# Class 11: One Son Born Tuesday & Relevance. Update: Homework Added!

Uncertainty & Probability Theory: The Logic of Science

# Video

Rumble

Bitchute

HOMEWORK: RELEVANCE. A psychic is guessing cards from a standard poker deck of 52 cards. She must guess the exact card, suit & value. When she guesses right, she receives a chocolate. When she is wrong, she gets nothing. In the first 10 cards, she got 4 right and 6 wrong. What is the probability P = “She gets the 11th correct”? The exact number is good if you can get it, but it’s much, much more important to explain your reasoning. I mean, what is your Q?

# Lecture

## Tuesday’s Child

Last week’s homework was this: You meet a guy who says, “I have two children, one of whom is a boy who was born on a Tuesday.” What’s the probability, given this and only this Q, that both this guy’s children are boys?

Let me not answer that right away. Instead, look at this, which is everything that can happen with two kids, recalling that part of our Q is always that there are two kids:

First Boy, Second Girl
First Boy, Second Boy
First Girl, Second Boy
First Girl, Second Girl

Given only that, and not some host of tacit premises about human births, what’s the probability of both boys? Obviously (well, we haven’t yet justified other numbers for probability, which we do next week, but pretend we have), it’s 1/4. Because there’s only 1 out of 4 ways to get both boys.

Now suppose you add to Q that the first child is a boy. What is the new probability? Right: it’s 1/2. Because we only consider those times where the first child was a boy, which can only happen two ways, one of which results in a second boy.

Understand, it would not matter if these were children or doubloons, a B one and a G one. Or whatever. This question assumes nothing about biology. It only mentions certain states—just two of them as possibilities. That’s it. You must always be wary of tacit and implicit premises!

Now suppose instead of assuming the first child is a boy, you assume the first or the second can be a boy. We must now consider only those instances in which the first or second child was a boy. This is three possibilities, only one of which resulted in two boys. That gives a probability of 1/3.

So we have three different probabilities for two boys, depending on which information we condition on.

Which brings up the most magnificent point: probability says nothing about cause. Do not think about birth order causing anything. There is no cause. The timing is only relevant because it opens up possibilities. In the last Q, we open more possibilities for girls. That’s it. Nothing more.

Let’s expand the example. Guy says one of his two children is a boy born on the weekend.

Now this is where the big trouble starts. Weekend is irrelevant in any causal sense to what happens to the other birth. Or even the first birth. But we naturally think of cause. We can see there is no causal relevance, so we give a probability with that Q in mind. But not this Q.

The big step is to supply the tacit premise “Well, if one is born on a weekend, the other might be born on a weekday.” This is now two possibilities. If we supply that premise, then we get a whole different probability. Here are the possibilities. To save writing, the order is birth order, B and G, and Y is for weekday and D is for weekend.

BY, BY
BY, BD
BY, GY
BY, GD
BD, BY
BD, BD
BD, GY
BD, GD
GY, BY
GY, BD
GY, GY
GY, GD
GD, BY
GD, BD
GD, GY
GD, GD

If I next ask what’s the chance for two boys, given all this, and only all this, we have 4 out of 16 possibilities, for a probability of 1/4.

If I ask, what’s the chance for two boys given the first was born on a weekday (Y), then we count only those tines in which a first boy was born on a weekday. That’s 4 possibilities 2 of which were both boys for a probability of 1/2. If I ask, what’s the chance for two boys given the first was a boy, that’s 8 possibilities out of which 4 are both boys, again a probability of 1/2.

But if I ask, what’s the chance for two boys given either the first or the second was born on a weekday, then we have 7 possibilities, only 3 of which result in two boys. For a probability of 3/7!

The boy born Tuesday is exactly like that, but with implied (tacit) premises of the other seven days of the week being possibilities.

It turns out we can generalize this probability to a formula. I haven’t seen anybody do this before. But it turns out to be:

(2n – 1) / (4n – 1),

where n is the number of possibilities we grant for birth situations. When only the order mattered, n = 1. In the weekend/weekday example, we have n = 2. In the Tuesday example, we have n = 7.

That gives probabilities:

n | Pr = (2n – 1) / (4n – 1)
————————————
1, 1/3
2, 3/7
7, 13/27

And so on. Obviously, in the limit, the probability goes to 1/2, as our intuition wanted it to be. In the limit, there are infinite possibilities for birth possibilities, which then turns out to be irrelevant to our calculation.

