Solomon Wisdom Probability Problem


I saw this problem in a Probability course offered by the Harvard Extension program. I highly recommend it if you have the time. Here’s the problem:

In Solomon’s reign there were two types of prophets: true prophets, who spoke the truth 9 out of 10 times, and false prophets, who only spoke the truth 5 out of 10 times. Solomon was the only prophet who always spoke the truth.

One day the queen of Sheba arrived at Jerusalem bringing spices, gold and precious stones. She told Solomon that she would give him all those items in exchange for a true prophet.

Solomon then asks for a servant to go fetch three prophets. The servant returns after a while and says: “Here are three prophets. Two of those are true prophets, and one is a false prophet, but I forgot who.”

At this point the queen of Sheba says: “Solomon, if you give me one of those prophets I’ll have a 1-in-3 chance of getting a false prophet, and that’s not acceptable to me. A chance of 1-in-10 would be acceptable.”

Solomon then walks to prophet 1 and asks: “Is prophet number 2 a true prophet?”. Prophet 1 replies, “Yes, he’s a true prophet.”

Solomon then holds prophet number 2 by the hand and says “There you go my queen, the chance of this prophet being a false prophet is even lower than 1-in-10, so take him without worries.”

Question: What’s going on, and what’s the probability that prophet number 2 is a false prophet?

Solution

First of all let’s outline the three possible arrangements of prophets. Let’s say that each arrangement is an event, so:

     Prophet 1     Prophet 2      Prophet 3
A1     True         False           True
A2     True         True            False
A3     False        True            True

Now let’s define what we are trying to find. We want to discover the probability of prophet 2 being a false prophet GIVEN that prophet 1 says he’s a true prophet. Therefore we have the following events:

Event A1: Prophet 2 is a false prophet
Event B: Prophet 1 says prophet 2 is a true prophet

Therefore we want to find the probability of A1 given B, or P(A1|B). Using the conditional probability rule we know that:

P(A1|B) = P(A1∩B) / P(B)

Each arrangement has a probability of 1/3, so we can say that:

P(A1) = 1/3

That is, the probability of prophet 2 being a false prophect is 1/3.

Now we need to find P(B), which is the probability of prophet 1 says prophet 2 is a true prophet. Event B will happen in A1 if prophet 1 tells a lie, so the probability is 1/3 * 1/10. Event B will happen in A2 if prophet 1 tells a truth, so 1/3 * 9/10. Finally event B will happen in A3 if prophet 1 tells a lie, and the probability here is 1/3 * 5/10. So:

P(B) = 1/3 * 1/10 + 1/3 * 9/10 + 1/3 * 5/10 = 15/30 = 1/2

Now to find P(A1∩B) we need to analyze the problem. What’s the probability that A1 and B will happen at the same time? That is, prophet 2 will be a false prophet AND prophet 1 will say he’s a true prophet? The probability is:

P(A1∩B) = 1/3 * 1/10 = 1/30

Notice that P(A1∩B) is not equal to P(A1)*P(B) because those events are not independent. Another way to find P(A1∩B) is this:

P(B|A1) = 1/10

That is, the probability that prophet 1 will say prophet 2 is a true prophet GIVEN prophet 2 is in reality a false prophet is 1/10 (i.e., prophet 1 would need to tell a lie). But we also know that:

P(B|A1) = P(A1∩B) / P(A1)

so

1/10 = P(A1∩B) / 1/3

so

P(A1∩B) = 1/3 * 1/10 = 1/30

Now we can plug P(A1∩B) and P(B) into our conditional probability rule to get the final result:

P(A|B) = P(A1∩B) / P(B) = 1/3 * 1/10 / 1/2 = 1/30 / 1/2 = 1/15

Therefore the probability that prophet 2 is a false prophet GIVEN prophet 1 said he’s a true prophet is 1-in-15, which is therefore acceptable by the queen. Notice that if prophet 1 had said prophet 2 was a false prophet Solomon would just have grabbed prophet number 3 by the hand and delivered him to the queen with the same probability of being a true prophet (i.e., 14/15).

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