<s>[INST] Contradicting probabilities? 
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?
 [/INST] 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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