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The Magic Cafe Forum Index » » Puzzle me this... » » Best strategy? (0 Likes) Printer Friendly Version

Scott Cram
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Here's an interesting challenge for you.

You hand a friend nine blank, opaque index cards. You tell him that you're going to look away (or otherwise make yourself truly unaware of the numbers he's writing), that you want him to write down a different whole number of his choice on each of the cards, with the stipulation that no two numbers can be consecutive. (If he writes 12 on one of the cards, then none of the other cards can have 11 or 13 on them, for example.)

Once he's done this, he's to turn the cards with the number side down, and begin mixing them. At this point you can turn around, and emphasize that he's to mix them so well, that even he doesn't know what order the cards are in.

At this point, he is to gather the cards in a single stack, and then start turning them over one at a time. Your challenge is to try and stop him on the one number that is the highest of all of them.

Without knowing what each of the numbers is, and what the highest of them all, this would merely seem to be a matter of luck. The question, then, is what strategy could you employ that maximizes your chances of winning?
leonard
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Scott,

While the friend doesn't know the order of the cards, he would know the largest value. Are you looking for a "cold reading" type of solution, or merely a mathematical approach?

leonard

P.S. I assume we stop when we see the highest value (versus the next card).
landmark
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Well,

My genius brother emailed me his solution. I'm not sure I understand it totally, but I'll give others a chance to respond, and then I'll post his answer.

Jack Shalom
Nir Dahan
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This was also in one of gardeners books—really a nice puzzle. Even though there seem to be to less information, there is a best strategy...

I will also let others respond.
Mushu
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This sounds like something for the mentalism forum. There's a method of pencil reading that works by sound, and perhaps a muscle twitch or a darting glance may give it away.
pxs
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My guess is that you hear the first three numbers and then go for the largest next number. Could be first four.

This is intuition, not sure what proof is.
Bill Hallahan
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Write google (10 to the 100th power) on the back of one of the cards, and put a small mark that you will notice later on the other blank side where the spectator will write their number. Stop the spectator when you see the mark you put on the card. Have him turn the card over to show the largest number (a google!). You have to hope he doesn’t write a larger number than that!

Seriously, it’s irrelevant who puts the numbers on the cards, or what their values are. There is only one largest number. The question is really how many numbers you should look at before you decide that you probably have seen the largest one.

There is only a one in nine chance that the first number is the largest number. Let’s say you skip that.

A strategy such as saying you will pick the fifth card every time (for example) is obviously no better than picking the first card. It will still only be a one in nine chance of being correct, so if this problem has any good solution, it will involve changing strategy as you see whether the value of each new card is lower or higher than previous cards.

When he turns over the second card, then you can only be sure you didn’t make a mistake letting the first card pass if the number is larger. Otherwise, all subsequent cards might be smaller.

Without considering the nine! orderings, I am not sure, but I would guess the best strategy is something like you should keep taking cards until the card you see has four (or three?) smaller predecessors. When I get out of work tomorrow, I will work on this more. (I have a SAM meeting tonight! Yay!)
Humans make life so interesting. Do you know that in a universe so full of wonders, they have managed to create boredom. Quite astonishing.
- The character of ‘Death’ in the movie "Hogswatch"
landmark
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Okay, here is my genius brother's comments:

Quote:
I assume that the problem is this: after each card is turned over, I must either say "this is the highest card" or else "turn over another". That is, I assume that I don't have to say "the card about to be turned over will be the highest card."

Then my solution as to the best strategy is to wait until the 8th card is turned over. If it is higher than any of the previous 7, then I declare that it (the 8th) is the highest card. If the 8th card is not the highest of the first 8, then I will opt for the 9th card. I will then win in all cases where the highest card is in place eight (1/ 9), and in every case where the highest card is in place nine, except when the next highest is in place eight (7/8*1/9). SO 1/9 plus 7/72 equals 5/24.

My brother's analysis of the probability of his strategy seems correct to me, however, I don't think he has proven that his is the best strategy. Any takers?

Jack Shalom
pxs
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The correct strategy must be to wait for the nth card and pick it if it is the highest or otherwise pick the highest card that comes next.

n=9
You have a 1/9 chance of winning and an 8/9 chance of losing

n=8
You have a 1/9 chance that number 8 is the highest. You win.

There is a (1/9 x 7/8) chance that number 9 is the highest and that number 8 is lower than the previous numbers. You win.

The remainder of the time: You lose.

Thus the chance of winning on strategy if n=8 is 1/9 + 1/9 x 7/8

n=7

You have a 1/9 chance that number 7 is the highest. You win.

There is a (1/9 x 6/7) chance that number 8 is the highest and that number 7 is lower than the previous numbers. You win.

There is a (1/9 x 7/8 x 6/7) chance that number 9 is the highest and that number 8 is lower than the previous numbers and that number 7 is lower than the previous numbers. You win.

Thus the chance of winning on strategy if n=8 is 1/9 + (1/9 x 7/8 x 6/7) + (1/9 x 6/7)

Etc.

Now you just need to do the math and see which total is highest.

Am I right???
Scott Cram
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Quote:
On 2004-01-21 03:28, pxs wrote:
My guess is that you hear the first three numbers and then go for the largest next number. Could be first four.

This is intuition, not sure what proof is.

This is the correct answer!

Using this strategy, you will win a little over 40% of the time, which is far better than the approximate 11% mere chance would dictate.

In short, look and watch the first three numbers. If the fourth one is bigger than anyone shown so far, stop them there. If not, wait for the fifth one. If it's bigger than any number shown so far, stop there, and so on.
landmark
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Mathematical analysis of the above strategy, anyone?