As to relevance this excerpt.

Like with Monty Hall, if you doubt all this, you can play the game yourself and see that the probabilities are correct. Do just the birth order because that’s easiest.

This is an excerpt from Chapter 4 of Uncertainty.All the references have been removed.

We need the idea of relevant and irrelevant evidence. Irrelevance occurs when the probability of some proposition, given whatever evidence we already have, is not changed in the face of different and new evidence. The probability of P would obviously not change it we merely repeated evidence in Q which was deducible from extant evidence. That is, the probability of P given Q is identical to the probability of P given “Q & Q”; and if Q logically implies W, then the probability of P given “Q & W” must equal the probability of P given Q—but it is not necessarily the case that $Pr(P|Q)=Pr(P|W)$.

Causality and its lack also plays a role. For instance, the proposition “This cargo ship is over 100,000 tons” has high probability given the evidence, “Most cargo ships are over 100,000 tons, and this ship is a cargo ship.” If I add the true premise “Einstein enjoyed tobacco” the probability doesn’t change: the evidence is irrelevant because we cannot identify any causal connection, no matter how complex, between the propositions. This supposition falls short of proof, naturally, but that’s because we are in the realm of the contingent. I cannot prove to you that Einstein’s hobby is unrelated to cruise ship weights; nevertheless, induction demands it. There is no identifiable logical connection between cargo ships and Einstein’s smoking. Induction lurks here as everywhere. The reason we know Einstein’s smoking has no bearing on the question is because of induction.

The opposite of irrelevant is relevant. If I add the premise “This cargo ship is smaller than most”, the probability of our proposition changes: by how much is not known, of course, but it is obvious, given our (tacit) understanding of the English language, that this new evidence is probative, that it is relevant, and it is relevant because it is determinative.

It makes a difference when evidence is introduced. If P = “John was the killer” and we began with Q_1 = “The murderer was a dentist and John is a dentist” the additional minor premise Q_2 = “John graduated from Honest Ben’s Dental College” is irrelevant because we already know John is dentist. But if all we had was Q_3 = “The murder was a dentist”, adding Q_2 (and leaving out Q_1) is relevant. This point is taken up again in models (especially time series). Relevance is itself conditional on the accepted premises.

Evidential importance or weight to a fixed proposition is measured, when it can be measured, but how much the probability of a proposition changes on the addition of new evidence. Importance is thus also relative to the information already present in the premises. If we start with Q = “At least 6 of the 11 balls are orange, and this is a ball”, the probability of P = “This ball is orange” can be calculated (see below). If we add to Q the evidence “There are 6 orange balls”, the probability of P conditional on the augmented Q is 6/11. How much has the probability “moved”? Clearly by some appreciable and happy amount. We have reduced the choices from six to one. Weight will not always be quantitative, just like probability is not always quantitative. For instance, we could start with Q = “Some of the 11 balls are orange, and this is a ball”. As before, we revisit this technique when we discuss modelling.

How relevancy, irrelevancy, importance, and weight are measured, using for instance the techniques of entropy and the like are, of course, of tremendous practical interest. But those techniques, while interesting, are incidental to the philosophical points made here. Jaynes is an ideal reference work.

Subscribe or donate to support this site and its wholly independent host using credit card click here. Or use the paid subscription at Substack. Cash App: \$WilliamMBriggs. For Zelle, use my email: matt@wmbriggs.com, and please include yours so I know who to thank.

1. Cloudbuster

Homework: If the previous ten cards are not being added back into the deck, and she is aware of the values of those cards, then the probability is 1 in 42 (~0.0238). Presumably she will know not to guess the suits and values of any of the cards that have already been revealed. Her previous guesses, success or fail, have no bearing on the success or failure of her next prediction other than to reduce the pool of unrevealed cards. If the cards are being added back into the deck, the probability is 1 in 52 (~0.0192), as her knowledge of the previous cards then doesn’t affect the probability of her guesses. I am also assuming there is no cheating, methodological sloppiness or unconscious tester bias going on.

I discount that she is “a psychic” as there is no evidence that such can be the case. A better-than-chance outcome on a one-time series of 10 guesses is not “evidence.” It is expected that in long series’ of guesses, there will be streaks that are out of the range of the mathematical probability, but that the trend will regress toward the mathematical probability over time. Many a person at the craps table has been emboldened by a statistically unlikely streak into eventually losing all his winnings.

