Simulating the Monty Hall problem

[CaptainShrimps]

I wrote a program to simulate the Monty Hall problem to show my dad that the odds are only 50/50 if you randomly choose to stay with your original choice or switch to the other door. He said my code was wrong, so I'm posting it here to see what everyone thinks.

code:

%This program is supposed to simulate the Monty Hall problem. var winID : int winID := Window.Open ("position:top;right,graphics:1024;512") var correct:int         %Counts the number of correct choices var a,b,c :boolean:=false       %3 doors var d:int       %Represents the door the prize is behind var q:int       %1 is stay, 2 is switch, 3 is random var ch:string(1) var randm :boolean proc playerchoose     cls     randm:=false     put "Type 1 to always stay, type 2 to always switch, 3 for random."     put "There are 10000 trials. The number of times the correct door was chosen is displayed."     get q     if q=3 then         randm:=true     end if     put "Press any key to start."     getch (ch) end playerchoose proc prizegen           %Generates which door the prize is behind     a:=false     b:=false     c:=false     d:=Rand.Int(1,3)     if d=1 then         a:= true     elsif d=2 then         b:=true     elsif d=3 then         c:=true     end if end prizegen proc pickdoor                   %Randomy pick a door, then stay or switch based on input from the start     var d:int:=Rand.Int(1,3)     if randm=true then         q:= Rand.Int(1,2)     end if     if d=1 then         if a=true then             if q=1 then                 correct+=1             elsif q=2 then                 correct+=0             end if         elsif b=true then             if q=1 then                 correct+=0             elsif q=2 then                 correct+=1             end if         elsif c=true then             if q=1 then                 correct+=0             elsif q=2 then                 correct+=1             end if         end if             elsif d=2 then         if a=true then             if q=1 then                 correct+=0             elsif q=2 then                 correct+=1             end if         elsif b=true then             if q=1 then                 correct+=1             elsif q=2 then                 correct+=0             end if         elsif c=true then             if q=1 then                 correct+=0             elsif q=2 then                 correct+=1             end if         end if             elsif d=3 then         if a=true then             if q=1 then                 correct+=0             elsif q=2 then                 correct+=1             end if         elsif b=true then             if q=1 then                 correct+=0             elsif q=2 then                 correct+=1             end if         elsif c=true then             if q=1 then                 correct+=1             elsif q=2 then                 correct+=0             end if         end if     end if end pickdoor     proc try            %Repeat 10000 times     for i:1..10000         prizegen         pickdoor     end for         put correct," out of 10000 correct." end try loop     correct:=0     playerchoose     try     put "Press any key to restart or press ESC to quit."     getch (ch)     if ch=(KEY_ESC) then         Window.Close (winID)         exit     end if end loop

[evildaddy911]

the output IS accurate.

* when you switch the chance of being right is 2 in 3
* when you stay, the chance of being right drops to 1 in 3
* if you randomly choose stay or switch, the chance should be the the average of the 2 options, making the chance 1.5 in 3, or 1 in 2

[nullptr]

The Monty Hall problem is one of those problems that seems so much a common-sense question that people will very stubbornly defend their answers. Good luck convincing your dad -- your code looks good to me.

[DemonWasp]

In general, shorter code is easier to prove correct (or at least assure yourself it is correct). Here's my take on the code, which displays each individual trial. Even with trials = 20, the pattern is very likely to be obvious:

Turing:

const TRIALS : int := 20 var correct_stay, correct_switch : int := 0 put "DOOR    CHOICE   WINNING STRATEGY" for i : 1 .. TRIALS     var door : int := Rand.Int(1,3)     var choice : int := Rand.Int(1,3)         if door = choice then         put "  ", door, "        ", choice, "     stay"         correct_stay += 1     else         put "  ", door, "        ", choice, "     switch"         correct_switch += 1     end if  end for put "STAY:   ", 100 * correct_stay / TRIALS, "%" put "SWITCH: ", 100 * correct_switch / TRIALS, "%"

The other argument you could use is this:
1. You have a 1-in-3 chance of choosing correctly at first.
2. If you chose correctly (1 in 3), then staying wins 100% of the time but switching loses 100% of the time.
3. If you chose incorrectly (2 in 3), then switching wins 100% of the time but staying loses 100% of the time.

Because case #2 has only a 1-in-3 chance of occurring, staying only wins in 1-in-3 cases.