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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.2012-08-17
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    Are the red balls perhaps bigger?2012-08-17
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    After you pick one ball you know which jar has the red balls so keep picking from that jar.2012-08-17
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    @sTEAK. Do you mean that after choosing some number of balls, more than 50% of them should be red?2012-08-17
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    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?2012-08-17
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    @i.m.soloveichik, "You are blindfolded"2012-08-17
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    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!2012-08-17

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

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    how to ensure >1/2 probability2012-08-17
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    @sTEAK. "guaranteed" means the probability is $1$, and $1 > \frac{1}{2}$.2012-08-17
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    @Sasha Getting red from 2 jars is 1/2 .2012-08-17
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    @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.2012-08-17
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    @sTEAK. If you take one from each jar then the probability of getting a red is $1$.2012-08-17
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    I think that the OP means to pick many times and have a majority of red balls, although that is clear not what is stated. The question is gettinga lot of tongue in cheek answers because if you don't get feedback about the previous balls picked then you never have any information as to which jar has the red balls. So a strategy of systematically selecting alternately from the jars guarantees that in an even number of draws you will have exactly half red balls. Random selection of jars would lead to a long range proportion close to 1/2.2012-08-17
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    There is no strategy that you can use to guarantee more than 1/2 are red.2012-08-17
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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.2013-05-16
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    But you don’t know which jar has the red balls, so you can’t perform the desired rearrangement of the balls.2013-05-16