2. Charlie O

Ok, I’m pretty baffled by this homework, but I’ll show what I’ve got so far so the prof knows what some of us dopes at the back of the class are thinking.

I assumed that the pack was shuffled in the first place and the guessed cards are removed from the deck, because the question didn’t say anything about reshuffling the cards. Otherwise she could just take the top card every time (I know I would! In fact I’d lie 6/10ths of the time until some idiot bet me money that I couldn’t guess the next card!)

If the psychic was psychic she wouldn’t have got any wrong guesses.

If she’s guessing at “random” (yeah, I know) then the probability starts at 1/52 and declines towards 1/1 as the cards are removed.

If she continues her 4/10 average then the probability stays at 4/10ish until there aren’t enough cards to do that.

If eating a chocolate makes her get the next guess right, then there’s only one scenario where she can have 4 of the first 6 right, which is 6 wrong followed by 4 right, in which case the next guess MUST be right.

If eating a chocolate makes her get the next guess wrong, there are 4 scenarios where she can get 4 of the first 10 right. 1 of those means she MUST get the next one wrong, the other 3 don’t specify the next guess.

What’s my Q? The stuff about the chocolates and the fact that she’s got 4 out of 10 right so far, and it’s a normal pack of poker cards.

3. Phil R

I think there is at least one more option. What if the cards are removed from the deck (as Cloudbuster noted) but she is only shown the four cards she guessed correctly (or knows because she gets a chocolate), and the other six were placed face down in a separate pile? She wouldn’t know the value or suite of at least six of the cards.

4. Charlie O

After further thinking, I’ll leave all 52 cards in the deck because removing them makes the maths annoying without having a huge effect.

Intuitively I wanted to say the probability was 0.4, but I realized that multiplying by a constant is a linear function while correctly picking more cards gets exponentially harder so the probability must reflect that.

The probability of picking one card correctly would be 1/52, and wrongly 51/52.

I think the probability of picking 4 out of 10 correctly would be P(4,10) = (1/52)^4 *(51/52)^6

And P(5,11) would be (1/52)^5 * (51/52)^6

So can we say that the psychic has done X times better than expected so far, where X = 0.4 / P(4,10)?

Then would the probability of picking the next card correctly be X * P(5,11)?

As for the chocolate, we can set limits on the effect, from forcing the next card to be right, to forcing it to be wrong. There are C combinations which give 4 correct cards out of 10. Only one of them forces her to get the 11th card right, and one forces her to get it wrong. So the probability of either must be 1/ C, which must be a very small number.

5. Cloudbuster

Charlie O: “…correctly picking more cards gets exponentially harder so the probability must reflect that.”

No, it doesn’t get harder, it gets more unlikely. That’s an important distinction as harder implies that there is a skill element. Each pick is completely unrelated to any activity that has occurred before it, unless the previous picks add relevant information to the the choice — as does removing previously-guessed cards from the deck and letting the guesser know the values of them, thereby reducing the available options.

You can describe a streak of correct guesses as statistically unlikely — a far lower probability than a single correct, guess, but the fact that there were previous correct guesses does not actually effect the upcoming guess in any way. You’re mistaking the description of the probability for the reality. If she is picking one card from a deck of 52, the probability is 1 in 52. Always, no matter how many correct guesses she has had in the past.

Describing her as a psychic doesn’t make her one. Real world evidence suggests that the probability of a psychic being a fraud is 1. It is, in my opinion, irrelevant data. We’ve been given no info about this supposed psychic ability except her previous guesses and as I said in my original post, the existence of one unlikely guessing streak isn’t sufficient evidence. In a sufficiently large number of examples, unlikely streaks occur with some regularity. Every casino owner knows this. You can definitely hit one right out of the gate. Your calculations above for 4 out of 10 show, what, 1.217*10^-7? That sounds small, but in fact, streaks like that actually occur. I’m not willing to concede that it is relevant information. For all we know she could blow the next 100 guesses.

The chocolate, of course, is a complete red herring as it also suggests a skill component that doesn’t exist. It doesn’t matter whether you reward her for correct guesses, because you are not training her to guess better by doing so. Even if she is a psychic, we have no information suggesting that chocolate rewards increase success rates. It is irrelevant data, just like the label “psychic.”

