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Lets say I have 8 objects. From those 8 objects lets say, 2 of those objects are "optimal". I am allowed to pick an item at random- my objective is to pick atleast one of the "optimal objects". I have four tries/attemps to pick one. What's the probability of you picking atleast one "optimal" object from those 4 tries? Note that picking an item doesn't remove it from the choices the next try - the number of total objects to pick from remains the same. Picking an object does not remove it.

I approached this seemingly very simple problem in two ways, but each way gives a different probability:

  • Method 1: There's a $\frac{2}{8}$ chance of you picking any "optimal" object at each try. Repeating this $4$ times gives the probability: $P =\frac{2}{8} * 4 = 1$

  • Method 2: There's are two possibilities of outcome: either I pick an optimal object by the end or I don't. Although these outcomes obviously don't have equal probability, the existence of the possibility of not picking an "optimal" object and the fact that's the probability of not picking an optimal object isn't zero, then it would be impossible for the probability of picking an optimal object to be 1! Both outcomes can't be true and since the probability of not picking an optimal object is greater than 0, then the probability of picking an object has to be less than 1, which is a contradiction with Method 1.

What's the fallacy in my logic?

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I downvoted this question because it was posed as a paradox whenin fact the inconsistency was due to a gross error in calculating the probability. –  Michael Chernick Jul 18 '12 at 4:54

4 Answers 4

You have double counted the cases where you pick an optimal one twice, triple counted the ones where you pick an optimal one three times, and counted four times the 16 cases you pick an optimal one all four times. A simpler problem would be the chance of getting one head in two tries. The four possibilities are HH, HT, TH, TT, giving a 3/4 chance. Your calculation would give 1 the same way because it double counts HH. You are right that the chance of picking an optimal one never equals 1, no matter how many trials you make. What you have calculated is the expected number of optimal objects after four tries, which is in fact 1. Similarly, the expected number of heads in two tries is 1, but that doesn't mean you are certain to get at least one head.

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The probability that you pick an optimal object at least once is equal to 1 minus the probability that you never pick an optimal object on any of your 4 tries.

Thus, the probability that you pick an optimal object at least once is $$1-\left(\frac{6}{8}\times\frac{6}{8}\times\frac{6}{8}\times\frac{6}{8}\right)=1-\frac{81}{256}=\frac{175}{256}$$

(Apologies - earlier editions of my answer contained terribly silly arithmetic errors.)


You can also see that this is the correct answer through a more tedious approach:

On the first try, you have a $\frac{2}{8}$ probability of picking an optimal object. If you do pick an optimal object, the outcomes of your other three tries are irrelevant; if you do not pick an optimal object, you must press on. Thus, the probability of picking an optimal object on at least one of 4 tries is equal to $$\frac{2}{8}\left(1\right)+\frac{6}{8}\left(\small{\text{probability of picking an optimal object}\atop\text{ on at least one of the 3 remaining tries}}\right)$$ Repeating this reasoning, we see that this is equal to $$\frac{2}{8}+\frac{6}{8}\left(\frac{2}{8}+\frac{6}{8}\left(\small{\text{probability of picking an optimal object}\atop\text{ on at least one of the 2 remaining tries}}\right)\right)$$ $$=\frac{2}{8}+\frac{6}{8}\left(\frac{2}{8}+\frac{6}{8}\left(\frac{2}{8}+\frac{6}{8}\left(\small{\text{probability of picking an optimal object}\atop\text{ on the single remaining try}}\right)\right)\right)$$ $$\begin{align}&=\frac{2}{8}+\frac{6}{8}\left(\frac{2}{8}+\frac{6}{8}\left(\frac{2}{8}+\frac{6}{8}\left(\frac{2}{8}\right)\right)\right)\\ &=\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}\right)\right)\right)\\ &=\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}+\frac{3}{16}\right)\right)\\ &=\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{7}{16}\right)\right)\\ &=\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}+\frac{21}{64}\right)\\ &=\frac{1}{4}+\frac{3}{4}\left(\frac{37}{64}\right)\\ &=\frac{1}{4}+\frac{111}{256}\\ &=\frac{175}{256}\end{align}$$


