Red/Green puzzler solution is wrong

The Show
21

George, I agree that your experiment is in line with the theoretical probabilities,
and your reasoning for why, and how you noted that 25% of the time the green card shows up,
which some people tend to ignore when counting all the possible outcomes, and how that effects
the relative probability of seeing R/R vs R/G

At this point, rather than convince the people who believe the outcome is 50/50,
I’d like to meet with them for some actual wagers.
We could do three cards, or simplified two card drawing.

I could agree to slightly different odds, like 53:47 biased in the direction of your choice.
We could bet something interesting, any consistent amount per hand,
and do 100 trials.

I’m game.
Anyone want to meet in the Bay Area to take the other side?

22

The experiment i did was exactly as described in the puzzler. Three cards, one rr,one rg, and one gg. I put them in a bag and drew them at random without looking and recorded the results, which i am happy to report in detail if anyone is interested. If you use 2 cards like George, you get a different result for the reasons he explains clearly. Fred, my kids live in the bay area- I will ask them if they want to wager you, or we can get together next time I’m there. But we must use 3 cards. Just to be clear, i’m saying that once you see the upside color, the bottom side has even chance of red or green.

23

Ok, I went ahead and did the experiment in Excel. This gave me the opportunity to perform it 100,000 times (using the rand function).

Step 1, draw one of three cards (RR, RG, GG) (1/3 chance for each)
Step 2, determine the face up color (R for RR, G for GG, R or G for RG (1/2 chance for each))
Step 3, if the face up color is G, discard the draw. (I can do this because the initial statement of the problem says that there was a random draw, and the side shown happened to be Red.)
Step 4, count the number of times R is on the opposite side.

Results:
Step 1
RR = 33,392
RG = 33,255
GG = 33353
Step 2
R = 50,050
G = 49,950
Step 3
Discard 49,950 draws
Step 4
R = 33,392
G = 16,658
That gives a ratio of 2.00 to 1 that the opposite side will be Red if the initial side is Red.

Just for fun, I did it again and got these results:
Step 1
RR = 33,283
RG = 33,308
GG = 33,409
Step 2
R = 50,049
G = 49,951
Step 3
Discard 49,951 draws
Step 4
R = 33,283
G = 16,766
Ratio
1.99 to 1

24

@Anonymosity–good job. I didn’t think about simulating the probem with Excel or I would have done it. I wrote a program to simulate the Monte Hall Game Show. I did it once in Pascal and then rewrote it in JAVA. What amazed me was that some of my students had a hard time accepting the answer when we worked the solution out mathematically, but when we traced through the program line by line and then ran the program, they believed the results we obtained mathematically.
When I taught college mathematics, it was interesting to observe students in the probability class I taught. I had some students who were outstanding in my calculus classes that really struggled in probability theory. I had students that struggled with calculus that grasped probability theory very quickly and did quite well. There were, of course, students who grasped the concepts of calculus and probabilty right away.

25

I think I understand the problem. The 2:1 folks are calculating if the UP side is R, what is the probability that a R/R card was drawn and not a R/G card. The 1:1 folks are calculating that if the UP side is R, what is the probability that the DOWN side will also be R. There is an important difference in these two questions.

26

Criss. I don’t see the distinction you’re making.

you mention two cases:

  1. what is the probability that a R/R card was drawn
  2. what is the probability that the DOWN side will also be R

how is that really different?
It seems to me that given you’re seeing a red side,
asking if the down side is also R is the same
as asking if the R/R card was drawn.

27

fredfhome: right, there is no difference. The people who are wrong (in thinking that the solution is wrong) are mostly focusing on the number of cards rather than the number of sides, and/or on the presence of the G/G card. Since the bet is offered AFTER the G/G card is taken out of play, all one has to look at is the four sides of the R/R and R/G card.

Make each card one-sided, so you have three red cards and one green card. You see a red card, then draw another one. What are the odds of drawing a red card?

