The Fourth Coffeehouse Challenge

Here's the original problem, followed by the results. Enjoy!

The Fourth Coffeehouse Challenge:MAGGIE'S BET

After leaving Zeno's late one night, Maggie posed the following challenge to Charles, as they stepped out on to the long street stretching each way in both directions: "Charles, let me mark the place where I am standing in the street---right here---as Point 0, and you walk down the street 100 feet ahead of Point 0 to Point +100. So we will be in a position like this," she desribed, as Maggie sketched the scheme:
Maggie=Point 0------------------------------------------------------------------Charles= Point +100
"So what's the deal?," quipped Charles, impatiently.
"C'mon, Charles, hear me out," Maggie continued. "When you are in position, I want you to walk either further away from me (to the right, on the scheme), or toward me (to the left, on the scheme), depending on the following ground rules: you will shuffle randomly these five red cards and five black cards, to be laid face down, and selected by you, one at a time, and then discarded after each selection, until all ten are selected. That will help determine where you walk. So far, so good?"
"Yes, but go on," added Charles.
"OK, now comes the fun stuff. After each card selected, you are to move one-half the distance from where you are to Point 0, going right (away from Point 0) that many feet with every red card selected, and going left (toward Point 0) that many feet with every black card. So, for example, your first move from Point 100 would be 100/2=50 feet; if the first card is red, you move to Point +150 feet, and if the first card is black, you would move to Point +50 feet from Point 0. And depending on which card you drew first, your second move would be either 150/2=75 feet (given red) and 50/2=25 feet (given a black draw first). And so on until all ten cards are drawn. Now for my challenge: I bet I can specify WITHIN TWO FEET where you will end up, no matter what cards you select! Wanna bet??
Charles uncomfortably considers the offer, and looks back to any Zeno patrons who might help out. What advice can YOU lend Charles? Should he take Maggie's bet? Can she really know where Charles must ultimately stop? Why or why not?

Results are as follows:

Re: the current question at Zeno's coffeehouse ( I found out about it via a member of the THINK-L list-serv who publicized this delightful opportunity!
Charles should not bet against Maggie, since it is a certainty that Charles will end up somewhere around 23.73 feet away from her. Each of the five "red" card events alters the distance between Maggie and Charles by a ratio of 1.5 to the preceding distance (a 50% increase). Each of the five "black" events alters the distance by a ratio of 0.5 to the preceding distance (a 50% decrease). The black and red cards may be drawn in any order, and since all cards must be used, the series of ratios may be arranged in any order. For convenience, this may be expressed as (1.5)(1.5)(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)(0.5)(0.5) times the initial distance of 100 feet, or a final location of 23.73 feet from Maggie.
This can be expressed in a more general formula, using:
ko = initial distance apart
Kf = final distance apart
r1 = ratio used for one category (red)
r2 = ratio used for alternate category (black)
n = number of cards in a category
ko * (r1) to the nth power * (r2) to the nth power = kf
This can be rearranged to show that,when r1, r1, and n are held constant, the ratio of starting distance (ko) to final distance (kf) is a constant.
Thus, if the street was inconveniently shorter than she had remembered, Maggie could have had Charles start at a distance of 50 feet from her, and known in advance that he would have ended up a mere 11.865 feet away, or start at 25 feet and end up at around 6 feet away (5.9325 feet). This change might be necessary if you want to try this on someone, because, during their game, the maximum distance Charles might have gone from Maggie would have been 100 ft * (1.5)(1.5)(1.5)(1.5)(1.5), or 759.375 feet!!!
Joan E. Cope Savage jesavage@dreamscape.com

