Best...Puzzle...Ever....

[BryonD]

But I was wrong. it isn't a 50/50 chance. There is no trick, Monty will always give the option to switch even if I pick the wrong one, and by switching I would get the right one. He will offer even if he knows he will 'lose' the game.

He always shows an empty chest. and by doing so gives me a 2/3 chance of getting it right by switching.

The only way it becomes 50/50 is if Monty doesn't know which chest the treasure is in and when he opens one it turns out to be the treasure.

RX

Sorry, bad wording on my part.

I was certainly confused by this one. I get it now. But at the time you posted I THOUGHT you had to be right.

Anyway, I don't know if I would have ever figured it out or not, but I didn't.

 

[Saeviomagy]

OK, here's the version of the treasure chest dillema I remembered, from a book on paradoxes and probability:

From Aha! Gotcha by Martin Gardner, pages 100-101

Operator: Step right up, folk, See if you can guess which shell the pea is under. Double your money if you win.

After playing the game a while Mr. Mark decided he couldn't win more than once out of three times.

Operator: Don't leave, Mac. I'll give you a break. Pick any shell. I'll turn over an empty one. Then the pea has to be under one of the other two, so your chances of winning go way up.

Poor Mr. Mark went broke fast. He didn't realize that turning on empty shell had no effect on his chances. Do you see why?
[/size]

Spoiler

Start - assume that cup 1 holds the pea.

I pick randomly:
If I pick 1 (1/3)
then:

1) the operator picks 2 (1/3*1/2)
1a) I stay and WIN (1/3*1/2*1/2) =1/12
2b) I switch and LOSE (1/3*1/2*1/2) = 1/12

2) the operator picks 3 (1/3*1/2)
1a) I stay and WIN (1/3*1/2*1/2) =1/12
1b) I switch and LOSE (1/3*1/2*1/2) =1/12

If I pick 2 (1/3)
then:
1) the operator picks 3 (1/3*1)
1a) I stay and LOSE (1/3*1*1/2) =1/6
1b) I switch and WIN (1/3*1*1/2) =1/6

If I pick 3
then:
1) The operator picks 2 (1/3*1)
1a) I stay and LOSE (1/3*1*1/2) = 1/6
1b) I switch and WIN (1/3*1*1/2) =1/6

That covers all the possibilities, doesn't it?

Winning chance = 1/6+1/6+1/12+1/12 = 6/12 = 1/2

IOW, we've got a 50/50 chance of winning now.

Which means that your book is wrong...

 

[Pyske]

I switch and LOSE

You were not offered the option of switching.

Spoiler

And your odds if you were would be 2/3 for switching, since we would be right back to the Monte situation.

. . . . . . . -- Eric

 

[Nasma]

The probability of event 'A' occurring, given that event 'B' has occurred is equal to:
Probability of event 'A' AND event 'B' occurring
divided by
probability of event 'B' occurring

Thus:
Probability of a child being a girl given that one is a boy equals:

Probability of a boy and a girl being born
divided by
probability of at least one child being a boy

equals:

0.5/0.75
(bb,bg,gb,gg)/ (bb,bg,,gb,gg)

equals 2/3

This can be verified by experiment

 

[MerakSpielman]

Which means that your book is wrong...

Think of it this way: You select a cup. You estimate that this cup has a 1/3 chance of being the correct cup. At this point you know that, regardless of your choice, at least one of the other cups is empty. Once one of the cups is revealed to be empty, you have no more information about your chosen cup than you had before. The odds of it being the correct cup are still 1/3.

Edit:

Martin Gardner, the author of this book, has been writing Scientific American's Mathematical Games column for over 25 years, and this book has an extensive bibliography in the Probability section. He isn't likely to be wrong.

That said, I think I've found the difference between the treasure-chest puzzle and the cups puzzle from the book - nowhere in the bit I quoted does he state that you're allowed to switch your choice after the empty cup is revealed (so it's really a completely different puzzle). Off-hand, I would guess that the ability to change your choice increases the odds to 1/2, regardless of whether you take the option. Which makes your analysis correct.

