The puzzle had to do with three cards … with each side red-red, green-green and red-green respectively.

You pull one card out of a bag, place it face down. It is red on the side facing up.

What are the odds that the other side is also red?

The boys say 2-1, not 50-50. Because you know the card is not green-green, so that card is eliminated. And the three remaining sides you cannot see, the side underneath your card and the sides of the remaining card in the bag, are red, red and green. 2 chances to one.

Not so fast, kimosabe.

Yes, the three sides that you cannot see are red-red-green. But you KNOW that two sides of the remaining sides are red. The red side you see on the table facing you, and one of the red sides in the other card in the bag. So only TWO sides are in doubt, not three. The odds the card on your table is red on the remaining side is, in fact, 50-50.

If you’re feeling lucky, take the bet.

Personally, I don’t like puzzles. When I hear these puzzles, it’s in one ear and out the other.

Hmm, that’s puzzling.

Sorry jwood. Leaving aside the fact that in the real world, the well-dressed gentleman would be cheating, there are three possible other sides to the card and only one of them is green, so he’s offering even money when you have a 1 in 3 chance of being right. The green-green card is a red herring.

I think where you’re falling down is not realizing that just because two sides are red, it doesn’t make them the same thing. Imagine putting numbers on the red sides. How many red sides do you have?

Now imagine that the number on the red side that you can see is covered up. How many red sides do you have?

I don’t know if 2-1 is the same as 1 in 3 because as a good Christian gentleperson I know nothing about gambling.

I did it by listing out all the equally likely combinations. Like someone dealed you the three cards randomly. I ended up with a list of 8 possible sets of three cards. It only took me 3 or 4 minutes to list them all out. That’s 24 possible draws. Of those, only a certain number have the red card up. So you can elminate the draws with the black card up. From those draws with the red card up, from my count, there were twice as many with the other side red, compared to the other side black. (It it possible I’m reversing which is more likely, the red or the black, I don’t have the list in front of me. But one had twice as many as the other, so I’m leaning toward that the boys are right.)

Try it. List out all possible draws. See if you get the same result.

I agree with Jwood, the real odds increase in the con’s man favor if the bet was placed on the face of the NEXT card that came out the bag.

Suppose the card with two red faces has one face labeled R1 and the other face labeled R2. Let the same hold true for the card with two green faces: one face is labeled G1 and one face is labeled G2. The remaining card has faces labeled R and G. Suppose the first listing in each pair is the face of the card that is up and the second listing in each pair is the face of the card that is down. The possibilities are:

(R1R2) (R2R1) (RG) (GR) (G1G2) (G2G1). With no other information, each of these possiblities is equally likely and the probability of each possibility is 1/6.

Now you are given some information–specifically that the card selected has a red face up. You can then eliminate (GR), (G1G2) and (G2G1). This leaves the possibilities (R1R2), (R2R1) and (RG) each of which is equally likely. Now the faces aren’t marked, but (R1R2) or (R2R1) occur 2/3 of the time.

Bayes theorem just uses given information to reduce the number of outcomes. In this case, the face that is turned up is red is the given information. Here is another example: Suppose you have a pair of dice. There are 36 ways the dice could fall from (1,1) to (6,6). If the dice are fair, each of the outcomes is equally likely. The probability that you roll an 8 is the probability of (2,6) (3,5) (4,4) (5,3) (6,2) which is 5/36. Now suppose I told you that doubles are rolled. What is the probability that you rolled an 8?

If you rolled doubles, the possibilities have been reduced to (1,1) (2,2) (3,3) (4,4) (5,5) (6,6). Knowing the information that doubles are rolled makes the probability of an 8 equal to 1/6.

Bayes’ Theorem is all well and good, but is not needed because you *know* that the green-green card has not been drawn. Bayes’ Theorem is all about *two* events, rendering it a math-wank in a Puzzler about *one* event. Really all you need to know is that there are three possible other sides to the card and only one of them is green.

If this were a Puzzler about prisoners having to solve the red-green card Puzzler, jwood and Carlos would not be doing too good.

I look at this way, there are two possible choices for the card that is showing It is either the r/g or the r/r. we already know one side. we are placing a bet on the other side that has one of two possible choices 5/50. The difference with the dice is that the sides are attached. This eliminates guessing one side in both cards possible. We already know One side is red (of either card, the r/r or the r/g) . you see you are choosing between one of TWO possible cards that’s a 50:50.