Received: from hal.idiscover.net (root@hal.idiscover.net []) by grits.valdosta.peachnet.edu (8.6.10/8.6.9) with ESMTP id TAA18866 for ; Mon, 25 Dec 1995 19:39:47 -0500 Received: from korer.idiscover.co.uk (korer.idiscover.co.uk []) by hal.idiscover.net (8.6.11/8.6.9) with SMTP id AAA05406 for ; Tue, 26 Dec 1995 00:45:37 GMT Message-Id: <199512260045.AAA05406@hal.idiscover.net> Date: Tue, 26 Dec 95 01:01:48 -0800 The Answer is: 23.73 100 * 1.5exp5 * 0.5exp5 Here's a problem for you: 3 thieves A B and C steal some money. They each take a handful of cash. when they count what they have taken, B and C complain to A that he has taken most. A replies that she will give B the same amount that B already has, and the same for C. Now they count the money again and now B has most so he does the same for A and C. But now C has most so C does the same for A and B. Finally they all end up with the same amounts. How much did they all start with? Give your answer in the simplest possible ratio. E-mail me if you want the answer Thanks for a very interesting site

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Do not accept the bet! Charles will end up, regardless of the order of the cards, about 24 feet from Point 0. Proof: Let's look at the smallest case and see if there is something to exploit. Let's assume that there are only two cards of each kind. There are only two combinations: case 1: Red-Black Explanation of what happens in case 1: decrease the distance to point 0 by 50% (multiply by 1/2) then increase the current distance by 50% (multiply by 3/2) case 2: Black-Red Explanation similar to case 1. For case 1, Charles' positions are 50, 75 feet from point 0. For case 2, Charles' positions are 150, 75 feet from point 0. So Charles ends up at the same point regardless of the order of the cards. Why is this? Because multiplication is commutative: 1/2*3/2=3/2*1/2=3/4. Extending this result to ten cards we find his final position to be: 100 ft*(1/2)^5*(3/2)^5 = (about) 24 ft.

Received: from jackson.freenet.org (ERC.JSCC.CC.TN.US []) by grits.valdosta.peachnet.edu (8.6.10/8.6.9) with SMTP id XAA08850 for ; Thu, 28 Dec 1995 23:31:05 -0500 Date: Thu, 28 Dec 1995 22:38:26 -0600 X-Personal_Name: Steve Beitz No matter how many times he moves back and forth, he will always have to go back. There for, he will end up near the middle every time.

From: macron@xmission.com (Joe Schlimgen) To: rbarnett@grits.valdosta.peachnet.edu Subject: Maggie's Bet Date: Tue, 16 Jan 1996 20:46:58 GMT
Stating the problem a bit differently, you will start "x" feet from me. With each black card, you will multiply that distance by "1/2". With each red card, you will multiply that distance by "3/2". Since multiplication is commutative, the order you draw the cards makes no difference. You will multiply the original distance "x" by (1/2)^5*(3/2)^5 or 243/1024. If I know the original distance, I can predict *absolutely* (at least well within the 2' window) the ending position.

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He should not take the bet; the distance will be multiplied by 1.5 5 times, and multiplied by 0.5 5 times; the order doesn't matter. He'll end up about 24' away

From rbarnett@grits.valdosta.peachnet.edu Wed Jan 17 15:20:10 1996 Date: Wed, 17 Jan 1996 13:56:42 +0000 From: Dawn Butson
go ahead, take the bet....

Sender: stein@thalia.nta.no
Yep she can. Every red card multiplies the distance with 1.5, every black card multiplies with 0.5. Factors can be arbitrarily ordered, so the ordering of the cards do not matter. The final distance will be 100 * 0.5^5 * 1.5^5 ~= 23.

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No, Charles should not take the bet, since Maggie can pinpoint his exact location. Here's a series of equivalent scenarios: Original problem: There are 5 red cards and 5 black cards. Start at 100, and order the 10 cards in any sequence. For each red card, move half the distance you are from zero away from zero, and for each black card, move half the distance you are from zero towards zero. Repeat until all cards used.
Version 1: There are 5 red cards and 5 black cards. Start at 100, and order the 10 cards in any sequence. Suppose you are at X. If the next card is red, move to 3X/2, and if the next card is black, move to X/2. Repeat until all cards used.
Version 2: There are 5 red cards and 5 black cards. Start at 100, and order the 10 cards in any sequence. Suppose you are at X. If the next card is red, multiply the number by 3/2, and if the next card is black, mutiply it by 1/2. Repeat until all cards used. Evidently, the order of the cards don't matter since all they do is multiply your number by a constant!

