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Yes. As I recall it was a great intellectual city in Africa that had a magnificent library. However the library was burned down and not much remains of the once great city. Here’s more information about Timbuktu from MrDowling.com.

A woodchuck would chuck as much as he could chuck if a woodchuck could chuck wood. Now say the question and answer ten times really fast.

Before I answer that, let me review recessive disease genetics , which is the case with Cystic Fibrosis (CF). Humans have two copies of each gene, one from the mother and one from the father. When there is a mating, the offspring will randomly inherit one each from the father and mother, resulting in two genes of his/her own.

In the case of CF, it takes two positive genes to be positive. In the case of one positive and one negative gene, the negative one will dominate. In such an event, the person is a carrier, negative for CF, but has a 50% chance of passing on the positive CF gene. Two negative genes will result in being completely clean of CF.

Given that both parents are carriers, here is the probability of each possible outcome for their offspring:

Positive: 0.5×0.5= 0.25
Carrier: 0.5×0.5 + 0.5×0.5 = 0.5
Negative: 0.5×0.5 = 0.25

Given one carrier and one negative parent, here is the probability of each possible outcome for their offspring:

Positive: 0
Carrier: 0.5×1 = 0.5
Negative: 0.5×1 = 0.5

Given two negative parents, the offspring will be negative with 100% chance.

Let's define the probability of the three possible states as:

p = positive
c = carrier
n = negative

Given random parents, let's solve for each after one generation.

p = pr(two carrier parents)×pr(positive given two carrier parents) +
pr(one carrier parent)×pr(positive given one carrier parents) +
pr(zero carrier parents)×pr(positive given two carrier parents) =
c² × 0.25 + 2×c×(1-c)×0 + (1-c)²×0 = c²/4.

c = pr(two carrier parents)×pr(carrier given two carrier parents) +
pr(one carrier parent)×pr(carrier given one carrier parents) +
pr(zero carrier parents)×pr(carrier given two carrier parents) =
c² × 0.5 + 2×c×(1-c)×0.5 + (1-c)²×0 = c-c²/2.

n = pr(two carrier parents)×pr(negative given two carrier parents) +
pr(one carrier parent)×pr(negative given one carrier parents) +
pr(zero carrier parents)×pr(negative given two carrier parents) =
c² × 0.25 + 2×c×(1-c)×0.5 + (1-c)²×1 = c²/4 - c + 1

So the probability of being a carrier, given not positive is:

(c - c²/2)/ (1 - c²/4) =
(4c - 2×c²)/(4 - c²) =
[2c×(2-c)] / [(2-c)×(2+c)] =
2c/(2+c)

We were given that the carrier rate now is 4%, so in one generation it will be 2×0.04/(2+0.04) = 3.92%.

The following table applies this formula for 100 generations.

Cystic Fibrosis Carrier Rate
Generation	Rate
0	0.040000
1	0.039216
2	0.038462
3	0.037736
4	0.037037
5	0.036364
6	0.035714
7	0.035088
8	0.034483
9	0.033898
10	0.033333
11	0.032787
12	0.032258
13	0.031746
14	0.031250
15	0.030769
16	0.030303
17	0.029851
18	0.029412
19	0.028986
20	0.028571
21	0.028169
22	0.027778
23	0.027397
24	0.027027
25	0.026667
26	0.026316
27	0.025974
28	0.025641
29	0.025316
30	0.025000
31	0.024691
32	0.024390
33	0.024096
34	0.023810
35	0.023529
36	0.023256
37	0.022989
38	0.022727
39	0.022472
40	0.022222
41	0.021978
42	0.021739
43	0.021505
44	0.021277
45	0.021053
46	0.020833
47	0.020619
48	0.020408
49	0.020202
50	0.020000
51	0.019802
52	0.019608
53	0.019417
54	0.019231
55	0.019048
56	0.018868
57	0.018692
58	0.018519
59	0.018349
60	0.018182
61	0.018018
62	0.017857
63	0.017699
64	0.017544
65	0.017391
66	0.017241
67	0.017094
68	0.016949
69	0.016807
70	0.016667
71	0.016529
72	0.016393
73	0.016260
74	0.016129
75	0.016000
76	0.015873
77	0.015748
78	0.015625
79	0.015504
80	0.015385
81	0.015267
82	0.015152
83	0.015038
84	0.014925
85	0.014815
86	0.014706
87	0.014599
88	0.014493
89	0.014388
90	0.014286
91	0.014184
92	0.014085
93	0.013986
94	0.013889
95	0.013793
96	0.013699
97	0.013605
98	0.013514
99	0.013423
100	0.013333

