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As of October, 2016, Embarcadero is offering a free release of Delphi (Delphi 10.1 Berlin Starter Edition ).     There are a few restrictions, but it is a welcome step toward making more programmers aware of the joys of Delphi.  They do say "Offer may be withdrawn at any time", so don't delay if you want to check it out.  Please use the feedback link to let me know if the link stops working.

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Mensa® Daily Puzzlers

For over 15 years Mensa Page-A-Day calendars have provided several puzzles a year for my programming pleasure.  Coding "solvers" is most fun, but many programs also allow user solving, convenient for "fill in the blanks" type.  Below are Amazon  links to the two most recent years.

(Hint: If you can wait, current year calendars are usually on sale in January.)

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We'll each take a 52-card deck, shuffle it thoroughly and place our decks face down on the table.  Now we'll start turning cards face up, one from the top of each deck on each turn.  If we go through the entire deck without two identical cards being turned, player A wins.  If an identical pair is turned, player B wins.  Do you want to be player A or B?

A derangement of  a set of numbers is a permutation in which no number appears in its   natural order.  So, for example,  {5,4,1,2,3}  is a derangement of  {1,2,3,4,5}.   The number of derangements for a set of n objects is sometimes call the subfactorial of n and denoted as !n.

There are number of algorithms for evaluating !n.   The simplest is derived from the equation:  !n= n! x sumk=0 to n  ((-1)k/k!) where n! is n factorial, the product of integers 1 through n.    It just so happens that the series represented by   sumk=0 to n  ((-1)k/k!) approaches 1/e  as k approaches infinity  (yup, the same irrational number e (2.718281...)  that is the base of natural logarithms).

So the easy way to calculate !n is as round(n!/e)n factorial divided by e rounded to the nearest integer.

Delphi does not have e as a predefined constant (it does have pi, but discriminates against e for some reason).  It has the Exp(x)  function, e raised to the  x power, though.  So defining e as exp(1.0) is pretty trivial.

Back to the original card game problem.  The "natural order" of a set can be any predetermined order.  In this case we'll let deck number 1 define the natural order, and deck number 2 will be a derangement if no card in deck 2 matches it's position in deck 1.    Player A wins in this case, Player B wins in deck 2 is not a derangement of deck 1.    The probability of an event is the number of "successful" outcomes divided by the total number of possible outcomes.   In this case the total possible outcomes number 52!    and the successful (derangement) outcomes is round(52!/e) which will be very close  to 52!/e.  So the probability of no match will be close to (52!/e)/52! = 1/e or about 1/2.71828=0.36788.   We can therefore conclude that A has about a 37% chance of winning and, since somebody will win, B has a 63% chance of winning.     I did the proof, so I get to choose first whether to be A or B and you are probably going to lose!

According to our Big Factorials program 52! has 68 digits, so I used a smaller problem in the program to reach the same conclusion.  The program provided, calculates actual derangement counts for N in the range of 1 to 10 and points out that the probability approaches 1/e as N gets larger.

The program will also display the first 100 derangements for the selected N.  I could find no reference to a function that would generate derangements directly.  So I fell back on our old faithful UCombos unit to generate permutations and then just weed out those that are not derangements.

 Created: November 25, 2002 Modified: July 29, 2017