your equation gives 1/3.
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### #7 Avalanche5x

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Posted 14 February 2009 - 12:21 AM

On #3, can "consecutive" be n+1 then n, or does it have to be n then n+1?
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### #8 Largaroth

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Posted 14 February 2009 - 12:23 AM

Spoiler for I think

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### #9 Avalanche5x

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Posted 14 February 2009 - 12:26 AM

If consecutive is only a then a+1. Then...
Spoiler for Possibly for #3

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### #10 EventHorizon

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Posted 14 February 2009 - 12:38 AM

Largaroth: Remember, it is a probability. This means it needs to be between 0 and 1.

Avalanche5x: You answer fails for n=3. the probability is 1/2, and you have 1/3.
Spoiler for Here's n=3 as an example...

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