If you’re considering all possibilities, you’re counting sides, but you’re in denial. Cut each card in half so you have six one sided cards, and then draw pairs of cards. There’s your higher level concept.
Bayes’ Theorem just adds an extra layer of complexity which doesn’t serve the student.
You fail to read the puzzler question as posted, yes if you were drawing cards the odds are different, but I must remind you to only answer the question at hand AS POSTED IN THE PUZZLER.
"“I’LL BET YOU EVEN MONEY THAT THE OTHER SIDE OF THE CARD IS ALSO RED.”
No drawing cards from a bag, no nothing else, only 2 cards can satisfy the option of showing a red side! 50/50 chance.
Final answer?
Crisss wrote: "I did the experiment. I drew a card randomly 40 times while I was eating breakfast and recorded the results. I stopped at 40 because it was pretty boring. The top face showing was red 20 times. When the top was red, the other side was red 10 times. Therefore, the experiment shows the probability of red or green was even. I was amazed that the numbers were so precise for such a small sampling. "
Criss, I think you should do your experiment again, drawing more cards this time. While it is consistent with the probability averages that the red face was up 20 times in this 3-card experiment, if CarTalk is correct, the other side should have been red more than 10 times (on average). If the odds are 2:1 in favor of red, then the expected number of times the other side should be red in the 40 draw (with 3 cards) experiment is about 14 times. I think Criss’s experiment just happened to be on the low side is all, due to the luck of the draw, because he didn’t draw enough times. The Excel simulations seem to prove that with enough draws you get the 2:1 ratio.
Anybody know, jsut for the sake of curiosity, what’s the probability of the other side being red 10 times or fewer? Anybody stat expert know the forumula to figure this out? I’m guessing 10%? 5%? Something like that. I
Anonymosity wrote:
One last try, and then I’m just going to give up…
Don’t get too frustrated. Some people just aren’t going to get it, no matter how clearly the right answer is explained.
@lioncar–I agree with you. I think it is time to close this thread. I am going to now devote my energies to the thread concerning the 1940 Ford pickup truck and possible engine problems. The red/green problem or some variation has been in textbooks on probability theory for years along with the solution. We can all be of more benefit to civilization by helping the screen writer with his Ford truck problem. I will be posting a thread very shortly that is of extreme importance to mankind.
yes, getting close to closing this.
Barkydog, I’m not sure what distinction you’re trying to emphasize.
You keep commenting on “AS POSTED IN THE PUZZLER”.
I think we’re talking about that.
A card was picked, and the well dressed gentlemen asks if you’d like to bet even money.
we’re saying it’s not a 50-50 bet. it’s 2/3 chance for red.
You say “no drawing cards from the bag”, but a card was drawn.
That’s the card that is shown, with the red side up.
etc.
Look–the puzzler is right; where’s why.
Let’s say you play this game 3,000,000 times. Roughly 1,000,000 times, the G/G card will come up, as will the R/G and the R/R. The chance of drawing any card are equal.
BUT…since we know the “up card” is RED, you can discard ALL of the G/G draws (obviously). Note that the R/G card will show up with GREEN facing up 500,000 times…so discard them, too.
So, out of 3,000,000 draws, you get:
1,500,000 IRRELEVANT results (either G/G picked, or R/G picked with G showing)
500,000 times R/G with R showing (i.e. half of all R/G draws).
1,000,000 times R/R.
1,000,000/1,500,000= 0.667 probability of a R/R card.
500,000/1,500,000=0.333 probablity of a R/G card.
meanjoe75fan,
so you’re agreeing that it’s not 50-50, and the probability is 2/3 for red
but Barkydog keeps saying that something about "AS POSTED IN THE PUZZLER"
should make it 50-50.
I don’t see any nuance in how the puzzler is stated that creates a different scenario from
what most people have been saying so far:
“pick a card, see color X, the probability the other side is X is 2/3”
Go one step further. Well-dressed man with dark glasses shows you three cards as described, then puts them in a bag. Says “I’m going to pull one card out of this bag and place it on the table so you can’t see the other side. Bet you even money the side you can’t see is the same color as the side you can.” Suddenly you realize the guy is blind. Doesn’t know whether the card he’s going to pull has two matching sides, doesn’t know what color the face showing is.
Now do you take the bet?
Helpful hint for the mathematically challenged: problem stated this way, it doesn’t matter what color is showing, the odds are the same.
Dado … you are right. That’s the reason for the G/G card. As stated in the puzzler, the G/G card would result in half the draws being non-bets, just wasting time. By doing it the way you say, every draw yields a bet, increasing the con-man gambler’s cash flow.
The odds are 2:1 that it will be red. I muddied the discussion in trying to disprove the 1:1 folks.
The 1:1 people are trying to start the argument at the point where you make the bet, ignoring how the card was selected/placed. I’m saying the odds would be 1:1 if the well dressed man intentionally selected either the R/G or the R/R.
But he didn’t. It was a random pick from the bag which brings all the probabilities discussed above into play creating a 2:1 likelihood that the card is the R/R instead of the R/G.
