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Ask The Wizard #403

Imagine a rubber band measuring one meter in length, with an ant starting at one end. As it moves towards the other end at a speed of one centimeter every second, the rubber band simultaneously stretches at a rate of one meter per second. What time will the ant take to reach the opposite 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).

Consider an ant located on the circumference of a circle that has a diameter of one centimeter. Beginning at time t=0, it travels along the edge at a speed of 1/(1+t) cm per second. How long will it take for the ant to make a full lap around the circle?

Ace2

e^pi -1 = 22.140693 seconds.

The ant my cover a distance of pi.

To determine the total distance covered, one can integrate the speed function over time. Let's denote the total time as 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

In a shuffled deck of cards, each card is turned one by one until the first queen appears. What is more probable to be the next card drawn— the queen of spades or the king of spades?

anonymous

I must confess that my first response regarding this matter was incorrect.

The probabilities are the same.

In the table below, you can observe the likelihood of each card position being the first queen followed by the queen of spades. The value in the bottom right corner indicates that the probability of the card succeeding the first queen being the queen of spades is 0.019231, which translates to 1 in 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 table below illustrates the chance that any card position is the first queen, which is subsequently followed by the king of spades. The lower right cell indicates a probability of 0.019231, meaning that the probability of the next card after the first queen being the king of spades is also 1 in 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

Initially, I thought the king of spades held a higher chance of appearing next, as there is a 1 in 4 possibility that the first queen turns out to be the queen of spades, resulting in it not showing up again. However, the crucial reason their probabilities align is tied to the fact that upon revealing the first queen, numerous possible queens were excluded from the preceding cards, even if some kings remained.

The explanation provided in the Mind Your Decisions video (see the link below) clarifies this concept as follows.

There are 51 factorial ways to arrange all cards in the deck aside from the queen of spades. By inserting the queen of spades right before the first queen, you maintain the same 51 factorial arrangements. When you divide this by the total number of arrangements of 52 cards, the probability that the queen of spades follows the first queen simplifies to 51!/52! = 1/52.

The same procedure can be applied if you exclude the king of spades, placing it in front of the first queen, yielding a probability of 1/52.

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