6. Cloudbuster

Phil R: “(or knows because she gets a chocolate)”

That’s a good scenario and should be included, as it indirectly provides information. I wish the example had explicitly stated the fate of the guessed cards otherwise — put face down in a discard pile, face up, shuffled back into the deck, etc. I mentioned the two extremes — she knows all the previously selected cards or, effectively none (shuffled back into the deck) as those set the outer bounds of the possibilities.

Your interpretation would mean she’s guessing from a pool of 48 possible choices — she can only eliminate the four guesses for which she received chocolates.

7. Paul Fischer

First off I solved the Born Tuesday problem as you did by enumerating all the combinations and counting the BB combinations throwing out the one duplicate.
Now the psychic problem is a bit more complicated. To calculate the probability of just randomly guessing 4 cards out of 10 picked from a 52 card deck without replacement is a purely combinatorial problem and I think I calculated something like 2.1% or whatever. It doesn’t matter because our Q tell us 40%. She was 40% correct. Furthermore, we know there are 42 cards left in the deck. Also doesn’t matter because the psychic doesn’t obey counting arguments. So what are we left with…40%.

8. JH

Was my answer to the homework too short to be published? 1/42

9. TFBW

I assume game rules draw one card with replacement and reshuffling, so the zero-knowledge probability of a correct guess is fixed at 1/52 (~0.02). These rules weren’t specified, so I’m assuming what’s convenient, mathematically. I assume that past performance suggests future performance: I assume the psychic’s powers, if any, do not vary and are not influenced by chocolate rewards or prior success/failure. I want to converge towards P=1 if she guesses right all the time. Ten samples isn’t much to work with: if I use that data as-is, P for the next guess being correct is 2/5 (0.4, the average so far). Maybe I want to dial up the scepticism a little and dampen my expectations: I add, somewhat arbitrarily, the assumption that she scored 1/52 for 52 cards previously. It’s fake data, but it represents my sceptical expectations of zero knowledge and will become less significant with more data. Given those priors, my new estimate is 5/62 (~0.08).

10. Briggs

All,

No reshuffling. It’s as the description reads. A card is guessed, then set aside and not used again in the trial.

JH,

11. Paul Fischer

It also matters if the psychic knows whether or not if they got cards correct. As far as Q goes isn’t that important feedback to the psychic? Does the psychic know the cards are removed? What if the psychic hates chocolate?

12. If I assume that cards are drawn from the deck and not replaced and she Does not see the cards she gets wrong. Then it’s 1/48.

However, I have been thinking about the Monty Hall problem, and there’s an issue that I don’t understand. I believe [sort of] that it is not possible to write a simulation for this problem, that does not inherently beg the question. For example, the easiest way to do the simulation is To place the goat and then have your imaginary player pick a door. Scoring this as plus one for not switching if he gets it right and plus one for switching if he doesn’t get it right. But notice that this gives switching an automatic two chances out of three. Other approaches may be more subtle but amount I think to the same thing, so my question is, has anyone actually tried this with real people in a real environment?

P. S. I am dictating this to an iPad so the capitalization and missing words and other really odd stuff is an artefact of the iPad not I am quite sure my in ability to pronounce English properly smile.

13. Nate

Looking at the homework, I dived into Chapter 3 of Jaynes. He talks about the Urn problem:

“This suggests a new question. In finding the probability for red at the k’th draw, knowledge of what color was found at some earlier draw is clearly relevant because an earlier draw affects the number Mk of red balls in the urn for the k’th draw.”

In our card case, “correct” vs “incorrect” being the knowledge that we have, with 4 ‘corrects’ and 6 ‘incorrects’, I would expect this information to change the probability of our psychic predicting the next card.

What if we start like you said, Matt, and break this down to something easier-like picking numbers from a bingo cage, and only telling the psychic if she is correct or incorrect after she guesses for each pick.

If we stipulate 2 balls, B1 and B2, in the bingo cage, on the first selection, our psychic has a 1/2 probability of guessing correctly.

1GC = 1st Guess Correct
Pr(1GC | {B1,B2}) = 1/2

2GC = 2nd Guess Correct
1GI = 1st Guess Incorrect

Pr(2GC | (1GC + 1GI) {B1,B2}) is always 1, since we always know which possibility remains.

