And now, a solution.

[theweaselking]

Yesterday, is was That Time Of Year Again!

The question is there - go ahead, read it again if you want. I'll wait.

If you said "There is a 1/2 chance that the second ticket matches the first ticket", you are wrong! You win nothing.

If you said "There is a 2/3 chance that the second ticket matches the first ticket", you are correct! You also win nothing.

This is the Bertrand's Box Paradox, with a few minor changes to make it harder to google the solution.

The SIMPLEST way to explain the problem is that the drawers are a red herring. The *real* question, here, is "You pick a box at random. Do the tickets in the box match?" - and since two boxes match and one box does not match, the odds that you've picked a box with two matching tickets is 2/3.

The more complicated version of the solution writes out all the possibilities:

Marking a Disneyworld ticket as W and a Disneyland ticket as L, this gives the three boxes as:

B1: WW
B2: WL
B3: LL

Assume you picked a W. You cannot *possibly* be in B3, but you might have chosen:
W of B2, and so the other ticket is an L
W1 of B1, and so the other ticket is a W
W2 of B1, and so the other ticket is a W.

Since the choice of box and drawer was random, each of these three possibilities is equally likely.
Since they're equally likely and 2/3 of them result in the match being a W, you've got a 2/3 chance of your second ticket also being a W.

(If you picked an L at first, repeat the analysis while reversing W and L. And then hit eBay to unload those dogs.)

PS: A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?
(PPS: The birth ratio of beagles is 50/50 for the purpose of this problem)

Comments

[scifantasy.livejournal.com]

See, my problem was, I thought it sounded too easy. (And it seems I was right.)

As to your PS: Whatever the probabilities on beagle birth gender are, which is pretty much always assumed to be 1/2, no?

[theweaselking.livejournal.com]

Doesn't the PS seem a bit *too easy*?

[scifantasy.livejournal.com]

The question boils down to, "what is the probability that a given beagle is male?"

And we know the answer to that one.

[theweaselking.livejournal.com]

Is that your final answer?

[scifantasy.livejournal.com]

Yes, Regis.

[theweaselking.livejournal.com]

I'm sorry, you do not win a prize.

(Had you been right, you wouldn't have won a prize, either, but that's not the point.)

There are four possibilities of gender distribution, all equally likely:

MM
MF
FM
FF

However, since we know that at least one dog is male, we know that the 25% chance of FF didn't come up.

This gives us THREE possibilities, still equally likely:
MM
MF
FM

And in those three, the second dog is male only once.

Therefore, given two random spherical cow puppies, at least one of which is male, there is a 1/3 chance of the second dog being male.

[scifantasy.livejournal.com]

I saw that in the comments to the previous, but something doesn't add up.

[theweaselking.livejournal.com]

It's counterintuitive!

The trick is that you're eliminating part of your sample set. Instead of using the set of all puppies, you're biasing your sample set by filtering, post-generation.

[theweaselking.livejournal.com]

Sure, but then you need to acknowledge that FM is twice as likely as each of MM and FF.

Basically, you get 25% MM, 50% FM, 25% FF. And you're still disqualifying the FF case, and you're still left with 2-1 odds that the second dog will be female.

Any given selection from the set of all random puppies is 50% likely to be male. However, we aren't dealing with the set of all random puppies. We're starting with the set of all random *pairs* of puppies, and then, after generating the set, we're removing all pairs which do not contain a male.

We're then querying the resulting set, which is only 75% of the size of the original set.

[scifantasy.livejournal.com]

Mm. Thinking about it...OK, I'll buy it.

[glitteringlynx.livejournal.com]

Ignore my comment I deleted. I was misreading (I'm so tired).

If puppy 1 is male, there is a 50/50 chance the other will be male as well. By the determination that one puppy is certainly male, we eliminate one possibility out of three. Therefore the only two left are:

MM
MF

Therefore, 50/50, n'est-ce pas?

[theweaselking.livejournal.com]

No. You're assuming generating a random puppy, noting it's gender, then generating another random puppy.

We're not doing that. We're generating *two* random puppies, making an observation that eliminates a large part of the potential result set, and asking about odds *within that reduced result set*.

In a given pair of random puppies, the possibilities are:

MM (1/4)
MF (1/4)
FM (1/4)
FF (1/4)

yes?

Now, given that you *know* that one of the puppies is male, it means that your potential pool of results is now:

MM
MF
FM

They're all still equally likely, like they were before. Our observation didn't change the odds of each possibility - it just just us that one of the possibilities DIDN'T happen.

And in 2/3 of the cases, the second dog here is female.

[glitteringlynx.livejournal.com]

*brain breaks* And this is why I tell people math does not agree with me. My brain is still insisting that genetically any given puppy has a 50% chance of being one gender or another. :X

I get where your example is coming from, I think my brain just can't get past the genetics.

[theweaselking.livejournal.com]

Any given puppy has a 50% chance. The thing is, we're not selecting from the set of all random puppies. We're not even selecting from the set of all PAIRS of random puppies.

We're selecting from the set of pairs of random puppies where at least one puppy is male. It *was* random, and then we nonrandomly eliminated a section of the sample set.

Heuristics with coins

[frith]

My stats course is a bit too far in the distant past for this to gel, so I haxxored it. I tossed quarters, heads for the win, all tails-tails combos eliminated (not counted). Lo and behold, end result: heads-heads occurred ten times out of 30 rolls in which at least one coin was heads. As you said, one chance in three that pup the second would also be a male.

Sir, I salute you.

Re: Heuristics with coins

[theweaselking.livejournal.com]

It's counterintuitive - the trick is that you're generating your random function, then *filtering out a subset of the random possibilities*, then asking questions about it.

So your experiment was perfect.

*bzzt*

[chrisrw109.livejournal.com]

The question boils down to 'What are the odds that any two beagle puppies are both male if at least one of them is male'.

(And don't feel too badly, my first answer was 1/4th until I read the commentary and 'DOH'ed' that it was distinctly impossible for their to be two female puppies.

Unless the Male puppy we've identified is a transsexual Beagle puppy, which makes it a whole different problem.

[elffin.livejournal.com]

What if my answer is "I'd never visit Disney* again in my life even if I were paid to do so." - ?

I feel that is the /most correct/ answer.

[theweaselking.livejournal.com]

But that answer does not address the question.

[elffin.livejournal.com]

It does so - it means the chances of us drawing the same ticket are precisely 0, unless my wife also does not draw, and still, NULL ticket isn't a ticket.

[theweaselking.livejournal.com]

Read the question again! Not only have you *already* drawn your ticket, but the question also tells us that you *do* want to go!

Troubling, no?

[elffin.livejournal.com]

Clearly, I, Citizen, must report to re-education camp.