# Class 12: Psychic Card Guessing & The Proportional Syllogism

### Uncertainty & Probability Theory: The Logic of Science

Uncertainty & Probability Theory: The Logic of Science

**Video**

Links:

Bitchute (often a day or so behind, for whatever reason)

**HOMEWORK:** Q = “The Metalunan interocitor (a die) must take one of 6 states when tossed: s_1 = 1, s_2 = 2, and so on.” P = “It takes state s_6 = 6”. What is Pr(P | Q)? And then what is Pr(P | Q & “fair die”)? That is, we add “fair die” to the Q we accept as true. We always accept Q is true. Always as in always.

**NO CLASS NEXT MONDAY. CLASS RESUMES 15 JULY.**

**Lecture**

Last week’s homework was to imagine a psychic guessing cards. She receives a chocolate when right and nothing when wrong. In her first 10 guesses (out of 52), she got 4 right and 6 wrong. Given that as our Q, what is Pr(she guesses next card right | Q)?

Now there was no way (unless you were a trained colleague) you could have got the answer for optimal guessing. The math is beyond that which we have done so far. I was interested instead in your *reasoning*.

One person nailed it. Nate (his answer is here). My heart soared like a hawk. A teacher is always delighted when a student understands. Like I said in the lecture, when you’re confronted with a difficult question, do not try to jump to the right answer. Try to first solve a simpler but equivalent problem. If your solution works on the simpler, then expand it to the harder. This is, after all, how many mathematicians work out theorems.

So here’s a simpler problem. The deck has 2 cards, a king and queen. You hold one up and think it, and I try to receive the impression using ESP. I guess King. Alas, you say I am wrong. Given that, what is the chance, using an optimal guessing strategy, I get the next card right? Notice the *implicit premise*—these exist by the score in real problems, which everybody forgets!—that you tell me the truth.

Anyway, the probability is 1; it is certain. I remember I was wrong. If the first card was not King, then it must have been Queen, and so the second must be King. If I remembered that, then optimally I will guess King—and cannot be wrong.

But, if I instead was an NPR listener, not paying too strict attention, and forgetting what I was about, remembering only there was King and Queen, then the chance I get the second is 1/2.

Two different probabilities because two different Q. **Change the evidence, change the probability.** That is the **one real lesson** in this entire class. Never forget it.

There are more Q possible in this problem. Instead of telling the psychic right or wrong, we could show her the card after each guess (removing the implicit premise of the sender telling the truth). That provides even more feedback. And, if it is used optimally, the chance of each successive guess being correct, grows much higher than the naive, no feedback case, where the cards are gone through and no indication of right or wrong until the end.

The math has been worked out in a nice paper by Persi Diaconis (I give him a shout out in the video), Ron Graham, and Sam Spiro, based off work Persi did back when ESP was a hot topic in the 1970s. “Guessing about Guessing: PracticalStrategies for Card Guessing with Feedback“. (If you’ve studied any analysis, you’ll get it.) This has applications everywhere, including clinical trials.

Forgetting there was feedback, and could be used, led some early psychic researchers to over-estimate the ability of their subjects. Because, as should now be clear, you can easily get more right guesses using feedback, even if you’re not psychic.

Now, at long last, but slowly, we begin to assign more numbers beside 0 and 1 to probabilities! We start with the proportional syllogism.

*This is an excerpt from Chapter 4 of *Uncertainty.*All the references have been removed. If you see them because I forgot to remove them, the dollar signs around certain entries indicate, to LaTeX, that math is happening.*

**he Proportional Syllogism**

Given Q = “An Metalunan interocitor must be in one of n states, s_1, s_2, …, s_n and S is a interocitor” then the probability Q = “S is in state s_j” equals 1/n (a tacit premise here and throughout is that n is finite; happily, in real life n always is). This value arises from the *proportional* or *statistical syllogism*. Carl Hempel originated the idea in statistics, and it is stressed in philosophy in the works of Stove, Williams, Groarke, and Franklin. The proportional syllogism arises in the natural way from recognizing the equality of probabilities in equations like this, called the symmetry of logical constants:

Pr(S is in state s_j|Q) = Pr(S is in state s_k|Q); j,k = 1,2,…,n.

