In the latest episode of Survivor there were 6 people and they had to draw rocks from a bag. In the bag were 6 rocks, 5 white rocks and 1 black rock. First person drew a rock and kept it in his hand. Then the second person drew a rock. Last person took the only rock that left.
Then they all showed what they got at the same time.
Was the possibility to draw a black rock for each person 1/6 or what was it?
Yes.
For the first person it's obviously $1/6$.
For the second it is either $1/5$ if the first person drew a white ($5/6$), or $0$ if the first person drew black already ($1/6$). So $p($Person 2 draws black$)=5/6 \times 1/5+1/6 \times 0=1/6$.
Logic continues and gives $1/6$ for each person.
The answer is yes.
First point of view is they can all draw the rocks at the same time without changing the outcome.
Another way of seeing this is to compute the probability that the $k$th person picks the black stone. The $k-1$ persons before must have picked a white stone, and the $k$th picks the black. The probability is $$\frac56\times\frac45\times\dots\times \frac{6-(k-1)}{6-(k-1)+1}\times \frac{1}{6-(k-1)} = \frac16$$ (telescopic product).
Yes, the probability of drawing the black rock is 1/6 for each person. This is obvious for the first person. Intuitively "by symmetry," it should be clear that everyone has the same probability 1/6 of drawing the black rock.
One more formal way to look at it is that there are $6! = 720$ orders
in which the six rocks can be drawn. Let's number the rocks 1 through 6,
with #1 being the black rock. Some examples among the 720 ways are
123456, 321456, 135246, 654132, and so on.
Suppose you are third in line to draw. Then you would get the black
rock for outcomes such as 321456 and 541326.
How many ways are there to arrange the rocks so that the black rock is third? Put the black rock in the 3rd position, and then there are 5! ways to arrange the other five rocks. So the number of ways for you to get the rock are 5! = 120.
Then the probability you get the black rock is $5!/6! = 120/720 = 1/6.$
Addendum: While I have been typing this and taking holiday phone calls, I see that two other answers have appeared. I have up-voted them both, as being reasonable alternative methods to show the same thing.
