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Puzzler answer is wrong

The con man is not really betting on the color of the hidden side of the card, he is betting that it matches the revealed side. If the revealed side was green, then he’d bet the hidden side was green. This being the case, the only way you can win is if you have drawn the red/green card, and your chances of having done that are 1 in 3. So you win 1/3 of the time, and lose 2/3 of the time.

If, before you drew the card, he bet you that the hidden side would be red, then your odds would be 50:50, because you could win either by drawing the green/green card (1/3), or drawing the red/green card and placing it green side down (1/3 x 1/2, or 1/6). So your total chances of winning would be 1/3 + 1/6, or 1/2.

@Triedaq: you are jibberjabbering. The person in the Puzzler is presented with a situation. There is no previous event. Bayes’ Theorem is irrelevant, type as much as you will, this is not a conditional probability problem. Plus, I have elegance on my side. It’s so much simpler doing things my way.

@Rushen: :“Here is how I see it” is what the you say just before you lose your money. You are wrong.

@Dsorgnzd: You also miss the point that the mark is presented with a fait accompli, a mise en scene.

The puzzler is a variation of the Bertrand’s box paradox. Anyone looking for a more cogent explanation should go to:

The dealer bets on the odds of drawing a card with both sides the same color (2/3) while the mark bets on the odds of the unseen side being either red or green (1/2). The mark may win half of the time but the dealer will win two thirds of the time. It just illustrates the danger of playing with less than a full deck.

MTraveler–good post and thank you for the link. This link explains exactly what I was attempting to point out in my posts.

@littlemouse: My second paragraph was only to illustrate a situation (granted, not the situation given in the puzzle) where the odds would be 50:50, as many people incorrectly think they are. Since the con man will not bet what color the hidden side is until he has seen the revealed side, this eliminates the possibility of winning by drawing the green/green card.

This is similar to the Monty Hall puzzle: because Monty knows what is behind each door, he will never reveal the good prize. If he picked a door at random to open, then there would be the possibility of the contestant having chosen one “zonk” door, and Monty opening the door with the good prize - in which case the question of whether to trade doors becomes moot as it neither benefits nor harms the contestant. If Monty revealed a door at random, then the probability of winning by trading doors would be 1/3, the probability of losing by trading doors would be 1/3, and the probability that it wouldn’t make any difference (trading one zonk for the other) would be 1/3. Thus (as many people intuitively assume) there would be no advantage to trading doors.

I agree the answer is 1/3 but why make the bet at all? Even if you mistakenly think the odds are 50-50 why would you make a bet where on average all you are going to do is get your money back?

Obviously, PT Barnum was correct. People will always gladly put their money down to play roulette or buy a lottery ticket even though they know the house always gets their share.

MTraveler"The mark may win half of the time but the dealer will win two thirds of the time."
Only if the probability of somebody winning is 1.67.

The reason why I stopped taking higher mathematics is because too many people make it sound more complicated than it needs to be.

Of all the answers, THIS ONE, made it easily understood:

The con man is not really betting on the color of the hidden side of the card, he is betting that it matches the revealed side. If the revealed side was green, then he’d bet the hidden side was green. This being the case, the only way you can win is if you have drawn the red/green card, and your chances of having done that are 1 in 3. So you win 1/3 of the time, and lose 2/3 of the time.

If, before you drew the card, he bet you that the hidden side would be red, then your odds would be 50:50, because you could win either by drawing the green/green card (1/3), or drawing the red/green card and placing it green side down (1/3 x 1/2, or 1/6). So your total chances of winning would be 1/3 + 1/6, or 1/2.

THANK YOU. After reading that, it’s obvious to me, the con man wins.

Personally I would have answered the question thusly:

Q: Should you take the bet?

A: No. The fact that it is a con man tells me right away I should run for the hills. Besides, this sounds too much like 3 card monte where the con man always wins.

A correct answer. Not the mathematically correct one.

