Friday Brain Puzzle

albionmoonlight · 07-30-2004, 09:59 AM

There are three notecards on a table in front of you face down. On the flip side of each card is a number. No card has the same number as any other card. The numbers are completely random and unbounded--meaning that they can be as small or as large as anything. They may be 1 : 13 : 8 billion. Or they may be 1,874,598 : 8,743,983 : 1,234,984,234,745. You don't know and have no way of knowing (all of that is my attempt to demonstrate that the size of the numbers matters only relative to each other. You cannot get anywhere in this puzzle by noting how large the numbers seem relative to numbers that we encounter in everyday life.)

Your goal is to end up with the highest number. The rules are that you pick a card at random and look at its number. You can either keep that card or discard it and pick another. If you discard it, you cannot later reclaim it. If you pick a second card, you can either keep it or discard it. If you discard it, you are stuck with the third card.

The question--is there a way to increase your chance of winning to over 1/3? If so, what is it and how good can you make your chances?

QuikSand · 07-30-2004, 10:03 AM

First thought:

Discard your first pick no matter what
If card #2 is greater than #1 was, stick with that
If card #2 is less than #1 was, discard it and take #3

QuikSand · 07-30-2004, 10:05 AM

I think that will give you a 50% chance of winning. Not sure if that's the best method, though.

With only three cards, it doesn't seem too likely that this gets all that complex. I wonder if there's a more general solution... to a similar game with n cards?

Raiders Army · 07-30-2004, 10:05 AM

Quote:

Originally Posted by QuikSand

First thought:

Discard your first pick no matter what
If card #2 is greater than #1 was, stick with that
If card #2 is less than #1 was, discard it and take #3

Agree. Isn't this a variation of the Monty Hall puzzle?

QuikSand · 07-30-2004, 10:06 AM

I don't see any meaningful connection to the Monty Hall puzzle.

albionmoonlight · 07-30-2004, 10:06 AM

Well, QS just answered mine in less time than it took to write it.

Let's move on to his--a general soultion for n cards.

Raiders Army · 07-30-2004, 10:08 AM

Three doors, three choices. In both instances you pick one, and have the option of picking another choice. If you pick the other choice, your odds are 2/3 instead of 1/3. The twist to this puzzle is that you have to give up your first choice to see the second.

QuikSand · 07-30-2004, 10:08 AM

Quote:

Originally Posted by albionmoonlight

Let's move on to his--a general soultion for n cards.

I'm thinking this gets brutal. What about trying it for 4 cards instead? Thst seems more manageable. If there's a trend, we might pick it up by doing the first few iterations - 3, 4, 5...

I'll give some thought to the 4-card variation.

albionmoonlight · 07-30-2004, 10:10 AM

Raiders--the "key" to Monty is the presence of an outside party who imparts knowledge into the system (Monty always shows you a loser door). No such key exists here.

QuikSand · 07-30-2004, 10:16 AM

Okay, there are two models I can think of for a 4-card option:

-Discard the first one, and keep anything that's the highest you have seen
(My back-of-envelope math suggests this will win 5 out of 12, or 41.66%)

-Discard the first two, and keep #3 if it's greater than both of the first two
(My back-of-envelope math suggests this isn't as good - only 37.5%)

Any other structured aproaches to 4 cards?

QuikSand · 07-30-2004, 10:17 AM

If you have 100 cards, though... I can't imagine that the best solution is to still discard one, and then take anything that qualifies from there. I suspect there's a sort of "sweet spot" involved when you generalize to n cards... but I don't see how to calculate it easily.

I think it should be less than half of n, though.

albionmoonlight · 07-30-2004, 10:28 AM

Real Job is keeping me from working with you on this one--I'm following with interest, but can't really contribute new ideas.

Mr. Wednesday · 07-30-2004, 10:29 AM

Quote:

Originally Posted by albionmoonlight

Raiders--the "key" to Monty is the presence of an outside party who imparts knowledge into the system (Monty always shows you a loser door). No such key exists here.

However, at the time that it comes to decide whether or not to go with the third card, you have knowledge of both the first and the second card (assuming QS's system). Isn't that the same effect as Monty, in that you have now eliminated one of the first two choices?

