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Sun Feb 19, 2012, 05:21 PM

Can you solve this puzzle?

From Car Talk:

Take yourself back in time to California, the Gold Rush, 1849. You're prospecting for gold. You've had a pretty good run of luck. So you decide it's time to clean up and go into the big city to celebrate.

You stumble out of one of the saloons, having spent most of your money on women and wine -- and you're about to squander the rest -- when you hear someone call out to you.

From the inky shadows emerges a well-dressed gentleman who proposes a game of chance. He says, "I have this little silk bag. In it are three cards. One of them is green on both sides. Another one is red on both sides. And the third is red on one side and green on the other.

"I'm going to allow you to inspect the bag and put the cards inside. Without looking, I will let you pull one of the cards and place it on this little table in front of me without revealing what's on the bottom of the card."

You reach into the bag, deftly pull out one card, and put it on the table. You see a red face.

The con man says, "I'll bet you even money that the other side of the card is also red."

Should you take the bet?

44 replies, 3980 views

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Arrow 44 replies Author Time Post
Reply Can you solve this puzzle? (Original post)
pokerfan Feb 2012 OP
Systematic Chaos Feb 2012 #1
Chan790 Feb 2012 #5
LisaL Feb 2012 #7
pokerfan Feb 2012 #9
LisaL Feb 2012 #10
pokerfan Feb 2012 #12
LisaL Feb 2012 #13
pokerfan Feb 2012 #15
quakerboy Feb 2012 #16
pokerfan Feb 2012 #18
Curmudgeoness Feb 2012 #19
pokerfan Feb 2012 #22
pokerfan Feb 2012 #23
quakerboy Feb 2012 #20
pokerfan Feb 2012 #25
LisaL Feb 2012 #31
quakerboy Feb 2012 #34
pokerfan Feb 2012 #36
petronius Feb 2012 #37
Incitatus Feb 2012 #39
Motown_Johnny Feb 2012 #35
warrior1 Feb 2012 #2
Chan790 Feb 2012 #3
LisaL Feb 2012 #8
cbrer Feb 2012 #21
pokerfan Feb 2012 #26
cbrer Feb 2012 #29
LisaL Feb 2012 #30
petronius Feb 2012 #33
Newest Reality Feb 2012 #4
csziggy Feb 2012 #6
pokerfan Feb 2012 #11
csziggy Feb 2012 #14
joeglow3 Feb 2012 #17
Dembearpig Feb 2012 #24
LisaL Feb 2012 #32
petronius Feb 2012 #27
baldguy Feb 2012 #28
Art_from_Ark Feb 2012 #38
pokerfan Feb 2012 #40
SwissTony Feb 2012 #41
HopeHoops Feb 2012 #42
RedCloud Feb 2012 #44
RedCloud Feb 2012 #43

Response to pokerfan (Original post)

Sun Feb 19, 2012, 05:28 PM

1. I think there are two ways of looking at this, and either one is breakeven or better.

I don't know for sure which one is right, but the first consideration is that there are a total of three red faces on all the cards. One-third of the time the card with the red face you draw will be the card with the green back, but two-thirds of the time it will be the card with the red back. This would give you 66 2/3% odds of having the winning card.

Or I guess you could just say that if you draw a card with any red on it, then half the time you'll have a green back and half the time a red back. So that's like a coin toss.

Either way, in the long run you will either come out ahead or remain even. In the former case, take the bet all day. In the latter, it's just playing for fun and over a lifetime it won't hurt you.

So, which is it in reality?

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Response to Systematic Chaos (Reply #1)

Sun Feb 19, 2012, 05:33 PM

5. You've looked at this backwards.

Conman is betting it's red.

Conman wins if the back is Red. You win if it's Green. There's a 2/3-odds it's red.

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Response to Chan790 (Reply #5)

Sun Feb 19, 2012, 05:43 PM

7. I think the odds are 50 % that it's red.

One card is green on both sides so that card is out.

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Response to Chan790 (Reply #5)

Sun Feb 19, 2012, 05:54 PM

9. Check out the big brain on Chan

There are three possibilities: R/R, R/R & R/G so the probability that the other side is red is 2/3.