Received: from cts02port18.cc.monash.edu.au From: Rob Meredith
No Charles shouldn't take the bet, because the movements will converge on one particular point. The further Charles moves away, i.e. he selects a series of red cards, the larger his movements will be. So, when he selects the first black card his movement will be that much closer to point zero by a large amount, also the probability of selecting a run of black cards is increased, counteracting the series of red cards. The opposite is true if he first selects a run of black cards. In the case where there is no run of red or black cards then Charles will oscilate around a particular point. Therefore it is more than likely that Maggie does in fact know where Charles will end up. On the other hand, the computations required to calculate this point are very numerous. Is it possible for Maggie to have already worked this out? Possibly not. But it may also be possible that she has access to some super-computer and have worked it out. Ther are three possible outcomes for Charles; 1. He takes the bet and loses; 2. he takes the bet and wins or 3. He doesn't take the bet and stands even. It is less-likely that Maggie would offer the bet if she didn't know the point than if she did. Ergo, the most likely win situation for Charles would occur if he didn't take the bet.

Received: from gate.icl.fi (gatekeeper@gate.icl.fi []) by grits.valdosta.peachnet.edu (8.6.10/8.6.9) with ESMTP id JAA19559 for ; Thu, 18 Jan 1996 09:49:13 -0500 From: Tapani Lindgren Charles ended up about 23.73 feet away from Maggie.
Drawing a black card card is equivalent to multiplying the distance by 0.5, red card by 1.5. There are five of each and all are drawn; therefore, the final distance is 100 * (0.5 * 1.5) ^ 5 = 023.73046875. Tapani

Date: Thu, 18 Jan 96 16:57:57 -0600 From: Tom Maciukenas
Charles had better not take the bet. Each black card multiplies the distance by 0.5, and each red card multiplies the distance by 1.5. So the final distance is: 100 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 1.5 * 1.5 * 1.5 * 1.5 * 1.5 = 237 feet. Since multiplication is commutative, it doesn't matter what order the cards come in.

Date: Thu, 18 Jan 96 16:28:47 0000 Organization: Freelance HTML/Web Design I tried it two different ways. First I assumed that he walked closer to her each time and wound up with about 27 feet. Then I assumed that he alternated walking further, then closer the whole time and wound up with about 23 feet. So I'm guessing she'll usually be okay guessing about 24 or 25 feet.

From: burns@mmsi.com Date: Fri, 19 Jan 1996 11:10:41 -0700 Trivial. Multiplication is commutative! Start with 100 5 red cards each multiply by 3/2 5 black cards each multiply by 1/2. And 100 x (3/2)^5 x (1/2)^5 = 24300/1024, which will be pretty close to 24 feet (looks like somewhat under 24, because 24^2 would be about 575, which is more than 300.) He should always end up in exactly the same place, but there will be some error involved because he is human. If he got the 5 black cards first, he would be at 3 feet, 1.5 inches. He might think he was at 3 feet and go on from there to end up at 729/32 feet, which should be about a foot short because the 1.5" error is multiplied by 243/32, which is about 7.5. So Maggie is safe with her bet.

