Not exactly true, the simpler flip is to say your first choice was likely wrong. 2/3 of the picks will be Goat to start and thus wrong. Monty showing you a goat doesn't change that your orriginal pick has only 33% chance of being right.

The trap is in thinking that because there are two doors that it's a 50/50 guess again. A trap most people fall into being drilled on elementary statistics about a penny flip. If you flip a penny heads 10 times in a row, what are the odds you'll flip heads again? Yes, still 50/50 but those odds were always 50/50.

To simplify, if instead of opening a door he said to you: "Would you like to trade your one door for the other two doors?" you'd likely agree. You'd go from 1/3 to 2/3 chance of winning. Showing you the goat and making you labour on the two doors is simply a form of entertainment.

TM

BTW, in a discussion of this I rewrote the code for people who need more handholding: 
#!/usr/bin/perl
my $wins  = 0;
my $loses = 0;

print "\n";
for (my $i=1;$i<=1000000;$i++)
{
  #all goats
  @doors = [0,0,0];
  #add a car
  $doors[int rand(3)] = 1;

  #contestant choice (cute varname declaration)
  my $choice = int rand(3);

  #monty makes his choice
  my $shown = -1;
  while($shown == -1)
  {
    $shown = int rand(3);
    if
    (
      $shown == $choice    || # he won't show contestants choice
      $doors[$shown] == 1     # or show the car
    ) { $shown = -1 }
  }
  $doors[$shown] = -1; # out of the equasion

  #If your on the car now, a change is a loss
  if($doors[$choice] == 1)
  {
    $loses++;
  }
  # presuming you don't pick the exposed goat, else is
  # a win.
  else
  {
    $wins++;
  }
  printf("\r Wins:%8d Losses:%8d Win \%:%5.3f",
         $wins,$loses,($wins/($i))*100);
}

[download]

In reply to Re^4: Marilyn Vos Savant's Monty Hall problem by ThinMonk
in thread Marilyn Vos Savant's Monty Hall problem by mutated

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