Another god-damn riddle

[Zylon]

Without the mathematical apparatus to analyze such problems statisticians couldn’t do their jobs, because the math tells us how to interpret the information the real world is spitting as us.

On the other hand, the expression “Lies, damn lies, and statistics” does exist for good reason.

[skedastic]

I disagree. The child-gender question is no different than if I toss two coins and one of them shows Heads. What is the probability of the other one coming up Heads?

As I vaguely recall discussing in this thread, let’s try a very slight variant of your problem: We play a game in which I toss two coins, look at them, and show you one. Then you tell me the probability the other coin is heads. We play a round and I show you heads. What’s your guess?

If I just select a random coin to show you, you should place 1/2 probability on the other coin being heads. But suppose I follow the rule: Whenever I get a heads and a tails, I will always show heads. Now you should place 1/3 probability on the other coin being heads.

In the boy-girl problem, suppose you knew that social convention in your town holds that boys should never ever stand in windows alone unless they have a brother. Would you still think the other child is a boy with probability 1/2, given you see a boy in the window?

Suppose you have a drug which, unbeknowst to you, cures an otherwise uncurable disease with probability 1/2, independently across patients. You collect a random sample of patients and give half the drug and half the placebo. A month later you collect data on how many of the patients in the two groups are sick. In the sample you started with, you’d find that about 1/2 the treatment group are cured. But what if patients given the drug drop out of the trial if they don’t find they’re getting better, because of nasty side effects from the drug?

In both cases, the answer you get depends on the process which selects units into your sample (children from households / patients from treatment statuses). Analysts working with data with selection problems, which are ubiquitous in many fields, have to deal with such issues every day.

[Mightynute]

As I vaguely recall discussing in this thread, let’s try a very slight variant of your problem: We play a game in which I toss two coins, look at them, and show you one. Then you tell me the probability the other coin is heads. We play a round and I show you heads. What’s your guess?

If I just select a random coin to show you, you should place 1/2 probability on the other coin being heads. But suppose I follow the rule: Whenever I get a heads and a tails, I will always show heads. Now you should place 1/3 probability on the other coin being heads.

I’d still place 1/2 probability on it, because if you show me heads, the only possible outcomes are it being HEADS-heads or HEADS-tails. That’s one out of two.

[Bahimiron]

And if I roll a die then the probability of me getting a six is always 1/6. Therefore the answer to the Tuesday boy problem is 1/6.

Lightning never strikes the same place twice. So the probability of the other kid being a boy is 0.

I’d still place 1/2 probability on it, because if you show me heads, the only possible outcomes are it being HEADS-heads or HEADS-tails. That’s one out of two.

Or tails-HEADS. One out of three.

[Dave_Perkins]

Lightning never strikes the same place twice. So the probability of the other kid being a boy is 0.

Curses, you’re right. I’m going to have to use Bayes Formula!

[Mightynute]

Or tails-HEADS. One out of three.

I’m still not sold on that. I’ll put my money on a coin flip having a 50% chance, every time.

[Dave_Perkins]

Mighty, do you think the probability of rolling an 11 on two six-sided dice is 1/36?

[Hawkeye_Fierce]

Or tails-HEADS. One out of three.

Well, if that’s true, you need heads-HEADS too. Two out of four.

[Mightynute]

Mighty, do you think the probability of rolling an 11 on two six-sided dice is 1/36?

If by an 11 you mean a 1 and a 1, then yes. Otherwise, there are multiple combinations that can produce a result of 11. Not sure where you’re going with this.

[Dave_Perkins]

Sorry, heh… I mean eleven. You can get 6-5 or 5-6, as you allude to. Each die is different, then, yes. So what’s the probability of getting a total of ten? Do you agree that there are three ways: 4-6, 5-5, 6-4, so it’s 3/36? You don’t count 5-5 twice, right?

[Mightynute]

Sorry, heh… I mean eleven. You can get 6-5 or 5-6, as you allude to. Each die is different, then, yes. So what’s the probability of getting a total of ten? Do you agree that there are three ways: 4-6, 5-5, 6-4, so it’s 3/36? You don’t count 5-5 twice, right?

Sounds about right to me.

