There needs to be a heavy metal cover of Midnight Oil's "Blue Sky Mine".

[theweaselking]

It's That Time Of Year Again!

You have three boxes. Each box has two drawers. You and your spouse both want to go on vacation, together, to a Disney theme park, and you'd like to get free admission.

In the first box, there is a single admission coupon for Disneyworld in both drawers.
In the second box, there is a single admission coupon for DisneyLAND in both drawers.
In the third box, one drawer has a single admission coupon for Disneyland and the other has a single admission coupon for Disneyworld.

You open one drawer at random. You get a Disney coupon! You're going to one park, or the other!

What is the possibility that the second admission coupon in the other drawer of the same box is *also* for the same park as the first one?

Comments

[jagash.livejournal.com]

I will go for 66.6666666667% chance.

[frith]

66.6%

[wynnsfolly.livejournal.com]

50%
the box with the other two matching tickets is thrown out of the equation
in other words, it's a homozygous box, or a heterozygous box ;-D

[ronebofh.livejournal.com]

50%

[cacahuate.livejournal.com]

If you pick Box #1, Drawer #1, the other coupon is to the same park.
If you pick Box #1, Drawer #2, the other coupon is to the same park.
If you pick Box #2, Drawer #1, the other coupon is to the same park.
If you pick Box #2, Drawer #2, the other coupon is to the same park.
If you pick Box #3, Drawer #1, the other coupon is to the other park.
If you pick Box #3, Drawer #2, the other coupon is to the other park.

So, the chance is 2/3.

...I think.

Edit: Similarly, if the setup is, "You pick a drawer, it turns out to be a coupon for Disneyland. What's the chance that the other drawer is also a coupon for Disneyland?":

If you picked the Disneyland coupon in Box #2, Drawer #1, the other coupon is also to Disneyland.
If you picked the Disneyland coupon in Box #2, Drawer #2, the other coupon is also to Disneyland.
If you picked the Disneyland coupon in Box #3, Drawer #2, the other coupon is to Disneyworld.

So the chance here is still 2/3.

[ronebofh.livejournal.com]

Yup, you did this right. I tried, but i was busy with work and failed.

I remember not going to the Blue Sky Mining launch. Those were the days...

[drjon.livejournal.com]

What is the possibility that the second admission coupon in the other drawer of the same box is *also* for the same park as the first one?

That depends. Are you the Monty Hall who always lies, the Monty Hall who always tells the truth, or the Monty Hall who "owes a little favour in the family"...?

Re: I remember not going to the Blue Sky Mining launch. Those were the days...

[theweaselking.livejournal.com]

It's not a Monty Hall problem!

[kadath.livejournal.com]

The problem is formulated with three boxes to be confusing. The actual probability in question concerns the drawers in any given box--a binomial test, akin to a coin flip. "Heads" is matching coupons. "Tails" is unmatched ones. Any chosen box has a p = 0.5 of containing matching coupons.

That said, if either of them is to Disneyland, it's going on eBay. :P

[theweaselking.livejournal.com]

You're close! And by "close", I of course mean "wrong!"

(I need to be able to edit comments)

[theweaselking.livejournal.com]

PS: I *love* the icon.

[aor.livejournal.com]

That seems, um, wrong.

Math suggests that you're only making a single choice (which box), and so you're looking at what chance you have of picking a box where both boxes match. 2/3.

Here's a rough simulation:
http://jhereg.net/disney-boxes.txt

(That's me convincing myself I'm not missing anything vital.)

[scifantasy.livejournal.com]

This feels like a Monty Hall problem with a twist...and I'm assuming that you mean there's a single admission coupon for Disney* in each drawer in the descriptions for boxes 1 and 2.

There are six drawers. In drawer 1-1 (box one, drawer one) is a ticket for Disneyworld. In drawer 1-2, ditto. 2-1 has Disneyland, 2-2 ditto. 3-1 has Disneyworld, 3-2 Disneyland.

So, the question becomes, what are the odds that a randomly selected drawer is one of: 1-1, 1-2, 2-1, or 2-2?

Which reduces to 2/3, in my book. But it seems too easy. So I must be missing something.

What?

[kadath.livejournal.com]

Hint: The number of boxes is irrelevant to the contents of any given box.

[theweaselking.livejournal.com]

It's not a Monty Hall problem!

[mhoye.livejournal.com]

You're not picking a drawer-in-a-box, here, you're just picking a box. Two of the three boxes have matching tickets, so your odds of going to the same place are two out of three. The two-drawers thing is a red herring, because you don't switch boxes.

[mhoye.livejournal.com]

Also, yes, there does need to be a metal cover of Blue Sky Mine.

[harald387.livejournal.com]

Math is hard. Let's go shopping!

[theweaselking.livejournal.com]

When shopping, there are three boxes, each with two drawers....

[aeduna.livejournal.com]

argh that song
My whole year level gets twitchy when it comes on, even now, nearly 20 years later. It was incessantly on the radio when we were all studying for the last high school exams.

[falconwarrior.livejournal.com]

There;s usually a trick to this sort of question. But after re-reading it several times, I'm quite convinced that the trick here is that there is no trick. So, there's a 2/3rds chance of the second ticket being the same as the first, since the box and drawer was chosen at random and only one of the three boxes has mismatched tickets.

[metahacker.livejournal.com]

50-50. Having chosen the first ticket puts you in a subset of the original problem.

No subset

[frith]

No, there is only one action: choosing a box. The fact that the box you choose has two drawers is a misdirection: you cannot choose a different box after opening that first drawer, so for all intensive purposes both tickets might as well be in the same drawer. So, each box contains a set of two tickets and two out of three boxes contain a matched pair of tickets. Ergo, you have two chances in three of getting a matched pair.

[silveradept]

Pretty good. Only one of the boxes is a split ticket, so you have a 2/3 chance that you picked a box with the same ticket in it.

[utmoonbog.livejournal.com]

The other ticket is both for Disneyland AND Disneyworld.

http://whatis.techtarget.com/definition/0,,sid9_gci341236,00.html

Wasn't there a physicist working on this problem in an above thread?

[fearmeforiampink]

You have a coupon.

It's for DisneyLand or DisneyWorld.

If it's for DisneyWorld, you're in box 1 or 3, equal chance of either, therefore 50% chance you'll get the same ticket as what you've just pulled.
If it's for DisneyLand, you're in box 2 or 3, equal chance of either, therefore 50% chance you'll get the same ticket as what you've just pulled.

Therefore the chance of getting the same ticket as the other one is 50%.

As soon as you open a drawer you narrow the situation and remove 1 box from the possibilities.

[theweaselking.livejournal.com]

Nope!