## Uncertain Urns Puzzle

A familiar probability example, using urns, is adapted to illustrate ‘true’ (non-numeric) uncertainty.

## Simple situation

The following is a good teaching example:

Suppose that an urn is known to contain black and white balls that are otherwise identical. A subject claims to be able to predict the colour of a ball that they draw ‘at random’.

They ‘predict’ and draw a black ball. What are the odds that they are really able to predict?

From a Bayesian perspective, the final odds are the initial odds times the likelihood ratio. If there are b black and w balls and we represent the evidence by E and likelihoods by P( E | ), then P( E | Predict ) = 1 and P( E | Luck) = b/(b+w). Thus the rarer the phenomenon predicted, the more a correct prediction tends to support the claim, of reliable prediction.

## Common quibbles

There is, however, some subjectivity in the estimated probability that the subject can predict:

• In this case, the initial odds seem somewhat arbitrary, and Bayes’ rule seems not to apply. For example, have you considered that the different colours may result in different temperatures? Such a thought is not ‘evidence’ in the sense of Bayes’ rule, but might change your subjective estimate of the probability prior to their draw.
• If we do not know the proportions of black and white balls for sure then the likelihood is uncertain.

## Multiple urns

Here we introduce a different type of uncertainty:

Suppose now that the subject is faced with two urns and selects a ball from one. Given the number of black and white balls in each urn, what is the likelihood, P( E | Luck ), of a correct prediction due to luck?

If you think the question is ambiguous, please disambiguate it however you wish.

Suppose you know the total numbers of black and white balls in the two urns. Is the likelihood estimate P( E | Luck) = b/(b+w) reasonable? Could it be biased? How?

Dave Marsay

## Uncertainty Puzzle

Here’s a puzzle from understanding uncertainty:

“Multiple Choice: If you choose an answer to this question at random, what is the chance you will be correct? A) 25% B) 50% C) 60% D) 25%”

This suggests a variant:

“Multiple Choice: If you choose an answer to this question at random, what is the chance you will be correct? A’) 25% B’) 50% C’) 0% D’) 25%”

0% for the first, ‘none’ for the second. These aren’t in the list of options supplied, but (at least for British school exams) this is not a novelty. (It would be  a ‘paradox’ , as claimed in the original source, if there were some law of logic that said that one of the answers provided is true: but there isn’t. It is only normally a mistake.)

My reasoning is …

For the first is to consider A…D in turn and to show that each is wrong:

• If A were correct, then so would D be, and hence the probability of the correct answer being selected is 50%. Thus  A and D are both wrong.
• If B were correct, then the probability of the correct answer being selected is 25%. Thus B is wrong. Similarly for C.

For the second, I claim that the answer is ‘None’. Again I consider each possible answer in turn and show that each is wrong.

• If C’ is correct then the probability of the correct answer being selected is 25%. Thus C’ is wrong.
• Since the answers are all wrong, if the probability of the correct answer being selected is a number, then it is 0%, and so C’ must be true. But C’ is not true, so the required probability is not a number.

To put it another way, ‘chance’ in the question is meaningless since the structure of the question mimics Russell’s paradox. This puzzle seems more appealing than Keynes’ examples of non-numeric probabilities, which assume a richness of life experiences.