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  #16  
Old 12-15-2007, 01:34 PM
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Quote:
Originally Posted by Kuan View Post
Assuming the odds are exactly 1/2 and 1/2.

I was thinking there are only six combinations.

M F
5 0
4 1
3 2
2 3
1 4
5 0

Hence the odds are 1/6?
Think of the number of ways each combination can be achieved:

M F
5 0 M M M M M
4 1 M M M M F - M M M F M - M M F M M - M F M M M - F M M M M
and so on. The probability of 2M 3F is a lot higher than 5F.

It's the same problem as the chance of snake eyes showing up when you roll a pair of dice.

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  #17  
Old 12-15-2007, 01:35 PM
t walgamuth's Avatar
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Quote:
Originally Posted by Kuan View Post
Assuming the odds are exactly 1/2 and 1/2.

I was thinking there are only six combinations.

M F
5 0
4 1
3 2
2 3
1 4
5 0

Hence the odds are 1/6?
That's not right!

the 1/32 is the best choice.

Tom W
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  #18  
Old 12-15-2007, 01:36 PM
truckinik's Avatar
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OK, You lost me at .... M.F.
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  #19  
Old 12-15-2007, 01:54 PM
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Your 1/6 only works if it is a single trial e.g. 1 puppy or a single roll of a dice. In the puppy example there are 32 possible outcomes, because the order of the puppies makes that many combinations.

The answer is 1/2! (1 divided by 2 factorial) or 1/32

This works because you are sampling with replacement. Sampling without replacement is like a lottery where once a ball is picked it cannot be picked again. Of course with only 2 outcomes per trial, you cannot model something without replacement because you can only do 2 trials.



Quote:
Originally Posted by Kuan View Post
Assuming the odds are exactly 1/2 and 1/2.

I was thinking there are only six combinations.

M F
5 0
4 1
3 2
2 3
1 4
5 0

Hence the odds are 1/6?
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  #20  
Old 12-15-2007, 02:00 PM
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Quote:
Originally Posted by Emmerich View Post
The answer is 1/2! (1 divided by 2 factorial) or 1/32
No it isn't. 2! is 2.

The answer is (1/2)^5.
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  #21  
Old 12-15-2007, 02:03 PM
Emmerich's Avatar
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You are correct! My bad, I meant to say 5!

Quote:
Originally Posted by Matt L View Post
No it isn't. 2! is 2.

The answer is (1/2)^5.
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  #22  
Old 12-15-2007, 02:04 PM
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It's not 5! either. That's 120.
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  #23  
Old 12-15-2007, 02:23 PM
Emmerich's Avatar
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Quote:
Originally Posted by Matt L View Post
It's not 5! either. That's 120.
You are right. Again. I hate it when I know how to do the math but totally botch the description.
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  #24  
Old 12-15-2007, 06:38 PM
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Quote:
Originally Posted by kerry View Post
Isn't it 1/32? To get the odds of a series of independent events occurring together, you multiply the odds of each one by the others??
That's what I was thinking but I have a feeling there's another factor that's more tricky.
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  #25  
Old 12-15-2007, 06:42 PM
cmac2012's Avatar
Renaissances Dude
 
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Quote:
Originally Posted by Kuan View Post
Assuming the odds are exactly 1/2 and 1/2.

I was thinking there are only six combinations.

M F
5 0
4 1
3 2
2 3
1 4
5 0

Hence the odds are 1/6?
But the chances of having a 5 zip outcome are much smaller than having a 2 -3 outcome . . . . . I think.
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  #26  
Old 12-15-2007, 07:24 PM
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You're correct, Cmac. See my post #16.
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  #27  
Old 12-15-2007, 11:18 PM
cmac2012's Avatar
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Of course. Well done. It's sort of a bell curve thing. To reach the outside of the bell curve, every gate has to run to the outside whereas in the middle chutes, the balls can fall more flexibly, back and forth.
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  #28  
Old 12-16-2007, 05:52 AM
Jim B.'s Avatar
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Oh that one is easy

Well, the first thing you do, is ...

Hey WAIT!!!





BRB

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