There is no proof of this: its truth is intuitive, i.e. provided by induction. Now “intuitive” does not mean every person can be made to understand it (or any proposition!), only that some can. Notice carefully that there is no evidence whatsoever about the interocitor *except* that it can take certain states; especially lacking is any evidence about the *symmetry* of its workings. There may or may not be any; we have no clue. All we have is that the interocitor must take a state with one of n labels (Q insists on this). We have *no* idea how *any* of these states arise. We cannot argue from physical symmetry or indeed use any knowledge from physics or engineering; though some try. Except for the assurance each of the n states are possibilities, the workings of the machine are a complete mystery to us and, in fact, can be no better than imaginary because, I shouldn’t have to add, there are in reality no Metalunans thus there are no Metalunan interocitors. This is of zero importance, however, because logic is not concerned with reality *but with the relations between propositions*. Recall the French-speaking cat example. Given we know the machine has to be in one of *n* states, and that is all we know, then the probability it is any one of them just is equal.

Interestingly, Persi Diaconis and ET Jaynes attempted proofs of the statistical syllogism which, they thought, avoided the necessity of the symmetry of logical constants, but it will be shown below that the proofs are circular and rely on the symmetry of logical constants after all. David Stove has a proof which also works but which has a quirk, and it will also be shown.

The proportional syllogism is a deduced principle of probability from the equi-probability of logical constants, as in the equation above. Whether in specific instances the results which flow from it match the results given by some other calculation device or principle, such as maximum entropy, is a nice coincidence, but does not obviate it. These principles, whatever nice or desirable properties they have, are not strictly part of probability. Take for example the premise Q = “90% of Martians wear hats” together with “George is a Martian” then the probability P = “George wears a hat” is 90%, which is also so because of the proportional syllogism; but it is constructed differently. To see that, let GM = “George is a Martian”, GH = “George wears a hat”, and S = “Sally”, etc.; then

Pr(GH|Q and GM) = Pr(SH|Q and SM).

Here we have no idea how many Martians there are, except that there are enough to form an even 90%. Unlike in the original equation, the evidence also changes. Note that Pr(GH|Q and GM) ≠ Pr(SH|Q and GM)$ because Sally could be a human or Metalunan. In general, Pr(XH|Q and XM) = Pr(YH|Q and YM) where X and Y are names. The “symmetry”, to use that word loosely, comes in considering that any names (labels) X and Y can be used: there is no information to prefer any name over another, thus the probabilities are equal.

The probability “George wears a hat” given “50% of Martians wear hats and GM” is less than the probability “Sally wears a hat” given “90\% of Martians wear hats and SM”, a fact which also follows from the proportional syllogism. Just as “George wears a hat” has higher probability given the premise “Most Martians wear hats and GM” compared to the premise “Few Martians wear hats and GM.” This example requires the tacit premise that {\it most} is more than {\it few}. Numbers aren’t needed. The probability “George wears a hat” given “X% etc.” is equal to “Sally wears a hat” given “X% etc.”, with only the tacit premise, given our understanding of percentages, that X is somewhere in 0 to 100.

It is a simple principle of logic that if the argument from Q to P is valid, then the argument from “Q and T” to P is also valid, where T is any necessary truth. Intuitively, adding something which cannot possibly be false to Q adds “nothing”; we might say that it doesn’t change how P flows from Q; it is like multiplying a simple algebraic equation by 1. Given “All men are mortal and Socrates is a man” then “Socrates is mortal” is true; and nothing changes if we append to the premise a necessary truth such as “A is A”.

