Nota Bene: this is still very much a work in progress. I have not yet achieved that mental state at which I can indulge, at least for a while, in the delusion that I have achieved the maximum point of crystalline clarity.’ I am not responsible for any brain damage anyone reading this stuff may incur.
What is the point of the arguments that are about to follow? These arguments are one snippet in an attempt to get clear in my mind regarding the nature of probability. (Yes, I know, this is absurdly ambitious. You may be a bit less inclined, gentle reader, to break out in raucous laughter if you keep in mind I am just trying to arrive at the point at which, in a doubtlessly delusional state, I suffer from the strong conviction I have gotten clear in my own mind regarding the nature of probability. Once achieved, this strong conviction will doubtlessly evaporate like a mirage as I increase my knowledge of the field.)
The reason I want to get clear in my own mind about the nature of probability because I think this is necessary in order to uncover at least one relation that makes the antecedent relevant to the consequent in relevant indicative conditionals. I expect to be making changes to this post as time goes on.
What is the conclusion I am heading towards with all the verbiage below? This: the existence of a probability greater than 0 but less than 1 has as both its necessary and sufficient condition a ratio of ignorance/knowledge within a given perspective. Probability within these two limits is perspectival down to the very root for this reason; it could not exist within the “perspective” of an infinite mind that does not suffer any ignorance at all, partly because such a Mind would not enjoy any perspectives at all. Given a deterministic universe, this is the only way there can be probabilities between 0 and 1 noninclusive.
In the clearest cases, the role knowledge/ignorance plays in determining such a probability is easiest to see in the case of independent events; but dependent events, as in the case of the Monty Hall puzzle, can increase/decrease the probability of a given event.
The Scene. A Shell Game Is Set Up. Let me begin by describing the scene. In an apple and cherry orchard in Iowa, a table has been set up. The sky above is clear. Unknown to and hidden from the people in and about to enter the orchard, but within view should one occupy the right vantage point, a tornado is touching down intermittently across the Missouri River, in Nebraska. I describe the scene this way because it is a situation. A situation is partially defined by what is hidden from one and unknown to one, and by the information that is available to one. Situations will become important in later posts because some versions of Relevant Logic rely on them rather than on possible worlds. I describe this particular one now because I will be returning to it later.
Elizarraraz (although this is not relevant to the example, the name is Ladino for ‘poor king’. Ladino is the Sephardic counterpart to Yiddish, and in . Elizarraraz’ case the name, and his paternal ancestry, comes from Mexico. Although they were not officially allowed to, a number of conversos managed to emigrate to Latin America in order to place a more comfortable distance between themselves and the Spanish Inquisition. Just thought I would provide my made-up characters with concrete backgrounds. But I digress) sets up on the table a shell game with three shells and a single peanut. The shells are labelled in order 1, 2, and 3. Employing a randomizing device of some sort (say, he throws a die), Elizarraraz places the peanut under the shell selected by his randomizer. Naturally, he knows under which shell the peanut is hidden.
At least for now, I will leave the concept ‘randomness’ as an unanalyzed primitive, explicated, not by a real, concrete example, but by a (vaguely described) ideal one. A fair 6-sided die would be suitably random if, after a very large number of throws, the average ratio of the times each number came up, divided by 6, remained sufficiently close to 1/6. And yes, I will leave ‘sufficiently’ undefined.
Smith (although this is not relevant to the example, the name is English for ‘smith’ as in ‘blacksmith’. But you knew that already) enters the scene. He knows that there is a peanut hidden underneath one of the shells. (Elizarraraz, who is a reliable conduit of information, has told him this.) Smith is about to play what I will call, for reasons that are about to become clear, the ‘non-Monty-Hall shell game. Again, using a randomizer, Smith selects one of the shells, and turns it over. Naturally, either there is a peanut showing up, or there is not. I think it would be uncontroversial to say that the probability there is a peanut there is 1/3, and the probability that there is not is 2/3.
Suppose no peanut was lurking under that shell — say, shell #1. Smith now knows that there was no peanut under shell #1. In at least some sense of the term ‘certain’, he is now certain that shell #1 was not the one hiding the peanut. But he knows that there is a peanut lurking under one or the other of the remaining shells, #2 and #3. I have, and I think most people will have, the strong intuition that the probability the peanut is under shell # 2 (alternatively shell #3) is 1/2. Given that the original sample space [*] of three has now been reduced to two, surely the probability is now 50/50! But hold that thought for a few more paragraphs  while I discuss for a bit the notion of ‘a possibility’.
