hench man said: "V Winning Door = INT(RAND * 3) becomes V Winning Door = INT(RAND * 2) after being shown the door that does not have the car behind it. The code must reflect the recalculation of the probability of winning after the elimination of one of the three doors. Nice simulation though... "
hench man said: "This is flawed... for example if you choose door 2, and the host shows you there is a goat behind door 3, then asks you if you want to switch your choice from door 2 to door 1... based on this simulation you will suceed 20-40% when you don't switch and 60-80% of the time when you do switch. The theory behind that is you first choose 1 of 3 doors so you are going to be right 1/3 of the time, so if you switch you will be right the other 2/3 of the time since door 3 is now eliminated. WRONG, eliminating door 3 removes the 1/3 chance the car was behind door 3, but now you have a choice between door 1 and 2, they are both equally likely to have a car behind them, thus yeild a 50% probability each, no one door is correct more often than the other and in 100 simlations in the real world you will have a 50/50 chance of winning the car regardless of if you A) switch your door when the host asks OR if you B) Do not switch your door when the host asks you. Think outside the box... :)"
(5)
The Monty Hall problem is a probability puzzle based on the American television game show Let's Make a Deal.
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Play this simulation and find out!