It's analogous to the boy or girl paradox.

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Response to pokerfan (Reply #9)

Sun Feb 19, 2012, 05:56 PM

10. One card has two green sides. You didn't pick that one.

That means the card with both green sides is out. So the other side is either red or green. Your chance is 50/50 after you pick the card with one red side.

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Response to LisaL (Reply #10)

Sun Feb 19, 2012, 06:01 PM

12. you have to count both sides as independent events

because they are different. Imagine that the card that is red on both sides is actually, say purple on one side but for the sake of the puzzle, let's say that purple = red. Maybe the contestant is color blind or something. Then we have: R/P, P/R, R/G

Probability that the other side of the card is purple or red = 2/3.

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Response to pokerfan (Reply #12)

Sun Feb 19, 2012, 06:03 PM

13. You are not setting it up correctly since OP claims one card has both sides green.

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Response to LisaL (Reply #13)

Sun Feb 19, 2012, 06:27 PM

15. Do you agree that three cards results in six possibilities?

RR, RG, GG plus their respective reverses (RR, GR, GG) so all six possibilities (the letter in the first position indicates the visible side): RR, RG, GG, RR, GR & GG.

Throw out the ones where green would have been showing: RR, RG, GG, RR, GR & GG and you're left with RR, RG and RR.

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Response to pokerfan (Reply #15)

Sun Feb 19, 2012, 07:47 PM

16. But once the card is out

That is again an independent event. You have red showing. That means that GG is out. As if it never existed, for probabilities purposes.

So your choices are R/R or R/G. I do not believe that the RR gets counted twice, as it is only one card, and one side is known.

I put it at 50/50.

On edit, no you shouldn't take the bet. He's a con man. He marked it somehow. Why would you ever bet with a conman?

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Response to quakerboy (Reply #16)

Sun Feb 19, 2012, 08:20 PM

18. You can't combine the two independent R/R events

and that's the trick. Three different sides have red and they're all independent. Of those three red sides, two of them have red on the opposite side and one does not.

No, I would not bet a con man.

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Response to pokerfan (Reply #18)

Sun Feb 19, 2012, 09:00 PM

19. I suck at math, but I don't agree.

You do have three sides that have red, but one if a known. There are only two possibilities here----this is the red/green and green is hidden, or it is the red/red and red is hidden.

Since there are only 2 possible options, it would be a 50/50 odds guess.

But I do agree that you don't bet with a con man---using his cards and bag. Besides, 50/50 isn't a good enough bet for me.

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Response to Curmudgeoness (Reply #19)


Response to Curmudgeoness (Reply #19)

Sun Feb 19, 2012, 09:59 PM

23. It might look like there are just two options

but actually there are three. Try it with some pen and paper. Nothing like empirical evidence.

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Response to pokerfan (Reply #18)

Sun Feb 19, 2012, 09:22 PM

20. But you can eliminate one.

There is no way to know which one you have eliminated, but you know, because red is up, and there is a physical card in front of you, that it can only be one of the r/r combos, and not the other one. Because there is only one card, the two possibilities are not independent.

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Response to quakerboy (Reply #20)

Sun Feb 19, 2012, 10:06 PM

25. You can't combine the two (eliminate one)

without skewing the probabilities. Here's a more rigorous analysis of an analogous problem.

Alternatively, you can always fashion some cards and try it out experimentally and see if you wind up with 50% or 2/3 after a number of trials.

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Response to pokerfan (Reply #25)

Mon Feb 20, 2012, 12:48 AM

31. You've already eliminated one.

The way you set up this puzzle, the card with both green sides was not picked.
So your choices are the card with either 2 red sides, or one green side and one red side.

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Response to pokerfan (Reply #25)

Mon Feb 20, 2012, 01:12 AM

34. I did, just to confirm my hypothesis

I get 10/10 on 20 trials.

I stand by what I said. There are 2 cards with red tops. One is green below and one is red. only 2 possibilities. If it helps, imagine that they are 2 regular playing cards, one with a black ace and one red ace. The back of the card is already showing, so it is eliminated as a possibility.