At this point, Smith confronts two possibilities. A possibility is a possible outcome. Possibility #1: the peanut lurks under shell #2 and shell #3 is empty. Possibility #2: the peanut lurks under shell #3 and shell #2 is empty. To talk about ‘a possibility’ here is to say the following: because Smith knows there is a peanut under one of the shells (he just doesn’t know which one), there is a peanut under one of the shells. For if one knows that p, then p is a true proposition (or, better, a state of affairs that obtains [I follow Chisholm in identifying propositions with a proper subset of states of affairs]. From Smith’s point of view, the peanut could be under shell #2 or shell #3; that is to say, he doesn’t know which one. So, at least in cases like this one, [yes, I know, this needs to be more sharply defined] ‘a possibility’ requires a combination of knowledge and ignorance. Remove the ignorance, and the possibility no longer exists.
From Smith’s point of view, it is no longer the case that the peanut could be under shell #1. Its being under shell #1 is no longer a possibility for Smith. And the probability that it is under shell #1 is now 0. Were Elizarraraz to turn over the shell that does hide the peanut (say, shell #3) (and were Smith to see the peanut that had been hiding there, and were nothing at fault in Smith’s visual apparatus), it would no longer be the case that, from Smith’s point of view, the peanut could be under shell #3. It is under shell #3. Its being under shell #3 is no longer a mere possibility, but a certainty. Again, remove the ignorance, and the possibility no longer exists. From Smith’s point of view, the probability that the peanut is under shell #3 is now 1.
When Smith turned over shell #1 and discovered it to be empty, he decreased the size of the sample space from three possibilities (the peanut is under shell #1 and shells #2 and #3 are empty; the peanut is under shell # 2 and shells #1 and #3 are empty; the peanut is under shell #3 and shells #1 and #2 are empty) to just two (the peanut is under shell #2 and shell #3 is empty; the peanut is under shell #3 and shell #2 is empty). A sample space is a set of possibilities; the cardinality or “size” of the space is the number of possibilities it has as members. The metaphor of ‘a space’ is apropos here because a given space, a room, for example, can contain items, just as set “contains” its members. If a sample space contains n possibilities and each possibility is equality likely, then the probability of each event (subset of the sample space) must be expressible as a ratio with n as the denominator. If the size of the sample space is six, for example, the probability of each event must be expressible as 1/6, 2/6, 3/6, 4/6, and 5/6.
When the possibilities involve physical entities, such as a number of shells one of which hides a peanut, it is easy to think of the size of the sample space as equal to the number of those entities. Later, however, I intend to show that the sample space can include possible as well as actual entities.
Now Morgenstern arrives on the scene.
But maybe we are not entitled to be confident about this intuition. The Monty Hall paradox shows rather clearly that our intuition in these matters cannot always be accepted at face value. Let me briefly describe the Monty Hall paradox.
The name of the paradox comes from a television game show hosted by a certain Monty Hall. The show employed doors hiding cars and goats, but I prefer to stick with shells hiding either a peanut or empty air. The game proceeds as it does with the non-Monty-Hall shell game, but with this difference. After Smith has selected a shell, he does not turn it over to see if it hides the peanut. Instead, Elizarraraz turns over one of the peanuts. The peanut he turns over has to meet two criteria: first, it cannot be hiding a peanut; and second, it cannot be the shell (initially) selected by Smith. Elizarraraz then gives Smith the choice of either sticking with his initial selection, or switching to the remaining shell (that has not yet been turned over).
One can be forgiven for having the strong intuition that neither strategy has any advantage over the other. As one pictures the two remaining shells with the mind’s eye, may seem completely obvious that Smith’s chances of winning the peanut are 50/50 if he sticks with his initial selection, and 50/50 if he switches. The sample space, after all, would seem to comprise just two possibilities, just as does the sample space of the non-Monty Hall game. Possibility #1: the one shell either hides the peanut, in which case the other shell hides just empty air; or (possibility #2) the former shell hides empty air, and the latter shell hides the peanut. This is what could turn up, what could be very shortly in the near future.
But, as it will turn out, this is not the sample space of the Monty Hall shell game. And Smith’s chances of winning the peanut are not 50/50 regardless of his strategy, but 1 in 3 if he opts to stick with his initial selection, and 2 in 3 if he opts to switch. As if that were not (at least initially) counter-intuitive enough, it remains true that Smith’s chances of winning the peanut are 50/50 if he chooses by flipping a coin which of the remaining two shells to select; and his chances of choosing his initial selection |alternatively| choosing the shell that was not his initial selection are also 50/50. How can all of these propositions be true at the same time? How can the ‘2 in 3′ be true at the same time the ’50/50’ is true? And what can we learn about the nature of probability from the co-truth of these propositions?
