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Yet again - oldbloke's mutterings
September 1st, 2010
10:54 am
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Yet again
This generated a depressingly long thread on urs:

There are 3 cards.
One is red on both faces, one white on both faces, one red on one face and white on the other.
You are shown a card on a table with a red face showing.
What is the probability that its other face is red?


This isn't the wording the ursOP used, or the wording the OQuestioner used, but it's equivalent.

Having dealt with the side issues that (a) we're not told the card we see s from the set described, and (b) we're not told the card is (i) drawn or (ii) oriented randomly, all of which make it unanswerable, there was /still/ a long argument about the answer.

LJ users, being far superior, know what it is, of course.

(28 comments | Leave a comment)

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From:nalsa
Date:September 1st, 2010 09:58 am (UTC)
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What? it's 1 in 3. Where on earth does the argument come in?
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From:oldbloke
Date:September 1st, 2010 10:07 am (UTC)
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Check your working! That's not one of the 2 answers the urs mob are argiung about!
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From:sarahx
Date:September 1st, 2010 10:07 am (UTC)
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No. 1 in 2. Only two of the cards could have a red face showing as only two of them have a red face. So it's not the white-both-sides card. Therefore it's 50:50 whether the card's other face is red!
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From:nalsa
Date:September 1st, 2010 10:13 am (UTC)
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Oh! *slaps forehead* Of course! Foolish boy.
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From:sidheag
Date:September 1st, 2010 10:42 am (UTC)
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Nope. This is the Bertrand's Box Paradox.
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From:oldbloke
Date:September 1st, 2010 10:41 am (UTC)
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Come quickly. Bring money.
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From:sidheag
Date:September 1st, 2010 10:43 am (UTC)
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:-)
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From:sarahx
Date:September 1st, 2010 10:48 am (UTC)
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:-)

Curious what the argument's about, mind. Surely the answer is bloody obvious, once the discounting-white-only option is understood?
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From:oldbloke
Date:September 1st, 2010 10:54 am (UTC)
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The argument is whether it's 1/2 or 2/3.

It's 2/3. Some people won't accept it.

You have 3 cards: r1/r2 r/w w1/w2
Random selection and orientation gives 1 of 6 equally probable states:
r1(r2) r2(r1) r(w) w(r) w1(w2) w2(w1)
We know we're in one of the first 3 equally probable states becasue we can see r
2 of those 3 states have a hidden r.
2/3
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From:sarahx
Date:September 1st, 2010 10:55 am (UTC)
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Course it is...
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From:sarahx
Date:September 1st, 2010 10:56 am (UTC)
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Maths A-level was more than 20 years ago.....
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From:femsc
Date:September 1st, 2010 01:16 pm (UTC)
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Ah. I'd have gone for 1:2, because I would ignore the w/w card completely, leaving just the two cards r/w and r/r to be considered.
We can see r, so the other side is either r or w, so it's 1:2

You're looking at it from the point of view of someone who's used to calculating odds and who understands what he's doing. I'm looking at it from the simplistic POV of someone who isn't and doesn't. You've asked me to consider three cards, but one of them is discounted automatically so I see no reason to include it in my calculations.

I can't see why it makes any difference which way up the r/r card is; however you seem to have counted it twice.

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From:oldbloke
Date:September 1st, 2010 01:24 pm (UTC)
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I can't see why it makes any difference which way up the r/r card is

Because there are 2 ways up it can be, while still satisfying the r-showing constraint. The rw card can only be one way up for that constraint.
So the probabilities of the two cards are not equal.

Some people prefer to think of sides rather than whole cards.
Ther are 3 red sides. 2 of them have red on the back (the red/red card), one doesn't.

