Red Card - Green Card Game

The Game Room
21

The trouble with this example is that it really isn’t equivalent to what Zenbeam originally said. You can’t easily model more than two colors in this problem while still leaving in the notion of two-sided cards.

Try this one out for size instead.

There are 10 DECKS of cards. Each deck has 9 cards. Nine of the decks have just one color: One deck has only red cards, one only blue cards, one only orange cards, etc. The tenth deck has one card of each color.

We shuffle the decks one by one. Then we choose a deck at random. We turn over the top card. It’s red.

Now, what are the chances that the next card (the one below it in the deck) is also red?

By ZenBeam’s reasoning, there’s a 9 out of 10 chance of picking a DECK that is all one color. And in fact there is a 9 out of 10 chance that the next card will be red (10 red cards total, 9 of them are in the all-red deck). So, the reasoning was just fine. I doubt ZenBeam would have used that same thinking to solve the problem you put forward here; they’re not really equivalent.

22

BTW–My example above, with the 10 decks of cards, is actually related to a way of thinking about the Monty Hall problem. It’s counterintuitive to think that changing your answer after you’ve already given it could change your odds, and a lot of people get hung up on this. And reasonably so.

So people will generalize: “What if there were 10,000 doors? and you picked door #366. And then Monty opens door number 512 and asks you if you’d like to change your answer? How does that help you?”

Which is again reasonable–but it’s not the best generalization, which is instead:

“Suppose there are 10,000 doors, and you picked door #366. And Monty opens EVERY SINGLE DOOR except for door #42, and of course, for door 366. Would you like to change now?”

Well, yes, yes, you would. Because the only way you can win if you stand pat is if that 1 in 10K chance came through and you happened to pick the correct door first time out. Now the 9,999 out of 10,000 chance that you were wrong the first time has all been concentrated in door number 42. So the odds actually are more malleable than people think.

Not everybody agrees that the two situations are similar. But it’s just a matter of rephrasing the original question from “I open one of the other doors” to “I open ALL the other doors but one.” When I put it like that, I’ve been able to convince a few people that the odds actually CAN “switch”–which in turn helps people be more receptive to the correct answer.

23

You can also look at it this way:

Card 1 - Red Red
Card 2- Red Green
Card 3 - Green Green

At the beginning, my chances of pulling each card is 1/3. If I pull a card and see Red, I can eliminate the card 3 possibility, leaving equal chances for Red Red or Red Green.

I see why a red top will most likely have a red bottom, but I don’t see how you can use that to modify the probability that you drew a certain card.

24

It doesn’t, that’s the point. The probability of drawing a double or not is a red herring that doesn’t affect the overall chance, once you have drawn.

25

We’re missing important information. How much is the bet and can you afford to lose it?

If you would be betting the last of your money, it may not be worth the risk even if the odds were in your favor.

26

Analogous problem: Suppose that we were dealing with cards with 100 sides, instead of cards with only 2 sides, and suppose that our three options were an all-red, an all-green, and a card with 99 green sides and 1 red side. If you see red on your card, you can completely eliminate the all-green card, but you can also almost eliminate the mixed card.

27

According to the OP, you’re about to “squander the rest”, so apparently it’s money that will otherwise be used for frivolous purposes.

But that’s rather besides the point, which is posing a probability question, wrapped in a story to make it more palatable.

28

Presumably, though, the grifter will offer you the same bet whether your card shows a green face up or a red face up. (That is, he will bet that the down color is the same as the up color.) So the bet becomes, did you pick a card with the same color on both sides? And what are the odds of that? 2 in 3.

29

The chance is not equal. If you pick card 1, it has a 100% chance of being red side up, but card 2 only has a 50% chance of being red side up.

30

I don’t take the bet. There is a 2/3 chance that if he flips it, the other side will be red.

There are 3 red sides, call them R1, R2, and B1. R1 and R2 are the red faces of the all-red card and B1 is the red side of the card with both red and green. Given that one side is red, it’s one of R1, R2, and B1. Since 2 of those have red on the other side, there is a 2/3 chance that the flipped side will be red.

31

You’re missing a step that could’ve sent you down a different track- had you picked card 2, it wouldn’t be guaranteed that you’d see Red. You had a 50/50 shot, in that case, of seeing green.

