Ron Barnette's Zeno's Coffeehouse Challenge #52 Result

Dear Zeno's patrons: Here are the results of our previous challenge. I have listed the original problem, with three results featured below. Thanks to all Zeno's patrons who took up the challenge! Over 250 entries were received, and I have included results and analyses from three of over 175 winners. Thanks for your critical thinking, and please take up the next Zeno's Challenge:)

Ron Barnette

Charles' Random Walk

Reflecting on the successful years with the Coffeehouse, and 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. Looking right, and then left, Maggie remarked:

"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 feet

"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, a bit puzzled.
"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??" Maggie smiles.

Charles uncomfortably considers the offer, and looks 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?


Here are some positive responses from three Coffeehouse patrons, whose detailed comments I include. Thanks go to all authors' submissions. Nice work! ...Ron Barnette

1. Troy Williamson (a long-standing Zeno's patron!)

comments: Maggie gave herself a really wide margin of error here:  no matter how the cards are drawn, Charles will end up 23.73046875 feet from Maggie.  (If you want a "measurable" distance, use 23' 8.75"--that is within .02" of the actual answer.)

When Charles draws a red card, he will end up 1.5 times the distance from Maggie that he begins.  For example, if the first card drawn is red, then he will end up 1.5(100) = 150 feet from Maggie.  (I multiplied by 100, since that is where he begins.)  If he were to draw another red card, he would be 1.5(150) = 225 feet away from Maggie.

In the same way, whenever Charles draws a black card, he will end up .5 times the distance from Maggie that he begins.  So if the first card is black, Charles will be .5(100) = 50 feet from Maggie.  If the second card is also black, he will end up .5(50) = 25 feet from Maggie.

Thus, the process is a matter of multiplying by .5 whenever a black card is drawn and multiplying by 1.5 whenever a red card is drawn.  But we know that there are five black cards and five red cards, which allows us to arrive at a specific answer.

Suppose that all five red cards are drawn first, then the five black cards.  The calculation would be:
What if the five black cards are the first five drawn?
Notice that the values involved are identical!!!  Since multiplication is commutative--meaning that it doesn't matter the order in which values are multiplied (6*7 = 7*6)--it really doesn't matter what order the cards are drawn:  we will begin with 100, we will multiply by 1.5 five times, and we will multiply by .5 five times.  The order of multiplication is irrelevant.

Thus, regardless of the order in which cards are drawn, Charles will always end up 23.73046875 feet from Maggie.

(Just for grins, I listed all of the ways in which the five cards could be drawn in a spreadsheet, then created formulas to calculate the distances moved on each draw.   The spreadsheet verified the answer given above--for each of the 252 ways in which the ten cards could be drawn:  Charles always ends up at the same place.)

By the way, if you want an exact answer rather than the one I gave (which is rounded to eight decimal places), it is:  100 times (3 to the 5th power) divided by (2 to the 10th power), which reduces to 6075/256.


2. Mark Young

Each card changes Charles' position by either adding (red card) or subtracting (black card) 1/2 of what it was.  That is the same as multiplying his position by either 1/2 (for a black card) or 3/2 (for a red).  For example, if he drew the cards in the order red-red-black-black-red-black-black-black-red, then he'd end up at

 100 * 3/2 * 3/2 * 1/2 * 1/2 * 3/2 * 1/2 * 1/2 * 1/2 * 3/2

which is about 23 feet 9 inches from point 0. 

Note that it's just a multiplication problem, and so it doesn't really matter what order you do them in -- IOW, it doesn't matter what order Charles picks the cards in.  He will end up within one inch of 23' 9" from position 0 -- tho the two foot factor will allow for accumulated measurement/movement errors....

3. Ben Ceschi

Charles will stop at 23.73046875 feet from Maggie.

D  = current distance.
D' = distance after selecting a card

a red card will move you one half the current distance away from zero or, D' = D * 1.5

a black card will move you one half the current distance towards zero or, D' = D * .5

One possible sequence would be:
D' = 100 * 1.5 * 1.5 * 1.5 * 1.5 * 1.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 23.73046875;

And Since multiplication is commutative, the sequence of moves is irrelevent. Charles will always end up in the same place.