28

I guess I didn’t say it well. I agree that after you see a red side up, there are 2 red sides left and 1 green side left. That would lead to the 2:1 (R:G) conclusion. But, the sides are not independent. If they were on individual cards, I agree that the chance would be 2:1. But, when you drew the card that you placed on the table, you drew 2 sides at once. The bottom color is not an independent draw. I do not think you can consider the sides as being independent. In other words, the fact that the 2 sides are on one card means that you cannot just focus on the sides and ignore the cards. If you have 3 cards and you draw 90 times, you would predict that you get the RR card 30 times, the GG card 30 times, and the RG card 30 times. If you see a R side up, the GG card is out of consideration. The R side that is up is also out of consideration because we know what it is. The color of the bottom side, then, is actually the probability of drawing a RR card over drawing a RG card. If this is wrong, I just do not see the fault in my un-logic. :slight_smile:

29

Your extrapolating beyond what the puzzler was asking. Sure we can get into the whatifs, but that is not what is involved. One simple question with 2 possible alternatives Quit making it harder than it is supposed to be!

"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.”

Should you take the bet?"

30

What is left is Red red card or red green card, no other options or extrapolations needed! 50/50 chance.periodend!

31

OK, what about this?

I agree that the probability would be 2R:1G if we were using separate cards (3 R cards, 3 G cards). To simplify, let’s eliminate the equivalent of the GG card by throwing away 2 of the G cards. So, we start with 3R and 1G card. On the first draw, we get a R card. There are 2 R and 1 G left, so the odds are 2:1 for R. But, with 2-sided cards, the situation is complicated by the fact that the sides are not independent. When you draw the first card (the side you see), you have automatically drawn the second card at the same time. Therefore, the probability of drawing two red cards would be 3/4 on the first draw and 2/3 on the second draw, or 9/12 * 8/12 = 72/144 = 50%.

32

I think that the confusion arises because the color of the side that is down depends on which card was chosen. Compare this problem with tossing a fair coin three times. In the case of the cards, we are told that a card is randomly drawn from three cards. After the card is drawn, it is laid on the table and either side is equally likely to be up. We are told that the face that is up is red and we want to know the probabilty that the card that has two red faces was drawn. Now think about tossing the fair coin three times. Suppose you know that the first two tosses resulted in a Head. You want to know that the third toss is also a Head.
To solve either problem, go back to basic principles. List each outcome that could happen and assign a probability to the outcome. Each outcome has to be assigned either a zero probability or a positive probability and the sum of the probabilities of the outcomes must be 1. Let’s look at the coin problem first. The outcomes from tossing a fair coin three times are :{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} Each outcome is as likely to occur as any other outcome, so each outcome has a probability of 1/8. Now suppose the fair coin is tossed and the first two tosses are a Head. What is the probability that the third toss results in a Head, given that the first two tosses were Heads? The possible outcomes with this condition are (HHH,HHT} and the probability that the third toss is a Head is 1/2. The third toss is independent of the first two tosses.
Now go back to the problem of the three cards. A card is chosen at random and then the card chosen is laid on the table so that either side is likely to be face up. Think of the faces of the card where both sides are red as labeled R1and R2, the faces of the card where both sides are green as labeled G1 and G2 and the card where one face is red and one face is green as labeled R and G. Now list the possible outcomes that could happen: {(R1.R2),(R2.R1).(R,G),(G,R),(G1,G2),(G2,G1)} Each outcome has the probability of 1/6 of happening. However, you know that the card laid on the table has a red face that is up. Now the possible outcomes are reduced to {(R1,R2),(R2,R1),(R,G)} with each of these outcomes equally likely. Therefore, the probabilty of {(R1,R2) or (R2,R1)} is 2/3 and the probability of {(R,G)} is 1/3.

The difference between the problem of the cards and the problem of the tossing of the coin three times is that in the coin toss, the outcomes are independent. Each time, a Head or a Tail is equally likely. However, with the cards, each card is different. If the first card is chosen, the probabilty of the face that is down being red is 1, if card 2 is chosen, the probability of the face that is red being down is 1/2 and if card 3 is chosen, the probabilty of the face that is down being red is 0. The color of the side that is down being red depends on which card was chosen.

When you do a probability problem, if you list each outcome and the probability of that outcome occuring, you won’t go wrong.

33

And when you’re doing a gambling problem based on probability, you pay attention to when the bet is offered. The green/green card doesn’t matter. It might as well not exist. You’re dealing with one red/red card and one red/green card, and that’s all. Saves a lot of typing.

34

Triadaq presented it more formally than I did, but it seems to me that we agree that the probability that the down face is red is 1/2 because it is not independent of the other side of the card that is drawn. The puzzler answer is wrong. Right?

