If you have $8$ shells, and $2$ of them have coins, what is the probability of guessing which have coins if you have $4$ guesses? 
If you have $8$ shells, and $2$ of them have coins underneath them (you do not know which shells), what is the probability of guessing both the shells with coins underneath if you have $4$ guesses?

I got a little stumped on this one. It's the guesses that really throw me off. I originally thought 
$$\frac{2}{8}\cdot\frac27\cdot\frac26\cdot\frac15$$ would work, but that doesn't make sense to me because by that logic, if you had two guesses the probability would be 
$$\frac28\cdot\frac27.$$ Which would mean the probability with $4$ guesses is lower than the probability with $2$.
 A: There are 6 ways in which your two correct guesses can be arranged within your four guesses (eg first, second or first, third etc). For the first of these two guesses you have 2 ways to be correct and for the second just one. For the first of the incorrect guesses there are 6 possibilities and for the second 5. So in total you have $6\cdot2\cdot1\cdot6\cdot5=360$ ways out of a total of $8\cdot7\cdot6\cdot5=1680$. So the probability is $\frac{3}{14}$.
A: There are $\binom84$ ways to select $4$ shells out of $8$.
There are $\binom62$ ways to select $4$ shells out of $8$ under the extra condition that $2$ specific shells (those with coins underneath) are selected. You could interpreted this as selecting $2$ shells out of the $6$ shells that have no coin underneath.
So the probability of this event is: $$\frac{\binom62}{\binom84}=\frac3{14}$$
A: If you are familiar with the hypergeometric distribution, then you can think about it as follows.
Our scenario is like having $2$ two gold balls (which represent the coins), and $6$ silver balls (which represent empty shells) in a box. Notice that choosing four shells is like drawing from the box without replacement four times. There are two "good" balls, and six "bad" ones. There are $\binom22$ ways to choose the good ones, and we are forced to make two more choices. Hence there are $\binom{6}{2}$ ways to choose two bad ones. There are $\binom84$ ways to choose from the total. So, by the hypergeometric distribution, the probability is 
$$\frac{\binom{2}{2}\binom62}{\binom84} = \frac{3}{14}.$$
A: Look at it from the idea of correctly placing the two coins in the shells that you guessed. There are 8 shells. Four of them are guessed shells. Randomly place one of the coins under a shell.  You have a 4/8 probability of it being under a guessed shell. Assuming you were lucky, there are now 7 shells you can place the coin under and 3 that are guessed shells. to give you a 3/7 probability of placing the coin under a guessed shell. So the probability of both coins being under guessed shells is 
$$
\frac{4}{8}*\frac{3}{7} = \frac{3}{14}
$$
A: There are six pairs within the four you guessed, out of 28 pairs altogether,  So your chance of winning is $$\frac{4\choose2}{8\choose2}=\frac6{28}=\frac3{14}$$