Half the current 4% rate is 2%. You can see from the table that that will be achieved in 50 generations. Assuming 30 years per generation, that will take 1,500 years.

First, for those readers who are easily offended, I apologize if it seems I am making light of this tragedy. That is not my intent. This is how the question was asked to me, and I don't see any reason to sanitize it to dropping a brick from a rooftop. That said, here is my answer...

Let's define some variables first.

• c = constant of integration
• d = distance traveled (in feet)
• t = time (in seconds)
• v = velocity (in feet per second)

The rate of acceleration is 32.2 feet per second per second. As you may remember from beginning calculus, velocity is the integral of acceleration. So v=32.2t + c. The constant of integration c is 0. Why? At the moment an object is dropped it has no velocity. A way to visualize that is if you throw a ball straight up in the air at the exact moment it reaches its highest point or apex, it is just hanging there, velocity and gravity exactly canceling each other out.

The distance the object traveled is the integral of velocity. So d=16.1 t² + c. There is that pesky constant of integration again, but it is zero again here as well, because at the moment you drop a ball (t=0) the distance traveled is 0.

We know the unfortunate guy fell 20 feet, so we just need to solve for t.

20 = 16.1 t²
t² = 20/16.1
t = (20/16.1)^1/2 = 1.1146

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

Absolutely, I agree! I think giving the public a list of the most stolen cars by the total stolen is at best giving the reader useless information, and at worst leading him to falsely conclude that an old Honda is more likely to be stolen than any other car.

What is useful information is to know is auto theft rates, in other words, cars stolen by the number on the road. An article that provides such rates is The 10 Most Stolen Cars in 2011 . Most often stolen? The Cadillac Escalade. No Honda even makes their top 10.

Publishing a list of total stolen will unfairly worry mathematically challenged old Honda owners, when it is the Escalade owners who should be worrying. To every media source who quoted the list from the National Insurance Crime Bureau, and there were lots of them, I say shame on you!

Image courtesy of Brilliant Puzzles . The puzzle in the above image is priced at $15.

Yes, I saw that episode of Survivor as well, where the player was struggling with the puzzle. I was screaming at the television that I could have solved it in 30 seconds and then bored them back at camp for three hours about the mathematics of it.

The puzzle is called the Tower of Hanoi. For the benefit of other readers, the puzzle consists of three pegs and any number of pieces, each the same shape but a different size. The starting state should have all the pieces on one peg, in order, with the largest one at the bottom. The object is to move the entire stack to another peg. However, you may not move any piece on top of a smaller piece. Here is an online version of the game.

The strategy is a simple recursive one. In computer programming logic, here is the function how to do it. For example, to move 10 pieces from peg 1 to peg 2 you would call TowerOfHanoi(10, 1, 2, 3).

void TowerOfHanoi(int NumberPieces, int origin, int destination, int storage)
{
    if (NumberPieces == 1)
    {
        cout << "Piece 1 to peg " << destination << "\n";
    }
    else
    {
        TowerOfHanoi(NumberPieces-1, origin, storage, destination);
        
        cout << "Piece " << NumberPieces << " to peg " << destination << "\n";
        
        TowerOfHanoi(NumberPieces-1,storage,destination,origin);
    }
}

Here is the output for a five-piece game:

