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Please first have a look at the following little problem:

There are two indistinguishable light bulbs A and B. A flashes red light with prob .8 and blue with prob .2; B red with .2 and blue .8. Now with .5 prob you are presented with either A or B. You're supposed to observe its flash color to make a best guess (maximizing probability of correct guessing) which bulb it is. Before you start to make observations, however, you must decide how many times you want to observe it (say n times, then you observe it flashing n times and make your guess). Suppose flashes are independent.

Intuitively, one would think the more observations one makes, the better one's chances will be. Curiously though, it is easy calculation to show that n=2 doesn't improve upon n=1, and n=4 doesn't improve upon n=3. I didn't go further but I speculate n=2k doesn't improve upon n=2k-1. I'm not able to prove it for the general case. But is it true? If so, how can one intuitively understand the result?

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You are correct: $n=2k$ does not improve upon $n=2k-1$ in this symmetric case.

Clearly the optimal strategy is to look at the number of red and blue flashes and choose A or B according to which colour appears more. If the same number appear of each, it doesn't make any difference which you guess, as your chance of being correct is $0.5$ in that situation.

If there is a majority of one colour after $2k$ flashes then the majority must be even and at least 2, so that colour also had a majority of at least 1 after $2k-1$ flashes. If there is equality after $2k$ flashes, then choosing the colour with a majority after $2k-1$ flashes is as good as any other decision rule in this situation. So with an even number of flashes, the final flash does not help you improve your change of guessing correctly.

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  • $\begingroup$ @Henry:"If there is a majority of one colour after 2k flashes then the majority must be even and at least 2" I may have misunderstood your point, but why must it be even? For example if k=10 and red is observed 11 times and blue 9 times, where does evenness come from? $\endgroup$ – Eric Nov 25 '11 at 15:51
  • $\begingroup$ @Eric: $11-9=2$ which is an even number. If $a+b=2k$ then $a-b=2(k-b)$ which is even. $\endgroup$ – Henry Nov 25 '11 at 16:19
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To answer in a rigorous way, this problem boils down to observing the number of red flashes $X$ which is either a binomial $\mathcal{B}(n,.8)$ (A) or a binomial $\mathcal{B}(n,.8)$ (B), with probability $0.5$ for each. The probability of selecting bulb A is thus given by Bayes theorem $$ \mathbb{P}(b=A|X=x) = \dfrac{\mathbb{P}(X=x|b=A)}{\mathbb{P}(X=x|b=A)+\mathbb{P}(X=x|b=B)} $$ so this is $$ \mathbb{P}(b=A|X=x) = \dfrac{{n \choose x} 0.8^x 0.2^{n-x}}{{n \choose x} 0.8^x 0.2^{n-x}+{n \choose x} 0.2^x 0.8^{n-x}}=\dfrac{1}{1+4^{n-2x}} $$ Therefore A (resp. B) is chosen when $n-2x<0$ (resp. $n-2x>0$). Thus, when $n=2k-1$, the probability to correctly choose A is $$ \mathbb{P}(X>(2k-1)/2|b=A) = \mathbb{P}(X\ge k|b=A) =\sum_{x=k}^{2k-1} {2k-1 \choose x} 0.8^x 0.2^{2k-1-x}\,. $$

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  • $\begingroup$ :That's helpful. It's the same formula as here stats.stackexchange.com/questions/18975/… , only in slight different notations. But to complete this rigorous proof, you still have to show $n=2k$ and $n=2k-1$ entail same correct probability. $\endgroup$ – Eric Nov 26 '11 at 8:53
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    $\begingroup$ This is done in the answer to your other question, <a href="stats.stackexchange.com/questions/18975/… for a binomial equation</a> $\endgroup$ – Xi'an Nov 26 '11 at 10:15
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    $\begingroup$ Note also that this other question of yours only provides the very last formula, while my answer explains why we do reach this formula. $\endgroup$ – Xi'an Nov 26 '11 at 10:34

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