 

[der_kluge]

*ahem*

So, does anyone have any D&D related puzzles they would like to share?

 

[Saeviomagy]

Think of it this way: You select a cup. You estimate that this cup has a 1/3 chance of being the correct cup. At this point you know that, regardless of your choice, at least one of the other cups is empty. Once one of the cups is revealed to be empty, you have no more information about your chosen cup than you had before. The odds of it being the correct cup are still 1/3.

Oops. Missed the fact that you couldn't change.
In that case, nothing changes about the puzzle from knowing anything - even if he then told you what the correct cup was, you still wouldn't have better odds.

 

[Tuzenbach]

Spoiler

If they're the same thing, then there IS no separate probability for each.

It's like this: If you role a d20 & get a 20, it has ABSOLUTELY ZERO bearing on what your next role will be.

 

[orsal]

Spoiler

Start - assume that cup 1 holds the pea.

I pick randomly:
If I pick 1 (1/3)
then:

1) the operator picks 2 (1/3*1/2)
1a) I stay and WIN (1/3*1/2*1/2) =1/12
2b) I switch and LOSE (1/3*1/2*1/2) = 1/12

2) the operator picks 3 (1/3*1/2)
1a) I stay and WIN (1/3*1/2*1/2) =1/12
1b) I switch and LOSE (1/3*1/2*1/2) =1/12

If I pick 2 (1/3)
then:
1) the operator picks 3 (1/3*1)
1a) I stay and LOSE (1/3*1*1/2) =1/6
1b) I switch and WIN (1/3*1*1/2) =1/6

If I pick 3
then:
1) The operator picks 2 (1/3*1)
1a) I stay and LOSE (1/3*1*1/2) = 1/6
1b) I switch and WIN (1/3*1*1/2) =1/6

This is correct... if your strategy is

Spoiler

to decide randomly, with probability 1/2, whether or not to switch.

But that's a stupid strategy. (This is assuming the version of the puzzle in which you always have the option of switching. If you don't, none of this is relevant.) You can do much better, as many people on this thread have already demonstrated, by always switching. Remember, it's in your power to decide that.

Spoiler

Go back over your figures, and change the probability of staying to 0 adn that of switching to 1, and see what you get out of this.

 

[Tuzenbach]

I guess your post really has 2 "solutions." I was talking about the second one.

Merak, this is for you, "Piel", and who ever else gives a damn. The reason the whole girl/boy thing has two answers is due to a statistical construct known as "MUTALLY EXCLUSIVE". Now, being the genius that I am, I actually took Stat twice! Well, OK, it's because I failed it the first time. But ANYWAY.....

Do a web search for statistics and put in "mutually exclusive" with it. Basically, it's a rule that determines AHEAD OF TIME whether or not the result of the first trial influences the second trial. I'm no wiz on these things, but I'm going to go out on a limb and say that Merak's answer assumes that the variables ARE mutually exclusive (i.e., the first and second trials do not influence one another).

Piel's (sorry for the abbreviation!) answer assumes NON-mutual exclusivity. Basically, the whole AA, AB, BA, BB thing was shown with the help of Gregor Mendel and his genetics experiments with pea pods in the 1800's (?).

Now PRIOR TO the trials, there exists an EQUAL probability for boys and girls. That is:

.25 for a boy, then a girl
.25 for a girl, then a boy
.25 for a boy, then a boy
.25 for a girl, then a girl

Assuming that by having a girl we increase the chance of the gender of the next child to be a boy is silly UNLESS we state AHEAD OF TIME that the probabilities ARE NOT mutually exclusive, that is, that they ARE dependent upon each other.

I'm no great teacher of stats, so don't take my word for it. But please, do the research yourself and you'll hopefully stumble across a more comprehendable explanation.