>Hi,
>
>>Monty Hall asks you if you want to change the door you have chosen. Should you change? And why?
>
>Yes, change. The probability to pick the right one is 33.3% per door. This probability doesn't change for the first door when the other one is opened, because Monty didn't arbitrarily opened one of the remaining two, but the one that is guaranteed to have no boat behind it. That gives the third door a probability of 66.6% to be the right one.
>
>If you don't believe that, just run the following VFP program. *g*
>

>lnChange = 0
>lnNotChange = 0
>Do While Inkey() == 0
>	* This door contains the boat
>	lnDoor = Int(Rand()*3)+1
>	* This is your selected door
>	lnChoice = Int(Rand()*3)+1
>	* These doors are available after you made a choice
>	lcAvailable = ChrTran("123",Str(lnChoice,1),"")
>	* Trim out the right one. Remains the door that is opened by Monty
>	lcAvailable = ChrTran(lcAvailable,Str(lnDoor,1),"")
>	* You picked the wrong one. You win when you change
>	If Len(lcAvailable) == 1
>		lnChange = lnChange + 1
>	* You picked the right one. You win when you don't change
>	Else
>		lnNotChange = lnNotChange + 1
>	Endif
>	@ 1,0 Say "Win when change:"+Str(lnChange)
>	@ 2,0 Say "Win when don't change:"+Str(lnNotChange)
>Enddo
>

>
>Christof

Christof,

I'm amazed that both you and Jim Booth have it all wrong! :-) You've both fallen into the trap of making something more complicated than it is. Consider these incontrovertible facts:

1. There are 2 remaining doors.
2. Exactly one of them has a goat, the other a boat.

Discounting any clues in the tone of Monty Hall's voice, clearly the chance of either door having the boat is exactly 50%. It makes no difference whether you change your choice to the other door or not.