Mr. Wednesday · 07-30-2004, 10:32 AM

Dola, the key difference with the Monty Hall problem is that it doesn't necessarily make sense to switch to the third card. I haven't really worked out a concise explanation for the theory that underlies this difference.

QuikSand · 07-30-2004, 10:32 AM

In this puzzle, I still see everything as being random. The unknown quantity or quantities in the remaining card(s) is a completely random variable, no matter what you have seen in the one or more cards you have already seen.

The key element in the MH puzzle is that the options are, after a certain point, no longer random -- the presence of a knowledgeable decision-makes ensures that (and is the backbone to the MH puzzle).

QuikSand · 07-30-2004, 10:33 AM

Quote:

Originally Posted by Mr. Wednesday

Dola, the key difference with the Monty Hall problem is that it doesn't necessarily make sense to switch to the third card. I haven't really worked out a concise explanation for the theory that underlies this difference.

Seems we're farther apart than I might have imagined. Of course you switch.

I don't know if this is the thread, but I'm interested in your "concise explanation."

Mr. Wednesday · 07-30-2004, 10:39 AM

In the MH puzzle, the unknown quantities are also random variables. Monty reveals an undesirable choice, leaving you to choose at random between the unknown you initially selected and the unknown that remains.

In this puzzle, and undesirable choice is also revealed, although it is possible that the undesirable is the choice you initially made.

Mr. Wednesday · 07-30-2004, 10:40 AM

Quote:

Originally Posted by QuikSand

First thought:

Discard your first pick no matter what
If card #2 is greater than #1 was, stick with that
If card #2 is less than #1 was, discard it and take #3

Quote:

Seems we're farther apart than I might have imagined. Of course you switch.

In the MH problem, you switch. If I read you right above, in this problem, you don't necessarily switch.

Edit: The key to the MH problem is that you are offered your initial selection, which was made with a 1/3 probability of being right, or an alternate selection, which would be made with a 1/2 probability of being right. I'm less sure about what happens with our new problem.

QuikSand · 07-30-2004, 10:46 AM

I guess I misunderstood your earlier post-- I though you were arguing that in the Monty Hall problem, you shouldn't switch. Now, I see the pivotal sentence causing the misunderstanding.

Now, it seems we're debating whether the two problems are just derivatives. I'll stick with the side that says the similarities are superficial (they happen to involv three options, that sort of thing) -- but that the critical element in the MH puzzle remains the introduction of a decision by someone with knowledge, which makes the puzzle no longer about random probabilities. I see no such introduction in this puzzle... and in my mind, that makes them essentially unrelated puzzles.

QuikSand · 07-30-2004, 10:47 AM

Quote:

Originally Posted by Mr. Wednesday

Edit: The key to the MH problem is that you are offered your initial selection, which was made with a 1/3 probability of being right, or an alternate selection, which would be made with a 1/2 probability of being right. I'm less sure about what happens with our new problem.

False. As any FOFC regular could tell you, I'll stand firm that the chances of you being right if you switch in the MH problem are cleary, unabashedly, and forever: 2/3.

I can show you my 2/3 tattoo, if you like.

Mr. Wednesday · 07-30-2004, 10:54 AM

Oops, yes, I forgot that the overall probability in the MH problem is 2/3. I'm trying to figure out if I can weasel by saying I was only referring to the final choice, but I think I'm still wrong then.

Mr. Wednesday · 07-30-2004, 10:57 AM

Dola, I don't think the problems are derivatives, necessarily, but I do think that making a subsequent "decision" with more information than the initial selection is a common element between the problems. The nature of the decision and the additional information is the key difference.

QuikSand · 07-30-2004, 11:15 AM

Quote:

Originally Posted by Mr. Wednesday

I'm trying to figure out if I can weasel by saying I was only referring to the final choice, but I think I'm still wrong then.

I think so, too.

