Posted by eldragonfly
The "dealer" is offering bets on either a SS or SG card, provided that the silver side of the SG card is showing.
According to the wording of the problem, the "dealer" is making a wager on "the opposite side of the card".
The focus on the card to the exclusion of sides is your interpretation.
As far as the equivalency of the problem given in the Wikipedia link, if you were to change black to silver, and white to gold, then toss in some cosmetic Western language, you would end up with the problem on page one without affecting the odds in any manner whatsoever.
If you think the solutions given in that article are false, then why would you link it, and if you think the article legitimate, why argue against it?
I understand that Wikipedia isn't a final authority on anything, but thus far I've found it is pretty reliable.
I also understand that majority thought and consensus doesn't constitute proof, but it does mean that you ought to make doubly sure of yourself if you disagree with it.
One more time:
The bayesian solution counts sides not cards. The "dealer" is offering bets on either a SS or SG card, provided that the silver side of the SG card is showing. Remember the GG card is excluded in the stated scenario. Now assuming that he picks randomly between these two remaining cards, then it is a fair bet, i.e. he turns over the card he has selected at random and reveals the other side, which can only be silver or gold.
Let's let this represent a silver side of a card ---> 😕
Let's let this represent a gold side of a card ---> 🙂
the dealer shows three cards in his hat :
🙂🙂
😕🙂
😕😕
The 🙂🙂 card is excluded in this example by definition.
Now we are left with two cards, assuming the silver side of the SG is face up. Remember for both cards the silver sides are face up.
😕😕
😕🙂
The dealer then selects randomly between the two cards and makes his even odds wager. He then turns over that card.
only two things can occur:
🙂😕
-or-
😕😕
It is an even money bet in the stated scenario.
Once again, I reject your notion that the individual sides don't matter.
Once again, I point out that according to the wording of the problem, the "dealer" is making a wager on "the opposite side of the card".
It is a side we see, and not the entire card. It is the "other side" that the problem directly centers on.
Once again, you are ignoring the possibility that the SG card could have come up gold-side-up, choosing to read the information on the results as follows.
The card selected has a silver side.
We have more information than that. We also know the orientation of the card is such that the card has a silver side FACING UP.
It seems unimportant, but this additional fact explicitly excludes the possibility that the SG was selected with the orientation of gold-side-up.
The chances that the SG was selected gold-side-up was equal to the chances it was selected silver-side-up.
The sum of these two chances are equal to the chances of the SS being selected (in which case a silver side WILL be showing).
This means the chances of the SS being selected silver-side-up is double the chances of SG being selected SILVER SIDE UP.
The good news is that this gambling problem can also be simulated relatively easily in excel. Take 10000 cases, choose a random card, then choose a random face up. Ignore the ones where gold is showing and see how many have silver side up and silver side underneath...
I have run it 5 times, got the following results for the other side when we know that the visible side is silver...
Silver Gold
3327 1683
3349 1678
3309 1695
3298 1615
3338 1623
Originally posted by eldragonflyThe way I see it, there are basically three possibilities: You've got the problem right, we've got
One more time:
The bayesian solution counts sides not cards. The "dealer" is offering bets on either a SS or SG card, provided that the silver side of the SG card is showing. Remember the GG card is excluded in the stated scenario. Now assuming that he picks randomly between these two remaining cards, then it is a fair bet, i.e. he turns over the card hings can occur:
🙂😕
-or-
😕😕
It is an even money bet in the stated scenario.
the problem right, or both you and we are wrong. Using your system of probability, it's an
even 1/3 chance that one of these three scenarios are the case -- that is, the system that doesn't
take into account evaluating new information.
We, however, think that since we all have come up with the solution using various different
mathematical models, that we comprise a wide variety of backgrounds in mathematics, and that
the very site you cited contained the very answer we are giving, that the likelihood that you are
right is actually 0%, just like the G/G card, and that we are right is 100%.
As for whether you're a troll or just obtuse, I would say that it would be a bad bet to take
even money on that question; I'd say the odds are 3:1 on that question, though I would appeal
to the mathematicians with greater expertise than I for a more refined proof of that probability.
Nemesio
Originally posted by geepamoogleYou're right, thank you for your patience, i apologize. My solution was in error.
Once again, I reject your notion that the individual sides don't matter.
Once again, I point out that according to the wording of the problem, the "dealer" is making a wager on "[b]the opposite side of the card".
It is a side we see, and not the entire card. It is the "other side" that the problem directly centers on.
Once again, you are ig ...[text shortened]... selected silver-side-up is double the chances of SG being selected SILVER SIDE UP.[/b]
Originally posted by eldragonflyDon't worry, many people get confused in a similar way when they are asked about the fairness of this bet.
i looked at two ideas here, the only two possible combinations or ways that the cards in the hat could appear, either
SS SG GG or SS GS GG, that is only the SG card changes its orientation depending on which side is face up.
I think the problem is a little easier to think about if it's worded differently. Suppose you pull out the card and the face is the color x. what are the chances that the opposing side is also color x.
That way you don't get into the "the gold/gold card is irrelevant" bull***.
It then becomes obvious that:
1) you pull out the gold/gold card - the opposing side is the same
2) you pull out the silver/silver card - the opposing side is the same
3) you pull out the gold/silver card - the opposing side is different
therefore, 2:1 that the opposing side is the same.