Originally posted by LemonJello
Yeah, please take a look at the article. It mentions that this scenario has been named the "three-card swindle". Sorry, eldragonfly, but you are one of the many who would have been swindled.
The victim is convinced that the bet is fair, but the dealer makes money in the long run by winning 2/3 of the time.i have said nothing to contradict this, once again the problem was poorly worded. Look LemmonJelly read this sentence again if you don't believe me.
In what Martin Gardner has called the three-card swindle, a card is drawn from a hat, and if a red mark is shown...This is what was conspicuously missing from the original problem statement, not to rain on PBE6s parade or anything.
...the dealer bets the victim even money that the other side is also red.
Originally posted by AetheraelNo misconceptions, common or otherwise, fact is that you and the others are now just conveniently backpedalling. As a one shot one draw experiment, the odds are 50-50% as many here besides myself have stated.
exactly.. no one ever said that Bayes' was the ONLY valid method to solving the problem. Bayes' theorem is, however, an interesting and elegant method for solving this problem, in addition to being a method that avoids the common probabalistic misconception that eldragonfly fell into his first go around on the problem.
The probability (without considering the individual colors) that the hidden color is the same as the displayed color is clearly 2/3, as this holds if and only if the chosen card is black or white, which chooses 2 of the 3 cards.
I'm failing to see here how any reading of the original problem has the answer of "It is a fair bet, and the odds are 50%."
The closest I come to a reading of the problem with this conclusion is if the man with the hat intentionally chooses a silver side (if possible, perhaps) when drawing the card, in which case he would need to be aware of which card is selected and would thus know with certainty whether the bet was a good one to take or not.
And the problem with that scenario is that his knowing with certainty would destroy the randomness of the problem, because the one offering the bet knows! As I said before, it ceases to be a probability problem at that point.
Now, PBE6 later clarified this point specifically, indicating that the man did NOT know what the other side was. It has been pointed out several times that a gold side could just as easily have come up, and that there wasn't anything particularly special about it being the silver end (unlike the revealed door in the Monty Hall problem).
If you take these two points into consideration, then it becomes obvious that the problem as stated is an equivalent in every way to the Three-Card-Problem, which (as others have pointed out) turns out to be the "Three-Side-Problem" in terms of the reasoning.
At any rate, with any problem of this sort, these are the steps I take to arrive at a logical answer.
1) Identify the key components of the problem in as specific a manner as possible. For the original problem, this winds up being the card chosen, the visible side, and the hidden side.
2) Identify the odds for every distinct combination of these components in the set-up, ignoring any additional information given. Sometimes odds are given for one component compared to alternatives, such as the coin toss in the sports problem (which meant 50% chance for each city) or the rate of infection (1:1,000,000). These ratios define the set-up. Since the result is additional information, it isn't factored into this step, so the presence of the gold card is noted in calculations at this point, although it will be eliminated soon.
The odds table should meet the following conditions.
a) The sum of all values should equal 100%. (All cases accounted for)
b) The ratio of any specified subsets should match any ratios given for those subsets. (i.e. if city A is three times as likely as city B, then City A combos need to total 75%, and City B 25%..)
c) You can multiply the entire thing by the same constant to make tracking numbers easier if you like.
3) Note the additional information given about the result, and eliminate those combinations which do not match the pre-condition (in this case, silver face showing). All the remaining combinations should meet the pre-condition now. (The gold card is now eliminated and a moot issue at this point. So too is the gold side of the gold/silver card being turned up, which eliminates the gold/silver card half the times it turns up.)
4) Sum up both the odds of the pre-condition occurring (all the remaining combinations) and the odds for the event occurring that you're looking at. You'll be using the numbers you arrived at in Step 2 to determine these sums. Divide the odds for the event AND precondition both occurring by the odds of the precondition.
And you should have your valid answer right there. Any qualms with this technique for conditional probabilities?
Originally posted by eldragonflyNah, don't pretend. We all know you got swindled.The victim is convinced that the bet is fair, but the dealer makes money in the long run by winning 2/3 of the time.i have said nothing to contradict this, once again the problem was poorly worded. Look LemmonJelly read this sentence again if you don't believe me.[quote]In what Martin Gardner has called the three-card swindle, a card is dr ...[text shortened]... picuously missing from the original problem statement, not to rain on PBE6s parade or anything.
Anyway, moving on, here's another problem for you to keep the thread going.
I have a six-chambered revolver in which I have loaded two bullets into successive chambers. The other four chambers are empty. I spin the chamber, then I point the gun at your head and pull the trigger. No shot fires. I tell you that I am going to pull the trigger again, but I give you the option of whether the chamber is spun or not beforehand. What is your choice? Do you wish for me to spin the chamber or not before I pull the trigger a second time?