[Dave_Perkins]

If I rolled two dice and said my total was 10 and that one of the dice read 4, would you say that the other die had probability 1/6 of being 6?

[Omniscia]

Depends on how many sides the die has. And how many of those sides are sixes.

[Dave_Perkins]

Your face!

[skedastic]

I’d still place 1/2 probability on it, because if you show me heads, the only possible outcomes are it being HEADS-heads or HEADS-tails. That’s one out of two.

The problems people have with basic probability never cease to amaze me.

[Mightynute]

If I rolled two dice and said my total was 10 and that one of the dice read 4, would you say that the other die had probability 1/6 of being 6?

Assuming a standard six-sided die, I’d say it had a 100% possibility of being six, since there’s no other option, right?

[Zylon]

I’d still place 1/2 probability on it, because if you show me heads, the only possible outcomes are it being HEADS-heads or HEADS-tails. That’s one out of two.

No, he’s right. Look at it like this (using binary instead of quarters)-- the possible outcomes are:

Quarter #1: 1 1 0 0
Quarter #2: 1 0 1 0

Four possible combinations, all equally likely, so the odds of each combination coming up are 25%.

But if following the “Must show heads if any heads flipped rule”, on those particular rounds where the rule comes into effect, the possible outcomes are culled to:

Quarter #1: 1 1 0
Quarter #2: 1 0 1

Or from the victim’s perspective:

Hidden: 1 0 0
Shown: 1 1 1

So while your statement that “the only possible outcomes are it being HEADS-heads or HEADS-tails” is correct, the trick is realizing that HEADS-tails is twice as likely to come up than HEADS-heads.

As a corollary, and to put a bullet in the “two outcomes = 50% odds” notion, if you’re shown tails under this rule, the probability that the other quarter is tails is 100%.

[Houngan]

The exact opposite is true: probability is not very intuitive, and math helps us understand reality when our intuition fails us.

Devlin begins his following column by noting that a bunch of applied statisticians wrote to tell him that his answer was wrong. That doesn’t surprise me, because the logic of this puzzle commonly plays out in applied statistics.

Notice the boy-girl problem has “two phases” just like the Monty Hall problem: in phase one some process selects a child whose gender we learn, and in phase two we assign probabilities to the other child’s gender. Analytically, this looks just like: in phase one some process selects patients to complete randomized trials for a new drug, and in phase two researchers evaluate the probability the drug works. Just like in the boy-girl problem, the answer hinges on how the stochastic processes in the two “phases” interact.

Without the mathematical apparatus to analyze such problems statisticians couldn’t do their jobs, because the math tells us how to interpret the information the real world is spitting as us.

Reading back, I see now that how he phrased it makes it a two phase problem, mea culpa. I still say it’s damned wanking and little more, but I get what’s going on now. Edit: Though I do stand by my observation that as you add variables it goes towards 1/2 mathematically. And since we can specify a near-infinite number of variables to extend the original question (boy/girl older/younger longer hair/shorter hair and any other binary <> aspect) it winds up at 1/2 in every instance except math problems.

H.

[Mightynute]

No, he’s right. Look at it like this (using binary instead of quarters)-- the possible outcomes are:

Quarter #1: 1 1 0 0
Quarter #2: 1 0 1 0

Four possible combinations, all equally likely, so the odds of each combination coming up are 25%.

But if following the “Must show heads if any heads flipped rule”, on those particular rounds where the rule comes into effect, the possible outcomes are culled to:

Quarter #1: 1 1 0
Quarter #2: 1 0 1

Or from the victim’s perspective:

Hidden: 1 0 0
Shown: 1 1 1

So while your statement that “the only possible outcomes are it being HEADS-heads or HEADS-tails” is correct, the trick is realizing that HEADS-tails is twice as likely to come up than HEADS-heads.

As a corollary, and to put a bullet in the “two outcomes = 50% odds” notion, if you’re shown tails under this rule, the probability that the other quarter is tails is 100%.

cue noise of “aaaaaah” I get what you’re saying now.

This is why I hate probability and its fallacies. If I have to guess the gender of a kid, I’m going to say “Hey, kid! You a boy or a girl?” and remove all doubt. Fucking Schrodinger and his damn cat.

[Zylon]

I still don’t get the Boy-Girl Paradox though.