The Paradox of Irrelevant Information

By John "Birdman" Bryant


October 26, 1997

To: Fellow Mensans, Fellow Libertarians and other Highly Intelligent People

From: Mensa's Resident Iconoclast (Tom Elliott's words: You did see Tom's review of my books in the October Mensa Bulletin, did you not?), aka John Bryant

Subject: An invitation to cast your vote on the Paradox of Irrelevant Information

For the last several weeks, about 20 of us (mostly Mensans, but with a scattering of Promethians and Libertarians) have been having a lively (and often acrimonious) exchange over a problem recently addressed by IQ- superstar Marilyn vos Savant, a problem which I have named the Paradox of Irrelevant Information (PII). Because I and most of my opponents disagree with each other just as strongly as we did at the outset, at my invitation (to all opponents), two of them have agreed to write refutations (or, in my opinion, "refutations") of my position, with the view that the Mensa email membership should have an opportunity to read our essays and then vote on whether they thought my position was right, or whether that of my opponents (including Marilyn, as it happens) is. You are hereby invited to read these three essays and cast your vote accordingly.

It may be noted that Marilyn took up this problem for a third time in her Parade Magazine column of 19 Oct, due apparently to the particularly large amount of mail (and interest) generated by this topic. Marilyn, as it turns out, did not have the benefit of my input, or she might have responded differently (she basically just restated the Standard Analysis (SA), given below.) She did, however, report the result of a statistical survey which seemed to support her analysis. In view of this, I should make it clear that I am not arguing that the statistics she cites are "wrong"; rather I am arguing that the SA leads to a contradiction. What that means is that the SA may give the right answers sometimes, but that because it leads to a contradiction, its results must always be suspect. But if I am right about the SA being wrong, then that fact would seem to shoot a big hole in probability theory. For me, the point in arguing my case is simply to find either a confirmation that I am right or a convincing argument that I am wrong, by presenting my arguments to the smartest and best informed people I can find. My thought is that if Mensans find my arguments convincing, perhaps the Mensa Bulletin editor will run a feature to let all Mensans hear the arguments, and maybe even Her Royal High-IQ-Ness Marilyn will deign to confront them.

As an aside, it may be noted that the explanation of the PII given in my first letter to the Mensa email membership has undergone considerable evolution due to the arguments thrown at me by opponents. But useful as those arguments have been in helping me clarify my thinking, they have nonetheless convinced me even more strongly that my position is basically right, since these counterarguments have been -- in my view, at least -- so totally ineffective in putting any real dent in my position. But whatever your view on the subject, the essays in this letter ought to convince you that the PII is not a simple problem which is answerable -- as HRH Marilyn would have us believe -- in a single paragraph or column.

So don't forget: After you finish reading, remember to send me your vote! Comments and criticisms -- to say nothing of a solution -- will also be appreciated.


The Paradox of Irrelevant Information

The standard situation: A Man has exactly two children, the older a boy. A Woman has exactly 2 children, one of whom is a boy. What is the probability of the Man's younger child being a boy, and what is the probability of the Woman's "other" child being a boy?

The SA: We assume that the probability of a boy (or girl) at any given birth is 1/2. This means that the probability of the Man's younger child being a boy is 1/2. Another way to characterize this is to say that the "event space" for the Man is the set of equally probable events {BB, BG}, and the probability of BB is the ratio of the number of "successes" (BB) to the number of possible events (BB, BG) in the event space, namely, 1/2.

As for the Woman's case, because we do not know whether it is the Woman's older or younger child who is the boy, the event space is the set of equally probable events {BB, BG, GB}, and the probability of the event "both boys" is the ratio of successes (BB) to possibilities (BB, BG, GB), ie 1/3.

Explication of the SA in terms of coin-flipping: Coin-flipping can be made to model the Man/Woman situation as follows: (1) Represent the event of the Man's younger child as a coin flip, with H for boy, T for girl, so that the event space becomes {HH, HT}; (2) represent the Woman's event space similarly as flipping two coins simultaneously, and eliminating all {TT} events from consideration, so that the event space becomes {HH, HT, TH}. Everyone agrees that the probabilities represented by the coin flips are the same as in the SA for the Man/Woman situation.

