## Disease

“You are suffering from a disease that, according to your manifest symptoms, is either A or B. For a variety of demographic reasons disease A happens to be nineteen times as common as B. The two diseases are equally fatal if untreated, but it is dangerous to combine the respectively appropriate treatments. Your physician orders a certain test which, through the operation of a fairly well understood causal process, always gives a unique diagnosis in such cases, and this diagnosis has been tried out on equal numbers of A- and B-patients and is known to be correct on 80% of those occasions. The tests report that you are suffering from disease B. Should you nevertheless opt for the treatment appropriate to A … ?”

My thoughts below …

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If, following Good, we use

P(A|B:C) to denote the odds of A, conditional on B in the context C, Odds(A1/A2|B:C) to denote the odds P(A1|B:C)/P(A2|B:C), and LR(B|A1/A2:C) to denote the likelihood ratio, P(B|A1:C)/P(B|A2:C).

then we want

Odds(A/B | diagnosis of B : you), given
Odds(A/B : population) and
P(diagnosis of B | B : test), and similarly for A.

This looks like a job for Bayes’ rule! In Odds form this is

Odds(A1/A2|B:C) = LR(B|A1/A2:C).Odds(A1/A2:C).

If we ignore the dependence on context, this would yield

Odds(A/B | diagnosis of B ) = LR(diagnosis of B | A/B ).Odds(A/B).

But are we justified in ignoring the differences? For simplicity, suppose that the tests were conducted on a representative sample of the population, so that we have Odds(A/B | diagnosis of B : population), but still need Odds(A/B | diagnosis of B : you). According to Blackburn’s population indifference principle (PIP) you ‘should’ use the whole population statistics, but his reasons seem doubtful. Suppose that:

• You thought yourself in every way typical of the population as a whole.
• The prevalence of diseases among those you know was consistent with the whole population data.

Then PIP seems more reasonable. But if you are of a minority ethnicity – for example – with many relatives, neighbours and friends who share your distinguishing characteristic, then it might be more reasonable to use an informal estimate based on a more appropriate population, rather than a better quality estimate based on a less appropriate estimate. (This is a kind of converse to the availability heuristic.)

My notes on Cohen for a discussion of alternatives.

Other, similar, Puzzles.

My notes on probability.

Dave Marsay

## Cab accident

“In a certain town blue and green cabs operate in a ratio of 85 to 15, respectively. A witness identifies a cab in a crash as green, and the court is told [based on a test] that in the relevant light conditions he can distinguish blue cabs from green ones in 80% of cases. [What] is the probability (expressed as a percentage) that the cab involved in the accident was blue?” (See my notes on Cohen for a discussion of alternatives.)

For bonus points …. if you were involved , what questions might you reasonably ask before estimating the required percentage? Does your first answer imply some assumptions about the answers, and are they reasonable?

My thoughts below:

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If, following Good, we use

P(A|B:C) to denote the odds of A, conditional on B in the context C,
Odds(A1/A2|B:C) to denote the odds P(A1|B:C)/P(A2|B:C), and
LR(B|A1/A2:C) to denote the likelihood ratio, P(B|A1:C)/P(B|A2:C).

Then we want P(blue| witness: accident), which can be derived by normalisation from Odds(blue/green| witness : accident).
We have Odds(blue/green: city) and the statement that the witness “can distinguish blue cabs from green ones in 80% of cases”.

Let us suppose (as I think is the intention) that this means that we know Odds(witness| blue/green: test) under the test conditions. This looks like a job for Bayes’ rule! In Odds form this is

Odds(A1/A2|B:C) = LR(B|A1/A2:C).Odds(A1/A2:C),

as can be verified from the identity P(A|B:C) = P(A&B:C)/P(B:C) whenever P(B:C)≠0.

If we ignore the contexts, this would yield:

Odds(blue/green| witness) = LR(witness| blue/green).Odds(blue/green),

as required. But this would only be valid if the context made no difference. For example, suppose that:

• Green cabs have many more accidents than blue ones.
• The accident was in an area where green cabs were more common.
•  The witness knew that blue cabs were much more common than green and yet was still confident that it was a green cab.

In each case, one would wish to re-assess the required odds. Would it be reasonable to assume that none of the above applied, if one didn’t ask?