That is why I think the real bet would be of the face on the NEXT card that the con man would pull out of the bag. In this case, if there is RED showing he knows out of the two cards left in the bag there is one with g/g and one r/r OR r/g. His odd are in this case in his favor ⅔ or ½. depending on the card he drew first

Littlemouse–there are two events here. One event is that you select a card. The other event is that the face that appears is red. Suppose card 1 is red on each face, card 2 is red on one face and green on the other and card 3 is green on both faces.

The probability that you select card 1 is 1/3; the probability that you select card 2 is 1/3 and the probability that you select card 3 is 1/3.

The probability that the top face is red if you selected card 1 is 1 (it is certain)

The probability the the top face is red if you selected card 2 is 1/2

The probability that the top face is red if you selected card 3 is 0 (it definitely can’t happen)

The probability that the face that appears is red is 1x 1/3 + 1/2 x 1/3 + 0 x 1/3 = 1/2

What one wants to know in this case is the probabilty that card 1 was selected given that the top face is red.

Applying Bayes’ theorem

P(card 1 selected | top face red) =[ P(top face red | card 1) x P(card 1 selected)]/ P(top face red)=

[1 x 1/3]/(1/2) = 2/3

No. The first event is that you are offered a choice, which has a probability of 1 because the Puzzler *says* you are offered a choice. That is why Bayes’ Theorem is superfluous. Imagine you realize that under these shady circumstances the well dressed man is probably cheating you, and at that moment, young Woody from Cheers walks up, having just got off work. The well dressed man describes the situation and offers Woody the same choice he offered you. There are no previous events, there is only the setup. As the Puzzler says,

"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.’"

He does *not* say, *before* you draw a card, that he will bet you that *if* the card you draw shows red, that the other side is red. The bet is only offered once there is a *fait accompli* (a lot of people think that’s Latin).

Keep It Simple.

"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.”

So the card is the red red card or the red green card. Even money, 50/50 chance, do ya feel lucky?

“No. The first event is that you are offered a choice, which has a probability of 1 because the Puzzler says you are offered a choice”.

The puzzler says that you draw a card from the bag and lay it face down on the table. The first event is that you draw the card with each face red. The probability that you drew this card is 1/3. The second event is that the card drawn has the red face up. There are an equal number of red faces and green faces, so this probability is 1/2.

What one needs to calculate is the conditional probability that the card with both red faces is drawn, given that the card laid on the table has the red face up. In otherr words, you draw a card, lay it on the table and the red face is up. He wants to bet that the other face is red with the knowledge that the top face is red.

You are doing exactly the same thing whether you list the possibilities and then eliminate those possibilities which can not occur by the given condition that the face that appears on top is red or you use Bayes theorem. If you look at the proof of Bayes theorem, you are doing exactly that–eliminating the events that could not occur under the given condition.

The con man sees the face on the card that is laid on the table. Had the face that appeared been green, he would bet that the other face would also be green. Again, the odds would be in his favor.

Bayes’ theorem is not needed to solve any conditional probability problem. One can go through and eliminate the outcomes that are not possible from the given condition. The theorem is used to make the calculations easier.

Here is a conditional probability problem where you can solve it very simply without Bayes theorem: A pair of dice is rolled. You are told that the roll resulted in an odd number. What is the probability doubles were rolled?

If you list all thirty six possibilities and eliminate the rolls that resulted in an even number, you will see that there are no doubles in the possibilities that remain.

"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.”

Is the quote from the website. There are no more drawn cards, only a bet on the other side of the card according to what was posted. Granted the odds would favor the gambler if one had to draw another card but that was not the question.

So I have to stand by as previously posted "So the card is the red red card or the red green card. Even money, 50/50 chance, do ya feel lucky? "

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

That is correct–one card and only one card is drawn. You put it down on" the table and see a red face. The con man also sees the red face and says I’ll bet you even money that the other side of the card is also red."

The con man is betting that the card drawn has two red faces knowing that one face is red. Imagine that the faces of the all red card are marked RA and RB, this is card 1.The faces of the all green card are marked GA and GB, this is card 3. Card 2 has a red face R and a green face G.

