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  #1  
Old 02-09-2007, 05:23 PM
jay_shark jay_shark is offline
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Default A bag full of jelly beans

Here is a question that seems interesting .

A bag contains a white and b black jelly beans. Jelly beans are pulled out of the bag according to the following method :

1) A jelly bean is chosen at random and is discarded.
2) A second jelly bean is then chosen . If its color is different from that of the preceding jelly bean , it is replaced in the bag and the process is repeated from the beginning . If the color is the same , it is discarded, and we start from step 2.

In other words , jelly beans are sampled and discarded until a change in color occurs, at which point the last jelly bean is returned to the bag and the process starts anew . What is the probability that the last jelly bean in the bag is white ?
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  #2  
Old 02-09-2007, 06:20 PM
Tom1975 Tom1975 is offline
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Default Re: A bag full of jelly beans

I'll guess a/(a+b)
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  #3  
Old 02-09-2007, 07:05 PM
jay_shark jay_shark is offline
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Default Re: A bag full of jelly beans

Interesting enough , the answer is 1/2 .
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  #4  
Old 02-09-2007, 09:54 PM
PairTheBoard PairTheBoard is offline
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Default Re: A bag full of jelly beans

[ QUOTE ]
Interesting enough , the answer is 1/2 .

[/ QUOTE ]

I can see how this might be true. As usual I am puzzled to see the nifty way to show it.

PairTheBoard
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  #5  
Old 02-11-2007, 03:48 AM
Siegmund Siegmund is offline
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Default Re: A bag full of jelly beans

Any time there are two jelly beans of the same colour in the bag, there's a possibility (increasingly likely as fewer of the other colour are present) that one of them will be removed.

However, the last jelly bean of either colour can never be removed. So the process will continue until the bag as one jelly bean of each colour remaining in it.
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  #6  
Old 02-11-2007, 11:43 AM
PairTheBoard PairTheBoard is offline
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Default Re: A bag full of jelly beans

[ QUOTE ]
However, the last jelly bean of either colour can never be removed. So the process will continue until the bag as one jelly bean of each colour remaining in it.

[/ QUOTE ]

I don't think so. You could have a bag with 4 white, 3 black beans. If the first 4 beans you pick happen to be white you will have removed all white beans and be left with nothing but 3 black beans.

PairTheBoard
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  #7  
Old 02-11-2007, 01:29 PM
AaronBrown AaronBrown is offline
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Default Re: A bag full of jelly beans

We need to know that a > 0 and b > 0.

Consider the event that you pull all the white beans out of the bag in one sequence. This will happen 1 time in C(a+b,a). That is equal to C(a+b,b), the chance of pulling all black beans out of the bag in one sequence.

Let P(a,b) be the probability the last bean in the bag is white, if there are a white and b black beans in the bag before a sequence of moves. Assume that P(a,b) is not equal to 1/2. The chance that the bag is all white after one sequence is equal to the chance that the bag is all black. After the sequence of moves, there will be fewer than a+b beans in the bag. Since P(a,b) is not equal to 1/2, there must be at least some P(a',b') not equal to 1/2, where 0 < a' <= a, 0 < b' <=b and a' + b' < a + b.

Repeating this logic, we discover that P(1,1) must not be equal to 1/2. Since this is false, the initial assumption that P(a,b) is not equal to 1/2 must be false.
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  #8  
Old 02-11-2007, 05:51 PM
PairTheBoard PairTheBoard is offline
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Default Re: A bag full of jelly beans

C(a+b,a) = C(a+b,b)

cool


PairTheBoard
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  #9  
Old 02-12-2007, 02:39 AM
DWarrior DWarrior is offline
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Default Re: A bag full of jelly beans

[ QUOTE ]
We need to know that a > 0 and b > 0.

Consider the event that you pull all the white beans out of the bag in one sequence. This will happen 1 time in C(a+b,a). That is equal to C(a+b,b), the chance of pulling all black beans out of the bag in one sequence.

Let P(a,b) be the probability the last bean in the bag is white, if there are a white and b black beans in the bag before a sequence of moves. Assume that P(a,b) is not equal to 1/2. The chance that the bag is all white after one sequence is equal to the chance that the bag is all black. After the sequence of moves, there will be fewer than a+b beans in the bag. Since P(a,b) is not equal to 1/2, there must be at least some P(a',b') not equal to 1/2, where 0 < a' <= a, 0 < b' <=b and a' + b' < a + b.

Repeating this logic, we discover that P(1,1) must not be equal to 1/2. Since this is false, the initial assumption that P(a,b) is not equal to 1/2 must be false.

[/ QUOTE ]

That's so baller.

What Stats course teaches that kind of stuff? I'm in Stat 2 now, but I don't think we go past distributions. Stat 1 only teaches the very basic stats and dips into distributions.

I've never heard of the Baye's Theorem until I came to 2p2.
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  #10  
Old 02-12-2007, 08:21 AM
jason1990 jason1990 is offline
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Default Re: A bag full of jelly beans

[ QUOTE ]
I've never heard of the Baye's Theorem until I came to 2p2.

[/ QUOTE ]
So you learned Bayes Theorem from 2p2? I'm just curious. What do you think Bayes Theorem is?
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