Other Puzzles.

My notes on probability.

Dave Marsay

## Coin toss puzzle

This is intended as a counter-example to the view, such as Savage’s, that uncertainty can, in practice, be treated as numeric probability.

You have a coin that you know is fair. A known trickster (me?) shows you what looks like an ordinary coin and offers you a choice of the following bets:

1. You both toss your own coins. You win if they match, otherwise they win.
2. They toss their coin while you call ‘heads’ or ‘tails’.

Do you have any preference between the two bets? Why? And …

In each case, what is the probability that their coin will come up heads?

Dave Marsay

## Clarification

In (1) suppose that you can arrange things so that the trickster cannot tell how your coin will land in time to influence their coin, so that the probability of a match is definitely 0.5, with no uncertainty. The situation in (2) can be similar, except that your call replaces the toss of a fair coin.

Other uncertainty puzzles .

## Which Car?: a puzzle

Here’s a variation on some of my other uncertainty puzzles:

You are thinking of buying a new car. Your chosen model comes in a choice of red or silver. You are about to buy a red one when you learn that red car drivers have twice the accident rate of those who drive silver ones.

Should you switch, and why?

Dave Marsay

## Football – substitution

A spanish banker has made some interesting observations about a football coach’s substitution choice.

The coach can make a last substitution. He can substitute an attacker for a defender or vice-versa. With more attackers the team  more likely to score but also more likely to be scored against. Substituting a defender makes the final score less uncertain. Hence there is some link with Ellsberg’s paradox. What should the coach do? How should he decide?

A classic solution would be to estimate the probability of getting through the round, depending on the choice made. But is this right?

Pause for thought …

As the above banker observes, a ‘dilemma’ arises in something like the 2012’s last round of group C matches where the probabilities depend, reflexively, on the decisions of each other. He gives the details in terms of game theory. But what is the general approach?

The  classic approach is to set up a game between the coaches. One gets a payoff matrix from which the ‘maximin’ strategy can be determined? Is this the best approach?

If you are in doubt, then that is ‘radical uncertainty’. If not, then consider the alternative in the article: perhaps you should have been in doubt. The implications, as described in the article, have a wider importance, and not just for Spanish bankers.

Other Puzzles, and my notes on uncertainty.

Dave Marsay

## The origins of Bayes’ insights: a puzzle

In English speaking countries the Rev. Thomas Bayes is credited with the notion that all kinds of uncertainty can be represented by numbers, such as P(X) and P(X|Y), that can be combined just as one can combine probabilities for gambling (e.g. Bayes’ rule).

You are told that one of these is true:

1. Bayes was in the  habit of attending the local Magistrates Court and making an assessment of the defendant’s guilt based on his appearance, and then comparing it with the verdict.
2. Bayes performed an experiment in which he blindly tossed balls on to a table while an assistant told him whether the ball was to the right or left of the original.

Assign probabilities to these statements. (As usual, I’d be interested in your assumptions, theories etc. If you don’t have any, try here.)

More similar puzzles here.

Dave Marsay

## The Sultan’s daughters

The IMA website has the following puzzle:

A sultan has 100 daughters. A commoner may be given a chance to marry one of the daughters, but he must first pass a test. He will be presented with the daughters one at a time. As each one comes before him she will tell him the size of her dowry, and he must then decide whether to accept or reject her (he is not allowed to return to a previously rejected daughter). However, the sultan will only allow the marriage to take place if the commoner chooses the daughter with the highest dowry. If he gets it wrong he will be executed! The commoner knows nothing about the distribution of dowries. What strategy should he adopt?

You might want to think about it first. The ‘official’ answer is …

One strategy the commoner could adopt is simply to pick a daughter at random. This would give him a 1/100 chance of getting the correct daughter. [But] the probability of the commoner accepting the daughter with the highest dowry is about 37% if he rejects the first 37 daughters and then chooses the next one whose dowry is greater than any he’s seen so far. This is a fraction 1/e of the total number of daughters (rounded to the nearest integer) and is significantly better than just choosing at random!

My question:

Given that the sultan knows what dowry each daughter has, in which order should he present the daughters to minimise the chance of one of them having to marry the commoner? With this in mind, what is the commoner’s best strategy? (And what has this to do with the financial crisis?)