2008-10-27 by Nathaniel Bluedorn in Mind Twist, 817 views
1 • Sonny Hall • November 16, 2008 • 10:10 AM
Both choices (stick or swap) have the same 50/50 odds at the time the question is asked. The question is “Which choice would give you the greatest chance to win the car? There are two doors to choose from whether you stick or swap. Even if you stick with your original choice, you have not stuck with your original odds of winning. Those odds changed as soon as Monty opened the door. The correct answer is NOOOOOO DIFFERENCE!!!!!! Sonny Hall
2 • Katie • November 24, 2008 • 11:51 AM
So is the answer switch right? I am really confused. I remember this but i do not remember.
3 • shawndumas • February 27, 2009 • 4:12 PM
I wrote a simulation using Ruby and ran it 100,000 times the results where:
Stay: 33.26%, Switch: 66.73%
I’ll post the source if anyone wants me too.
4 • Renee • August 13, 2009 • 12:30 AM
shawndumas, please do post your source.
I think, if I am understanding this correctly, what is described as being assessed is the FIRST CHOICE made—the 1 out of 3, 33% door choice—and how that 33% ORIGINAL CHOICE (door #3)stacks up to the other unknown doors (the door #1 and door #2 group) of 67%. That FIRST CHOICE, 1 out of 3, is what is being analyzed. If you had only ever had two choices, the odds of your FIRST CHOICE would be 1 out of 2, but that is not the original case. // On the other hand, if each door is worth 33% the 50/50 people see this as choosing between “33%” & “33%”, since the 3rd (door #2) 33% door has been revealed. However, the revealed door was not revealed randomly, which is the other pivot point in the arguement for 67% “falling on” the—as yet—unchosen, unrevealed door (door #1). // Is that “falling on” assumption logical? // I think it is. // Anyway, we are planning to run the test and prove the given answer to ourselves! This is fun!
5 • Renee • August 13, 2009 • 1:19 AM
Accepting that statistically the “switch” is correct about 67% of the time, could it be that this is because “statistics” is blind to what is revealed (door #2 in this event). Revealed or not, the statistical value of each door is still 33%, although we would never choose the revealed goat door, once revealed (unless we wanted a goat!). In a quarter toss, for example, if “heads” comes up, there is still a 50/50 chance of “heads” not coming up, even though we see clearly that it did. In the 3-doors scenerio, our brain discounts the revealed, unwanted goat door. Remember, a goat door will always be revealed purposefully—the partner of the UNchosen, unrevealed door. This at-first-UNchosen door GROUP (door #1 & #2 in this event) carries 67%. Because we, in reality, would never choose the revealed goat door, the UNchosen, unrevealed door ends up statistically “carrying” the 67%. Does that seem to make sense?