Jack
TomasB
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Pxs seems to be absolutely right in his calculations. Great work! Just for fun I made 1000000 simulations of each of the cases and found that the percentage of successes in each case (starting with card number 1) is:

30.1
38.1
40.6
39.3
35.3
29.0
20.9
11.1

I have not managed to understand why the condition that no two numbers are allowed to be consecutive was given. By saying that you in effect make a new set of consecutive numbers so that for example 5 follows 3.

I'm pretty sure that the optimal strategy makes use of actually analyzing the numbers. If there's a 1 among the first three you are sure that's the lowest number in the batch. If there is a 7 and a 9 you know that there can be no number between them. This surely could be used to some advantage.

The strategy of waiting for the highest after the first three would be the optimal if any number could be the highest or the lowest, so it'd be nice to rephrase the problem to have them write any real number...so they can write negative numbers, fractions, numbers with decimals, etc. This seems to be a harder problem to the person you are doing it for.

To make this into a real bar bet I think you should say that you can pick out the highest number at least once if you do the experiment twice with the Mark (with him picking new numbers for the repeat). You'll win that bet 65% of the time.

/Tomas
sashain
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To: ThomasB, Scott Cram, or whoever wants to pipe in.

The problem is stated for 9 cards, N=9. The solution as given is the next highest number that turns up after observing the first 3 cards, n=3. What is the generalization for other values of N. How is n related to N for N>9? Is there an analytical relation? Or is ThomasB's brute force method required for each N?

Steve
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Jonathan Townsend
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This sounds a lot like a mentalist thing. If you add the instructions that the first number is the largest of all, and...anyway it's the kind of thing you do with a marked card.

Let's consider the case where they chose 5, 10, 15, 20, 25...and the cards are shuffled. What are the outcomes then?
...to all the coins I've dropped here
dersmith
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This is actually called the secretary problem. Here is a link.
landmark
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Thanks for the informative link. Guess my genius brother ain't such a genius after all!

Jack
Mike Powers
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If you go to the "Secretary Problem" link given above, you'll see another link to the "Monte Hall's Revenge" puzzle and solution. This formulation of the puzzle seems to be a generalized version of the puzzle presented in this thread. Here's the puzzle and solution taken from:

===

Monty Hall's Revenge
Puzzle by Philip Newman
You are once again a contestant on a game show. This time, however, you are presented with 100 bags.

"These bags," the host explains, "each contain a different amount of money. I'll let you open a bag, and then you may decide to pick that bag, or choose another one. Once you have passed up a bag, you may not go back to it. You may keep the money in the bag you choose. If, however, you manage to choose the bag with the most money, you will also win this!"

You watch, along with the rest of the crowd, as a door opens revealing the car of your dreams (a 1981 Chrysler Station Wagon).

What strategy should you take to give yourself the best chance of winning the car?

===

Answer to Monty Hall's Revenge
At first glance, it appears that no matter which bag you pick, your chances are the same: 1/100. However, once you have opened some bags, you know something. Since you obviously won't pick the second bag if it is less than the first, or the third if it is less than either of the first two (etc.), since you know you can't win, you now have information you can use to improve your odds. So, this best strategy is to open some number of bags, and then pick the next bag that has more money than all the previous ones.

How many should you pass on? By calculating the probabilities for each number of bags passed, you can find the highest probability (around 37%) for passing on 37 bags. This number can also be found by noting that it is the highest number of bags that can be passed on such that the odds of the most money being in one of the first x bags is still greater than the odds that the most money will be found in the last 100-x bags *and be picked*. This amounts to finding the smallest x with:

x/n > x/n * (1/(x+1) + ... + 1/(n-1)), where n=100 in this case.

This solution is a simplification of the puzzle; it is actually not so obvious that this is the best strategy. Also, when dealing with discrete values of money, we have additional information. For example, in the case of two bags, knowing how much money is in the first bag actually allows us to choose the richer bag with > 50% probability. However, these considerations do not appear to affect the solution to a great extent.

===

If you apply the above formula to the case at hand i.e. n = 9, it predicts what Tomas' computer analysis shows - x = 3 is the key number. For n = 100, x = 37 is the cut off point. The formula gives you a way to determine how to maximize your chances for n = any number. Note that there is no restriction on consecutive numbers here. It's not clear how this restriction affects the solution.

===

Here's another question, how does the number of cards used affect the probability of success?? Tomas indicates that betting that you can do it in two trials with 9 cards gives you a 65% chance. I guess, less cards gives you a better chance, you just need to know the value of x. What is the probablility with, say, 7 cards instead of 9? 10 cards seems like a more natural number of cards to use. Does the probability go down noticeably with 10 cards? Or up noticeably with 7 cards?

BTW the formula shows that x = 2 for 7 cards while it's still 3 for 10 cards.

Mike
TomasB
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Quote:
On 2004-01-28 08:07, Mike Powers wrote:
Here's another question, how does the number of cards used affect the probability of success?? Tomas indicates that betting that you can do it in two trials with 9 cards gives you a 65% chance. I guess, less cards gives you a better chance, you just need to know the value of x. What is the probablility with, say, 7 cards instead of 9? 10 cards seems like a more natural number of cards to use. Does the probability go down noticeably with 10 cards? Or up noticeably with 7 cards?

BTW the formula shows that x = 2 for 7 cards while it's still 3 for 10 cards.

Mike

For strategy 10:3 the probability is 39.8%, 8:3 gives 41.0% and 7:2 gives 41.5%.

So you could let the Mark have a, to you, unknown number of cards from 8 to 10 and you can still use the strategy of skipping the first three cards.

/Tomas
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