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There are $B$ blue balls and $R$ red balls in a bag.
Now, randomly two balls are removed .
If both of them are of different color , then blue ball is added to the bag .
If both of them are of same color , then red ball is added to the bag .

Also the two drawn balls are discarded.

This is done until one ball is left.

What is the probability that the last ball is blue?

Now, I could not even start a solution on this , so could anyone provide help in solving this ?

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marked as duplicate by Marc van Leeuwen, TZakrevskiy, Daniel Rust, O.L., William Aug 18 '13 at 10:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Have you tried performing this procedure with e.g 2,3 or 4 balls of arbitrary color? In that case you can quiet easily compute the probabilities depending on the number of initially given red and blue balls. Can you generalize this to more balls? –  Quickbeam2k1 Aug 18 '13 at 7:42
    
Yes, the main problem I am having is generalizing it. –  Mod Aug 18 '13 at 7:52

1 Answer 1

up vote 3 down vote accepted

Claim: The parity of the number of blue balls is invariant.

So if $B$ is initially odd, then it will always be odd, so it must be the last ball to be drawn. Otherwise, if $B$ is initially even, then it will always be even, so it will never be the last ball to be drawn.


To see this, just consider each possibility.

Case 1: Suppose that both selected balls are blue. Then both will be removed and one red ball will be added, leaving us with $B-2$ blue balls and $R+1$ red balls. Subtracting $2$ from $B$ did not change its parity.

Case 2: Suppose that both selected balls are red. Then both will be removed and one red ball will be added, leaving us with $B$ blue balls and $R-1$ red balls. The parity of $B$ did not change.

Case 3: Suppose that one of the selected balls is blue and the other is red. Then both will be removed and one blue ball will be added, leaving us with $B$ blue balls and $R-1$ red balls. The parity of $B$ did not change.

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This is because either 2 or 0 blue balls are removed by any move ? Ok understood :) –  Mod Aug 18 '13 at 7:54
    
Yep, exactly. =] –  Adriano Aug 18 '13 at 7:57
    
So probability will always be 1 or 0 , right ? –  Mod Aug 18 '13 at 8:00
    
Yes, depending on the initial parity of $B$. –  Adriano Aug 18 '13 at 8:00

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