~Six years ago, I wrote a blog entry about The Monty Hall problem called The Classic Monty Hall Problem Gets Goatsed and I continue to get comments on it. Recently, the entry was linked by a commenter on the Dilbert blog by fiftyfiftyman AtlantaDude who later recanted his statement and appears to have turncoated to side of the switchmen, and a new influx of switchmen have arrived to proclaim my profound insanity.
After ~six years of comments, I have held my ground and have gained even further insight as to why the authentic reality of the situation is 50/50, rather than 33/66.
To establish the ground rules, the Monty Hall problem in question (of which there are several) is strictly as follows:
You are a contestant on a gameshow hosted by Monty. Monty gives you the option of selecting one door from an available indistinguishable three. Behind one door is a new car, behind another is a goat, and behind the remaining door is also a goat. You select one from among the three indistinguishable doors. Monty now reveals one of the doors you have not selected to be a Goat Door. What is the probability of selecting the Car Door?
For ease of reference, I refer to those who believe that the elimination round is relevant as Switchmen, while those like myself who believe the elimination round as irrelevant as Fiftyfiftymen.
A switchman will advise to always switch, presuming that the elimination round impacts the stay/switch round, which is untrue. The final question is a diversionary tactic to create a false situation in which a certain branch of math appears to be applicable, but does not genuinely apply.
The explanation goes that, when conducted with a cup-and-ball experiment in which a ball is placed under one of three cups without your knowledge of which cup conceals the ball, and you play out the game an arbitrarily sufficient number of times (perhaps as with this simulation), that by empirical, statistical observation of wins and losses, wins by switching will gradually approach 66% of the time, whereas wins by staying will gradually approach only 33% of the time, so the safest bet would be to switch.
In order to arrive at that conclusion, however, the elimination round must be taken into account, even though the elimination round is irrelevant. The acceptance of the irrelevant elimination round as weight upon a 50/50 choice is the switchman’s error, when it genuinely bears no weight.
Allow me to offer another example of such diversionary questioning that creates a similarly false mathematical circumstance:
Three men require lodging, and at the chosen hotel, management asserts the cost for one room’s rent for the trio is $30, so each man pays his fair $10 share for the single room. Management is later crunching the numbers, and realizes that the price for a room rental for three men is actually $25. The next morning, management confronts the trio and offers a partial refund for the difference, giving back each man $1 each and proposing that $2 be offered to the bellhop as a tip, to which the trio agree. The trio therefore only paid $9 each for the room for $27 total, and $2 goes to the bellhop, which only totals $29. Where did the other dollar go?
The diversion within the question is that the $2 tip is added, rather than (properly) subtracted from the $27 total, to make $25. The question attempts to suggest that the original $30 total is still relevant, rather than the new total of $25 that is the only relevant total for application.
Similarly with the Monty Hall problem, the previous elimination round bears no effect on the most recent choice, to stay or switch. The math involved for Switchmen, like those fooled by the lodging problem, requires the now-irrelevant prior circumstances to remain within the body of evidence, when in reality both prior matters are newly and entirely irrelevant.
Further evidence to the contrary of the Switchman’s claim is when the figures are tallied for a situation involving four doors and three goats. Analysis of past wins indicates that switching wins 5/8ths of the time (according to this article) — but all of those require the belief that the prior choices matter in a new situation in which a 50/50 situation is now present.
There are several ways to explain how elimination rounds no longer matter to the final Stay/Switch round.
Imagine a large, blank, perfect-circle dartboard that can plane-rotate on its center point when spun. With a fine marker, precisely draw three radius lines from the center so that each radius intersection with the outer edge of the board create 3 pie-shaped regions of the board equal in surface area. Each region corresponds to a specific door, perhaps labeled 1, 2 and 3. The board is spun, the contestant throws a dart, and the mark made indicates which door is chosen in the elimination round, and that door is removed from play.
The elimination round dartboard’s layout is now wholly irrelevant and is no longer an acceptable instrument for selecting between whether one should stay with the door chosen by dart’s mark or switch to the other door, because of the presence of region that still represents the now-eliminated door.
A new dartboard is instead brought forth, which has a single bisecting diameter line that passes through center, dividing the board into two equal regions of Stay and Switch. The contestant still has no idea which door contains a goat door or a car door, and the new board is strictly a matter of Stay or Switch. Regardless of how many previous elimination boards that came into play before the 3-way board, such as a 4-way board, 5-way board, or 100-way board or more, the final decisive moment is a matter between Stay or Switch.
The elimination round, itself, is an illusion, because no choice has genuinely been made. The host’s selection of elimination-round doors could have been entirely nonsensical — the question of, “What is the capital of Kentucky?” could be posed and depending on the answer right or wrong, one or the other of the two goat doors could be removed. An insider-audience vote of which person, either Monty or contestant danced best to the People’s Court theme song, could be the deciding factor whether which of the two goat door would be removed. The final option would still, in each above case, result in a single 50/50 choice remaining.
The doors could be arbitrarily named Heads, Tails, and Wildcard, whereas the Wildcard will take the name of the door which is eliminated:
Regardless of what door the contestant selects, all three situations result in a Heads or Tails event.
The technique to choose a door is decided by the roll of a 1D3 die — perhaps in the form of an imperfection-free, equilateral triangle extruded at a distance four times the length of one of the triangle’s lines, with ends that equally force a final resting position with one of the three triangle’s sides face down without the possibility of landing on the end (perhaps better visualized as a 3-sided pencil with both ends sharpened). The end facing down is the door that the contestant selects to keep, and one goat door is removed from among the two unselected doors. The 1D3 is now no longer a valid instrument for determining whether to Stay or Switch, so a 1D2, or simply a coin flip would be an accurate way to make the next decision since the prior options of the 1D3 are no longer relevant.
There is a 100% chance that the car door is a car door, and a 0% chance that either of the goat doors is are a car door. One of the two goat doors (each with zero percent chance being a car door) is removed. The contestant must now decide whether to select one of two options — one door which is 100% the car door, or another door which is 0% a car door. Regardless of the choice the contestant makes in any elimination round, or how the choice was made, or how many doors that had 0%-car doors were previously been removed in how many elimination rounds existed previously, the final choice will be between a door which is 100% a car door, and another which is 0% a car door.
In an effort to resolve the dispute, I propose that the final question of the Monty Hall Problem be reworded:
Does the empirical evidence that the win/loss figures change based on the increase in number of elimination rounds preceding the final stay/switch round, substantiate or invalidate the belief that switching, when presented with the final stay/switch choice, is the most reasonable option?
Is a half-cup of water in a cup that could hold 1 cup, half empty, or half full? Math doesn’t lie, after all. Statisticians, however..