# Ask The Wizard #403

There is a rubber band that is one meter long. An ant is at one end of it. The ant travels to the other end at a speed of 1 centimeter per second. Starting from the time the ant starts moving, the rubber band expands at a rate of 1 meter per second. How long does it take for the ant to reach the other end?

anonymous

The answer is e100 – 1 =~ 26,881,171,418,161,400,000,000,000,000,000,000,000,000,000 seconds.

Here is my solution (PDF).

There is an ant on a circle of diameter 1 centimeter.  Starting at time t=0, the ant moves along the circumference at a speed of 1/(1+t) cm/sec.  How long does it take him to complete a revolution?

Ace2

e^pi -1 =~ 22.140693 seconds.

The ant my cover a distance of pi.

A way to get the total distance traveled is to integrate the speed over time. Let the answer be T.

The integral from 0 to T of 1/(1+t) dt = pi.

Integrating, we get:

ln(1+T) - ln(1+0) = pi

ln(1+T) = pi

1+T = e^pi

T = e^pi - 1

Cards are turned over in a shuffled deck one at a time until the first queen appears. What is more likely to be turned over as the next card, the queen of spades or king of spades?

anonymous

I admit my initial answer to this one was wrong.

The probabilities are the same.

The following table shows the probability that any given position in the deck is the first queen followed by the queen of spades. The lower right cell shows the probability that the card following the first queen is the queen of spades is 0.019231 = 1/52.

### Next Card Queen of Spades

Position of
First Queen
Probability
First Queen
Probability Next
Card Q of Spades
Product
1 0.076923 0.014706 0.001131
2 0.072398 0.001086 0.001086
3 0.068054 0.001042 0.001042
4 0.063888 0.000998 0.000998
5 0.059895 0.000956 0.000956
6 0.056072 0.000914 0.000914
7 0.052415 0.000874 0.000874
8 0.048920 0.000834 0.000834
9 0.045585 0.000795 0.000795
10 0.042405 0.000757 0.000757
11 0.039376 0.000720 0.000720
12 0.036495 0.000684 0.000684
13 0.033758 0.000649 0.000649
14 0.031161 0.000615 0.000615
15 0.028701 0.000582 0.000582
16 0.026374 0.000549 0.000549
17 0.024176 0.000518 0.000518
18 0.022104 0.000488 0.000488
19 0.020153 0.000458 0.000458
20 0.018321 0.000429 0.000429
21 0.016604 0.000402 0.000402
22 0.014997 0.000375 0.000375
23 0.013497 0.000349 0.000349
24 0.012101 0.000324 0.000324
25 0.010804 0.000300 0.000300
26 0.009604 0.000277 0.000277
27 0.008496 0.000255 0.000255
28 0.007476 0.000234 0.000234
29 0.006542 0.000213 0.000213
30 0.005688 0.000194 0.000194
31 0.004913 0.000175 0.000175
32 0.004211 0.000158 0.000158
33 0.003579 0.000141 0.000141
34 0.003014 0.000126 0.000126
35 0.002512 0.000111 0.000111
36 0.002069 0.000097 0.000097
37 0.001681 0.000084 0.000084
38 0.001345 0.000072 0.000072
39 0.001056 0.000061 0.000061
40 0.000813 0.000051 0.000051
41 0.000609 0.000042 0.000042
42 0.000443 0.000033 0.000033
43 0.000310 0.000026 0.000026
44 0.000207 0.000019 0.000019
45 0.000129 0.000014 0.000014
46 0.000074 0.000009 0.000009
47 0.000037 0.000006 0.000006
48 0.000015 0.000003 0.000003
49 0.000004 0.000001 0.000001
Total 1.000000 0.019231 0.019231

The following table shows the probability that any given position in the deck is the first queen followed by the king of spades. The lower right cell shows the probability that the card following the first queen is the king of spades is 0.019231 = 1/52.

### Next Card King of Spades

Position of
First Queen
Probability
First Queen
Probability Next
Card Q of Spades
Product
1 0.076923 0.019231 0.001479
2 0.072398 0.019231 0.001392
3 0.068054 0.019231 0.001309
4 0.063888 0.019231 0.001229
5 0.059895 0.019231 0.001152
6 0.056072 0.019231 0.001078
7 0.052415 0.019231 0.001008
8 0.048920 0.019231 0.000941
9 0.045585 0.019231 0.000877
10 0.042405 0.019231 0.000815
11 0.039376 0.019231 0.000757
12 0.036495 0.019231 0.000702
13 0.033758 0.019231 0.000649
14 0.031161 0.019231 0.000599
15 0.028701 0.019231 0.000552
16 0.026374 0.019231 0.000507
17 0.024176 0.019231 0.000465
18 0.022104 0.019231 0.000425
19 0.020153 0.019231 0.000388
20 0.018321 0.019231 0.000352
21 0.016604 0.019231 0.000319
22 0.014997 0.019231 0.000288
23 0.013497 0.019231 0.000260
24 0.012101 0.019231 0.000233
25 0.010804 0.019231 0.000208
26 0.009604 0.019231 0.000185
27 0.008496 0.019231 0.000163
28 0.007476 0.019231 0.000144
29 0.006542 0.019231 0.000126
30 0.005688 0.019231 0.000109
31 0.004913 0.019231 0.000094
32 0.004211 0.019231 0.000081
33 0.003579 0.019231 0.000069
34 0.003014 0.019231 0.000058
35 0.002512 0.019231 0.000048
36 0.002069 0.019231 0.000040
37 0.001681 0.019231 0.000032
38 0.001345 0.019231 0.000026
39 0.001056 0.019231 0.000020
40 0.000813 0.019231 0.000016
41 0.000609 0.019231 0.000012
42 0.000443 0.019231 0.000009
43 0.000310 0.019231 0.000006
44 0.000207 0.019231 0.000004
45 0.000129 0.019231 0.000002
46 0.000074 0.019231 0.000001
47 0.000037 0.019231 0.000001
48 0.000015 0.019231 0.000000
49 0.000004 0.019231 0.000000
Total 1.000000 0.019231

I admit my initial reaction was that the king of spades was more likely, because there is a 1/4 chance the first queen is the queen of spades, in which case there would be zero chance to see it again. However, the simple reason the probabilities are the same is when the first queen is seen, the deck was rich in queens. In other words, a bunch of random cards were removed before that first queen, that could have been kings but not other queens.

The way it was explained in the Mind Your Decisions video (see link below) is as follows.

There are 51! ways to arrange all the cards except the queen of spades. Then put the queen of spades directly in front of the first queen and you still have 51! orders. Divided that by the 52! possible orders and the probability the the queen of spades follows the first queen is 51!/52! = 1/52.

You could do the exact same thing except omit the king of spades and then put it in front of the first queen and still get 1/52.

This question was taken from the Mind Your Decisions YouTube channel.