So yes, you can eliminate Card 3 as a possibility, but you can also eliminate half of Card 2. What you’ve done is take it for granted that you’re seeing Card 2’s red side.

So let’s suppose it was granted that you’d see Card 2’s red side. Suppose if you picked Card 2, the grifter always turned it before you looked at it so you saw the red side. That is, 100% of the “I picked Card 2” scenarios result in you seeing Red. In that case, and that case only, you’d be fine taking the guy’s bet…your odds would be at parity once again.

32

I’m pretty sure you’re wrong about that. But maybe I’m misunderstanding what you say.

Before I present my analysis, please clarify two points:

  1. What you mean by 'symmetric ’

  2. You say the odds are N/(M+N). That’s the odds of what event, exactly?

33
  1. Interchange of any two colors leaves the situation unchanged. You have the same number of each type of card before and after the interchange.
  2. N/(N+M) is the chance of the underside of whatever card was (randomly) picked being the same color as the top of the card.
    (I shouldn’t have said “Odds”; that might not be the correct term, as it is used in betting. I’ve never been able to keep that straight.)
34

Okay, I’m still not sure what you mean by that. Suppose you have 8 colours: Red, Yellow, Green, Blue, Pink, Black, White, Grey. You make one card for every possible combination, 64 cards in all.

Is that symmetric? If not, why not?

And as for the chance that the underside is the same as the top, lets calculate them by your method
N single-color cards, and M two-color cards, the odds will be N/(M+N).

given that N=56 and M=8 then by your method odds are 56/64 or 7/8 chance that the bottom will match the top. Clearly this is wrong.

Even if it’s the other way around, and you meant to say that is the chance that the bottom won’t match the top, it’s still wrong.

The actual chance of the bottom matching the top are 2/9.

35

I believe your mistake is that you miscounted the cards. There aren’t 64 distinct cards here, there are 36. You’ve counted each two color card twice.

Fixing that, ZenBeam does give the 2/9 you claim.

36

In addition to the error Cabbage pointed out, Peter Morris also switched M and N. N is the number of single-color cards, which is 8 in his example, so the result should be 8/64 = 1/8 chance. This is correct for the problem he actually gave.

37

As for symmetry. Count the number of each type of card: RR, RG, RB, GG, … Then pick two colors, say red and blue. Put a blue sticker over all the red sides, and put a red sticker over all the originally blue sides. Then count the number of each type of card again, and see if you have the same number of each type of card. If you will always get the same number of each type of card no matter which pair of colors you swap, then it’s symmetric WRT color.

Maybe someone has some better way of explaining it.

38

Ah, yes. So I did.

39

This is actually more restrictive than it needs to be. Not sure how best to describe the less restrictive condition, so I’ll just give an example.

Use four colors, RGBY, with one card of each type: RR,GG,BB,YY,RG,RB,RY,GB,GY,BY. That satisfies the requirements above. But if you add one more RB card, and one more GY card, it’s still symmetric in that the situation looks the same for each of the four colors, but it doesn’t satisfy the above requirement. Maybe there’s an accepted name for this symmetry that someone can point to.

40

ZenBeam wrote:

> 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.

yes, but that’s not the question.
If I was asked before picking the card; before looking at it;
“I’ll bet you 50/50 that you pick a card with the same color on both sides”,
then I’d agree. 2/3 of the time the card you pick will have the same color on the other side.

But that’s not what happens.
Like the Monte Hall problem, as #22 02-20-2012, 09:54 AM Ulf the Unwashed

so wonderfully described; what you’re doing is picking a card,
looking at a side, and given that additional information you’re trying to decide if the
other side is the same color, or not.

That’s a different dependent probability.
If you pick a card, and see that one side is red,
then you can ask yourself what is the probability that I picked a card with red somewhere.
That would be 2/3

Given that you’ve picked a card with red on at least one side,
what is the probability that the other side is red?
That would be 1/2

you have the multiply the two probabilities together: 2/3 * 1/2 = 1/3

So, given the additional information of seeing the color of one side of the card,
it’s like Monte opening some of the other doors.
You’re now specifically looking for red, not some card with red or green that may have the same
color on the other side.

see the difference?