Common necessary truths are logical tautologies. Examples, “Either Mars now is 12 parsecs from Earth or it isn’t,” “If it is sunny then I will go swimming which implies I will not go swimming if it is not sunny” (we take the entire proposition here), “Either unicorns like chocolate or they don’t” (this is true whether unicorns exist or not), and the standby “P or not-P” (which we know is necessarily true based on tacit premises of logic). That last proposition means if Q to P is valid “Q and ‘P or not-P'” to P is also valid. Adding the necessary truth “P or not-P” did not change the argument in any way.

The same conditions hold in probability. If on the argument from Q to P we deduce the probability of P given Q to be some value, then adding a tautology or other necessary truth to Q does not and cannot change this value. Thus

Pr(P|Q) = Pr(P|Q and T).

And this holds even if Pr(P|Q) doesn’t have a numerical value or if it is an interval; it also holds if T = “P or not-P”. A concrete example. As above, given *only* Q = “A Metalunan interocitor must be in one of n states, s_1, s_2, \dots, s_n and S is a interocitor” then the probability P = “S is in state s_j” equals 1/n. Take the tautology T = “S is in state s_j or it isn’t.” T is necessarily true whether Metalunan interocitors exist or not, and true regardless which state any of them might happen to be in. Adding T to our list of premises thus does nothing to the conclusion that P takes probability 1/n. T, because it is a tautology, does not suddenly change our notion of P, nor does it add any information about it. (*NOTE NOT IN BOOK: This T does not even tell us how many states are possible!*)

Consider the probability of P = “S is in state s_j” given only T = “S is in state s_j or it isn’t.” Since again T is a tautology it gives no information about P. In particular it is false, though it is often asserted, that Pr(P|T)=1/2. This is more easily believed when any necessary truth can take the place of T. Let T’ = “There are 82 millions ducks in the world or there aren’t.” Then obviously Pr(P|T’) ≠ 1/2, but even stronger

Pr(P|T) = Pr(P|T’);

and this is so even though neither side produces a unique, single number. The closest Pr(P|T) comes to a number is the interval (0,1); note that this does not include the extremes, which is just another way for saying P is contingent. In this sense there is some information in T and T’, or rather the words and grammar of T, T’, or P, which is usually that P is contingent, and which is enough to exclude complete certainty, but that’s all; information in tautologies is thus of (literally) infinitesimal value (when it has any value).

Another version of the tautology is T” = “S might be in state s_j.” Implicit in this is that S might not be in state s_j, thus T = T”. Still another version is T^{3} =”S might be in state s_j or s_k” with the same tacit inclusion (though it can be argued this version says only these two states are possible; words matter!). Another: T^{4} = “S might be in state s_1 or s_2 or, …, or s_n”, again with the tacit admission they might not. Note carefully that T^{4} is *not* equivalent to Q = “A Metalunan interocitor must be in one of n states, s_1, s_2, …, s_n and S is a interocitor” because T^{4} *never* asserts that any interocitor must be in *any* state, only that it might be in one or another state, or none at all.

It turns out that Metalunan interocitors are actually that race’s version of dice, that n=6, and that each “state” is a side upper most upon tossing. Thus Q really equals “A Metalunan die when tossed must show one of 6 sides, labeled 1 through 6, and S is the side that shows on a toss” then the probability P = “S is 3” equals 1/6. This also follows from the proportional syllogism. Adding tautologies, I hope is now plain, changes nothing.

*We finish the die next class!*

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edited Jul 15So, my attempt at the homework:

1. Pr(P=It takes state s_6 | Q as stated) = 1/6. (Q - "must take one of 6 states" implies this is a fair die, no other outcomes are possible.

2. Pr(P | Q & “fair die”) = 1/6; "fair die" is an irrelevant addition since the definition of Q excludes any other possible outcome than 1 through 6.

3. If "must take one of 6 states" is excluded from the definition of Q, then we may not have a fair die, so there may be a 7th state which is not 1 through 6. In which case Pr(P=It takes state s_6 | Q as stated) = 1/7.

How am I doing?

edited Jul 1At the end of this lecture, how far along should we be in your text and in Jaynes at this point? The lectures are not difficult, but the text has been a bit of heavy going at times.