Here are the possibilities when you reach in the bag, deftly pull out one card and lay it on the table:

Side Up Side Down Card number

RA … RB … 1

RB … RA … 1

R … G … 2

G … R … 2

GA … GB … 3

GB … GA … 3

If you did not know that the side up was red, then you would have an even game. However, you and the con man know that the card deftly pulled out and laid on the table has a red face up. Therefore, you can eliminate the bottom three possibilities. Since the faces aren’t marked A and B, what you have is this situation with the possibilities that haven’t been eliminated:

Side Up Side Down Card number

R … R … 1

R … R … 1

R … G … 2

Therefore, the chances are 2 out of 3 that if the face that is up is red, when the card is turned over, the other face will be red. If the con man and the other fellow play this game, say 90 times, the con man will win 60 times and lose 30 times while the other fellow will win only 30 times, but lose 60 times.

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!

Make yourself the three cards and go play the game a number of times. Deftly draw the card from the bag with your eyes closed. Lay the card on the table and then open your eyes. Whatever color face you see, note whether or not the other side is the same color. I am certain that you will find that 2/3 of the time the face that was down is the same color as the face that is up. (The con man would make money from playing with you).

You try it. whatever color comes up, there are only 2 cards it could be. So turning the card over it is one card or the other. A same color or other color. No other options. 50/50 chance explain any other non existant possibilities.

@Triedaq: “The first event is that you draw the card with each face red.”

A wise man once said: “Once a magician starts distracting you, the trick is already done.”

Good Lord. No. The first event is that you are offered a wager. You are not offered the wager *until* you draw the card. You would *not* be offered the same wager (even money that the other side would be whatever color) if the side showing was green. The card is there for the next mark to see. Drawing the card is irrelevant in this problem, which is why Bayes’ Theorem is irrelevant in this problem. It doesn’t mean Bayes’ Theorem is wrong for this case, just that it’s too much tool for the job.

@Barkydog: No. Wrong. You are focusing on cards, not sides.

I should not have used “first” which implies order. An event is a subset of the possible outcomes.

Assume for the moment that the faces are labeled RA and RB on card 1 which has a red face on each side. The faces on card 3 are labeled GA and GB which has a green face on each side. The faces on card 2 which has a green face and a red face are R and G. The possible outcomes are

{(RA,RB), (RB,RA), (R,G), (G,R), (GA,GB),(GB,GA)} where the first element in each pair, for example (RA,RB), is the face of the card that is up, in this case RA and the second element in the pair,in this case RB is the face of the card that is down.

Let E be the event that the face of the card that is up is red. E = (RA,RB), (RB,RA), (RG)}. F is the event that card 1 is chosen, so F= {(RA,RB), (RB,RA)}. The con man wins if card 1 is the one laid out on the table, which means that event F occurs. The con man sees that the face of the card laying on the table is red, so he knows that event E occured. He then makes the wager that card 1 must have been drawn or that event F occured. Event F is contained in event E as the outcomes in the brackets within the braces: {[(RA,RB),(RB,RA)],(R.G)}. The wager is not an event as it is not one of the possible outcomes. Since the outcomes are all equally likely, the probability that the face of the card that is down is red is 2/3.

Littlemouse-- you are correct in saying that Barkydog is wrong for focusing on the cards and not on the sides. Now the answer is the same whether one reasons it out directly from the set of all outcomes as I just did or whether one uses Bayes’ theorem which states P(F|E) = (P(E|F)xP(F))/P(E) = (1 x 1/3)/1/2 = 2/3.

Chapter 5 of William Feller’s classic book “Probability Theory and Its Applications Volume 1” gives a very thorough explanation of conditional probablility and Bayes’ theorem.

Here is how I see it.

The fact that the side facing up is red is a red herring.

The magician is winning more often than losing because you have a two in three chance of selecting a card with matching sides and he will always bet that the reverse matches the revealed color. If green is revealed, he will bet on green. There is only a one in three chance that the reverse color will be opposite the revealed color.

Once the color red has been revealed however, the odds of the reverse side matching or being opposite become 50/50. There is one red card with a red back and one red card with a green back.

It never mattered to the puzzle as presented that there are three red sides. Once you reveal the red side faced up, that side is no longer in contention for being the side that is concealed. The concealed side can only be one of two options red or green.