Then, extend the bingo cage to 3 bingo balls, B1, B2, B3.

Pr(1GC | {B1,B2,B3}) = 1/3

Pr(2GC | 1GC {B1,B2,B3}) = 1/2
This one’s easy, since we got the first right, the next ball is either of the remaining.

Pr(2GC | 1GI {B1,B2,B3}) = 1/3…
I enumerated the possibilites for this.
At first glance, getting the first guess wrong appears to be irrelevant for the 2nd guess.
However, if the psychic guesses wrong, she could always choose to only ever repeat the same number for the next guess:

R = Repeat same if previous guess is incorrect
Pr(2GC | 1GI R {B1,B2,B3} ) = 1/2

Also, this assumes she is not told the value of the prior number. If she is told the value of the prior number, then she obviously doesn’t need to guess it and has the same Pr as when she guessed correctly (1/2).

Finally, for the 3rd guess:

Pr(3GC | 1GC 2GC {B1,B2,B3}) = 1
Pr(3GC | 1GC 2GI {B1,B2,B3}) = 1
Pr(3GC | 1GI 2GC {B1,B2,B3}) = 1

When there’s at least one correct guess in the mix, getting one wrong isn’t a problem for guessing #3, because now our psychic is certain of the others.

Pr(3GC | 1GI 2GI {B1,B2,B3}) = 1/3
If she gets both the 1st and 2nd numbers wrong she’s still left with a 1/3 shot at guessing correctly.
Unless, of course, she always follows the repeat strategy.
In which case:
Pr(3GC | 1GI R1 2GI R2 {B1,B2,B3}) = 1

So our psychic has a path to always get the 3rd number correct, no matter what happens on the previous guesses.

So I expect that we could, in theory, extend this process to a deck of cards, with 52 items.

I imagine there’s some formulas that could simplify this, for the math part at least,

With 52 cards, we’d have to compute something like:
Pr(11GC | 1GC 2GC 3GC 4GC 5GI 6GI 7GI 8GI 9GI 10GI | {52 unique cards})
We’d have to add this to the other Pr() for the other guessing order possibilities to come up with a final Pr(11GC|All Q).
We’d also need to determine if there’s more strategies like ‘always repeat the next guess on an incorrect guess’, that further improves her odds.

I expect there’s math using permutations/combinations that we can use to compute the exact Pr()?

14. Briggs

Nate,

My heart soars like a hawk. I am very proud.

There is a method (not easy!), yes. It will be revealed Monday.

15. TFBW

I have a controversial take on the Boy Born on a Tuesday problem. It’s been doing my head in for the last couple of days, but I’ve finally sorted out why my intuition is screaming at me that the 13/27 answer is wrong. I’m going to walk through my solution in the style of another Monty Hall problem in the hopes that this makes it easier to follow, then discuss the key difference between my solution and the standard solution. Feedback is invited.

The standard solution is summarised as follows. There are 2 * 2 * 7 = 196 possible sex-day combinations for a first and second child. Of these, 27 contain at least one boy born on a Tuesday, and 13 of those combinations are both boys. P(both boys) is therefore 13/27.

Now consider the following variation on the Monty Hall game show.

Let us annotate sex as B or G, and days as numbers from 1 (Monday) to 7 (Sunday). A boy born on Tuesday is “B2”. Let us suppose that all 196 possible combinations for two children are written on one Scrabble tile each. Monty Hall has chosen one tile—we do not know which one—and placed it behind one of three doors marked “A”, “B”, and “C”. If the tile was both boys, it went behind door A; if it was both girls, it went behind door C; mixed pairs go behind B.

With no further information, we compute the probability of the tile being behind these doors as 1/4 for A, 1/2 for B, and 1/4 for C, as those are the proportions of the conditions across all tiles. In formal notation, using “Q” for our priors and “~” for negation, we have the following calculations.

P(A|Q) = 1/4; P(B|Q) = 1/2; P(C|Q) = 1/4;
P(~A|Q) = 3/4; P(~B|Q) = 1/2; P(~C|Q) = 3/4;

In the original Monty Hall problem, we would now get to choose. Instead, he tells us, “one of the children is a boy born on a Tuesday”. In other words, there’s a “B2” on the tile. This new information changes our assessment of the probability: Monty has effectively just opened the “C” door, revealing that the tile is not behind it. We revise our probability estimates in light of the new information using Bayes theorem. For Bayes theorem to apply, something has to go from “possible” to “known true”. In this case “~C” is now known true; its prior probability was 3/4. We also note that P(~C|AQ) = 1: given door A, “not door C” follows with certainty. The same applies for P(~C|BQ). Thus, given ~C, we can compute new probabilities for P(A|~CQ) and P(B|~CQ) as follows.

