0
$\begingroup$

I am doing a research paper about this topic. It has really puzzled me and although I seem to have found a way to calculate it, my answers are rather weird.

I assumed that since 1300 AD a total of approximately 1.546×10^20 times a set of cards have been shuffled (700 years x every second since then).

The probability that such permutation has never been seen before is then (1- 1/52!)^(1.546×10^20 ) and that it did happen is the answer - 1.

However, I always got the answer being equal to 1.

What am I missing? Any more tips and tricks concerning this topic, PLEASE do. I need to cover 12 pages.

Also: how long would it take until the chances increase of us seeing a permutation ever again?

$\endgroup$

marked as duplicate by AlexR, Travis, Daniel Fischer, Mark Fantini, mookid Oct 23 '14 at 17:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Couldn't find a definite answer though $\endgroup$ – Janeliza Oct 23 '14 at 13:29
3
$\begingroup$

Let us assume: 52 card deck, shuffled 7 times so now random, one player dealt a total of 52 cards when order matters. In this case, we use the formula for permutations:

$P$($n$, $r$)$ = $P$($52$, $52$)$ =

$80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000$

possible permutations. Next, assume 365 days per year, 24 hours per day, 60 minutes per hour, and 60 seconds per minute. Then, over 700 years, the number of time a deck has been shuffled would be:

$700*365*24*60*60$ = $22,075,200,000$

Consider then that the chance it has happened is then

$\dfrac{1}{P(52, 52)}$*$22,075,200,000$ = $ 0.0000000000000000000000000000000000000000000000000000000003$

Using the complement rule then yields the probability is has not occurred. Does this reply help?

$\endgroup$
1
$\begingroup$

$52!\approx 8*10^{67}$, so $1-1/52!$ =0.999999... beginning with 67 nines.
Raise it to a power around $10^{20}$ only removes the last twenty of those nines, so there are still 47 nines to begin with. It is that close to 1, that most calculators will just call it 1.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.