Now suppose instead that I initially pull the trigger two times in succession. Neither results in a shot fired. I give you the choice of whether or not the chamber is spun before I pull the trigger a third time. What is your decision?
Originally posted by eldragonflyCan you quote where any of us define the original problem as anything BUT the following?
This is insane. You keep changing the problem definition around as it suits you, there is neither rhyme nor reason to your overblown remarks and tedious assumptions.
1) The hat has three cards in it. One gold, one silver, and one gold/silver.
2) The man draws randomly from the three cards, with no single card being more likely than any other card to be drawn out of the hat.
3) That card is placed random side up, without anyone examining the card until it is placed on the table. The opposite side remains unknown to anyone.
4) The side we see is silver.
5) The man bets us even odds the other side of that card is silver, too.
Could you also tell me which if any of the above statements you believe are not accurate representations of the problem as given (or are ambiguous)?
Originally posted by geepamoogleI also posed this one back on page 21. It's also a relatively simple pair of problems.
Suppose a family has two children. The children are not twins, and each child is either a boy or a girl, and equally likely to be either.
What are the chances the family has a girl if...
1) the oldest is a boy?
2) they have at least one boy?[/b]
(Please note that there are two different questions here, neither question referring to the other.)
EDIT: I also pm'd you my answer to Russian Roulette question.
Originally posted by eldragonflyBut my response leads to the 2/3 answer that you objected to in the first place. And my response is equivalent to those of the people you are still arguing with. Very confusing. Anyway, we seem to be agreeing about some bits and going round in circles about the rest, so I will try and drop it this time!
Thanks for a half-way coherent response. This is what i have been saying all along.
Originally posted by mtthwI think eldragonfly is using the Posers and Puzzles community to work out the practical aspects of the 'three card swindle.' Picture this:
But my response leads to the 2/3 answer that you objected to in the first place. And my response is equivalent to those of the people you are still arguing with. Very confusing. Anyway, we seem to be agreeing about some bits and going round in circles about the rest, so I will try and drop it this time!
Instead of an all silver card he has one with the command, "Argue vociferously with these idiots, they know nothing!" on both sides. Instead of the all gold card he has one which instructs him, "Agree with the conclusion but claim to have been right all along and, or, the original wording of the problem is flawed" on both sides. Of course the third card is a combination of the two.
This would explain why we see outright denial 2/3 of the time and sardonic acceptance 1/3 of the time.
Originally posted by eldragonflyStop saying that everything that is against your opinion is worthless. Prove it!
Yet another unfortunate redefinition of the cards in the hat problem. First you say it's not a 3 card problem, then you say it is. Make up your mind kbaumen, your explanation is worthless.
And read the article again.
Originally posted by mtthwmtthw more to the point...i disagreed with your reasoning, that is all. The wikipedia article was succinct and lucid, many of the proffered "knee-jerk" explanations and "proofs" clearly were not.
But my response leads to the 2/3 answer that you objected to in the first place. And my response is equivalent to those of the people you are still arguing with. Very confusing. Anyway, we seem to be agreeing about some bits and going round in circles about the rest, so I will try and drop it this time!
There are currently two sets of questions posted here without a revealed answer.
LemonJello posed the following...
I have a six-chambered revolver in which I have loaded two bullets into successive chambers. The other four chambers are empty. I spin the chamber, then I point the gun at your head and pull the trigger. No shot fires. I tell you that I am going to pull the trigger again, but I give you the option of whether the chamber is spun or not beforehand. What is your choice? Do you wish for me to spin the chamber or not before I pull the trigger a second time?
Now suppose instead that I initially pull the trigger two times in succession. Neither results in a shot fired. I give you the choice of whether or not the chamber is spun before I pull the trigger a third time. What is your decision?
I have proposed the following to complete the trifecta of classic probability 'paradoxes'.
Suppose a family has two children. The children are not twins, and each child is either a boy or a girl, and equally likely to be either.
What are the chances the family has a girl if...
1) the oldest is a boy?
2) they have at least one boy?
(No further comment on the original problem until the opposing side tells me substantively and clearly where we are either changing or inaccurately stating the problem. Alternatively, I will respond to their restatement of the entire problem in their own words if it is clearly worded.)
Originally posted by geepamoogle[/i]1. Don't spin. There is only one chamber that will have a bullet in it next and three that will be empty. So these 1 in 4 odds are less than the initial 1 in 3 chance.
There are currently two sets of questions posted here without a revealed answer.
LemonJello posed the following...
[i]I have a six-chambered revolver in which I have loaded two bullets into successive chambers. The other four chambers are empty. I spin the chamber, then I point the gun at your head and pull the trigger. No shot fires. I tell you espond to their restatement of the entire problem in their own words if it is clearly worded.)
2. No difference. The odds are 1 in 3 whether you spin or not.
I know the answer to the boy/girl problem.