P(A|~CQ) = P(~C|AQ) P(A|Q) / P(~C|Q)
= 1 * 1/4 / 3/4
= 1/3
P(B|~CQ) = P(~C|BQ) P(B|Q) / P(~C|Q)
= 1 * 1/2 / 3/4
= 2/3

In plain English, the probability is now 1/3 for door A, 2/3 for door B, and (of course) zero for C. Door A is P(both boys). Note that “Tuesday” never factored into any computation: it was never relevant.

BUT NOW, what if we changed the game slightly? What if Monty tells us BEFORE he chooses and places the tile, “I will select a tile that has a B2 on it.” There are only 27 such tiles from which he can choose, 13 of which have both boys. Our initial probability calculation is now 13/27 for A, 14/27 for B, and 0 for C. No further information is revealed, so this is also our final calculation. This is the standard solution.

What if we go back to the original scenario and Monty plans ahead of time to pick a B2 tile, but he doesn’t tell us? Well, then our initial probability calculations are off due to lack of knowledge about reality, but they’re still the best we can do with the information available to us. However, we could ensure that Monty isn’t choosing ahead of time by making him pull the tile out of a bag, blind. This would mean that our initial calculations are good.

Now back to the ACTUAL original problem: the guy who tells us he has two children, one of whom is a boy born on a Tuesday. Did he select “a boy born on a Tuesday” at will, or were the parameters determined by circumstances beyond his control? One assumes the latter, no? Therefore this is like Monty pulling the tile out of a bag, not like him choosing the tile beforehand.

The other child is thus a boy with probability 1/3 unless the boy who was born on Tuesday was artificially selected for those properties. In this day and age, I can’t rule it out, but I’d rather assume it’s not so.

16. Charlie O

@Cloudbuster, thanks for your comment, I’m thinking out loud on this so I appreciate these little corrections.

As for the psychic, are we seriously saying that the psychic is tracking an entire deck of cards in her head to rule out possibilities? In which case she’s clearly a magician, which means that she’s clearly cheating and almost certainly switched the decks when we weren’t looking and statistics have nothing to do with it. May as well play Find the Lady.

17. Charlie O

Thinking about it a different way, If she guesses the first card right, then she has 51 cards remaining and is choosing from a pool of 51 possibilities. But if she guesses the first card wrong, there are 51 cards remaining but she’s still choosing from a pool of 52 cards as she doesn’t know the identity of the discarded card.

In our scenario, there are 42 cards remaining but she’s choosing from a pool of 48 cards as she can’t identify the 6 that were discarded unseen. There’s a 6/48 chance that her chosen card isn’t even in the deck, let alone the correct choice.

What if we started off with 48 cards in the first place? Let’s say we agreed with the psychic that we’d take all the aces out of the deck. If she gets the first 6 guesses wrong, she’s still choosing from a pool of 48. So what difference does it make?

In that case, our Q would include only the number of cards remaining and the number of wrong guesses.

18. Paul Fischer

All we can say about your psychic is that she got 4 out of 10 correct. We don’t know if she was shown the cards either wrong or correct. So we can’t even speculate (induce) if she has an edge by knowing what cards are not left in the deck. Since the cards are not replaced this problem lends itself to a hypergeometric distribution. So what do we have…we have these parameters for the hypergeometric
N=52 total cards
K=1 specific card to guess
n=11 number of draws
k=1 guess the 11th card right
P(X=k)=(1C1)X(51C10)/(52C11)=11/52

19. Paul Fischer

Nate, Who is Matt? A friend of yours?

20. Nate

Matt Briggs, the proprietor of this site?

21. Darin Johnson

In the weekend example, it’s true there are seven possibilities three of which have two boys, but they aren’t all equally likely because there are only two weekend days and five week days.

In the extreme, this leads to the “Girl named Florida” version, where the relevant probability approaches 1/2.

(I’m a week behind, so mea culpa if this has been discussed and I missed it.)