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This Problem is already asked but not answered properly...

Grimmet Book Problem

i believe there must be a tricky short answer for it, but even though i couldn't solve it through induction or recursive formula...

Can someBody Help Me To find either short or long answer for it?

How do I solve this probability problem of randomly drawing balls from a urn?


it could be a miss Understanding about this problem: notice that the problem said "Restarted" it means if after a run of Azure You pick a Carmine, if the next ball was an Azure it will be acceptable. i think the answer in the above link is wrong because of this miss understanding

Thanks...

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    $\begingroup$ The other question may or may not have been intended to pose the problem you're posing, but that doesn't make the answer wrong. The answer is an answer to the problem posed, not to the problem you think may have been intended. $\endgroup$ – joriki Jul 2 '16 at 11:40
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The probability is clearly $\frac12$ by symmetry if the number of azure and carmine balls is equal, and in particular if there are $2$ balls, one azure and one carmine

Suppose as an inductive hypothesis that the probability that the last ball drawn is azure is $\frac{1}{2}$ if you start with $n-1$ or fewer balls not all the same colour

Then if you start with $n=a+c$ balls, you might

  • draw all the carmine balls initially, so the last ball drawn will be azure, and the probability this happens is $\dfrac{1}{{n \choose c}}$
  • draw all the azure balls initially, so the last ball drawn will be carmine, and the probability this happens is $\dfrac{1}{{n \choose a}}$
  • some other case, in which case you restart with fewer balls not all the same colour, and we know from the inductive hypothesis this gives a probability of $\frac12$, and the probability this happens is $1-\dfrac{1}{{n \choose c}} -\dfrac{1}{{n \choose a}}$

So the probability that the last ball drawn is azure is $$1 \times \dfrac{1}{{n \choose c}} + 0 \times \dfrac{1}{{n \choose a}} + \dfrac{1}{2} \times \left(1-\dfrac{1}{{n \choose c}} -\dfrac{1}{{n \choose a}}\right) = \dfrac12$$ since ${n \choose a}={n \choose c}$

This means that by strong induction the assertion is true

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    $\begingroup$ Oh My God... Tnx $\endgroup$ – MR_BD Jul 2 '16 at 13:16
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    $\begingroup$ Very nice solution. $\endgroup$ – drhab Jul 2 '16 at 13:36
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    $\begingroup$ The "quick answer" may be that, given that a particular round of draws is the second last round, that round must then draw all the carmines or all the azures, and those events are conditionally equally likely no matter what the pattern of balls is going into that second last round $\endgroup$ – Henry Jul 2 '16 at 14:59

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