This problem is amazingly obvious to understand once you analyze it correctly and remove certain ambiguities from the problem statement. Here's my analysis and some Mathematica simulations to add some weight (as if any is needed).

Okay, we all can agree that the probability of NOT picking the DREAM VACATION is 2/3, right? There are two GOATS and one DREAM VACATION.

Now, when Monty shows you the remaining door with a GOAT he just beamed you some very significant information.

*He's told you that if you picked a GOAT then the probability of getting a DREAM VACATION is 1 if you switch!*We already know the probability you picked a GOAT is 2/3 so after he gives you this new info your probability of winning is now 2/3. So switch for GOAT's sake!! If you don't switch, your probability is just 1/3.

Here is a Mathematica program for the non-believers.

GOAT = 0; (* Goat worth zero *)

VACATION = 1; (* Vacation worth one*)

makePrizes[] := Module[{},Switch[RandomInteger[{1,3}],

1,{GOAT,GOAT,VACATION},

2,{GOAT,VACATION,GOAT},

3,{VACATION,GOAT,GOAT}]]

randomPick[doors_List] := Module[{},RandomInteger[{1,Length[doors]}]]

strategy1VS2[trials_Integer] :=

Module[{winnings1=0, winnings2=0, firstPick, secondPick, doors, doors2},

SeedRandom[];

Do[doors = makePrizes[];

firstPick = randomPick[doors];

(*winnings of person who keeps first pick*)

winnings1+= doors[[firstPick]];

(*delete first pick from choices*)

doors2 = Drop[doors,{firstPick}];

(*delete goat from remaining*)

doors2 = Drop[doors2,Position[doors2,GOAT][[1]]];

(*Always pick remaining prize *)

secondPick =doors2[[1]];

(*winnings of person who switches*)

winnings2+= secondPick,{trials}];

{winnings1,winnings2}](*Run simulation 10000 times. *)

strategy1VS2[10000]

{3356,6644}

The result {3356,6644} means keeping first choice only paid 3356 over 10000 runs but switching paid 6644!

Now, there are ASSUMPTIONS here (there always are). One assumption is that on each run the position of the prize changes. It turns out that keeping the prize always in any particular door for the entire simulation does not matter (as long as the contestant does not have the information, obviously!)

strategy1VS2A[trials_Integer,init_List] :=

Module[{winnings1=0,winnings2=0,firstPick,secondPick,doors,doors2},

SeedRandom[];

Do[doors = init;

firstPick = randomPick[doors];

(*winnings of person who keeps

first pick*)

winnings1+= doors[[firstPick]];

(*delete first pick from choices*)

doors2 = Drop[doors,{firstPick}];

(*delete goat from remaining*)

doors2 = Drop[doors2,Position[doors2,GOAT][[1]]];

(*Always pick remaining prize *)

secondPick =doors2[[1]];

(*winnings of person who switches*)

winnings2+= secondPick;,{trials}];

{winnings1,winnings2}]

strategy1VS2A[10000,{GOAT,GOAT,VACATION}]

{3316,6684}strategy1VS2A[10000,{GOAT,VACATION,GOAT}]

{3267,6733}

strategy1VS2A[10000,{GOAT,GOAT,VACATION}]

{3382,6618}

The other assumption is that your not forced to switch before seeing the goat. This IS important!!

strategy1VS2B[trials_Integer] :=

Module[{goatPositions,pos,winnings1=0,winnings2=0,firstPick,secondPick,doors,doors2},

SeedRandom[];

Do[doors = makePrizes[];

firstPick = randomPick[doors];

(*winnings of person who keeps first pick*)

winnings1+= doors[[firstPick]];

(*delete first pick from choices*)

doors2 = Drop[doors,{firstPick}];

(*Randomly choose from remaing*)

secondPick =randomPick[doors2];

(*winnings of person who switches*)

winnings2+= doors2[[secondPick]];,{trials}];

{winnings1,winnings2}]

strategy1VS2B[10000]

{3373,3295}

So information has value, Duh!

So now that you have this information, become a believer, make the switch!

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