35

No. The puzzler answer is correct in that if the card face that is up is red, the probability that the face that is down is red is 2/3.
Let’s suppose that the chances are even (the probability is 1/2) that if the top face is red, the bottom face is also red. Again, thinking about the problem, one randomly selects a card. Therefore the probability of the card with two red sides being chosen is 1/3; the probability of the card with a red side and a green side being chosen is 1/3 and the probability of the card with two green sides being chosen is 1/3. Label the faces on the all red card as R1 and R2; the faces on the red and green card as R and G and the faces on the all green card as G1 and G2. After a card has been randomly drawn, one side is equally as likely to be face up on the table as the other card. Listing the possible outcomes we have
{(R1,R2),(R2,R1),(R,G),(G,R),(G1,G2),G2,G1)}. Now since we see that when the card is chosen and a red face appears, we can reduce the outcomes to {(R1,R2), (R2,R1), (R,G)}. Now if the odds when we turn the card over of having a red face are even, then the probability of (R,G) is 1/2; the probability of (R1,R2) is 1/4 and the probability of (R2,R1) is 1/4.
If we now think back to orignally selecting a card before we knew that the face that is turned up is red, we then have the assignment of probabilities as follows:

R1,R2 is 1/8
R2,R1 is 1/8
R, G is 1/4
G,R is 1/4
G1,G2 is 1/8
G2,G1 is 1/8

Therefore, the probability of initially choosing the car with both red faces is 1/8 + 1/8 = 1/4
The probability of choosing the card with one red face and one green face is 1/4 + 1/4 = 1/2
The probability of choosing the card with both green faces is 1/8 + 1/8 = 1/4.

Then the probability of choosing the card with the red and green face was twice as likely as choosing either of the other cards. This, though, contradicts the original assumption that each of the three cards was equally likely to be chosen. Therefore, the odds that when we see that the face turned up is red that the other side of the card is red can not be even.

36

Barkydog, and Criss,
let’s meet and try it… for real…

Criss, see your car talk inbox.

I have an another idea for what Criss (and maybe Barkydog) are missing.

Criss, in your note from March 11

But, the sides are not independent.

there is a correlation to consider, but in this case it doesn’t seem to matter.
Perhaps there’s some commutative property in effect.

In any case, my post on March 4’th attempts to explain in terms of picking cards,
then picking a side. This is more in line with how it appears you’d like to think of the problem.
I got 2/3 for red, etc.

If you see a R side up, the GG card is out of consideration.

true.

The R side that is up is also out of consideration because we know what it is.
The color of the bottom side, then, is actually the probability of drawing a RR card over drawing a RG card.
If this is wrong, I just do not see the fault in my un-logic. :slight_smile:

I think what you’re missing is how to count the case when the G side of the RG card shows up.

It may be easier to consider this when drawing 2 cards: R/R and R/G
As I did on March 4’th.

You can work with 3, but it gets a little confusing since the cases you “throw out” will depend on the color
of the side you see.

With two cards, what would you do with the case when the G side shows up?
not count it?
you have to consider it somehow.
it is a possibility.

With 2 cards, the chances of picking the R/R card is 50:50,
and picking the R/G card is 50:50,
but if you “throw out” the cases when G shows up, you’re left with
two out of three times; when you see red, then the other side is red.

all the enumerations presented by Triedaq look correct to me.

37

Criss, yes, you and Triedaq agree, in the sense that 2/3 = 1/2.

Fred: offering the bet after the red side is visible is the same as using only two cards. The G/G card might as well not exist.

It’s sides not cards. Once the red side shows up there are three possible sides left, and two of them are red.

38

Littlemouse, I’ve always been a big advocate of approximation :slight_smile:
I wish i could understand this!

39

One last try, and then I’m just going to give up…

How many ways can you draw a card with a red side showing? Three (R/R side 1, R/R side 2, R/G side 1). Of these three possible draws, how many have a red opposite side? Two (R/R side 2, R/R side 1). How many have a green opposite side? One (R/G side 2).

The mathematics, the simulation, the simplification, the willingness to do the actual game for cash, all these give the same answer. At a certain point, further discussion isn’t worth the time.

40

littlemouse

Fred: offering the bet after the red side is visible is the same as using only two cards.

agreed.

that is simpler, but you get the same answer using Bayes’ theorem, etc.
as many people have noted.

It’s sides not cards.

it appears possible to approach the problem from either direction. using cards, or sides.
I get the same answer with either approach.
The key is to consider all the possibilities.

I do wonder if some higher level concept/rule/theorem applies that can show the two approaches are identical (apart from simply getting the same answer).