Taking my cue, first from Judea Pearl, then from Luis Jorge Borges, I will prove the ‘1 in 3’ vs. ‘2 in 3’ probabilities for sticking with the initial choice vs switching. Then, after proving the 50/50 cases, I will show how these are compatible with the 1 in 3 and the 2 in 3.
Computer simulations of Monty-Hall-type games (for example, the one available online here or here) show definitively that Smith’s chances of winning the peanut are 1 in 3 if he sticks with his initial choice and 2 in 3 if he switches. One of the simulations I linked to repeats the game ten million times. Few, I think, would dispute that these simulations show that the chances are 1 in 3 | 2 in 3. But they won’t suffice to give one any intuitive sense why those are the chances. No Aha Erlebnis will be coming from just observing the simulations.
A table listing all of the possibilities, all the possible cases, goes some way, I think, towards giving one this intuitive sense. As shown in the table below (a modification of the table presented by Judea Pearl in his BOOK OF WHY (BOOK OF WHY, p. 191), which in turn is taken from Marilyn vos Savant’s column from the 90’s), there are nine distinct possibilities, nine possible cases. Each of the nine cases is equally likely. One can then start to see why the computer simulations would give Smith a 1/3 chance of selecting the shell with the peanut if he sticks with his initial choice, and a 2/3 chance if he chooses the remaining shell.
|Shell #1||Shell #2||Shell #3||If Same||If Different||Which Means That|
|peanut, initial selection||empty, not initial selection||empty, not initial selection||Smith wins||Smith loses||either shell #2 was turned over, leaving shell #3 to be select should Smith opt to change his selection; or shell #3 was turned over, leaving shell #2 to be selected should Smith opt to change … in either case, Smith loses if he opts to change his selection|
|empty, initial selection||peanut, not initial selection||empty, not initial selection||Smith loses||Smith wins||shell #3 is the only shell eligible to be turned over, which means that Smith will choose shell #2, and win, if he opts to change his selection|
|empty, initial selection||empty, initial selection||peanut, initial selection||Smith loses||Smith wins||shell # 2 is the only shell eligible to be turned over, which means that Smith will choose shell #3, and win, if he opts to change his selection|
|peanut, not initial selection||empty, initial selection||empty, not initial selection||Smith loses||Smith wins||shell # 3 is the only shell eligible to be turned over, which means that Smith will choose shell #1, and win, should he opt to change his selection|
|empty, not initial selection||peanut, initial selection||empty, not initial selection||Smith wins||Smith loses||either shell #1 was turned over, leaving shell #3 to be selected should Smith opt to change his selection; or shell #3 was turned over, leaving shell #1 to be selected should Smith opt to change. In either case, Smith loses if he opts to change his selection|
|empty, not initial selection||empty, initial selection||peanut, not initial selection||Smith loses||Smith wins||shell #1 is the only shell eligible to be turned over, which means that Smith will choose shell #3, and win, if he opts to change his selection|
|peanut, not initial selection||empty, not initial selection||empty, initial selection||Smith loses||Smith wins||shell #2 is the only shell eligible to be turned over, which means that Smith will choose shell #1, and win, if he opts to change his selection|
|empty, not initial selection||peanut, not initial selection||empty, initial selection||Smith loses||Smith wins||shell #1 is the only shell eligible to be turned over, which means that Smith will choose shell #3, and win, if he opts to change his selection|
|empty, initial selection||empty, initial selection||peanut, initial selection||Smith wins||Smith loses||either shell #1 was turned over, leaving shell #2 to be select should Smith opt to change his selection; or shell #2 was turned over, leaving shell #1 to be selected should Smith opt to change. In either case, Smith loses if he opts to change his selection|
The table, however, is not perfect as a device for generating the desired Aha Erlebnis giving one to see that Smith’s chances are only 1 in 3 if he sticks with his initial choice. One may want to see rows 1, 4, and 7 in the table as each comprising two possibilities, not one, rendering problematic the math that gives us the 1/3 and 2/3 probabilities. One would be wrong, of course; nonetheless, it remains true that the table is burdened as an Aha-Erlebnis-generating tool by this complication. Also, the table does not show why the 50/50 chances (initially and perhaps even non-initially) seem so powerfully intuitive.
Listing out all the possibilities in the form of a tree, gives us a picture, another way of showing the 1/3 and 2/3 probabilities without the burden of this complication. We can picture repeated plays of the Monty Hall shell game as a trunk branching off into a number of branches. Doing so will nail down the 1/3 and 2/3 probabilities quite conclusively, though perhaps without generating an Aha Erlebnis, a concrete intuition.