Piece 1 to peg 2
Piece 2 to peg 3
Piece 1 to peg 3
Piece 3 to peg 2
Piece 1 to peg 1
Piece 2 to peg 2
Piece 1 to peg 2
Piece 4 to peg 3
Piece 1 to peg 3
Piece 2 to peg 1
Piece 1 to peg 1
Piece 3 to peg 3
Piece 1 to peg 2
Piece 2 to peg 3
Piece 1 to peg 3
Piece 5 to peg 2
Piece 1 to peg 1
Piece 2 to peg 2
Piece 1 to peg 2
Piece 3 to peg 1
Piece 1 to peg 3
Piece 2 to peg 1
Piece 1 to peg 1
Piece 4 to peg 2
Piece 1 to peg 2
Piece 2 to peg 3
Piece 1 to peg 3
Piece 3 to peg 2
Piece 1 to peg 1
Piece 2 to peg 2
Piece 1 to peg 2

In English this code means that to move any piece you follow these three steps:

1. Move all the pieces above it, if any, to the storage peg.
2. Move the desired piece to the destination peg.
3. Move all the pieces that were moved from step 1 to the destination peg.

Where I say to "move all the pieces," just follow the same instructions above but for the substack you want to move. Eventually you will get down to moving just a single peg.

If you follow these instructions, you will move the stack in the minimum number of moves, which is 2^NumberPieces-1.

There are other ways to express a solution in English. Here is a clever one where the initial stack is on peg A. The destination stack will be B for an odd number of pieces, and C for an even number. Keep repeating these steps until finished:

1. Make legal move between A and B.
2. Make legal move between A and C.
3. Make legal move between B and C.

Finally, here is an outside link to a binary number solution .

To be honest, I'm not entirely sure. They used to be known as Knaves, but at some point they became known as Jacks. Perhaps about the time they started to put numbers and letters on cards (they didn't used to) and it would have been confusing because both King and Knave start with a K. So that brings up the question of what is a Knave? dictionary.com gives us these definitions, aside from the playing card usage:

1. an unprincipled, untrustworthy, or dishonest person.
2. male servant (archaic).
3. man of humble position (archaic).

Given the company the Knave keeps with kings and queens, you would think the Knave is a male servant. Playing cards were certainly around during "archaic" times. However, the fact that the Knave turned into a Jack argues for a "man of humble position." dictionary.com says one of the various meanings of the word jack:

• fellow; buddy; man (usually used in addressing a stranger): Hey, Jack, which way to Jersey?

The expressions "Jack of all trades" and "That's the fact, Jack" also spring to my mind. Wikipedia mentions that the Knave was called a Jack in the game All-Fours before cards were printed with a J, but the term may have been slow to catch on because it was considered "vulgar." I don't always trust Wikipedia, so take that with a grain of salt, as with everything in this answer.

To help us further, let's look at a deck of French cards I happen to have from the Casino De Montreal. In that deck they use an R for Roi (king), D for Dame (lady), and V for Valet. It should be noted that the French word for queen is reine, so I suspect they went with a lady instead, to avoid two ranks that begin with an R. So, what is a Valet? www.french-linguistics.co.uk gives this definition:

• manservant; je ne suis pas ton ~ I'm not your slave

That would seem to go along with the English "male servant." Still, I'm not entirely comfortable with that, because if that is the meaning, how did they go from Knave to Jack? I suggest that the better translation, to avoid words that start with K and Q, would have been to follow the French and go with Valet, which has a similar meaning in English. Here are the dictionary.com usages for Valet:

1. a male servant who attends to the personal needs of his employer, as by taking care of clothing or the like; manservant.
2. a man who is employed for cleaning and pressing, laundering, and similar services for patrons of a hotel, passengers on a ship, etc.
3. an attendant who parks cars for patrons at a hotel, restaurant, etc.

In closing, let me be the first to suggest that we replace the J on English playing cards with a V, and call them valets.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .