The solution to the red/green puzzler is incorrect. The probability is 50/50, because before you draw the card you have 6 possible bottom faces, three greens and three reds. As soon as you draw and a red face is showing, you eliminate several of those possibilities. You eliminate the two green faces on the G/G card, you eliminate the red face of the R/G card (because the green face isn’t showing), AND you eliminate one of the red faces of the R/R card. Which one? You eliminate the face which is showing since the top face can’t also be the bottom face ( it can be the same color, but not the same face). That leaves two possibilities, one red face (on the R/R card) and one green face (on the R/G card).
There’s already a thread about this that’s closed…I read most of the comments, but didn’t see my thought…
There are TWO events that must be considered, the initial drawing of the card AND the bet. You cannot just eliminate the first event as a “given” in the statistical sense. If you think backward, knowing that the face came out red, the red/red card was twice as likely to have been the card chosen as red/green. Therefore, the odds are 2:1 that the face on the other side is also red.
Because the first event was a random choice, the probability of the card being the R/R instead of R/G must be factored into the probability of the bottom being red.
If the first event was a deliberate selection, THEN the probability of the face down being red would be 1:1.
There have been two threads on this already, and the puzzler answer is correct. People just have a hard time grasping why. This is a variation of the Monty Hall problem.
Yes, you are looking at the Red/Red or the Red/Green card. But now you have to forget the concepts of cards, and think of it as sides. So add numbers to distinguish the sides. So we have a Red1/Red2 card and Red3/Green1 card.
You are looking at a red side, but are unable to distinguish which side (and thus cannot eliminate it as a possibility). So you are either looking at…
…Red1 in which case the other side is Red2, so the well dressed man wins the bet
…Red2 in which case the other side is Red1, so the well dressed man wins the bet
…Red3 in which case the other side is Green1, so the well dressed man loses the bet
Given the information available to us, those are the three possible outcomes of the game, and you lose 2 of them.
Another way to think about it is that the well dressed man is essentially betting that you pulled a card with the same color on the other side (he’d say that the other side is green if you showed a green side). The odds of choosing a card with the same color on the other side is, obviously, two out of three.
Statesboro — "There's already a thread about this that's closed..."??? Why was the thread closed? Considering all the other crap that gets thrown against the wall and sticks on here, I thought it was an interesting discussion. I hate posting on forums where the moderator terminates an interesting discussion on the basis of a personal whim. There are a qadzillion other automotive forums ...
I’m not sure that’s it Kitty…I don’t think sides actually matter - at the point of guessing the side that’s face down.
For one, it matters not which side of the R/R card is up, whether you call it R1, R2 or R3, there is still a 50/50 possibility that the card on the table is either the R/R or the R/G.
Let’s look at it two ways. First, if you let someone purposely place one card face on the table, and he puts a red face up, then, it is 50/50 whether the face that’s down is R or G. Because he could have chosen either of the cards with a red face with NO RANDOMNESS. At that point, we would all agree, the bet is 1:1 whether he used R/G or R/R to put a red face up.
But, adding a random placement to the problem, because two cards have same colors on each side, it’s twice as likely that the card with a red face picked was the R/R than the R/G, therefore it’s twice as likely that R/R is the card on the table. Obviously same outcome if it’s a green face.
I think the odds come in in the initial placing of a red face up. A random, blind picking which puts a red face up means it’s 2:1 that the card is the R/R. But if we play the game with a deliberate placement of one of the three cards, I think the odds of guessing the face that’s down become even money.
Write down all the possibilities, as if you were dealing the three cards randomly. You can assume without affecting that you always deal the r/r one first, the r/g one next, and the g/g last. Now write down the entire set.
Because each card has two faces, there’s 2^3 = 8 ways these three cards can be dealt. So there’s 8X3 = 24 equally likely possible draws. You next eliminate those draws among these 24 that are the wrong color, as you know the color of the card face. That eliminates 8 of the 24 possibilities. So there’s 16 left, all equally likely. If the bet was fair, 8 opposing faces would be one color, and 8 would be the other. But inspection of the 16 will show that’s not the case.
As I said a week ago. You also spelled out why Bayes’ Theorem is an unnecessary math-wank, because the color of the card face is a fait accompli. Remember you aren’t offered a bet until the card is drawn.
The probability is 50/50
to those who think the probability is 50/50,
and those focussed on the cards vs the sides,
consider thinking of it this way.
Trim the problem to two cards.
P(picking card 1) = 1/2
P(picking card 2) = 1/2
P(G) = P(picking card 2) x P(picking the G side) = 1/2 x 1/2
P(R3) = P(picking card 2) x P(picking the R3 side) = 1/2 x 1/2
P(R2) = P(picking card 1) x P(picking the R2 side) = 1/2 x 1/2
P(R1) = P(picking card 1) x P(picking the R1 side) = 1/2 x 1/2
P(R1) or P(R2) = .25 + .25 = .5 // i.e. red on the other side
P(R3) = .25 // i.e. red side of card 2 picked
If you picked from two cards, how do you count the case when G shows up?
If you don’t, then P(R3) has only half the probability of picking R1 or R2.
Picking the G side is a possibility, with a probability of .25,
which you have to account for somehow…
Or, when you pick a card and see Red,
the probability of Red on the other side is .5 vs .25 for it being green
or 2/3 of the time, or some would say the chances are 2:1
With three cards,
the chances of picking a card with the same color on both sides is 2/3
because there are 2 out of 3 cards with the same color on both sides.
In the 3 card case, picking and showing G takes you down the path of cards:
which is the two card situation again.
That’s the subtle trick of the 3 card situation.
Depending on what color shows up,
how to count the other possibilities changes.
I found it a little difficult to get my head around this one,
but I do believe the answer is that is you pick and see color X,
the chances of X on the other side is 2/3
I hope this explanation helps someone.
Statesboro, yes, the sides do matter. They matter a lot, and that’s the whole point. (See above)
The simplest way to wrap your head around it is that the well dressed man is essentially betting you that you have pulled a double sided card out of the bag, which is two out of three in his favor. The point of making that bet after drawing the card is to trick you into not realizing that is what he has done.
Simplify. Once the card is drawn, the ones remaining in the bag don’t matter. The face we see is red. It is one of three possible red faces – the front of the red/red, the back of the red/red or the front of the red/green. In two of these cases the down face is red.
Mechaniker, the other discussion was closed not because it was censored, but just because there was a duplication of subject. It’s not a value judgment; it’s just a way of streamlining the threads.
I’m not saying the sides don’t matter. I’m saying that they do, just not when the guess is made, which to me, is the trick to why some people don’t understand it. Separate into two scenarios:
The well dressed man shows you 3 cards, you close your eyes and HE selects one and puts it face down. Your odds of guessing the hidden face are 50:50. It was not a random selection of the first card so the chances of it being R/R or R/G are the same.
The well dressed man chooses a card blindly and red faces up. Given that the showing face is red, there are 2 of 3 remaining sides it could be.
All I’m saying is the odds of this come in when the first card is selected, not when guessing the hidden face.
- The well dressed man shows you 3 cards, you close your eyes and HE selects one and puts it face down.
we’re discussing the probabilities. If it’s not a random selection then the probabilities don’t matter.
If the well dressed man is selecting a card on purpose, why doesn’t he always pick the one that will make
He doesn’t have to. the probabilities are in his favor. He can make it seem more fair by letting you pick
If the cards are picked randomly, the chances are 2/3 of getting a card with the same color on the other side.
Once a card is picked, the chances are still 2/3 that the opposite face is the same color
(for different reasons, but still in favor of the well dressed man, as discussed by me and others above).
- The well dressed man chooses a card blindly and red faces up.
are you agreeing that the chances are 2/3 for red on the other side?
All I’m saying is the odds of this come in when the first card is selected, not when guessing the hidden face.
I’m not sure which odds you’re talking about here.
with 3 cards, before picking randomly, the chances of getting the R/R card are 1/3,
for the G/G card it’s 1/3, so for picking one with the same color on both sides it’s 2/3.
Once a card is picked, it’s the same as picking randomly from two cards,
because some possibilities have been eliminated. Yet the choice is still random, and
you’d get the same answer whether you had two cards and kept picking 1,
or 3 cards as described.
What is different with 3 cards is that the color with 2/3 probability of being
on the other side keeps changing.
Sometimes red, sometimes green.
I did the experiment. I drew a card randomly 40 times while I was eating breakfast and recorded the results. I stopped at 40 because it was pretty boring. The top face showing was red 20 times. When the top was red, the other side was red 10 times. Therefore, the experiment shows the probability of red or green was even. I was amazed that the numbers were so precise for such a small sampling.
Criss … what you did is called the Monte-Carlo method to calculate probabilites. Seriously, that’s what mathematicians call it.
I’m surprised by your result though. Did you use three cards, two, or just one? Explain your method with some more details.
Did you read the puzzler? From the site as stated previously "It only says. “I’ll bet you even money that the other side of the card is also red.” all your hypothetical suppositions are null! "So the card is the red red card or the red green card. Even money, 50/50 chance, do ya feel lucky? " look at the question! Sure we could extrapolate into the what ifs, but as I tell anyone, read the question,and post answers pertinent to the question only. There is one card and the bet is on the other side of the card only. No mention of anything else! It is a red on top and red on bottom, or red on top and green on bottom. no other options! 50/50. Sure if you want to throw other assumed ideas other than the question posed, that is your right, but your answer will be WRONG! READ THE QUESTION ONLY!
this helped explain it for me.
" Barkydog March 7 Report
Did you read the puzzler? From the site as stated previously "It only says. “I’ll bet you even money that the other side of the card is also red.” all your hypothetical suppositions are null! "So the card is the red red card or the red green card. Even money, 50/50 chance, do ya feel lucky? " look at the question! Sure we could extrapolate into the what ifs, but as I tell anyone, read the question,and post answers pertinent to the question only. There is one card and the bet is on the other side of the card only. No mention of anything else! It is a red on top and red on bottom, or red on top and green on bottom. no other options! 50/50. Sure if you want to throw other assumed ideas other than the question posed, that is your right, but your answer will be WRONG! READ THE QUESTION ONLY! "
If you see red!, how do you know if! you’re looking AT! the top or the bottom?!
"Criss March 7 Report
I did the experiment. I drew a card randomly 40 times while I was eating breakfast and recorded the results. I stopped at 40 because it was pretty boring. The top face showing was red 20 times. When the top was red, the other side was red 10 times. Therefore, the experiment shows the probability of red or green was even. I was amazed that the numbers were so precise for such a small sampling. "
What are the chances of your results being so suspiciously even? 40, exactly 20, and exactly 10?
Ok, I did the experiment too. I think everyone already agrees the G/G card doesn’t affect the odds, so I dropped it from the experiment. I used two cards, one R/R, and one R/G. I randomly choose the card by shuffling the cards to & fro, flipping them at random, then drawing a card, thenfrom that card, selecting the face, 40 separate times. Here’s the results:
Out of 40 draws
- 11 draws showed a Green face
- 24 draws showed a Red face, and the other side was Red also.
- 5 draws showed a Red face, and the other side was Green.
The expected results of this experiment would be (assuming Tom & Ray are correct)
- 10 draws showing a Green face
- 20 draws showing a Red face, and the other side being Red also.
- 10 draws showing a Red face, and the other side being Green.
Summary: The results my experiment are consistent with the face that the higher odds are on the other side being Red.
When I did the experiment, it was clear why the other side will more likely be red than green. To end up with the other side being red, all you have to do is draw the R/R card, which you’ll do 50% of the time. It doesn’t matter which face you choose to show. To end up with the other side being Green, not only do you have to draw the R/G card (which you’ll do 50% of the time), but you also need to select the Red face showing, which you’ll only do 50% of the time you draw the R/G card. In other words, you’ll only end up with the Green as the other side (1/2)*(1/2) = 1/4 or 25% of the draws.