@tigerboy: I suspect the reason you stopped taking math is that they stopped letting you make up your own problems on the test. Your long post above addresses a problem with is not this particular Puzzler, and the fellow in the Puzzler is not a “con man” he is a “well dressed” man.

“Thusly”? Why don’t you take a pause and rewrite that carefullyly?

@littlemouse and others. You are not reading the puzzler, you are imputing conditions that do not exist. Use the constraints as defined. The puzzler only says. "I’ll bet you even money that the other side of the card is also red."
What part of it can only be the red green card, or red red card am I missing? Thus 1/2, 50/50 even money!

The part where you’re wrong. Think. Why would someone offer you even money if the odds were really 50/50? I’ve explained why you’re wrong above. I AM reading the puzzler, and I am NOT imputing conditions that do not exist.

The ongoing dispute revolves abound whether you are talking about the odds of winning the hand in front of you (1/2) or the odds of winning the game as it is being played (1/3).

A short quote from the Wikipedia article, http://en.wikipedia.org/wiki/Bertrand's_box_paradox

“Alternately, it can be seen as a bet not on a particular color, but a bet that the sides match. Betting on a particular color regardless of the face shown, will always have a chance of 1⁄2. However, betting that the sides match is 2⁄3, because 2 cards match and 1 does not.”

This is the reason casinos are so profitable. It is pretty hard to tell the guy that just won mega millions that he made a bad bet…

I suppose we all can blame T&R.

Despite the fact that there are three red faces on the cards. Your odds for the question asked are still 50/50.

There is a 0% chance that the concealed side is the red side that is on the reverse of the green card.

Therefore it can only be the red side that is on the reverse of the red card, or the green side that is on the reverse of the red card. From the point when the card is on the table, your chances are 50/50.

As DSorgnzd points out the dealer is betting on the odds that you chose a card with matching sides and will always bet on the color that is revealed.

MTraveler. yes. that’s a good way to summarize the ongoing dispute.
thanks for the reference.

Tom initiated the confusion for me by stating in the solution:

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

which isn’t true in general.
It’s not about red, it’s about the con man betting it’s the “same” color.

So Tom’s statement seems wrong to me.
It’s 2 to 1 in favor of “same”, in general. not red.

there’s 1/3 chance the card is red/red
there’s 1/3 chance the card is grn/grn
there’s 1/3 chance the card is red/grn

or, stated differently, there’s 2/3 chance the card you pick has the same
color on both sides, as Rushen and Dsorgnzd note.

Over the long run, the con man wins because he keeps switching the offer.
If red is shown first, the offer is that the other side is red.
If grn is shown first, the offer is that the other side is grn.
That will win 2/3 of the time.

The link from MTraveler also noted one of the confusions in this discussion:

Bertrand’s point in constructing this example was to show that merely
counting cases is not always proper. Instead, one should sum the
probabilities that the cases would produce the observed result;

i.e. some cases are weighted differently than others.

If you could convince the con man to keep betting that the other side is red,
but only when red is showing, and you win if green is showing,
then you’d win 2/3 of the time.

BTW.
I found littlemouse’s reasoning specious, at best, relying on the stereotype
of what motivates a con man, not the mathematics of why the scheme works
for the con man.

littlemouse even says:

You would not be offered the same wager (even money that the other side would
be whatever color) if the side showing was green.

I don’t believe that.
The offer will always be made in the real con (even though the puzzler doesn’t
talk about doing it again).
Given that color X is showing, the con man will bet that the other side is X.

Bayes’ Theorem works just fine here.

Triedaq, you said:

Littlemouse-- you are correct in saying that Barkydog is wrong for focusing
on the cards and not on the sides.

I was not so sure about that at first.
I thought since we’re picking cards, it was necessary to assign probabilities
to the cards, etc.
Apparently it doesn’t matter.

The reference from MTraveler seems to use either approach,
starting with cards, or sides, assigning probabilities,
and getting the same answer in either case.

from Rushen,

Despite the fact that there are three red faces on the cards. Your odds for
the question asked are still 50/50.

with a similar statement by Barkydog.

