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You are blindfolded and placed in front a table with two jars. One jar has $50$ red balls and other has $50$ blue balls.

What should be your strategy so that you pick up the red ball with more than $50\%$ probability.

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Bribe the person running the experiment to tell you which jar has the red balls. – Gerry Myerson Aug 17 '12 at 12:30
Are the red balls perhaps bigger? – clark Aug 17 '12 at 12:30
After you pick one ball you know which jar has the red balls so keep picking from that jar. – i. m. soloveichik Aug 17 '12 at 12:43
Your description of the experiment is incomplete. How many balls do you draw? Which feedback do you get? How exactly is the probability which shall be optimized defined? Over the balls drawn in the experiment? Or for the last drawn ball over repetitions of the experiment? Or maybe it's the probability that you've got at least one ball in each run of the experiment? – celtschk Aug 17 '12 at 12:56
Since the rules seem to be incompletely defined (nothing is said about what you can and cannot do after being placed in front of the table, as @celtschk notes), my strategy would be to remove the blindfold! – Henning Makholm Aug 17 '12 at 13:48
up vote 3 down vote accepted

Take one ball from each jar. You're guaranteed to get a red.

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how to ensure >1/2 probability – h4ck3d Aug 17 '12 at 12:35
@sTEAK. "guaranteed" means the probability is $1$, and $1 > \frac{1}{2}$. – Sasha Aug 17 '12 at 12:36
@Sasha Getting red from 2 jars is 1/2 . – h4ck3d Aug 17 '12 at 12:55
@sTEAK.: If you get one ball from each jar, then you obviously got one black and one red ball, unconditionally (that is, the probability of getting one black and one red ball is $1$). Of course if you got both a black and a red ball, then you got a red ball. So the probability of getting a red ball is $1$. That's at least one way to complete your incomplete description. – celtschk Aug 17 '12 at 13:15
@sTEAK. If you take one from each jar then the probability of getting a red is $1$. – T. Eskin Aug 17 '12 at 13:17

If the red and blue balls are identical in shape and size and picking one ball from two jars which are also identical should result in the probability of getting a red ball as .5. Now if the experiment is biased in some way, then the probability of getting a red ball may be greatr than .5.

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I don't think moving the balls makes any difference.

The question is what is your strategy to improve from 50% of the balls. Not 50% out of one jar.

Ultimately there are always 100 balls in front of you and you have a 50/50 chance. The only way to change that is to take two balls out, or throw away one balls so that N changes.

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take 49 red balls from one jar and put them in the other jar wit the blue balls,

now its 100 % he will pick red from the first jar, and 50% red from the mixed jar.

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Not quite $50\%$ for the second jar, as it is $\frac {49}{99}$, but the approach does guarantee more than $50\%$ overall as requested. – Ross Millikan May 16 '13 at 14:27
But you don’t know which jar has the red balls, so you can’t perform the desired rearrangement of the balls. – Brian M. Scott May 16 '13 at 16:46

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