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The Magic Cafe Forum Index » » Puzzle me this... » » More annoying than Monty Hall! (1 Likes) Printer Friendly Version

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TomasB
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Landmark,

That only works if the questions you ask don't fill up the probability. If you get a "No" on "Is there at least one head before 1940?" and then ask "Is there at least one head at or after 1940?" you gain no information hence the probability is 1 of that being a property.

Also the combined answers to several earlier questions tell you how much information you gain with your new question, so the easiest is to keep the questions totally unrelated. For example the correlation of the shinyness of a coin and its date is high.


LobowolfXXX you had a really hard time convincing me that 1/3 didn't have to be the correct answer. It took several days to convert me as I remember.


I just thought of a variation on this problem:

I shuffle a deck and deal two cards until I find a pair with at least one Ace, re-shuffling each time. What's the probability that there are two Aces in that pile?

Repeat but look for a pair with a Black Ace. What is the probability of it containing two Aces?

Repeat but look for a pair with the Ace of Spades in it. What's the probability that it has two Aces?


Does it make any difference if I instead use this procedure: Shuffle the deck and deal 26 two-card piles and look through them until I find the first pair meeting my criteria? It would be much quicker.

/Tomas
landmark
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Before we move on, now consider:

So I now ask a series of a billion independent such questions (those with a probability of about 1 in a million). It would be unusual not to get a yes answer. If I take the set of ALL such possible questions, eventually there should be a yes, with an infinitesimal probability of all nos. But that means then that I don't need to ask those questions in the first place since I know that somewhere in that set there must be a yes! And, by our prior agreements, if there's a yes answer to such a question, then the odds change . . .
TomasB
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You write "unusual to not get a yes answer" so as long as that holds true you still have to ask the questions.

If I understand it all correctly you should be able to keep asking questions if they are totally uncorrelated, or you can ask correlated questions but have her flip new coins each time taken at random among all coins in distribution.

I don't follow the reasoning that you would not need to ask the questions. The probability is not 1 of there being a yes among them.

/Tomas
LobowolfXXX
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The best way to convince someone that the answer to the Monty Hall problem isn't 50-50, by the way, is to take a deck of cards, tell him to pull out a card and not look at it, and you'll give him a prize if he pulls out the ace of spades. Then take the rest of the deck and look through it...turn up 50 cards that aren't the ace of spades, so you and he each have a face down card, and ask him if he wants to switch, or if he thinks it's 50-50.

The "extra questions" thing sort of reminds me of the black raven proposition. I have a red magic book (Let's say it's Dusheck's Card Magic, since that's red). Curiously, this helps support (prove?) the proposition that all ravens are black, at least inductively. The logical equivalent to "All ravens are black" is "No non-black object is a raven" or "All non-black objects are non-ravens." By finding an instance consistent with these latter versions of the proposition, I incrementally support their equivalent, too. This red thing COULD have been a raven, but instead it's a magic book...making it a teeny tiny bit more likely that yes, all ravens are in fact black.

S2000Magician should be credited (blamed?!) in this thread, too. He's (correctly) pointed out to me in similar threads that I was importing the unspoken presupposition that a choice between two objects would be random (e.g. when I would say, without comment, something like, "If he had a male and a female dog, there's only a 50% chance he would have told you about the female dog.") Although, to paraphrase (or maybe quote) poker guru David Sklansky, "In the beginning, all bets were even money," i.e. in the absence of outside information, you probably SHOULD assume that a binary proposition is 50-50.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley.

"...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us."
landmark
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The chances of getting a billion nos (assuming independence) is .999999^(10^9). This # is near zero. (Asking 10^8 questions gives a probability of 3.7 x 10^-44 of all nos) I can increase the number of questions I ask arbitrarily, and approach as close to zero as much as I like. I have 10^20 questions in my pocket. I will take the chance that IF I asked them I wouldn't get all nos.

Jack
TomasB
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Lobowolf, this got me thinking of the black raven too. What are the odds? Smile

Hmmm, I got an idea on how to visualize it. I think you could formulate your 10^20 questions as a single question with AND between each question. Then you'd have a single question which would be answered "yes" to with probability p (which is pretty darn close to 1 now). In other words, it'd not increase the 1/3 probability much.

That tells me that I was wrong in thinking that the question (of many questions) that got a "yes" would work with full force with its probability even if they are uncorrelated. There are more layers of deception in this puzzle than I could have imagined.

/Tomas
landmark
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Yes the black raven occurred to me too.

But I like your solution of having it equivalent to one question. It makes it all consistent with what's been said before.

I think I'm ready to go on to the next round.

Jack
landmark
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Quote:
On 2009-07-25 01:46, TomasB wrote:


I just thought of a variation on this problem:

I shuffle a deck and deal two cards until I find a pair with at least one Ace, re-shuffling each time. What's the probability that there are two Aces in that pile?


/Tomas


Do you mean two Aces in that pair ? (You wrote "pile" there. Just making sure.)
TomasB
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Yeah, sorry, I must have been thinking about the follow-up question with 26 piles. I mean the pair that was just dealt.

/Tomas
LobowolfXXX
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Quote:
On Jul 21, 2009, landmark wrote:


1) The question as posed above, has an answer of 1/3. Consider that given that we know there is at least one daughter, we have three kinds of families {gb, bg, gg}. Only one of those families has two daughters, hence 1 in 3. The classic answer.



In the first two situations, you're betting on a parlay - The apparently-as-likely-scenario of one boy & one girl (as compared to two girls) is offset by the 50-50 probability that a family that could have told you about either child chose to tell you about the girl, making the GG and BG cases 50-50 (as one would intuitively expect).

The classic 1 in 3 answer is problematic, because, as the analysis would be the same if you were told the first child was a boy, then regardless of what you were told about the sex of a child, the apparent probability of a mixed-sex family would be 2/3, which we know it isn't. It's ignoring the fact that the choice of which child's sex to reveal, which carries a nonzero/noncertain probability, that makes it appear to be a paradox.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley.

"...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us."
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