The reasoning in your first method can be seen to be fallacious as follows: first of all, probability cannot work the way you use it there, because it would mean that taking 5 tries would give you a 125% probability of choosing an optimal object, which makes no sense. You can also consider the simpler case of two tries: $$\begin{array}{c|c|c|} & \text{first try success} & \text{first try failure} \\\hline {\text{second try}\atop\text{success}} &\frac{2}{8}\times\frac{2}{8} & \frac{6}{8}\times\frac{2}{8}\\\hline {\text{second try}\atop\text{failure}} & \frac{2}{8}\times\frac{6}{8} & \frac{6}{8}\times\frac{6}{8}\\\hline \end{array}$$ You have picked an optimal object at least once in all situations except the one in the bottom right square, so the probability you have picked an optimal object after two tries is $$\left(\frac{2}{8}\times\frac{2}{8}\right)+\left(\frac{6}{8}\times\frac{2}{8}\right)+\left(\frac{2}{8}\times\frac{6}{8}\right)=1-\left(\frac{6}{8}\times\frac{6}{8}\right)=1-\frac{9}{16}=\frac{7}{16}$$ and not $$\frac{2}{8}+\frac{2}{8}=\frac{1}{2}$$ Your "method 2" is not really a method, as it is not producing an answer, but your reasoning there is correct, in that the probability of picking an optimal object can never be one after any finite number of tries because the probability of not picking one is positive and independent of the number of trials or previous outcomes.

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I see. What exactly is wrong with my listed methods though? –  user26649 Jul 18 '12 at 2:52
    
@Riddler: I've updated my answer. –  Zev Chonoles Jul 18 '12 at 3:16

Zev is correct but he didn't explain it or tell you why you are wrong. The first answer treats independent events as though their probability sums. This confuses the probability law for disjoint events with the rule for independent events. In fact they multiple.That is why 1-p where p is the probability of getting none is the probability of getting at least 1 and p=(3/4)$^4$ becuase each time you fail with probability 3/4 and the independent probabilities multiply.

The reasoning in the second case is of ocurse correct but only qualitative. It does not tell you how much less than one the probability is.

Zev did something clever by computing 1-p and getting the result in an indirect way. Looking at the complementary event is simpler because it only occurs one way and give a simpler calculation than the direct approach. The direct approach would be to compute and sum the four disjoint outcomes that result in at least one optimal object picked.

You can get exactly 1, exactly 2, exactly 3 or exactly 4. For 4 it can happen only one wayand the probability of 4 is (1/4)$^4$. You can get 3 with probability (1/4)$^3$ (3/4) for each sequence but there are 4 ways that the draw where the non-optimal object occurs. for 2 it is (1/4)$^2$ (3/4)$^2$ and there are 6 ways this can happen. This is a binomial distribution and we are determining the binomial probability of k successes for k=1 to 4 where success is defined as the probability of drawing an optimal object on any particular draw.

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As I worked out the tedious approach Zev edited his answer to inclue it. –  Michael Chernick Jul 18 '12 at 3:17

1-P(not picking an object). This is $1-(6/8)$ $^4$ since 6/8 is the probability of a non optimal object being chosen on any draw.

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You have a LaTex error because part of your answer is invisible to us. –  Michael Chernick Jul 18 '12 at 3:18
    
@MichaelChernick: Though a LaTeX error can cause things to appear to be missing, I'm not seeing anything further in Andrew's answer (you should be able to see an "edit" button that will let you look at the contents of any post). –  Zev Chonoles Jul 18 '12 at 3:32
    
@ZevChonoles Regardless he should fix his post even though I could guess at how to fix it if I know what teh LaTex error is. –  Michael Chernick Jul 18 '12 at 3:45
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@MichaelChernick Out of curiosity, are you actually getting a LaTeX error appearing in your browser? Because the equation shows up fine for me (even though Andrew separated 1-(6/8) from ^4 for some inscrutable reason). | Edit: Wait, it seems you're the one that separated the two for some odd reason. Now I'm even more curious. –  anon Jul 18 '12 at 3:53
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@Andrew: I don't understand. To get a 4 in an exponent, one writes $(6/8)^4$ with only two dollar signs, as Andrew has it. If this is not what you see, then somehow the error is on your side. –  anon Jul 18 '12 at 3:57

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