[Maths/Logic] Probability and Cards

This is not a question for school or anything, just one of general interest.

Three cards are in a hat. One is red on both sides, one is white on both sides, and one is red on one side and white on the other. I draw a card from the hat and drop it on the table. The upward-facing side is red. What are the odds that the downward-facing side is also red?

My answer was one-in-two, or fifty per cent. Apparently the correct answer is two-in-three, or sixty-six per cent.

Here is my working.

First of all, we have three cards. They are: R/R, R/W and W/W. The card that is drawn from the hat has a red side, so the W/W card is immediately eliminated. That means that there are now only two cards: R/R and R/W. As we know that one side of the card is red, there should be two possible outcomes for the other side of the card. Either the other side of the card is red (R/R), or the other side of the card is white (R/W). These have an equal probability of occuring, thus there is a 50% chance of getting a R/R card. Yet, the correct answer is 66%. I've drawn the decision tree (or whatever it is called, I can't remember) and listed the outcomes and it seems that using formal probability (i.e. P(R then R/R)) I can arrive at my answer. It is strange.


The correct answer and its justification can be found here. After reading through the allegedly correct answer's justification once or twice I can comprehend how they came about their answer, but it still seems counter-intuitive to me. Can someone double-check it please? I think you'll find it rather interesting.

EDIT: Sorry, when I first tried to post this topic the page returned as an error instead of a confirmation, so I tried again. Apparently it had actually posted it though, so this thread can be deleted, plz mods. Thanks.

yeah... it´s 66%. The thing is that you didn´t take out a card from the hat, you took one side of one card.... originally you had 3 red sides and 3 white sides. 2 of those 3 belong to the all red card... so if you took a red "side" out of the hat, the probability that it belongs to the all red card is 2/3.

Imagine the cards in the hat are not cards, but "double cards". Now separate them in "simple cards" which are still paired... now you have to take one of those simple cards out... the rest is simple.... 2/3

Thanks for that, Rock Lee. After re-reading the justification with your comments in mind I see how the correct answer is 66%. It is because I was reading the question with the notion of "three cards, equal probability" in mind that I made that mistake. I think this is an interesting little logic puzzle. ^__^

I believe it is because there are three cards total, and there are two chances of it being red on the other side. That's all. ^^

Confusing, but yes. And Rock Lee, that's a good explanation and reson! =O

Thanks LM.
By the way, skate_mate, do you have any other confusing mathematical riddles?
...

Alrighty, just for you Rock Lee I've got one that I recently did at another forum I visit.

There are two hour glasses. The duration of one is eight minutes, the duration of the other is four. Assuming you cannot measure partial times with either of the hour glasses, how do you determine when nine minutes have past?

Enjoy.

are you sure it's not 4 and 7 minutes?... I have heard before the 4-7 one...not particularly difficult...
I'll try this 4-8 configuration, which seems a lot more difficult...

Err, it is actually meant to be four and seven minutes! Sorry about that. It was early in the morning, so I made a bit of a mistake in typing that >_>

But anyway, if you can work out an answer for four and eight minutes that'd be pretty awesome too, lol!

I think I have it for the 4-7 configuration.

1. Start both hourglasses.
2. When the 4-minute hourglass runs out, turn it over immediately.
3. When the 7-minute hourglass runs out, turn it over immediately.
4. When the 4-minute hourglass runs out, turn the 7-minute hourglass over immediately.
5. When the 7-minute hourglass runs out, 9 minutes have passed.

8 is a multiple of 4, so I don't think the second configuration's possible. You're welcome to try, though.

Snarwin, you are spot-on with your answer there. Good work!

actually there are infinite answers for the problem. Many of them can be found by solving any of these equations (just the positive integer solutions):

7x - 4y = 9
or
4y - 7x = 9

x is the number of times you have to turn around the 7 minutes hourglass
y is for the 4 mins one.

let's take the first equation for example, one of the solutions is X=3 and y=3. if you start both hourglasses at the same time, and turn them when they run out (just the one that ran out), 9 minutes will pass between the third time you turn the 4 mins hourglass and the third time you do it with the 7 mins one.

These are called diofantine equations (at least in spanish, hope the translation is ok), integer variable equations, quite useful in recreational mathematics eh.

as for the 4-8 problem, the diofantine equations have no solutions, however this doesn't imply there is no way to do it, maybe you have to do weird stuff, like opening them and mixing the sand in one hourglass (actually that's not a good idea), but anyway, something like that.

(please correct any grammar or spelling mistake you find... still learning english)...

PS: I believe this thread is starting to diverge from the purpose of this forum... however it's very interesting, let's go on until it gets moved.

Here's a fairly simple puzzle involving Tetris:



Using any set of pieces you like, and starting from this position, clear the screen.

Edit: If you can't see the image, try this link.

I can't see the picture...