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pokerfan

(27,677 posts)
Sun Feb 19, 2012, 06:21 PM Feb 2012

Can you solve this puzzle? [View all]

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?

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Can you solve this puzzle? [View all] pokerfan Feb 2012 OP
I think there are two ways of looking at this, and either one is breakeven or better. Systematic Chaos Feb 2012 #1
You've looked at this backwards. Chan790 Feb 2012 #5
I think the odds are 50 % that it's red. LisaL Feb 2012 #7
Check out the big brain on Chan pokerfan Feb 2012 #9
One card has two green sides. You didn't pick that one. LisaL Feb 2012 #10
you have to count both sides as independent events pokerfan Feb 2012 #12
You are not setting it up correctly since OP claims one card has both sides green. LisaL Feb 2012 #13
Do you agree that three cards results in six possibilities? pokerfan Feb 2012 #15
But once the card is out quakerboy Feb 2012 #16
You can't combine the two independent R/R events pokerfan Feb 2012 #18
I suck at math, but I don't agree. Curmudgeoness Feb 2012 #19
This message was self-deleted by its author pokerfan Feb 2012 #22
It might look like there are just two options pokerfan Feb 2012 #23
But you can eliminate one. quakerboy Feb 2012 #20
You can't combine the two (eliminate one) pokerfan Feb 2012 #25
You've already eliminated one. LisaL Feb 2012 #31
I did, just to confirm my hypothesis quakerboy Feb 2012 #34
Amazing pokerfan Feb 2012 #36
You're not doing your experiment with only two cards, are you? petronius Feb 2012 #37
Think of it this way. Incitatus Feb 2012 #39
R/R, G/G or R/G Motown_Johnny Feb 2012 #35
yes warrior1 Feb 2012 #2
No, he wouldn't be a very good conman if the odds were even. Chan790 Feb 2012 #3
The other side is either red or green. So that's 50 % odds. LisaL Feb 2012 #8
Ding Ding Ding cbrer Feb 2012 #21
I have a lottery ticket pokerfan Feb 2012 #26
Riiiiiiiight... cbrer Feb 2012 #29
The odds of it being a winning ticket are not 50-50. LisaL Feb 2012 #30
That would be true if he had to pick his color before you drew, petronius Feb 2012 #33
If the other guy Newest Reality Feb 2012 #4
Is this a variation on the Monty Hall door conundrum? csziggy Feb 2012 #6
sort of pokerfan Feb 2012 #11
Yeah but my family is an illustration that probabilities mean nothing csziggy Feb 2012 #14
Hell no joeglow3 Feb 2012 #17
Correct! Dembearpig Feb 2012 #24
your chances are 50-50, but his chances are better than yours LisaL Feb 2012 #32
Of course I take that bet! But then, I'm stumbling drunk at the time petronius Feb 2012 #27
There's two different events being sampled here. baldguy Feb 2012 #28
Correct Art_from_Ark Feb 2012 #38
Answers to last week's puzzler pokerfan Feb 2012 #40
The con-man wins if a one-colour card is chosen. SwissTony Feb 2012 #41
Fuck no. Cold cock punch him and steal his wallet. HopeHoops Feb 2012 #42
We often think very much alike! RedCloud Feb 2012 #44
dang nab it. I accidently shot him out of excitement! RedCloud Feb 2012 #43
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