Picturing the game this way will also provide at least a start at an explanation why the conclusion that the chances are not 50/50 seems so paradoxical. The idea of treating the game this way came to me in a flash of insight after reading Jorge Luis Borges’ short story THE GARDEN OF FORKING PATHS. (“You are so smart!” at work, though sometimes I suspect they mean this in a ‘you have a wonderfully intuitive sense for the blindingly obvious’ way), but, of course, essentially the same idea has occurred to other people, as one can see here and at numerous other places on the internet. I would like to think, however, that I have my own twist on the idea. Anyway, onto the chart shown below and an explanation of what it shows.
The Monty Hall Shell Game Considered As Conceptual Sleight Of Hand: In the chart shown below, Elizarraraz (employing a randomizing device) chooses which shell to place the peanut under (tanned orange). In order to make the chart readable, I show just Elizarraraz’ choice of shell #1. The possible choices that ensue from the “space” that would open up if Elizarraraz placed the peanut under this shell are, I claim, canonical. That is to say, they comprise a piece (shell #1) of the larger picture that enable one to draw conclusions about the larger picture (all three shells).
A moment later, Smith comes into the scene and, employing a randomizing device, makes his initial selection of a shell (pink). Elizarraraz then turns over one of the shells, employing, not a randomizer, but his knowledge of which shell Smith has selected and which shells are empty (baby-aspirin orange). Those shells Elizarraraz cannot turn over are crossed out by red lines.
Finally, using a randomizer, Smith decides either to switch shells or stick to his initial choice. The decision to switch is shown (for reasons that will become clear when I get to the ‘forking paths’ metaphor) by the bolded arrow. The winning shell (Smith gets the peanut) is shown by the darker viridian or “sea-glass” green color of the oval symbol picturing the shell. The losing shell is shown by the lighter viridian green, which looks like a light blue.
[Each oval represents a possible outcome (for example, Smith initially selects shell #1). Until we get to the culminating possibilities (represented by the green ovals), each possible outcome opens up (and sometimes closes down) what I will call a ‘possibility trail’, i.e., a “trail” in which one possible outcome follows another. Smith’s initial choice of shell #1, for example, opens up a path in which Elizarraraz turns over shell #2, which in turn forks into two paths, one leading to Smith’s winning the peanut and the other leading to his losing the game; and opens up another path in which Elizarraraz turns over shell #3, which path in turn forks into…; and results in a dead end, in which Elizarraraz is constrained by the rules of the game from turning over shell #1. ]
[Each fork opens up what I shall call a “cone” of possibility paths. Elizarraraz placing the peanut under one of the shells opens up three such cones, not labelled here. Smith’s choosing a shell opens up three cones, which I label A, B, and C. The paths in cone A culminate in four different possible outcomes; the paths in cone B and cone C each culminate in two possible outcomes. ]
[Cones A, B, and C match with rows 1, 2, and 3 respectively in the table shown previously. Each cone/row constitutes a wider sample space whose “places” or “slots” are themselves narrower “sample spaces” whose “places” are still narrower samples spaces defined by the forks and, ultimately, by the possible ending outcomes. These narrower sample spaces would (note the subjunctive mood) succeed one another in time; one such sample space, one set of possibilities would open up for example were Smith to initially select shell #1. There are two final sample spaces in cone A. These sample spaces begin, respectively, at Elizarraraz’ possibly turning over shell #2, or his possibly turning over shell #3, and include their ending “leaf” possibilities: shells #1 or #3; or shells #1 or #2 respectively. Both of these final sample spaces are included as places in the sample space comprising cone A. The sample space that is cone A is defined by the fork that gets generated by Smith’s possibly making the initial selection of shell #1. Cone A in turn, along with cones B and C, are included in the sample space that is generated by Elizarraraz’ possibly placing the peanut under shell 1.]
If Elizarraraz has placed the peanut under shell #1, then of course Smith has only a 1 in three chance of winning if he sticks by his initial choice. For in this case he will win the peanut only if that initial choice was shell #1. But the chances shell #1 was his initial selection are just 1 in 3. So his chances of winning by sticking with his initial choice are also just 1 in 3. It follows that his chances should he switch will be 2 in 3. If this conclusion is not already already intuitive to you, gentle reader, I think it will become more intuitive once I start laying out the forest of forking paths picture.