It does seem like 50/50 at first, and it took me awhile to get my head around
this part, which was helped by reading the ref by MTraveler
i.e. http://en.wikipedia.org/wiki/Bertrand’s_box_paradox

yet, the chances are 2/3 for red.

A = given that red is showing, what’s the chance of the other side being red

P(A) = (combinations with red on the other side) / (total combinations)

You might also think of this with two cards, and eliminate the confusing
dependant probability cases.
Think of picking one of two cards again and again.
When you pick “green”, pick again.

The cards can show up as these combinations:
R1 / R2
R2 / R1
R3 / G

so P(A) = 2 / 3

with red showing, there are 2 cases where red is on the other side,
and only one with green.

To say there’s only 2 choices, so the odds are 50/50 is incorrect because
some of the possibilities are not being counted.

It’s like asking:

“What are the chances I have $1 Million in my pocket?”

either I have $1 Million in my pocket, or I don’t. two choices.
Therefore the chances are 50/50.

The fallacy is that not all possibilities are accounted for.

This is different than asking what is the probability of picking R/R
from 3 cards, sight unseen. That’s 1/3.

Here’s a crazy thought…

Why not try it? Take three playing cards and label them. Then mix them up under the table, and pull out a card and place it on the table. Note the result, and try it over and over. You will find that the whenever the face you see is red, about 2/3 of the time the opposite face will also be red.

Just try it! Probability works.

“with red showing, there are 2 cases where red is on the other side,
and only one with green.”

This cannot be true.

If red is showing, the red side of the red/green card cannot be on the other side (or you would be seeing green).

The other side can only be red if you chose the red/red card (one case). And can only be green if you chose the red/green card (one case).

There are six ways to draw a card. Three of these ways have a red side up. Of these, two of the ways have the other side being red.

Two out of the three ways you can have a red side showing, the other side is red.

Look at it this way…

If, as you say, it’s a 1/2 chance that then if you see a red side, then there must be a 1/2 chance that it’s the Red/Red card. Since there’s a 1/2 chance that you will see a red side on the draw (6 sides, 3 are red), there must be a 1/4 chance that the card is Red/Red. The same logic means that also a 1/4 chance that any card you draw is Green/Green.

3 cards. 1/4 = Red/Red, 1/4 = Green/Green. What’s left? 1/2 chance that you drew the Red/Green card? What is it about the Red/Green card that would cause you to draw it half the time?

It may seem counterintuitive, but it’s true. Seeing a red side gives you a 2/3 chance of the other side being red.

Try it!

"his cannot be true.

If red is showing, the red side of the red/green card cannot be on the other side (or you would be seeing green)."

No. You’re adding information that you do not yet know and changing the odds. You need to not think of it as drawing a card, but looking at a side. You basically want to know what is the probability of seeing a particular side.

When you draw a card, you see a red side. So that means you pulled either the red/red card or the red/green card. Now, pretend each side of the cards are labeled R1, R2, R3, and G. Those are the only possible sides available.

Now you need to start thinking in terms of the SIDE of the card you are looking at, rather than cards themselves. You are looking at a particular SIDE of a card, you just don’t know which one.

We know we are not looking at G, so we need only look at the cases of looking at an R. If you are looking at…
… R1, then the other side is R2 so the nicely dressed man wins
… R2, then the other side is R1 so the nicely dressed man wins
… R3, then the other side is G so the nicely dressed man loses

Those are the three, and ONLY three possible outcomes of the game given the information available to you. And the nicely dressed man wins two out the three times. Play the game yourself, over many games you will see the odds work out.

Another way to look at it - the nicely dressed man is essentially betting that you pulled out a card that has sides of the same color (he’d make the same bet if you showed a green face rather than red), which is, obviously, two out of three.

Aha!

I see where I have been mistaken when I look at it this way: If I have selected a card with a red face (which we know), there is a 75% chance that I will lay that card with red faced up (100% for R/R, 50% for R/G).

This is how there are two opportunities for the reverse to be red, as several commenters have explained.

Mea Culpa.