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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  They do say "Offer may be withdrawn at any time", so don't delay if you want to check it out.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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### Problem Description

Calculate the probabilities of winning or losing for each outcome for the dice game of Craps.

### Background & Techniques

Craps is a dice game with scoring based sum of rolling 2 dice. A game ends in a single roll if the player wins if he rolls a 7 or 11 (a Natural), or loses if he rolls a 2, 3, or 12 (Craps).

If the first roll is one of the other possibilities, 4, 5, 6, 8, 9, or 10, that sum is called the "point". and the game continues with additional rolls until the point is rolled again (a win) or a 7 is rolled (a loss).

This program calculates the theoretical chances of winning or losing for each of the eleven initial roll possible outcomes (2 through 12). It also has a page which simulates a million games to verify that the theoretical results are valid.

The math is mostly "Probability 101" level.  By definition, the probability of an particular outcome for an event is the number ways that outcome can occur divided by the total number of outcomes.  When rolling two standard six sided dice, there are 36 possible outcomes.  For each of the 6 outcomes for die A there are six possible outcomes for die B.  The number of occurrences for each sum are shown in this table:

 Sum of dice values 2 3 4 5 6 7 8 9 10 11 12 Number of outcomes (36 total) 1 2 3 4 5 6 5 4 3 2 1

So, for example,  the probability of rolling a 11 sum in a single roll is 2/36 or 1/18.   That takes care of the winning or losing probabilities for the naturals (7,11) and the craps (2,3,12) outcomes.

For the "point" outcomes (4,5,6,8,.9,10) things get slightly more complicated.  To win with any given point, two events must occur, we must roll the point on the first roll, then we must roll the point again before we roll a 7.  Each roll is an independent event so the probability that both will occur is the product of the two events.  The first event is as described above but for the second event, the probability depends on the relative frequency of the point and 7.  So for example if the point is 4, it can occur in 3 ways, but 7 can occur in 6 ways.  While rolling for point, any outcome that is not the point or a 7 is ignored so on our example, there are 9 outcomes which count.  Three of these are winners (the point) and six are losers (the 7s). Therefore the probability of winning with a 4 is 3/36 * (3/9) = 1/36.

If you follow that, the rest is a piece of cake.