User talk:Craigkb

Think of it binary numbers.

There are 3 possibilities:

100 (car is behind door 1), 010 (car is behind door 2), 001 (car is behind door 3),

The first part of the game - you picking the door - is pure showmanship. The real game begins when Monty removes a goat (or a zero in our simulation), which he is obliged to do by the rules.

Thus:

100 becomes either 10 or 10, 010 beomes either 01 or 10, 001 becomes either 01 or 01

Note that there are three 10's and three 01's

Thus a random sequence of the three binary digits representing the start of the game could be say:

010 100 100 010 001 001 001 100 010

and becomes a sequence of two binary digits after Monty removes one of the zeros, and as we have shown, with equal probability of a 10 or a 01 occuring.

It should be intuitive at this point that the odds of the real game i.e. the part after he removes a goat or a zero, is the equivalent of a toss of the coin. It matters not whether you switch or keep your original selection.

The vast majority of the discussion I have read on this problem is flawed because it fails to recognize that there is a change to the game once Monty removes a goat. All the probabilities are calculated when there are three choices (thus you see references to "one in three" or "two in three"), but the real game - the part after one of the three doors is removed - has only two choices. Three is not a valid denominator in the probability calculations.

In simpler language, you start off choosing one of three, Monty removes one of the two remaining choices and returns your selection to "the deck" and effectively gives you a choice of one in two.

Craigkb (talk) 06:49, 26 June 2013 (UTC)

The famous paradox
Thank you. Please consider the special scenario that lets the paradox arise. The famous "paradox" (odds staying:switching is not 1/2:1/2 but is 1/3:2/3), arises because in 2 out of 3 the host is not free to open one of his two doors at random, but on the contrary in 2 out of 3 is bound to show "the second goat only". In 2 out of 3, when the guest selected goat A or goat B, the host is extremely biased because the only door he can and may open in that case depends on the guest's first selection. In 2 out of 3, "by showing the second goat" (!), the host annunciates and promises that the door offered to switch on, actually does indeed hide the price for sure. Regard, Gerhardvalentin (talk) 20:26, 26 June 2013 (UTC)

No Bias

Thank you Gerdhardvalentin. Rather than re-iterating someone else's flawed logic, can you please disprove my arguments? Just one?

In the case where you correctly select the door at your first guess, the host has a choice of which door to remove to reduce the game to a coin toss. In any case where you select the wrong door, the host has only one choice to reduce the game to a coin toss. There is no argument here. That is a logical description of all the possibilities. But there is no "bias". Bias is a word that indicates a pre-condition exists that will affect the outcomes and thus affect the observations. There is no pre-condition here.

What about this variant of the problem to highlight the contention that the first "choice" is irrelevant?

Monty shows the contestant and the audience the three doors and states truthfully that there is one car and two goats behind the three doors. Before the contestant can make a choice, they are led away to a soundproof booth where he or she cannot see anything happening on stage.

Monty then shows the audience one of the goats and it is taken off stage, one door dismantled and removed, and the contestant is then brought back out and asked which of the two doors the car is behind.

This is exactly the same as the problem as stated. Can someone please prove that it is not? But here, there is no choice to change. We have removed that. So anyone quoting "one in three", "two in three", 33.3% or 66.6% in their probability calculations is wrong.

Back to the original problem, there are two events:



Event 1: Selection by the contestant of one of three doors. Outcome: No prize awarded. One of the three goats is eliminated. Event 1 ends.

Event 2: Two doors remain. Contestant given the choice of keeping their original selection or changing to the other single remaining door. Outcome: Contestant has 50% chance of selection the winning door regardless of decision taken in Event 1, because Event 1 is statistically independent of Event 2.



Disproving arguments please.

Craigkb (talk) 21:53, 26 June 2013 (UTC)


 * Huh, thanks. I thought we were talking about the scenario of the famous paradox, where always swapping definitely has a 2/3 chance of getting the car. Gerhardvalentin (talk) 22:35, 26 June 2013 (UTC)

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Thank you. --SineBot (talk) 23:15, 26 June 2013 (UTC)