Suppose that Smith, compulsive gambler that he is, plays the Monty Hall Shell Game ten million times. At the end of each game, he is presented with just two shells. One was initially selected by him, the other not. Now suppose that the shell that was initially chosen is marked as such; ditto the shell that was not initially chosen. If Smith sticks to a strategy of of chosen the shell he did not initially select, he will win 2/3 of the time and lose 1/3 of the time. Conversely, if he sticks to a strategy of sticking to his initial choice, he will lose 2/3 of the time and win 1/3 of the time.
Now suppose the markings ‘initial choice’ and ‘not initial choice’ are removed from the shells — and, because the shells looks so similar, Smith cannot remember which one he had initially selected. No labels ‘shell #1’, ‘shell #2’, ‘shell #3’ have been applied to help guide him. Smith has to flip a coin to decide on which shell to select. I think it is clear from the chart that Smith will win the peanut 1/2 the time by flipping a coin. This 50/50 probability is, I think, what makes the Monty Hall Shell Game so drastically counter-intuitive. One looks at the two shells, each of which could be hiding the peanut, and (correctly) sees a 50/50 chance should they flip a coin.
But notice that in the game, Smith is not asked to flip a coin to decide between the two remaining shells. Instead, he is asked either to stick with his initial choice or to switch. That is the Monty Hall Shell Game, which presents Smith with a 2/3 (alternatively, 1/3) chance of winning. He is not asked to flip a coin to decide between the two shells. That is a different game altogether, one that results in a 50/50 chance of winning. Let me call this other shell game the ‘Monty Hall Shell Game With A Final Coin Toss Added In At The End For Good Measure.’
So which game is being played — and what the rules are for each — matters for what the probabilities are. The rules of the Monty Hall Shell Game require that Smith, the player, know which of the two shells remaining in the penultimate step was his initial choice, and which was not. The rules require keeping track of what happened in the past — there has to be a trail, a path, so to speak, leading from the past to the present. If Smith loses this trail — say, all shells have the tendency to look alike to him, and Elizarraraz does not bother to inform him which is which — then Smith has no available evidence to base his choice on except for flipping a coin.
[Since in both these games the designations ‘shell #1’, ‘shell #2’, ‘shell #3’ drop out of the picture, one may get the feeling that these are similar to the shell game as traditionally played, in which a slick operator switches the peanut between hard-to-distinguish shells by slight of hand. But here, of course, one is not trying to force their eyeballs on three actual shells in an attempt to keep from getting fooled within a single playing of the game. Shell stays the same; peanut surreptitiously moves. Instead, one is dealing with labels which stay the same even as the shells they apply to change with each new playing of the game. [How come 2/3 probability when only 2 shells remaining?]]
Under one description for the shells, the chances of winning the peanut are 50/50. Under another description (shell not initially chosen; shell initially chosen), the chances are, respectively, 2 in 3 and 1 in 3. But these are (at least at any given time) the same shells. What accounts for the difference? The difference, I think, lies in the history of how the shells got there. And in explaining this, Borges short story THE GARDEN OF FORKING PATHS will prove useful.
Enough of the shell games. Let me now apply a completely different picture, one inspired by Borge’s short story THE GARDEN OF FORKING PATHS. This picture will be of a forest containing within it a multitude of forking paths. It will, I propose, make it easier to articulate certain aspects of the paradox I am trying to make sense of.
The chart above was originally drawn as a graphic tree depicting the Monty Hall Shell Game, but it can also be interpreted as a map depicting several forking paths in a forest. Smith will be walking the paths ten million times (he is an indefatigable hiker).
These paths are in a parallel universe which mirrors our universe, in which Smith is playing the Monty Hall Shell Game. The ovals in the chart above, which used to represent choices (Smith’s or Elizarraraz’), now represent clearings in the forest. The arrows, which used to represent ‘go on to the next step’ now represent paths leading from one clearing to the other. Which clearing Smith ends up in, and which path he takes, is determined by the choices he and Elizarraraz take in the shell game in our universe. So the forking paths picture will be a bit science-fiction-y; nonetheless, my hope is that it will result in a gain in intuitive clarity (certain points will be easier to make) which will make up for its contrived character. Think of it as like the Mercator projection which serves as the standard in maps of the world. In this projection, certain features are captured at the expense of distortions in the areas of the land and water masses mapped.
