Red/Green puzzler solution is wrong

The Show
61

MTraveler you said there are two different bets, one being the dealer bets on the drawn card having the same colored sides 2/3. That’s true while there are 3 cards to be considered, but as soon as the green/green card is eliminated, it’s a two-card game (as Barkydog said) - with a common color on one side and different color on the other. Once the color of the common side is revealed, it essential takes two of the four sides out of contention, so odds are 50/50.

Little mouse is on to something also – if there were multiple trials, or a drawing game. I would change my answer. If the game were “what are the odds of drawing the red side of a card from a bag that had 3 red sides and 1 green side and currently one of those red sides revealed?” I would say 2/3. But when you are simply looking at one card and having to anticipate the unseen side, and there is one opportunity for it to be red, and one opportunity to be green, I don’t see how it can anything but 50/50.

Fredfhome asked what is the probability of me having $1 million dollars in my pocket? Well, if your options were any number of dollar bills up to a million, then obviously the odds are hugely different. But if you had, say, a check for a million and it was either in your pocket or not, that would be the same as a two-card game. 50/50.

62

Enginerd, you’re confusing the issue by thinking about “3 cards.” What there really are, is “6 sides”: (3 cards)X(2 sides on each card)=6 sides.

When we know there’s a “red” side showing, there are three SIDES (not cards) that are equally likely to occur. Since there are two of these sides on the R/R card, and only one on the G/R card, the R/R card is twice as likely to be the card in question.

63

“I don’t see how it can anything but 50/50.”

That’s really not much of an argument. It’s certainly not math.

64

The dealer always offers the same bet no matter what card is drawn. Even money if the unseen side is different. Of the three cards in the bag only one will win for the player. The other two win for the dealer. If the player only considers the card in front of him he believes his odds of winning are always 50/50. The dealer knows he will win 2 out of three draws.

You can argue the puzzler doesn’t give you enough information to reach the right answer, but that is the essence of the con.

65

MTraveler, you said:

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.
and
if a player only plays one round his chances of winning are 1/2

I don’t believe this is true.
I thought so at first, but this idea doesn’t correspond to results counting the
possibilities, or the simulations by Anonymosity on March 9

I think it’s 2/3 for the same color on the other side,
whether the first draw or the N’th draw.

Although the puzzler didn’t ask about a two card situation, I think it can
serve to illustrate why the probability is 2/3, and once you see it for two cards,
you can extend to 3 cards.

as Enginerd says:

That’s true while there are 3 cards to be considered, but as soon as the
green/green card is eliminated, it’s a two-card game (as Barkydog said)

So let’s focus on 2 cards for a moment.

MTraveler, Enginerd, and Barky Dog,

If you had two cards: R/R and R/G,
and you picked one card randomly, and looked at a side,
how would you count the case where the G side shows up?

What are the other cases?
I think there are 3 other cases.
There are 4 cases total.
2 cards x 2 sides.

What probability do you assign for each of these 4 cases?

Enginerd, you’re getting close with your previous response.
I think if you answer this question, you might see why some of
your statements are not true.

66

Fred. You are correct.

67

Probability analysis is very confusing. Even mathematicians trained in it can make mistakes when faced with this kind of problem. Usually when there is a question whether the analysis is correct, mathematicians resort to trial by fire: Monte Carlo analysis. They program thousands of trials on their computer. So I wonder, for those saying it is an even bet (1:1), how do you explain the computer simulation results posted above by multiple posters, where thousands of hands were simulated, and the results showed 2:1 rather than 1:1?

Someone is wrong. Is it some mis-understanding of the problem by the person who programmed the computer? Is it the computer itself that is faulty?

If you think it is the fault of the computer or its programmer, have you tried the “trial by fire” method yourself? Either program your own computer and do the simulation, or is is probably faster and easier just to sit down with three cards and play the game exactly as stated in the original puzzler 300 or 400 times. It shouldn’t take more than an hour or two, and you can do it while you watch tv. Then let us know the result.

68

Monte Carlo analysis. They program thousands of trials on their computer.

The Monte Carol method of integration!! I haven’t thought about that in over 30 years. I Taught that method in college to figure out the area under a curve. This was before they had math libraries you could use that could actually do Integration (for you young-ins : Anti-derivatives). One of those cool things taught in college that you never ever used in real life…at least no my life.

69

If one has to use simulation to solve a math problem, one has not solved the math problem.

70

Mechaniker Good point. But some math problems in probability can’t be solved with basic pen and paper analysis. They have to be solved trial and error, usually by some form of Monte Carlo methods or something similar. A good example is the probabilities of the casino game Blackjack (i.e. 21). Professor Thorpe (1962) was the first to publish the basic probabilities in his bestseller “Beat the Dealer”. Card counters have been thankful to him ever since! Professor Thorpe went on to do the basic probability ciphering for hedge funds, using the same Monte Carlo computer techniques to analyse the stock market.