From yesterday’s Car Talk:

I say that you should not take the bet but I will leave my reasoning for later in the thread. As usual, googling is cheating.

From yesterday’s Car Talk:

I say that you should not take the bet but I will leave my reasoning for later in the thread. As usual, googling is cheating.

Okay, I usually misread these, but I’ll swing:

No you should not bet. There’s a two/three chance that the other side is red.

That’s what I came up with. Six possibilities (three cards x two sides), of which three involve a red side but only one of those have a green alternate side.

I thought of it in terms of, there’s a 2/3 chance of picking a card with one color on both sides, so 2/3 of the time, the color on the bottom matches the color on top, whichever color that happens to be.

That’s true in this case but not generally so. Suppose there were also cards that were blue on both sides, black on both sides, etc. (but no other mixed cards), your chances of picking a card with one color on both sides would go way up, but the probability of this problem would remain 2/3.

Even simpler than that, you can refuse the game on the Nathan Detroit principle, no math needed. The dude wouldn’t be offering you the bet if it didn’t favor him.

Seems to me most of the probability calculations here fit only if they’re done before a card has been drawn. But that’s not the situation.

There are three cards. There’s a zero percent chance that the card we’re looking at is the one that’s green on both sides. There’s a fifty percent chance that what we’re looking at is the card that’s red on both sides, and a fifty percent chance that it’s the one that’s red on one side and green on the other.

If you had to put your money up before the cards are touched, it’s a bad bet. It’s 50/50 now, isn’t it?

Is it cheating that I once read a detective story where the hook was this particular con?

There are three red faces in the game, call them R1, R2, R3.

There are three green faces in the game, call them G1, G2, G3.

The cards are

G1-G2

R1-R2

R3-G3

If you see a red face, there are three equally likely possibilities

R1 with R2 underneath

R2 with R1 underneath

R3 with G3 underneath

Two out of three possibilities have a red face underneath.

Good explanation, that makes sense.

It’s not surprising that a different problem will have a different probability.

The point is that your method of calculating the odds was flawed. On this particular occasion, by chance, it gave the right answer. But in other conditions following your method will give the wrong answer.

I guess not. I’ve been a puzzle fan all my life so I’m rarely surprised by any of them. Do you remember the author/title of the story?

It wasn’t flawed, and it wasn’t by chance. In the OP, it’s symmetric whether a green card or a red card turns up, so the color you see doesn’t give you any more information. My approach holds generally for any such symmetric case.

Look at another symmetric case, one with three colors, red, green, and blue. In the bag put the six possible cards: RR, GG, BB, RG, GB, and BR. Pick one card and look at the top color. Say it’s green. What’s the chance of the bottom being green? 50%, the same as the chance of picking a single color card at the start.

As long as it’s symmetric WRT colors, if you have N single-color cards, and M two-color cards, the odds will be N/(M+N). In the OP’s case, N = 2 and M = 1. The case in the previous paragraph has M = N = 3. For a more complicated example, if there two each of RR, GG, and BB, and seven each of RG, GB, and BR, the odds would be 6/(21+6) = 2/9.

It wasn’t chance that **ZenBeam**’s method got it right. He used a method that is perfectly valid for this problem. The method might or might not also be valid for the problem that **OldGuy** presented, but we can’t tell because he didn’t actually completely specify his problem

*The Percentage Player*, from the collection *The Saint To The Rescue *by Leslie Charteris.

I drew the same conclusion (although for me, it’s more of an Admiral Akbar principle).

I don’t think so. But let’s take a concrete example, following OldGuy’s suggestion.

The cards are as follows: Red-Green, double-Red, double-green, double-White, double-Black, double-Grey, double-orange, double-yellow, double-blue, double-pink.

That’s 10 cards, 9 of which are double coloured.

You pick one at random. It shows Green face up. What are the chances that the other side is red?

According to Zenbeam’s method, there is only one chance in 10 of picking the red-green card, and 9 chances of picking a double colour. Therefore, the chances that the other side is red are 1 in 10, or 9 to 1.

I’m going to bet on Red. Will you give me odds of 7 to 1, that should favour you, right?

If that’s too steep, then maybe 6-1 would be preferable to you?

That’s not a symmetric problem. Read my post again.

Your post said:

*"I thought of it in terms of, there’s a 2/3 chance of picking a card with one color on both sides, so 2/3 of the time, the color on the bottom matches the color on top, whichever color that happens to be. "*

So, according to you, if there’s is a 9/10 chance of picking a card with one color on both sides, a red face showing will have red on the other side 9 times out of 10, if your logic is right.

:rolleyes:

Read my post **#13** again. That’s the one you quoted in your post #17. Here, I’ll just quote it for you, with the parts you completely ignored in bold, so there’s no confusion:

Your example is not symmetric WRT colors.