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Response to quakerboy (Reply #34)

Mon Feb 20, 2012, 02:13 AM

36. Amazing

Still, I don't know if twenty trials is really enough to confirm a difference between 50% and 67% (a 16.7% margin). I ran sixty trials with some scraps of paper and wound up with a 36-24 distribution in favor of red.

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Response to quakerboy (Reply #34)

Mon Feb 20, 2012, 02:34 AM

37. You're not doing your experiment with only two cards, are you?

Because that won't prove anything - the trick is in the extra info the con-man gets when he sees the color you've pulled before he has to declare a choice...

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Response to quakerboy (Reply #34)

Mon Feb 20, 2012, 03:41 AM

39. Think of it this way.

There may be only two cards to choose from so it seems like a 50/50 chance, but you have to consider all 4 sides. Only 1 side will be showing.

3 sides are red and 1 is green. On card #1 the top is red and the bottom is red and on Card #2 the top is red and the bottom is green. So if the face is red on the card pulled, there are 3 possibilities.

1 that the other side is the bottom of card #1 (red)

2 that the other side is the top of card #1 (red)

3 that the other side is the bottom of card #2 (green)

That gives the con man a 2/3 chance of being right if he chooses red.



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Response to pokerfan (Reply #9)

Mon Feb 20, 2012, 01:45 AM

35. R/R, G/G or R/G

and you know that the card is either the R/R or the R/G because the G/G would not have a red side facing up.


This leaves only R/R or R/G and you have a 50% chance of either being true

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Response to pokerfan (Original post)

Sun Feb 19, 2012, 05:29 PM

2. yes

You have 2/3 chance of being right.

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Response to pokerfan (Original post)

Sun Feb 19, 2012, 05:31 PM

3. No, he wouldn't be a very good conman if the odds were even.

We can know definitively that the G/G card is in the bag still.

That means of the remaining two cards (R/G, R/R) 3/4 of the faces are red. Since we know the face-side is red, that there is a 2/3 chance that the hidden-face is also red.

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Response to Chan790 (Reply #3)

Sun Feb 19, 2012, 05:44 PM

8. The other side is either red or green. So that's 50 % odds.

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Response to LisaL (Reply #8)

Sun Feb 19, 2012, 09:51 PM

21. Ding Ding Ding

 

We have a winner!

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Response to cbrer (Reply #21)

Sun Feb 19, 2012, 10:07 PM

26. I have a lottery ticket

It's either a winning ticket or it's not. Therefore, it's 50-50.

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Response to pokerfan (Reply #26)

Sun Feb 19, 2012, 11:27 PM

29. Riiiiiiiight...

 

2 colors, as opposed to millions of number combinations.

Same thing!

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Response to pokerfan (Reply #26)

Mon Feb 20, 2012, 12:47 AM

30. The odds of it being a winning ticket are not 50-50.

But the odds of the other side being either green or red are, since those are the only two choices.

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Response to LisaL (Reply #30)

Mon Feb 20, 2012, 01:11 AM

33. That would be true if he had to pick his color before you drew,

but by showing him the card before he specifies a color, he gets an edge. What he's really betting is that you'll draw a same-color card, and he does that by naming the color you show. There are only three possibilities:

You draw R/R - he says R - you lose
You draw G/G - he says G - you lose
You draw R/G - he says whichever color shows - you win! OMG I won!? Let's play again! $$$$$$$ $$$$$$$

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Response to pokerfan (Original post)

Sun Feb 19, 2012, 05:32 PM

4. If the other guy

is a con man and his face is red, I think I'd pass

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Response to pokerfan (Original post)

Sun Feb 19, 2012, 05:42 PM

6. Is this a variation on the Monty Hall door conundrum?


Where they deal with the mathematics starts at about 12:30.

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Response to csziggy (Reply #6)

Sun Feb 19, 2012, 05:56 PM

11. sort of

but I think it's closer to the boy or girl paradox.

