And now, the answer to That Time Of Year Again!

[theweaselking]

Last time, you were asked a question:

Given a pair of randomly generated animals, each with a 50/50 chance of being male/female, you investigated and discovered that one of them was male. What are the odds that the other is male?

If you said "Dice don't have a memory, babies aren't entangled, any random baby is male so the answer is 50%", you are wrong. You win nothing!
If you said "1/3", you are correct. You also win nothing!


The trick here is that, had the babies been born separately and considered separately, people who thought the probabilities weren't related would be correct. However, we're not serially generating random Velocibabies. We've generated A PAIR of random babies, and then filtered the possible results of random pairs of babies, and THEN looked at possibilities.

When you generate two random babies, the possibilities are:
1. Female/Female
2. Female/Male
3. Male/Female
4. Male/Male

Each of those possibilities is equally likely. With me so far?

By examining one baby (you don't know which baby), and determining that the baby is male, you *eliminate* some of your sample set. You KNOW you didn't get Female/Female. However, you still don't know which of F/M, M/F, or M/M you're in.

And, at the same time, those three results are equally likely. And in two of the three results, the other baby is Female. In only one of the possible sets of results is the other baby Male.

As such, the real question being asked is "given what we know, what are the odds that we're in the sample set with both babies being Male". And that is 1/3.

Comments

[scifantasy.livejournal.com]

So all that "you got it wrong last time, are you sure?" was pure Regis Philbin on your part. Thought so.

[dreamshade.livejournal.com]

Are you sure about that?

[stormfeather.livejournal.com]

Is that your final answer?

[kafziel.livejournal.com]

Just the tip?

[jaundicedaye.livejournal.com]

So because you looked at the gender of one of the probability of the other being male changed to 1:3 from the coin toss if you hadn't looked. Does that make it Schrodinger's Velociraptor?

[theweaselking.livejournal.com]

Not exactly. The probability of the other never changed - the problem is that the generation of the pair is a different thing.

[dscotton.livejournal.com]

Given the pairs you described:

1. Female/Female
2. Female/Male
3. Male/Female
4. Male/Male

Isn't a random observed male equally likely to be in position 2b, 3a, 4a, or 4b?

[twilightbloo.livejournal.com]

As velociraptors are reptiles it is plausible that the gender of the hatchlings is dependent on the temperature of the eggs during incubation - as with turtles and crocodiles.
Therefore if one hatchling is male and the eggs were both incubated at the same temperature then the odds of the other hatchling being male are greatly increased.

[It can still be true that exactly 50% of all cloned velociraptor babies are male, when averaged across multiple clutches of eggs incubated at varying temperatures.]

[corruptedjasper.livejournal.com]

But the one being examined was either number one or number two, and *unlike* the three doors thingy, the examiner does *not* have special info and (and this is an assumption) does *not* deliberately pick one that he knows is male.

Therefore if he examined hatchling 1, the odds are 50% on it being male, and the remaining two options show a 50% chance for hatchling 2, and if he examined hatchling 2, the remaining odds for hatchling 1 are also 50%. He presumably has a 50% chance of picking up hatchling one or two, which makes the total odds 25%+25%=50%.

At least, the Monty Don thing I've always seen described as the critical issue being that Monty Don does not pick a door at random, but deliberately picks one that is Not The Prize, and that that's where the extra information enters the system.

[theweaselking.livejournal.com]

Yes, in the Monty Hall Game the extra info enters because Monty chooses a non-prize.

In this case, the extra information is added in that we *know* that we're not in the F/F case. To make it more similar to Monty Hall, imagine if Monty didn't know where the prize was and was opening doors at random, but *in any case where he revealed the car, the results were ignored and the game restarted*.

[theweaselking.livejournal.com]

I didn't even see this question until Jasper responded to it!

And no, not exactly. You don't have 8 hatchlings in 4 pairs of 2 and are "picking a random male" - you have a pair of randomly generated hatchlings, and, post-generation, you learn that at least one of them is male.

The set of all random pairs of hatchlings is 25% FF, 25% MF, 25% FM, 25% MM.
We have generated a random pair, meaning we're in ONE of those 4 situations, and we don't know which one.

We then learn that we are *not* FF, but learn nothing else about the situation. This eliminates the FF option, but leaves the other options as all equally likely. In those other options, 1/3 has two males and 2/3 have only one male.