Probability Confusion/Clarification Two pizzas are each divided into eight slices and placed inside separate boxes with the
lids closed, and left in an empty room. People randomly arrive at the room to take a slice.
Each person who arrives randomly chooses a box, opens it, takes a slice of pizza and closes
the box. If all eight slices from a box are taken then the lid is not closed.
The first person to discover an empty box opens the other box. Calculate the probability
that it contains all eight slices.
The answer says $\left(\dfrac{1}{2}\right)^8$. This is what I intuitively thought, until my friend posed the solution $\left(\dfrac{1}{2}\right)^7$ as he is looking at the problem from the standpoint that all the others have to do is pick the same box as the first person, and it doesn't matter which box the first person chooses. I have a feeling that perhaps his situation is more conditional probability, but I'm struggling to get my head around what is correct! Can anyone clarify?
The difference between $\left(\frac{1}{2}\right)^8$ versus $\left(\frac{1}{2}\right)^7$ depends on how we interpret the problem.
Problem 1:

We have two pizzas, each with eight pizza slices.  Each time someone enters the room, they open one of the two boxes uniformly at random, independent of each other persons' choices, and will take a slice if able and then close the lid except in the case where they have taken the last slice in which case they throw out the box.  We ask what the probability is that the first person to come in the room after a box has been thrown out will find a full eight slices of pizza in the remaining box.

For this problem, let us label the two pizza boxes $H$ and $T$ and let the people make the decision which box to eat from based on a coin flip.  Here, we have the two sequences HHHHHHHH and TTTTTTTT both result in the ninth person coming in and seeing one of the boxes missing and the remaining box having all of the slices of pizza in it, with a probability of $\left(\frac{1}{2}\right)^8 + \left(\frac{1}{2}\right)^8 = \left(\frac{1}{2}\right)^7$
The calculation could have been made simpler as your friend had noted by not caring which of the two boxes was chosen by the first person and then for the next seven all of them choosing the same box, yielding the same $\left(\frac{1}{2}\right)^7$

Problem 2:

We have two pizzas, each with eight pizza slices.  Each time someone enters the room, they open of of the two boxes uniformly at random, independent of each other persons' choices, and will take a slice if able and then close the lid... even in the case that they closed the lid on a now empty box.  We ask what the probability is that the first time that someone comes into the room and opens an empty box (that they did not cause to be empty themselves), that the remaining box has all eight slices still in it.

For this problem, similarly to the last, we describe via a sequence of coin flips.  Here however, after the first box becomes empty, we require that the empty box be discovered before slices from the still full box be taken.  In this case we required nine heads in a row or nine tails in a row.  This occurs with probability $\left(\frac{1}{2}\right)^9 + \left(\frac{1}{2}\right)^9 = \left(\frac{1}{2}\right)^8$

Which of these was the intended problem?  That is up to the person who asked the problem, but I interpret it as the first as per the included phrase &quot;If all eight slices from a box are taken then the lid is not closed.&quot; which implies to me that no random event needs to occur for the ninth person in the event that the first eight people all ate from the same box since the contents of the empty box are plainly visible due to the lid remaining open.
As for why the linked Matchbox Problem gives the other answer, that is because it follows the style of problem 2 where we do not remove the newly emptied box once it becomes empty and we leave it until we attempt to retrieve a match from it and discover there are none to be had.
The only possible way for a person to discover an empty box and a full one, is if  eight slices have been taken (not more). The second condition is, that every slice was taken from the same box. This adds up to your result $(\frac{1}{2})^8$ since every person must take a slice from the same box.