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Response to pokerfan (Reply #11)

Sun Feb 19, 2012, 06:18 PM

14. Yeah but my family is an illustration that probabilities mean nothing

Especially for the boy or girl paradox.

My grandmother had two sons, Dad and his brother. Dad had four daughters. My uncle had four daughters, one son, then another daughter.

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Response to pokerfan (Original post)

Sun Feb 19, 2012, 07:51 PM

17. Hell no

Look at all three cards. No matter what side is showing, he will offer the same bet (that the other side is the same color). This ensures he has a 67% chance of being right before you even walk up to him.

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Response to joeglow3 (Reply #17)

Sun Feb 19, 2012, 10:01 PM

24. Correct!

 

An easy con because most mistakenly assume it must be 50/50.

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Response to Dembearpig (Reply #24)

Mon Feb 20, 2012, 12:54 AM

32. your chances are 50-50, but his chances are better than yours

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Response to pokerfan (Original post)

Sun Feb 19, 2012, 10:13 PM

27. Of course I take that bet! But then, I'm stumbling drunk at the time

And, when I lose 2 time out of 3 for a dozen or so plays, I pull out my six-gun, pistol-whip him unconscious, get arrested, and wind up sold by the corrupt sheriff into the crew of a China-bound clipper ship.

My sober, mathematically-competent self, would just walk on by...

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Response to pokerfan (Original post)

Sun Feb 19, 2012, 11:15 PM

28. There's two different events being sampled here.

The first is "From a selection of three cards: R-R, G-G and R-G, what are the odds of drawing one card with one red face?"

The second is "Of the cards that have at least one red face, what are the odds of the other face being red also?"

The puzzle only involves the second event, since the first event - that the card has at least one red face - has already been stipulated.

So, given that you have a card with one red face, what are the chances of the other face being red also? You could have either R-R or R-G. So the odds are exactly even.

The question is: Do you feel lucky? Well, do you punk?

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Response to baldguy (Reply #28)

Mon Feb 20, 2012, 02:40 AM

38. Correct

By this time, the odds are 50-50 that the other side is red.

However, since a con man is involved, no doubt he has one card marked in some way (the RG card). That's why he wants to see the card. If he sees no tell-tale mark, he knows it's RR. By the same token, if the green side is up and there's no discernible mark, then he knows it's GG.

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Response to pokerfan (Original post)

Sun Feb 26, 2012, 01:16 PM

40. Answers to last week's puzzler

RAY: Here's the answer. When the game started, before anyone did anything, there were equal chances of getting either red or green because there were three green faces and three red faces.

TOM: Right.

RAY: But now you know that card you turned over isn't the "green, green" card. So, there is really only one green face left. And there are two red faces left!

TOM: So you're saying the chances of winning are two to one in favor of red?

RAY: Believe it or not. You can believe it because what you're dealing with is not cards-- you're dealing with the sides. When you see the red side up, you could be seeing one or the other face of the red card. That's what most people don't grasp-- and that's why this guy became a millionaire playing this game in California. Do we have a winner?

TOM: Yes we do. The winner this week is Kelly Deal from Hattiesburg, Mississippi. Congratulations, Kelly!

http://d2ozqge6bst39m.cloudfront.net/CT120805.mp3

http://www.cartalk.com/content/red-card-green-card-0?answer

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Response to pokerfan (Original post)

Mon Feb 27, 2012, 11:20 AM

41. The con-man wins if a one-colour card is chosen.

If the first card shows red, he bets on red. Likewise, if it shows green, he picks green. The punter only wins if the mixed colour card is picked.

Since there are two same-colour cards and one mixed card, the chance that the con-man wins is 2/3.

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Response to pokerfan (Original post)

Mon Feb 27, 2012, 11:44 AM

42. Fuck no. Cold cock punch him and steal his wallet.

 

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Response to HopeHoops (Reply #42)

Mon Feb 27, 2012, 07:09 PM

44. We often think very much alike!

Scary, huh?

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Response to pokerfan (Original post)

Mon Feb 27, 2012, 07:06 PM

43. dang nab it. I accidently shot him out of excitement!

I was going to bet him "odd" money just to see if it exists!

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