07-30-2004, 10:06 AM	#5
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	I don't see any meaningful connection to the Monty Hall puzzle. Last edited by QuikSand : 07-30-2004 at 10:09 AM. Reason: responding to comments below

07-30-2004, 10:16 AM	#10
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	Okay, there are two models I can think of for a 4-card option: -Discard the first one, and keep anything that's the highest you have seen (My back-of-envelope math suggests this will win 5 out of 12, or 41.66%) -Discard the first two, and keep #3 if it's greater than both of the first two (My back-of-envelope math suggests this isn't as good - only 37.5%) Any other structured aproaches to 4 cards? Last edited by QuikSand : 07-30-2004 at 10:28 AM.

07-30-2004, 10:17 AM	#11
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	If you have 100 cards, though... I can't imagine that the best solution is to still discard one, and then take anything that qualifies from there. I suspect there's a sort of "sweet spot" involved when you generalize to n cards... but I don't see how to calculate it easily. I think it should be less than half of n, though. Last edited by QuikSand : 07-30-2004 at 10:18 AM.

07-30-2004, 10:03 AM	#2
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	First thought: Discard your first pick no matter what If card #2 is greater than #1 was, stick with that If card #2 is less than #1 was, discard it and take #3

07-30-2004, 10:05 AM	#3
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	I think that will give you a 50% chance of winning. Not sure if that's the best method, though. With only three cards, it doesn't seem too likely that this gets all that complex. I wonder if there's a more general solution... to a similar game with n cards?

07-30-2004, 10:06 AM	#6
albionmoonlight Head Coach Join Date: Oct 2000 Location: North Carolina	Well, QS just answered mine in less time than it took to write it. Let's move on to his--a general soultion for n cards.

07-30-2004, 10:08 AM	#7
Raiders Army Coordinator Join Date: Oct 2000 Location: The Black Hole	Three doors, three choices. In both instances you pick one, and have the option of picking another choice. If you pick the other choice, your odds are 2/3 instead of 1/3. The twist to this puzzle is that you have to give up your first choice to see the second.

07-30-2004, 10:10 AM	#9
albionmoonlight Head Coach Join Date: Oct 2000 Location: North Carolina	Raiders--the "key" to Monty is the presence of an outside party who imparts knowledge into the system (Monty always shows you a loser door). No such key exists here.

07-30-2004, 10:28 AM	#12
albionmoonlight Head Coach Join Date: Oct 2000 Location: North Carolina	Real Job is keeping me from working with you on this one--I'm following with interest, but can't really contribute new ideas.

07-30-2004, 10:32 AM	#14
Mr. Wednesday Pro Starter Join Date: Jul 2003 Location: South Bend, IN	Dola, the key difference with the Monty Hall problem is that it doesn't necessarily make sense to switch to the third card. I haven't really worked out a concise explanation for the theory that underlies this difference.

07-30-2004, 10:39 AM	#17
Mr. Wednesday Pro Starter Join Date: Jul 2003 Location: South Bend, IN	In the MH puzzle, the unknown quantities are also random variables. Monty reveals an undesirable choice, leaving you to choose at random between the unknown you initially selected and the unknown that remains. In this puzzle, and undesirable choice is also revealed, although it is possible that the undesirable is the choice you initially made.

07-30-2004, 10:46 AM	#19
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	I guess I misunderstood your earlier post-- I though you were arguing that in the Monty Hall problem, you shouldn't switch. Now, I see the pivotal sentence causing the misunderstanding. Now, it seems we're debating whether the two problems are just derivatives. I'll stick with the side that says the similarities are superficial (they happen to involv three options, that sort of thing) -- but that the critical element in the MH puzzle remains the introduction of a decision by someone with knowledge, which makes the puzzle no longer about random probabilities. I see no such introduction in this puzzle... and in my mind, that makes them essentially unrelated puzzles.

07-30-2004, 10:54 AM	#21
Mr. Wednesday Pro Starter Join Date: Jul 2003 Location: South Bend, IN	Oops, yes, I forgot that the overall probability in the MH problem is 2/3. I'm trying to figure out if I can weasel by saying I was only referring to the final choice, but I think I'm still wrong then.

07-30-2004, 10:57 AM	#22
Mr. Wednesday Pro Starter Join Date: Jul 2003 Location: South Bend, IN	Dola, I don't think the problems are derivatives, necessarily, but I do think that making a subsequent "decision" with more information than the initial selection is a common element between the problems. The nature of the decision and the additional information is the key difference.

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