Received: from aix2.uottawa.ca (aix2.uottawa.ca []) by grits.valdosta.peachnet.edu (8.6.10/8.6.9) with SMTP id MAA20350 for ; Sat, 20 Jan 1996 12:37:48 -0500 id AA102717; Sat, 20 Jan 1996 12:45:04 -0500 Date: Sat, 20 Jan 1996 12:45:04 -0500 X-Personal_Name: jasoneddo From: s632144@aix2uottawa.ca Subject: problem
She's got you there, you will invariably end up 23 187/256 feet from where she stands. This comes from calculating 100 * (3/4)^5. The problem is easy to see if you simplify the problem to having only one red and one black card then you get to 75 = 100 * 3/4 regardless of wwhich colour is chosen first. If you have two of each you end up at 56 1/4 = 75 * 3/4 ... and so on until we get to ten. The reason she said to within two feet is likely that she would prefer to say you are at 25 feet. hope you like my answer.

Date: Sat, 20 Jan 96 09:56:34 -0800 Sender: kend@rosemail.rose.hp.com From: Ken Duisenberg
Since Charles increases his distance by 50% five times and decreases it by 50% five times, it's equivalent to his multiplying his distance by 0.5 and 1.5 five times each. Since multiplication is transitive, these operations can happen in any order to arrive at the same answer of: 100*[(1.5)(0.5)]^5 = 100*(3/4)^5 = (2^2)(5^2)(3^5)/(2^10) = (5^2)(3^5)/(2^8) = 25*243/256 = 23.73 feet Ken Duisenberg

From: Thomas Erlebach Hi, just came across the "Maggie's Bet" challenge in Zeno's Coffeehouse. Charles should not take Maggie's bet. She knows exactly where Charles will ultimately stop, no matter in which order he selects the cards. On the selection of a red card, the distance between Maggie and Charles is multiplied by 1.5. On the selection of a black card, it is multiplied by 0.5. Since multiplication is commutative, Charles will always end up at a distance of 1.5^5 * 0.5^5 * 100 = 23.7 feet. Correct? Regards, Tom
-- Thomas Erlebach (e-mail: erlebach@informatik.tu-muenchen.de)

Date: Mon, 22 Jan 96 16:49:56 EST From: mclean@itd.nrl.navy.mil (John McLean) If Charle's is n feet from Maggie, then a black card will move him to n x .5 feet from Maggie and a red card will move him to n x 1.5 feet from Maggie. Since he starts off 100 feet from Maggie, his final location will be 100 x z1 x ... x z10, where 5 of the zi are equal to .5 and the other 5 are equal to 1.5. Since multiplication commutes, Charle's final location will be 100 x 1.5^5 x .5^5 = 23.730469 feet to the right of Maggie. Hence, if Maggie can mark off such a distance within 2 feet (e.g. 25 feet), she'll win the bet. John McLean (mclean@itd.nrl.navy.mil)

From: "Robert F. Rorschach" Date: Fri, 26 Jan 96 13:32:44 -800 If Maggie can compute 100' * 3^5 /2^10 to within 2' she can win the bet. Each black card multiplies the current distance by 3/2, each red card multiplies it by 1/2. Since order of multiplication doesn't change the result, the final distance is unaffected by the order in which the cards are drawn.

From: "Paul E. Sochocky" No matter what order he draws the cards, whether red or black, the fact that there is an even number of them will ultimately bring him back to the point from which he started, which is +100 from zero. Thank you. Paul Sochocky- pes109@psu.edu

Date: Thu, 01 Feb 1996 00:23:36 -0600 From: Pete Bender Charles ought to turn on Excel and he will find that he is 23.73046875 ft away from point 0.

From: "Mark A. Young" Charles will end up 24 feet from Maggie. DON'T TAKE THE BET!
Each red card takes Charles 50% further away from Maggie. (Position := Position*1.5) Each black card brings Charles 50% closer to Maggie. (Position := Position*0.5) After the ten cards have been drawn, Charles' position has been increased by 50% five times (1.5^5) and decreased by 50% five times (0.5^5). Take the product, and you get about 23 feet, 9 inches. (12 inches in a foot, 24 feet in a flook, and 220 fleek in a mile: Hooray for Imperial Measures!) ..mark young

Date: Sat, 03 Feb 1996 08:23:58 -0800 Listen, Maggie, next time you have the decaf, ok?