Nail-Tinted Glasses

Monty Haul Doors revisited

Once again tripping over a misunderstanding of this puzzle, I've worked out another possible illustration which may confern enlightment on the reader.

Consider first a different version of the puzzle, which works like this: a prize is hidden behind one of three doors. The player, with no information as to which door hides the prize, makes a choice from the three available doors. The host flips a coin. He now opens a door, subject to the following rules: the direction of the door shall be determined by the outcome of the coin toss (heads for left, tails for right - order the doors like a ring so that going off the end to the left brings you to the right, and vice versa), the host shall open the first door encountered (starting from the door chosen by the player) that does not include a prize.

To keep things simple, consider the cases where the player has chosen door number 1 (the other two work exactly the same way - this just saves bookkeeping). Given that the location of the prize and the coin flip are independent of one another, there are six distinct cases to consider.

Player   Prize   Coin   Host opens
1        1       H      3
1        1       T      2
1        2       H      3
1        2       T      3
1        3       H      2
1        3       T      2

The contest conditions tell us that these are equally probable outcomes. Now, the player is ignorant of the position of the prize, but has seen the coin flip. So collect those terms by the information that the player has.

Player   Prize   Coin   Host opens
1        1       H      3
1        2       H      3

1        1       T      2
1        3       T      2

1        2       T      3

1        3       H      2

So of the first two cases, there is one where StandPat is the correct strategy, and one where Switch is correct. The same is true of the second pair. But the two straglers at the end favor the switch category - there is no corresponding case where StandPat wins that produces the same behavior in the host. So both of these cases favor the Switch strategy - giving us 4/6 which favor the switch and 2/6 which favors StandingPat. Switch wins twice as often.

The game we just analyzed differs from the original only by showing the coin to the player. If we hide the coin, then the sample breaks into two groups.

Player   Prize   Coin   Host opens
1        1       H      3
1        2       H      3
1        2       T      3

1        1       T      2
1        3       T      2
1        3       H      2

So when the host shows door 3, there are still two cases that favor Switch vs. one that favors StandPat; the same is true when the host opens door number 2. So Switch is still successful more often than StandPat.

March 14, 2004 1:45 PM | TrackBack