Monty Hall Marathon

Monty Hall offers you the choice of three doors. Behind one is $1000. Behind the others are nothing. He'll let you choose one door, then he'll open one of the remaining doors to show nothing. He'll then allow you to choose if you want to keep your original door selection or switch to the remaining door.

Monty has played this game so often he decides to make it a little different. Before you make your choice, he tells you that if you don't find the money, he'll let you play the same game again, but increase the amount to $2000. Perhaps it would be better to try to lose the first game.  Should you switch doors?

On another day, Monty lets you know before you make your first choice that if you don't find the money, you'll play again for $2000, and if you don't find that money, you'll play one more time for $3000. When offered your first choice, should you switch doors, or stay with your first choice?

What are your best decisions to maximize your potential winnings?

Source: Original.


Solutions were received from Jayavel Sounderpandian, Alan O'Donnell, Joseph DeVincentis, Nick McGrath, Jackson, and Jeremy Galvagni.  All found it better to not switch (try to lose) if the potential winnings on the next round were greater than 1.5 times the current round.  An example solution (from Nick McGrath) is below.
 
In the first case, our expectation is given by :
E=p(win first game)*1000 + p(lose first game/win second)*2000
If we "try" to win both then:
E=2/3 * 1000 + 1/3 * 2/3 * 2000 =~ 1111
If we try to lose the first and win the second then:
E=1/3 * 1000 + 2/3 * 2/3 * 2000 =~ 1222
 
So clearly we should NOT switch on the first game and switch on the second (if we get that far).
 
In the second case we have 3 possible strategies(assuming we should always try to win the last game if we get there)
a) switch,switch,switch
b) stay,switch, switch
c) stay,stay,switch
 
E(a) = 2/3 * 1000 + 1/3 * 2/3 * 2000 + 1/3 * 1/3 * 2/3 * 3000 =~ 1333
E(b) = 1/3 * 1000 + 2/3 * 2/3 * 2000 + 2/3 * 1/3 * 2/3 * 3000 =~ 1600
E(c) = 1/3 * 1000 + 2/3 * 1/3 * 2000 + 2/3 * 2/3 * 2/3 * 3000 =~ 1600
 
So we should not switch on the first game and we are indifferent to switching on the second game.

 


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