1) [1 - Pr(none)] seems quite a bit easier.
2) What's the point of randomly shuffling BOTH decks?
Okay, there is a lot of debates about the answer to this question in class. Most of the proposed solutions are plausible, but I'm not going to list them because I don't want to give ideas.
But here is the question:
If you have two randomly shuffled deck of cards of 52 cards per deck, and put both decks side by side.... then you flip the 1st cards from both decks and compare them, and you do the same for the 2nd, 3rd, and subsequent cards in the deck until you compare all the cards.... what is the probability that there is at least 1 EXACT match out of 52?
What do you think is the answer? And why? THanks
Hasn't the answer got something to do with ? I vaguely remember this from my own studies, and Martin Gardner mentioned this one in one of his famous columns in Scientific American.
It's the same as the shuffled letters problem - the incompetent secretary who stuffed letters at random into the envelopes. The proportion of the number of people getting their correct letters by chance tends to as the number of people tends to infinity.
The professor said the answer was 1-(51/52)^52 = 0.6357
I told him I thought it was 0.6321 by generalizing with smaller sets, and then computing the answer making a huge mess, but I'm 99.8% it's correct.
Of course, the two answers are so CLOSE to each other. What do you think about the professor's answer? Do you think it's correct?
BTW, what program do you use to conduct the Monte Carlo simulations?
Does anyone what's the best way to tell the professor why 1-(51/52)^52 isn't the correct answer?
That is the number of permutations in which every term is active.
Line numbers up so that none is in its correct place.
You may have seen ‘hat check’ or ‘letter stuffing’ problems.
Those numbers are given as .
So as Matt Westwood noted above we can approximate .
So in this problem we use .
You could simply show your professor what you have found. There are multiple methods here, but one answer.
Euler but any of the Matlab like systems will do, though in some you may have to provide your own random permutation generator.
(for problems like this a Monte-Carlo estimate is always useful as a reality check on other calculation methods, also it takes only a few minutes to write and only a little more time for the odd million replications).
Let's say that is the event of a match on the ith pair. Then , and . So the probability that there are no matches in 52 pairs is .
This is wrong. The fallacy is that it assumes independence of events among etc. To see that the events are not independent, suppose we have matches on 51 of the pairs. What is the probability of a match on the 52nd pair? 1, because it is forced.
Plato's solution is correct.
You may find interesting to have a look at Euler's paper (translated to English).
Thanks for that, Laurent - it's the first time I've ever read a paper by Euler. I never realised that he was available online. I'd always heard he was an excellent writer - now I know for sure. If every textbook was written as clearly as that, the world would have more mathematicians in it.