Each oval represents a clearing in the forest. Each arrow represents a path leading from one clearing to the next. There are three different starting clearings which map to Elizarraraz hiding the peanut under shell #1 alternatively shell #2 alternatively shell #3; above, only the clearing corresponding to his hiding the peanut under shell #1 is shown, since I take this to be canonical. Three paths fork of, or, more precisely, trident off from the starting clearing. If Smith takes the path to the left, These of course map onto Smith’s initially selecting one of the three shells. If Smith takes the path on the left, hink of the arrows in the chart above as depicting Let me first describe the forking-path interpretation in just enough detail to let me state the two points I want to make. Then I will lay out the interpretation in more adequate detail. We will be having Smith walk the paths…maybe ten million times would be cruel and unusual punishment, but enough times that a frequency becomes apparent. The paths end in a forest clearing which contains something stupendous which I will leave to the reader’s imagination. Maybe it is a glorious vision of a topless Channing Tatum clearing brush. Maybe it is seeing Edward in full shining resplendent crystalline display. Maybe it is seeing a gorgeously feral Jacob — another graceful son of Pan! Or maybe it is just an extra-special peanut that outshines any other peanut. Whatever.
When Smith, walking down the path for the x number of times, comes to the final fork in the path, he can do one of two things. First, he can select the path by flipping a coin. Or, second, he can adhere to a right-hand/left-hand strategy: always choose the path on the right (alternatively the left).
I think it is plan from the graph that if he chooses by flipping a coin, he will arrive at the clearing with the special prize (a view of Channing Tatum, or the extra-special peanut) one half the time. If he adheres to the right-hand/left-hand strategy, he will arrive at the clearing with the special prize two thirds of the time if he always takes the path on the right, or just one-third of the time if he always takes the path on the left. Always taking the path on the right corresponds, in the Monty Hall Shell Game, to Smith’s switching, and always taking the path on the left corresponds to his sticking to his initial choice.
The different strategies lead to different probabilities. In a short while, I will relate these differing probabilities to those of the Non-Monty-Hall Shell Game played by Smith and Morgenstern. I intend to show that just as knowledge (or lack of knowledge) accounts for the difference in probabilities in the Smith and Morgenstern case, the related concept of evidence (or lack thereof) accounts for the difference in probabilities in the forking path case (and in the Monty Hall Shell Game).
But given the difference in the probabilities established by the different strategies, one can explain why the Monty Hall Shell Game seems so paradoxical to about everyone, at least at first. For when one imaginatively confronts the choice faced by Smith (stick to the initial choice of shells or switch), one surreptitiously thinks of the choice in terms of a ‘let’s flip a coin’ scenario. This scenario is, after all, easy to picture imaginatively. The alternative is to have the choice guided by something like the graph above. This graph is, naturally, not at all easy to picture.
Let me now turn to a fuller explanation of the above chart, interpreted either as a tree (the Monty Hall Shell Game) or as a set of forking paths.
I think I have fulfilled my promise to use the forking paths picture to nail down even more firmly the 1/3/2/3 stick with the initial choice/switch probabilities. Now let me show how this picture helps explain why this result seems, at least initially, so counter-intuitive.
Now after Smith has traveled down one or another of the paths in one or another of the three possibility cones, he is presented with two shells (in cone C, for example, either shell #1 or shell #3). The peanut could be under either of those shells. At the time of this writing (September 8, 2019 — I note the date because particular pieces of my autobiography have in the past turned out, somewhat surprisingly, to be philosophically fruitful), it seems absolutely clear to me from looking at the chart that Smith’s chances of winning the peanut are 50/50. Later I may try to nail this intuition down more firmly by coding my own simulation of the Monty Hall shell game.
But note that what I am ascribing a 50/50 chance to is the peanut’s being under (for example) shell #1 or shell #3. I am not ascribing a 50/50 chance to the peanut’s being under the Smith’s initial choice of shells or his switched choice. The descriptions ‘initial choice shell’ or ‘switched choice shell’ have no meaning in this narrow sample space delimited by what could be, i.e., by the present and the potentialities of the (presumably) near future.
To get these descriptions, we have to go deeper than what could be and move into what could have been. We have to move into the past. Smith could have chosen shell #2, but he has chosen shell #3, which in turn made shell #2 the only possible choice of shells for Elizarraraz to turn over, which in turn left Smith with a final choice of shells #1 and #3. Were Smith to go back in time multiple times to his initial choice of shells but with his randomizer determining different choices — or, less science-fictionally, were he to repeat the Monty Hall shell game a large enough number of times, he would end up winning the peanut 1/3 of the time by sticking to his initial choice, and 2/3 of the time by switching.
The probabilities are determined by the sample space. When the descriptions ‘initial selection shell’ and ‘switched choice shell’ make sense, the sample space embraces three possibilities, the three possibility cones, one of which culminates in Smith’s winning the peanut should he stick to his initial choice, and two of which culminate in his winning the peanut should he switch choices. That’s the sample space that counts when those descriptions are meaningful. When those descriptions don’t make sense because we are restricted to what could be, that is, to the present because the sample space is restricted to the present, to what is facing Smith now, and to a narrow snippet of the near future, the sample space comprises only two possibilities: the peanut is under this shell or under that other one.
Were Smith told, when confronted with the two shells, to choose one of two strategies: switch or stick with the initial choice, neither strategy would make any sense at all unless he had access to enough of the past to let him identify which shell was his initial choice; or unless someone who was keeping track told him. And even then his adopting one strategy or the other would be incompletely rational unless he had plotted out all the cones with the possible paths that could have been, including both the paths that led to the present situation and the paths that ended up as dead ends. He would be better off not worrying about which shell was his initial choice and just flipping a coin.
What the sample space is, and therefore what the probabilities are, depends upon which game is being played — flip a coin, or stick-with-the-initial-choice-or-switch. Different sample space, different game; different game, different sample space. Although Pearl’s point in the following may be a bit different from what I have just described, his actual words still fit with my point. (Maybe there is another Borges story about something similar.) Pearl notes:
The key element in resolving this paradox is that we need to take into account not only the data … but also … the rules of the game. They tell us something about the data that could have been but has not been observed.BOOK OF WHY, p. 192
When confronted with just the two remaining shells in the present, it is easy to forget that these are two different games.
Thinking about the the different cones containing different possible paths requires a certain amount of time, patience, and wetware power and bandwidth. Considering the possibilities when confronted (perceptually or imaginatively) with just two shells requires much less time, patience, and wetware power and bandwidth. This fact, plus the fact that it is perhaps not so obvious when staring at the shells that the descriptions ‘initial choice’ and ‘switching choice’ cannot be applied to the shells if one’s time horizon (and the resulting sample space) are too narrow are, I submit, at least one reason the actual probabilities of the Monty Hall shell game seem at first so drastically counter-intuitive.
As Pearl notes, there are probably 10,000 different reasons, one for each reader, why the actual probabilities of Monty Hall game seems so counter-intuitive. To return for a moment back to cars, goats, and doors:
Even today, many people seeing the puzzle for the first time find the result hard to believe. Why? What intuitive nerve is jangled? There are probably 10,000 different reasons, one for each reader, but I think the most compelling argument is this: vos Savant’s solution seems to force us to believe in mental telepathy. If I should switch no matter what door I originally chose, then it means that the producers somehow read my mind. How else could they position the car so that it is more likely to be behind the door I did not choose?BOOK OF WHY, pp. 191-192.
The specter of mental telepathy is doubtlessly one reason the result seems so counter-intuitive; one’s tendency, resulting from the limitations on human mental power, to be perceptually/imaginatively restricted to what could be as opposed to what could have been is another. I won’t try to judge here whether one is more compelling than the other, especially since I have not yet wrapped my head around Pearl’s account of causality.
Now back (finally!) to the point of bringing up the Monty Hall puzzle in the first place. Regarding the non-Monty-Hall shell game, I asked what makes us so sure the probability is now 1/2 that the peanut is under one of the remaining shells after Smith has turned over one of the shells which turned out to be empty. Why should we trust our intuition in this case, when our intuition regarding the Monty-Hall case were initially so far off? Well, let’s provide a table of the possibilities.
|Shell #1||Shell #2||Shell #3||Shell Uncovered by Smith||Former Possibility Converted to Actuality|
There are two independent events a work here: Elizarraraz randomly placing the peanut under one of the three shells, and Smith’s randomly turning over one of the shells. Neither event affects the probability of the other. If we then eliminate the rows in which Smith happened to turn over the shell containing the peanut (as marked by ‘yes’ in the column ‘Possibility (that the shell hides the peanut) turned into actuality (yes, the shell did hide the peanut), we get 6 rows. Each of the three pairs of rows describes a probability: if Smith finds that shell #1 was hiding nothing except empty air, then row 2 (the peanut is under shell #2) and row 3 (the peanut is under shell #3) describe the situation. Since both rows describe equally likely possibilities, the chances are 50/50 that shell #2 hides the peanut, and the chances are 50/50 that shell #3 hides the peanut.
Our initial intuition is therefore vindicated. Smith’s turning over one shell and finding it empty changes the probability the peanut is lurking in any one of the remaining shells from 1 in 3 to 1 in 2. (It sure is nice to have a wonderfully intuitive sense for the obvious.) The probabilities changed because the sample space changed, just as changing the Monty-Hall game from ‘switch or stick with the initial choice’ to ‘flip a coin’ changed the probability of winning the peanut from 2/3 (if Smith switches) to 50/50. The probabilities in the Monty Hall case changed because the sample space relevant to the game Smith was playing changed. Having the ability to describe one of the remaining shells as ‘the initial choice’ expanded the sample space needed to support this description from two possibilities regarding each shell’s hiding or not hiding a peanut to three possibility cones each containing one or more possible paths to the current situation.
Now Morgenstern (German for ‘morning star) enters the scene, after Smith has put back the shell he turned over. (Say, this is shell #1) She does not know that shell #1 turned up empty. The peanut is still under one of the remaining shells. Elizarraraz points to shell #2 and asks both Smith and Morgenstern what are the chances the peanut is under that shell. For Smith, surely, the answer is 1 in 2. For Morgenstern, the answer has to be 1 in 3. For Elizarraraz, who knows where he put the peanut, the answer has to be either 0 or 1. Were Elizarraraz to point to shell #1, the answer for both him and Smith would have to be 0. What the probabilities are differs from the perspectives of each of the three because the sample space differs for each given what each knows.
From Elizarraraz’s perspective, there is no hiddenness, no ignorance given how things stand with regard to the peanut under shell situation, because his knowledge is complete regarding that situation. Obtaining within that perspective is certainty: either a probability of 1 or of 0. I will go out on a limb and say that within that perspective there is no sample space at all.
Uncertainty, a probability greater than 0 but less than 1, can exist only given a particular ratio of local ignorance and local knowledge. If one’s local knowledge of the peanut under shell affair is 0 (one does not even know if there is a peanut under one of the shells) and even Elizarraraz has forgotten if he has placed a peanut under one of them or not, one can appeal to a (possibly hypothetical) infinite (or at least extremely large) Mind that does know, in which case the probability is either 0 or 1. Or one can appeal to a brute, currently unknown fact of the matter, in which case, again, the probability that the peanut is under any given shell is either 0 or 1.
But if there is to be a probability greater than 0 or less than 1 within anyone’s perspective — including the Infinite (surely impossible for that one) or at least Extremely Large Mind’s — there has to be some ignorance, some hiddenness as well as some knowledge. For an omniscient God, everything has either a probability of 1 or 0. Ignorance/knowledge is a necessary condition for such probability in between 0 and 1.
It is also a sufficient condition for there being, within a particular perspective, for there being such a probability. All that Morgenstern needs to know is that there is a peanut under one of the shells, and all she needs to be ignorant of is which one, for there to be, within her perspective, of a probability of 1 in 3 that the peanut is under this shell, or that one, or the one remaining one. The probability is 1 in 3 within this perspective because Morgenstern’s ignorance/knowledge determines the sample space.
Knowledge/ignorance suffices for the existence of a probability between 0 and 1. But other factors help determine what exactly that probability is. In the non-Marty-Hall shell game, we need only to take into account the increase in Smith’s knowledge in determining the size of the sample space when he turns over one of the shells and discovers it to be empty. The probability the peanut is under one of the shells increases from 1 in 3 to 1 in 2 because the two events — the placement of the peanut under one of the shells and Smith’s turning over one of the shells — are both random and independent.
But in the Marty Hall shell game, Elizarraraz’s turning over one of the shells doubles the probability that switching will win the prize from 1 in 3 to 2 in 3. It therefore constitutes evidence that the peanut is likely to be under the shell that wasn’t Smith’s initial choice, whether Smith is in a position to utilize this evidence for not. Since, prior to the final step in the Monty-Hall shell game, the only difference between it and the non-Marty-Hall shell game is that in the former Elizarraraz’s turning over one of the shells is, because of his knowledge, not random and is independent of neither his placement of the peanut under one of the shells nor of Smith’s initial selection of one of those shells, it follows that this lack of independence is another factor in addition to Smith’s knowledge/ignorance helping to determine the specific probability of Smith’s finding a peanut if he switches (sticks with the initial choice). By itself, all his knowledge/ignorance does by itself is guarantee a probability of at least 1 in 2 should he switch (stick with the original choice) ; given the additional factor of a lack of independence in the event of choosing which shell to turn over, that probability increases to 2 in 3 (decreases to 1 in 3) should he switch (stick with his initial choice).
At the time of this writing, however, I am unable to say anything more succinct and more sophisticated regarding why this should be so other than ‘look at the chart shown above; given the all the ovals crossed out because Elizarraraz’s choice of shells to turn over was neither random nor independent of the other events, this is how all the possibilities panned out — all three of the possibility cones, and all of the possible trails within those cones. Stay tuned.
Today’s homage to Plato’s SYMPOSIUM is Channing Tatum. Again. Who would want anything more?