The intuitive objection to the SA: (Note: This is not a precise objection, but rather what is known in mathematics as a "motivation".) Consider the following statements: (1) Of x's 2 children, the older is a boy (2) Of x's 2 children, one is a boy The SA says in effect that (1) implies a higher probability for x's second (other) child being a boy than does (2). This, however, is intuitively unsettling, because olderness is not a quality which impacts on the probability of sex (ie, sex and age are independent), and hence olderness is irrelevant to the probability of being a boy. The supporters of SA have two basic responses to this: (a) Statement (1) -- and in particular, the assertion of olderness -- gives "more information" (b) Any category matters, hence is not irrelevant. (To be understood, this point needs explanation; but because it is irrelevant to our arguments below, we merely state it here for the sake of completeness.) Because the above arguments are all somewhat indefinite, they settle nothing. They merely point to the field of battle.

Using an additional category with SA: Note that virtually any comparative quality whatsoever (eg, fatter, smarter) could be substituted for "older" and the SA would not be essentially changed. Now let us suppose that we know a comparative quality of the Woman's boy (or one of them), eg, suppose we know that the Woman's fatter child is a boy. Then we can use this information in the SA to show that the probability of the Woman's second child being a boy is 1/2, not 1/3, in exactly the same way we use olderness to show the probability that the Man's younger child is a boy equals 1/2. We can use the two bits of information (olderness, fatness) separately or simultaneously: If we use them simultaneously, the event space for the Man -- using F/S for fatter/slimmer and O/Y for older/younger -- can be represented as (2 boys) {BOF, BYS} + {BOS, BYF} + (1 boy, 1 girl) {BOF, GYS}, {BOS, GYF}; while the Woman's event space can be represented as (2 boys) {BOF, BYS} + {BOS, BYF} + (1 boy, 1 girl) {BOF, GYS} + {BYF, GOS}. (Note that the Man's case for (1 boy, one girl) excludes the events {BYF} GOS}, {BYS, GOF}; while the Woman's case for the same excludes {BOS, GYF}, {BYS, GOF}.) This event space yields a probability of 1/2 for the second boy for both the Man and the Woman, just as does using the bits of information (O/Y, F/S) individually.

Why the SA is contradictory: We have already noted how virtually any comparative category will yield a higher probability for the second child being a boy. While this is counterintuitive against the SA, the situation is actually much worse, for rather than a comparative category (eg, older/younger, fatter/slimmer, smarter/dumber) we can simply use the category "one/the other". This is counterintuitive to the contention that the SA is valid not only because this new category gives no new information whatsoever (recall that one justification for SA was that the older/younger category gave "new information") but it is actually a non-category category. Or to put it another way, we can raise the probability of the Woman's other child being a boy without any new information whatsoever by simply recognizing that "one/the other" may be used in SA as a category. Which means that SA has two possible interpretations, one of which yields a  probability of 1/2 for the Woman, and one which yields a probability of  1/3; ie, SA is contradictory! 

Arguments against using "one/the other": SA supporters will say that "one/the other" does not make a unique designation for the children, unlike the other comparative qualities (fatter/slimmer, etc), since either child could be "one" (or "the other"). But as far as I can see, this carries no more weight (and in fact, much less weight) than the objection against the SA that using irrelevant information to determine probabilities is illegitimate. But in any event, as I noted in earlier letters, the nature of using comparative qualities in the context of the SA constitutes nothing more than slapping an arbitrary label on the objects of the event space (in the present case, the children), so "one/the other" does quite as well for labeling as anything else, altho it is different from the other labels in the sense that we can only imagine that they uniquely designate the children, since we cannot know "which" child each designates. However, it is my contention that this fact doesn't matter, because even if we knew which was "one" and which "the other", it would have absolutely no effect on the analysis. The above response is reinforced by a completely different approach, namely, the fact that we know that there always exists a comparative quality Q (and in fact there exist infinitely many) which uniquely distinguishes each child, eg, "noisier/quieter at time t". Thus if Q exists (and we know it does), then it is -- unlike "one/the other" -- a real quality; hence this overcomes the potential objection against "one/the other" that it is empty of content. In fairness, it should be noted that, in general, we do not know what Q is; but since we know it exists, this seems adequate for use in SA.

Responding to a final objection: The major argument in the minds of several of my correspondents in favor of SA appears to be that, because the coin- flipping situation described earlier seems to represent the Man/Woman problem precisely, and since the SA seems to correctly describe the coin- flipping situation, it follows SA must also correctly describe the Man/Woman problem. To this I admit that I do not have a conclusive response, except in the sense that I have discovered a contradiction in the SA. Perhaps the problem is that the coin-flipping does not properly represent the Man/Woman situation, as suggested by my discussion of labels. If all else fails, we can simply say that this is the crux of the PII, namely, that we seem -- like Buridan's ass -- to be drawn in two opposite directions at the same time. To which the supporters of SA will undoubtedly respond that when an ass is drawn in two opposite directions, the result can only be the exposure of an asshole.