I was only trying to demonstrate the flaw in their logic. The probability/odds are created at the time of picking the card, not making the bet and have to be considered when making the bet.
It’s not so much the answer, but maybe the puzzler question is wrong!! The odds are 50/50 for the reason BarkyDog stated on March 13, when you specifically answer the puzzler question asked. I think most of you are giving weight to red as the favored outcome because you can see variations on this game such as what you draw, what you see, what’s left in the bag, etc. But the question isn’t about how odds change with the next card to drawn (you only play the one card drawn). It isn’t about sides of cards because you don’t turn a card over and continue playing or drawing. Erroneously the puzzler answer gives favorable odds to red because two red sides and one green side have yet to be revealed (the face down side and the two sides in the bag), However, you will not draw or turn over other cards so that is irrelevant.
It simply boils down to: what color is on the back of this specific card? Let’s walk through it: based on the all the info given (number of cards and colors) and process of elimination (what we see and don’t see), we know the card’s unseen side is 1 of 2 choices: either red or green. It doesn’t matter how many cards - or colors - there were originally. Now that the green/green isn’t a possibility, both remaining cards have a red side, and only one card has a green side. So that “common” side acts as single side, just for the purposes of guessing the flip side of the red table card. We have effectively combined two red-side possibilities into one because it is the common side of both cards, and it’s the color that’s been revealed.
We have reduced our possibilities to only the two “uncommon”, unknown sides as choices. One red, one green. 50/50.
Another way to think of this common side is to think of the remaining two possible cards as two red cards, one with a green backside and one with a red backside. When the table card is showing a red side, we know it’s analogous to the “common” side. This concept is easy to grasp if the common side was a third color. With the card on the table showing us a red side, and its known the card in the bag has at least one red side, we can deduce both cards have at least one side in common. Let’s imagine those “common” sides turning white. So now we have a white side facing us on the table, and one side of the card left in the bag is also now white to represent it has at least one side in common with the table card. So essentially we now have a white/red and white/green card, with one card on the table showing white. What color is the unseen side of the table card? That is specifically what the puzzler asked. Either red or green, 50/50.
Typing a lot doesn’t make you right.
“What color is the unseen side of the table card? That is specifically what the puzzler asked.”
No. Read the puzzler. It asked “Should you take the bet?”
And you are wrong.
@littlemouse explain how he is wrong.there are 3 cards. Green and green, green and red or red and red.
Draw one card, place it on the table, it is red.
The puzzler as posted. “”“I’LL BET YOU EVEN MONEY THAT THE OTHER SIDE OF THE CARD IS ALSO RED.”
All Possible cards = 2, so the card on the table is red on top and bottom, or red on top and green on the bottom, thus 50/50 chance.
1 of 2 possible cards.
Should you take the bet? it is an even wager, you can win or loose. so tell us how we are wrong that you have a 50/50 chance of winning or loosing.
Barkydog and Enginerd,
I don’t want to convince you otherwise. just visit in the Bay Area and we can wager with each other
based on our beliefs. I’ll even give you 52:48 odds.
That said, I’ll try to explain.
It took me a little while to get my head around this.
One thing that’s difficult is how to count the number of cases.
Barkydog and Enginerd,
you emphasize that there are two cards possibly in play.
true enough.
Let’s say you played with two cards, say R/R and R/G, randomly picking.
How would you count the case of the G side showing?
Perhaps if you try to answer this question you may see what we’re talking about.
See my post on March 3.
If you’re picking from 3 cards, the two “in play” will change.
Sometimes you’ll see R, and it’s either the R/R or R/G card.
Sometimes you’ll see G, and it’s either the G/G or R/G card.
This makes it tricky to count the cases.
A side related, somewhat facetious question is this.
What is the probability of me having $1 million dollars in my pocket?
By your reasoning, it’s 50:50.
Either I have $1 M dollars in my pocket, or I do not.
two choices. 50:50
But, I’ll suggest there are more possibilities to consider,
as with the card problem,
and mis-counting the cases appears to lead to the confusion.
The ‘‘red/green’’ puzzler means only ONE thing in New Mexico.
And that is the most often asked question in most restaraunts here and is the New Mexico official question.
When you place your order for your food there remains one constant variable…
red or green ?
( chili on your food )
I prefer green chili on my
brakfast burrito
steak
stuffed sopapilla
cheese burger
etc
I finally got my head around this problem by realizing two things:
There are actually two different bets being made.
The player bets on the color of the unseen side 1/2
The dealer bets on the the drawn card having two same colored sides. 2/3
The odds of the player winning the bet in front of him are different from the odds of the dealer
winning the game plated over multiple trials.
So, if a player only plays one round his chances of winning are 1/2. As soon as the player plays another round his chances of winning the game drop to 1/3.
Hope that helps.
There’s only one bet being made.
The player’s bet IS one of the dealer’s multiple trials. The puzzler doesn’t suppose multiple trials. What it does suppose is that you are offered an even money bet when the odds aren’t 1/2.
“@littlemouse explain how he is wrong.”
See my earlier post where I explain it. Finding it is your task.