If that doesn't help, some people get it if we number all the sides - even replaces red, odd replaces white
2:6 4:3 1:5
You can see an even number. What's the chance the hidden number is even?
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From:femsc
Date:September 1st, 2010 01:52 pm (UTC)
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OK, I'm giving this more thought after I've had a cuppa. Your explanation above is helpful; I think I can see where you're coming from and it's not often that happens when Hard Sums are involved. Thanks.
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From:sidheag
Date:September 1st, 2010 02:16 pm (UTC)
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Here's another way that helped C. I take the three cards; you pick one of them, and look at only one side (both at random). We ask: what is the probability that the other side is the same colour as the one you can see? The answer to that is obviously 2/3, because it will be definitely true for two of the three cards, and definitely false for the third, and the three cards are equally likely. If we change our language if necessary so that the word for the colour we can see is "red" - and changing language obviously doesn't change probability! - this answers the original problem.
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From:femsc
Date:September 2nd, 2010 05:14 pm (UTC)
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I can see your logic I think, but I can't see why the two ways up of the r/r card don't cancel each other out to become one.
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From:femsc
Date:September 2nd, 2010 05:21 pm (UTC)
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Scrub that. Getting there. Slowly, but getting there.
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From:oldbloke
Date:September 2nd, 2010 05:22 pm (UTC)
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Why would they? The card has two sides, it's not a moebius strip. They happen to both be red, but they're still different sides.

Suppose instead you marked the cards with numbers, and the red/red card was marked 2/4 - you'd see a 2 or a 4, but they'd both be even.
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From:jestersong
Date:September 1st, 2010 11:09 pm (UTC)
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Thanks! I've never understood before why it's 2/3 not 1/2, but now I get it.
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From:oldbloke
Date:September 1st, 2010 11:29 pm (UTC)
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My work here is done
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From:drdoug
Date:September 1st, 2010 11:28 am (UTC)
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:-)

Reminds me of this gem of an exercise from a statistics textbook:

Q 7.1
Laplace argued that when two coins were tossed, there were three possibilities, namely two heads, two tails, or one of each, so the probability of each must be 1/3. How would you convince him that he was wrong?

(flip to back of book)

A 7.1
Do not try to convince him that HT and TH are different using logic - he will run rings round you. (For particles obeying Bose-Einstein statistics he would be right!) Invite him to toss pairs of coins, you pay him 2 francs if they give the same result, he pays you 3 francs if they are different, and see how long his convictions last.
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From:sidheag
Date:September 1st, 2010 11:30 am (UTC)
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Sweet! Do you happen to remember which textbook?
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From:drdoug
Date:September 1st, 2010 11:37 am (UTC)
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From:sidheag
Date:September 1st, 2010 11:58 am (UTC)
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Thanks!
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From:drdoug
Date:September 1st, 2010 10:32 am (UTC)
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We also need to know the distribution of the 3 types of card in the set our challenge card is drawn from. In these sorts of questions one tends to assume that it is uniform (i.e. equal numbers of each type), but in real life situations that assumption is often not true.

My current-favourite puzzle on those lines is this one by Paul Wilmott, which goes:
You are in the audience at a small, intimate theatre, watching a magic show. The magician hands a pack of cards to a random member of the audience, asks him to check that it's an ordinary pack, and would he please give it a shuffle. The magician turns to another member of the audience and asks her to name a card at random. "Ace of Hearts," she says. The magician covers his eyes, reaches out to the pack of cards, and after some fumbling around he pulls out a card. The question to you is what is the probability of the card being the Ace of Hearts?
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From:drdoug
Date:September 1st, 2010 10:33 am (UTC)
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Duh. Just read your puzzle properly, which states clearly that there are 3 cards, not that there are 3 types of card. Ignore me.
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From:oldbloke
Date:September 1st, 2010 10:44 am (UTC)
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Depends on the magician, but usually 1. Sometimes it's some other card but the magician then discovers the Ah in the audience member's wallet or somesuch.
But since this is posed as a puzzle, presumably that's wrong.
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From:drdoug
Date:September 1st, 2010 11:24 am (UTC)
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It's more an exercise in thinking about probabilities in the real world than one with a right answer, I think.

The only sure thing to my mind is that the simple analysis which says it's definitely 1/52 is wrong.

I'm not even sure that that figure (about 2%) is the wrong ballpark, i.e. it's possible but not very likely. On its face it's an absurdly simplistic and boring trick for a stage magician without some twist. It could be that the selected card isn't the Ah, but now all of the others are, or it's not in the pack at all but in a sealed envelope inside a glass box that has been on the stage since the start of the show, or it turns in to confetti before anyone can look at it, or any number of other possibilities.

The thing about good magic tricks is that they are supposed to surprise you, so the actual answer in a given situation is unlikely to be one I've thought of. Luckily real life isn't deliberately designed to be so capricious!
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