Randomizing The Order Of Cards

Probability Level 4

A magician takes a deck of $52$ playing cards and records their position, labelled $i$ for a card $C$ (i.e. the Queen of Spades is on top ( $C_1$ ), the $5$ of Diamonds is beneath ( $C_2$ )... the $10$ of Spades is on the bottom ( $C_{52}$ )). Then, he gives the cards to his assistant, who shuffles them in a completely random manner. The magician then records the cards' resulting positions, $C'_{i}$ .

What is the probability that no cards are in their starting position? In other terms, what is the probability that for any card $C_i$ , $C_i\neq C'_{i}$ ? Write your answer as a percentage.

Note: The assistant doing nothing is one possible configuration: in other terms, $C_i=C'_{i}$

The answer is 36.787944117.

1 solution

Andrei Li
Aug 21, 2018

Relevant wiki: Derangements

The probability that the event described occurs is the number of configurations with $C_i\neq C'_i$ , divided by the total number of possible configurations. From the information in the problem, we see that the total number of configurations is $52!=80658175170943878571660636856403766975289505440883277824000000000000$ , quite a lot. Now, the possible configurations satisfying $C_i\neq C'_i$ can be found by the formula for subfactorials/derangements:

$!n=\lfloor{\frac{n!}{e}+\frac{1}{2}}\rfloor$

Using this formula, we see that $!52=29672484407795140000000000000000000000000000000000000000000000000000$ . Then, the probability of the magician finding a configuration satisfying $C_i\neq C'_i$ is

$\frac{29672484407795140000000000000000000000000000000000000000000000000000}{80658175170943878571660636856403766975289505440883277824000000000000}\approx0.36787944117=36.787944117\%$

It is worth noting that another way of writing the number for your solution is $\large \frac{!n}{n!}=\frac{\lfloor{\frac{n!}{e}+\frac{1}{2}}\rfloor}{n!}$ and since $n!$ is very large this answer is very close to $\frac{1}{e}$ . In fact as decimals they differ only in the $10^{th}$ decimal place.

Jeremy Galvagni - 2 years, 9 months ago

Thank you! As a matter of fact, $\lim_{n\to\infty}\frac{!n}{n!}=\frac{1}{e}$ . As a matter of fact, I am curious about from where $e$ comes from.

Andrei Li - 2 years, 9 months ago

(this isn't a confirmed answer, this is just me asking a question) wonder if it is possible to use inclusion exclusion to solve this like [52!-(52C1)51!+(52C2)50!-(52C3)49!+...+(52C52)1!]/52! and so it would be 1/0!-1/1!+1/2!-1/3!+1/4!...+1/52! which is super close to the expansion of e^x with x=-1

A Former Brilliant Member - 2 years, 9 months ago

Randomizing The Order Of Cards

The answer is 36.787944117.

1 solution

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