Additional reinforcement: Libertarian Eric Carroll came up with an independent analysis of the PII which reaches the same conclusion as mine. While he has been known to flip-flop on his positions, I looked his work over and it seems right to me, and in fact quite elegant. However, the last time I heard, he had begun to waffle. All I can say is, Eric, stiffen that backbone!

Final remarks: It is my suspicion that the resolution to the PII is in having better criteria for setting up problems. But because I am not well- enuf acquainted with probability theory to know what impact my work might have, I can only speculate on this point. But whatever implications the PII carries for probability theory, it is clear from our discussion that there are some rather confusing elements in this area which no one has yet succeeded in properly clearing up, Marilyn and my correspondents included. Which means that there is hope, after all, that Las Vegas can be beaten.

What follows below are two essays by Mensans who believe I am mistaken and who have agreed to write responses to my essay above for publication. The first is by Anthony C. Patzelt (; the second by Tom Serb ( I have not changed the text of either of their essays except to indicate with "note 1", "note 2" etc the places in the respective essays to which my comments at the end of each essay apply. My general reaction to these essays is that they are illogical and do not prove their point. Specific problems with these essays are noted in my comments following the essays.


Tony's essay:

I have attempted to relay this to John Bryant in private correspondence to allow him to correct himself. I have tried various illustrations in an attempt to point out his logical flaw. He seems to be ignoring this information while continuing to publicly bluster.

John seems to believe that by introducing what he terms "irrelevant information" into a probabilistic Standard Analysis that you generate a "paradox." I think "paradox" is far too strong a term and I might feel more comfortable with "counter-intuitive."

Now, we know that the statistics, the experimental data prove John wrong. So either we can believe that John has discovered a "paradox" that shakes the foundations of probability theory, or more likely there is a flaw in John's logic.

His logic is basically: "If SA predicts blah, blah using irrelevant information"

Well, lets stop right there. John has been very crafty in concocting this puzzle. He storms quickly through the SA, and guess what, he is using it incorrectly.

John has tried to bamboozle us into believing that the SA is based on THE NUMBER OF POSSIBLE OUTCOMES as opposed to the correct, NUMBER OF ALL POSSIBLE WAYS TO REACH ALL POSSIBLE OUTCOMES.

I will illustrate this with the more classic presentation of this problem, which contains no potentially "irrelevant" information. (note 1)

You have a black, velvet sack. Inside the sack are two blue balls and one green ball. The balls were all manufactured in the same way, so there is not way for you to tell which color a ball is when you reach into the bag to pull one out. So it is equally probable that you pick any ball.

The equivalent of the Man situation:

You reach into the bag and pull out a blue ball, what is the probability that the next ball you pick will be blue? Note that this is predicated on pulling out a blue ball first. If you pull out a green ball first, you have to try again.

John's (incorrect) analysis:

Event Space:

First Ball Second Ball

Blue Green Blue Blue

So, if the first ball you pick is Blue, the probability of the second ball being blue is 1/2 or 50%

Note that for the correct analysis we need to label the blue balls Blue(1) and Blue(2) to avoid the contradiction of retrieving Blue(1) first and then Blue(1) second or the contradiction of retrieving Blue(2) first and Blue(2) second. (note 2)

The correct analysis:

Event Space:

First Ball Second Ball

Blue(1) Blue(2) Blue(2) Blue(1) Blue(1) Green Blue(2) Green

So, if the first ball you choose is blue there is a 2/4 or 50% chance that the second ball will be blue.

The equivalent of the Women situation:

You reach into the bag and simultaneously extract two balls, what is the probability of both balls being blue?

John's (incorrect) analysis:

Event Space:

One Ball The other Ball

Blue Blue Green Blue

Therefore the probability of retrieving two blue balls is 1/2 or 50%.

Again, for the correct analysis, we need to label the blue balls Blue(1) and Blue(2) to avoid the contradiction of selecting Blue(1) and Blue(1) simultaneously or the contradiction of selecting Blue(2) and Blue(2) simultaneously.

The correct analysis:

Event Space:

One Ball The other Ball

Blue(1) Blue(2) Blue(2) Blue(1) Green Blue(1) Green Blue(2) Blue(1) Green Blue(2) Green

(note 2a) So the probability of you picking two blue balls 2/6 or one in three.

I know that John will argue that the labels on the blue balls are "irrelevant" because this lets him slide past the problem of THE NUMBER OF POSSIBLE OUTCOMES versus THE NUMBER OF ALL THE WAYS TO REACH ALL POSSIBLE OUTCOMES. (note 3)

Unfortunately, that is not "irrelevant." This leads to the contradictions highlighted above.

So, not surprisingly, John's misapplication of the Standard Analysis has led to his supposed "paradox."


The following are my comments on Tony's essay:

General remark: As is obvious, Tony did not actually bother to respond to my arguments except in an indirect way. You may judge for yourself as to whether this means he cannot understand them, or simply prefers to avoid dealing with them. In any event, I find his essay completely unconvincing.

(note 1) Using the balls-in-a-sack situation to represent the children situation is dubious at best, so any conclusion drawn from it is dubious. Furthermore, it adds unneeded complexity to a situation which can use all the simplification it can get.

(note 2) Tony blithely decides to slap labels on the balls without any consideration of the implications which this has for his analysis. It is the act of changing objects from unlabeled to labeled which I used to show SA to be contradictory, so if Tony justifies his labeling, how does he avoid my contradiction?

(note 2a) Unless Tony is using "One Ball" and "The Other Ball" as labels for the balls in addition to 1 and 2, then lines 1&2, 3&5 and 4&6 represent the same event, since the balls they refer to are supposedly not ordered in being drawn from the bag.

(note 3) I did not so argue: see note 2. I also find Tony's apparently- crucial distinction between possible outcomes and possible ways to reach them opaque.


The following is Tom's essay:

When I was a child, my grandfather told me a riddle: Three men walk into a hotel and ask for a room. The manager tells them that the room costs $30, so each man puts in $10. After they leave for the room, the manager realizes he gave them the key to a $25 room, and he sends the bellboy after them with the $5 overpayment.

The bellboy decides to pocket some of the money; he gives each of the men $1, and keeps $2 for himself. Now each man paid $10, and got $1 back, so each really paid $9. $9 times three men is $27, and the bellboy kept two, which makes $29. Where did the other dollar go?

This was somewhat confusing to me as a lad, until I wrote it out, and saw the error: the correct expression would be ($9 * 3) - 2 = $25, the cost of the room. Funny how words can get in the way of understanding.

With that in mind, let me propose my proof exposing the 'paradox' posed by John Bryant. John has challenged others to prove the validity of standard analysis in response to his approaches; I'll not do that -- it should be sufficient to prove he is wrong.


A man has two children, and the older one is a boy. A woman also has two children, and at least one is a boy. What are the probabilities that each of them has two boys, given that the likelihood of a boy or girl for any given birth is equal?


1. Probability may be expressed as a fraction: the number of desired outcomes divided by the number of total outcomes. 2. The chance of any given child at random being a boy is 0.5 3. The chance of any two unrelated events occurring together can be found by multiplying their probabilities together; in this case, the chance of two boys is (0.5) * (0.5), or 0.25. This becomes the numerator of the fraction, the outcome to be found in either case.

For the man: 4. The denominator will be all events which result in an older boy (1.0 by definition) times the events that result in a boy at random second birth (0.5). 5. This can be expressed: (0.5 * 0.5) / (1.0 * 0.5) which reduces to 1/2

For the woman: 4. The number of total outcomes that will satisfy her parameters is the sum of probabilities of two-boy occurrences (shown to be 0.25) plus the chance of a mixed boy-girl occurrence. 5. Mixed boy-girl probability can be found by taking all births (1.0) and subtracting the probability of two-boy and two-girl births. 6. The probability of two boys is expressed: (0.5 * 0.5) / ((0.5 * 0.5) + (1.0 - (0.5 * 0.5) - (0.5 * 0.5)) which reduces to 1/3.


John argues that this is counter-intuitive, because "this boy", "fatter boy", etc. can be substituted. Let's take three children, A= a 10 year old boy, B = an 8 year old girl, and C= a six year old girl (note 1). These result in the following combinations, with either child selected first; columns note whose parameters are satisfied, and mark those resulting in two boys with asterisks:


Note that of all possible outcomes, the man will have two boys 2/4, or 50% of the time; the woman will have two boys 2/6, or 33% of the time.



John argues that Standard analysis implies a higher probability for the man's second child being a boy than it does for the woman. Of course they do; so are her odds of having an older boy less than those of the man (0.67 versus 1.0). He finds this unsettling because olderness does not impact the probability of sex. (note 2)

This is certainly true in the man's case. However, if we accept the fact that the woman's older child may be a boy 50% of the time, we find the following: Boy-boy = 25% (50% of 50%) Boy-girl = 25% (50% of 50%) Girl-boy = 50% (100% of 50%) (note 3) This is by nature counter-intuitive. We now have twice the likelihood that a mixed-sex set will consist of an older girl. (note 4)

In the case of the woman, the second child's sex will be a dependent variable if the first child is a girl. John correctly defined an event space as one in which each outcome is equally likely; if we accept the fact that a boy may be older 50% of the time in the woman's case, we may not create a Standard Analysis event space to solve the problem: by definition, it will not work. Returning to the mathematical expression, the chance of either child at random being a boy is 2 in 3, and we get a correctly weighted space: Boy-Boy = 33% (1/2 of 2/3) Boy-Girl = 33% (1/2 of the other 2/3 of first boy) Girl-Boy = 33% (the remainder; note that independent of birth order, B/G and G/B must be equally likely)

The error here is the assumption that if all birth possibilities are equal, then the first child has an equal chance of being a boy or a girl. This is only true if we include the girl-girl possibility, which has been eliminated in the problem definition.


Will it? Let's restate this and say that one of the woman's children is fatter, and that child is known to be a boy. The fatter child will be shown in caps.

We now have, using the same A,B,C possibilities shown above:

Ab, Ac, bA, bC, Ca, Cb, all of which fulfill her parameters. 50% of the time the fatter child will be the older, and 50% the younger. What John is actually doing is finding that events (Ab, bA, and aC, Ca) are pairs of identical events, and therefore reduce the universe of possibilities.

For this we need to look again at weighting: in any given set of random boy-girl births, considered in order, any one ordered pair will occur 25 percent of the time. If we exclude Ab and bA as being identical, we have to ask why we are excluding the possibility of an existing Bc set (a boy- girl set with a fatter girl). John's addition of this label does in fact reduce the universe of possibilities; he has re-stated the problem in terms other than those originally proposed.

Recognizing that fat and chosen are independent identifiers, we would find: a random 25% chance of two boys with either fatter; a 12.5% chance of a boy-girl with fatter boy (the other 12.5% of the time the girl will be fatter), and a 12.5% chance of girl-boy with fatter boy. Once again, the outcomes he proposes are not suited to an event-space analysis, as they are either not equally likely.

To state that one child is fatter than the other makes no change to the numerator; however, associating this quality with sex WILL change the denominator. That's exactly why the man's and woman's probabilities differed in the first place.


To substitute "This child", and say that we've chosen "this child" from the woman's two, and found it a boy, we can easily see that 50% of the time the boy will be the older child, and 50% the younger. The odds don't change: although if the boy is the older one, 1/2 of the time he will have a sister, and 1/2 of the time a brother, the same is true of finding if we find a younger boy first. If we treat each case individually and then combine them, we must eliminate the duplicates from the resulting set: AC,CA,AB,BA + AC,CA,BC,CB = AC,CA,AB,BA,BC,CB

Considering each case separately, and then eliminating the true duplicates, will give us a proper solution for an event space analysis, but both possibilities must be considered simultaneously, and we must recognize those that are in fact the same event.


Any information attached to the sex of a child will change the odds. For the information to be truly 'irrelevant', it cannot restrict the variable to be found: in this case, sex. To state that the woman has at least one boy, and at least one fat child, will have no impact; to state that the woman has a fatter boy will change the nature of the beast.


My comments on Tom's essay: 

General remarks: While Tom is slightly better in attempting to address my actual argument, I nonetheless find his argument opaque, so that while there might be something to it, I am unable to figure out what it is.

(note 1) Like Tony, Tom cannot seem to avoid the temptation to complexify a situation which needs to be simplified (eg, why mention ages?). He also seems to have made a mistake in substituting "girl" for "boy" in referring to the six-year old.

(note 2) I'm not sure what Tom is saying, but it does not seem to properly represent anything I have said.

(note 3) I do not understand the basis for this statement

(note 4) The latter statement is irrelevant to the matter at hand as far as I can see.


Note (Jan 2001): Three years ago I burned out on this particular subject. However, I am posting the final essasy, and if there is some interest from the cybercommunity, I may actually get back into it.


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