Wackerly/Mendenhall/Schaeffer Problem 2.14: Perspective's Role in Probability

[Ackbach]

Problem 2.14. According to Webster's New Collegiate Dictionary, a dividing rod is "a forked rod believed to indicate [divine] the presence of water or minerals by dipping downward when held over a vein." To test the claims of a divining rod expert, skeptics bury four cans in the ground, two empty and two filled with water. The expert is led to each of the four cans and told that two contain water. He uses the divining rod to test each of the four cans and decide which two contain water.

• List the sample space for this experiment.
• If the divining rod is completely useless for locating water, what is the probability that the expert will correctly identify (by guessing) both of the cans containing water?

Answer.

• For each can, there are four possibilities:
  1. $E_1$: the can has water and the expert says it has water (WW),
  2. $E_2$: the can has water and the expert says it does not have water (WN),
  3. $E_3$: the can does not have water and the expert says it has water (NW),
  4. $E_4$: the can does not have water and the expert says it does not have water (NN).
  Theoretically, there are $4^{4}$ possible outcomes, since there are four possibilities for each can and there are four cans. However, there are many possibilities we can rule out, since many of them would have three cans with water, e.g. So, we could have:
  \begin{align*}
  &WW, WW, NW, NW \\
  &WW, WW, NW, NN \\
  &WW, WW, NN, NW \\
  &WW, WW, NN, NN \\
  
  &WW, WN, NW, NW \\
  &WW, WN, NW, NN \\
  &WW, WN, NN, NW \\
  &WW, WN, NN, NN \\
  
  &WN, WW, NW, NW \\
  &WN, WW, NW, NN \\
  &WN, WW, NN, NW \\
  &WN, WW, NN, NN \\
  
  &WN, WN, NW, NW \\
  &WN, WN, NW, NN \\
  &WN, WN, NN, NW \\
  &WN, WN, NN, NN.
  \end{align*}
  
  Re-arranging the actual cans would not change the structure of the probabilities, so we can assume, WLOG, that this is the sample space.
  
• For the first water can, there are two possibilities: $E_{1}$ and $E_{2}$. So the probability is $1/2$ that the diviner will correctly identify it. The same goes for the other can: a $1/2$ chance. Hence, the probability that the diviner will correctly identify both water cans as containing water is $(1/2)(1/2)=1/4$. This we can verify by looking at the table: there are four rows in which the diviner correctly identifies the two water cans, out of sixteen rows total. Hence, the probability is $1/4$.

My question is this: does the perspective matter? That is, you could look at this problem from the point of view of the diviner (Does he know how many cans have water and how many don't? Presumably not, if it's a good test.) Or you could view it from the perspective of the test designer, who knows there are two water cans and two cans without water. Does it make a difference in the probability? Or even in the sample space? Surely it makes a difference in the sample space: the test designer can reduce what he looks at, because of the reasoning I put forth before. But the diviner has to consider quite a few more possibilities: if the cans are "indistinguishable", then there could be anywhere from zero to 4 water cans. For each of those five possibilities, the diviner could be guessing water or no water.

 

[awkward]

Hi Ackbach,

I think you're making this too hard.
For (a), the sample space is simply all the subsets of size two taken from the collection of four buckets. We assume these are all equally likely if the divining rod is useless.
For (b), how many of these subsets contain the two buckets with water?

 

[Ackbach]

Hi Ackbach,

I think you're making this too hard.
For (a), the sample space is simply all the subsets of size two taken from the collection of four buckets. We assume these are all equally likely if the divining rod is useless.
For (b), how many of these subsets contain the two buckets with water?

Ah, it would also help if I read the problem correctly. The diviner is told that there are two cans with water (and presumably that there are two without).

Thanks much!