Comments on: Let’s Make a Deal http://www.damninteresting.com/lets-make-a-deal/ A collection of legitimately fascinating information culled from the past, present, and anticipated future. Sun, 20 May 2012 04:50:01 +0000 hourly 1 http://wordpress.org/?v=3.3.2 By: Perry Curling-Hope http://www.damninteresting.com/lets-make-a-deal/#comment-27095 Perry Curling-Hope Sun, 06 May 2012 10:57:01 +0000 http://www.damninteresting.com/?p=211#comment-27095 People make this ‘problem’ hugely complicated, running simulations and introducing all manner of irrelevancies to try to ‘prove’ that the odds prevail. They do, and always will, by simple logic, unless some metaphysical precognition is at work to confound them, but that is not the issue at hand here. If the contestant applies the strategy of not switching, the odds are a simple 1 in 3 of picking the car, whatever the host does is irrelevant as it will have no bearing on the result. If the contestant applies the strategy of switching, the odds are still a 1 in 3 of initially picking the car, and the contestant will lose, as the host will reveal the one goat, and the contestant will migrate to the other goat. The odds of the contestant initially picking a goat are of course 2 in 3, and the contestant will win, as the host will reveal the other goat, and the contestant will migrate to the car. It is only the initial selection by the contestant which has anything to do with randomness and probability, after that, it plays no part, and the outcome is determined. The host has to know where the car is, and is presumably constrained to revealing a goat after the contestant’s choice for the ‘Monty Hall Problem’ to prevail People make this ‘problem’ hugely complicated, running simulations and introducing all manner of irrelevancies to try to ‘prove’ that the odds prevail.
They do, and always will, by simple logic, unless some metaphysical precognition is at work to confound them, but that is not the issue at hand here.

If the contestant applies the strategy of not switching, the odds are a simple 1 in 3 of picking the car, whatever the host does is irrelevant as it will have no bearing on the result.

If the contestant applies the strategy of switching, the odds are still a 1 in 3 of initially picking the car, and the contestant will lose, as the host will reveal the one goat, and the contestant will migrate to the other goat.
The odds of the contestant initially picking a goat are of course 2 in 3, and the contestant will win, as the host will reveal the other goat, and the contestant will migrate to the car.

It is only the initial selection by the contestant which has anything to do with randomness and probability, after that, it plays no part, and the outcome is determined.
The host has to know where the car is, and is presumably constrained to revealing a goat after the contestant’s choice for the ‘Monty Hall Problem’ to prevail

]]> By: Evil1 http://www.damninteresting.com/lets-make-a-deal/#comment-27009 Evil1 Wed, 25 Jan 2012 18:01:22 +0000 http://www.damninteresting.com/?p=211#comment-27009 If there were 1 Billion doors to choose from, and I picked one randomly that I thought may contain the prize, I would have a 1 in 1 Billion chance of picking the right door first time (very little chance of success). If the host then opened 999,999,998 doors that HE knew did not contain the prize (leaving my original picked door and only one other door), that would mean that either my door was right (on first choice) - or the only remaining door he has left unopened is right and the odds would suggest I should swap to his door. Much better odds than the three doors but the principle is the same, unless someone sees a flaw in my reasoning? If there were 1 Billion doors to choose from, and I picked one randomly that I thought may contain the prize, I would have a 1 in 1 Billion chance of picking the right door first time (very little chance of success). If the host then opened 999,999,998 doors that HE knew did not contain the prize (leaving my original picked door and only one other door), that would mean that either my door was right (on first choice) – or the only remaining door he has left unopened is right and the odds would suggest I should swap to his door. Much better odds than the three doors but the principle is the same, unless someone sees a flaw in my reasoning?

]]> By: cornflower http://www.damninteresting.com/lets-make-a-deal/#comment-26501 cornflower Fri, 29 Apr 2011 01:02:35 +0000 http://www.damninteresting.com/?p=211#comment-26501 *rather than the first *rather than the first

]]> By: cornflower http://www.damninteresting.com/lets-make-a-deal/#comment-26500 cornflower Fri, 29 Apr 2011 01:01:47 +0000 http://www.damninteresting.com/?p=211#comment-26500 I'd like to point out what I think about the coin tossing question. The question "What are the odds that..." is the key here, and is what may make the problem misleading. It makes a difference when you say: "What are the odds that the coin will come up heads on the next (100)th flip?" and when you say: "What are the odds that you will get a series of 1oo coin flips that come up all as heads?" The illusion is in that we tend to think of the second formation of the question rather first, as our brain tends to recognize stark patterns more quickly, as the article says. I’d like to point out what I think about the coin tossing question. The question “What are the odds that…” is the key here, and is what may make the problem misleading.
It makes a difference when you say: “What are the odds that the coin will come up heads on the next (100)th flip?” and when you say: “What are the odds that you will get a series of 1oo coin flips that come up all as heads?” The illusion is in that we tend to think of the second formation of the question rather first, as our brain tends to recognize stark patterns more quickly, as the article says.

]]> By: Alucin Veritas http://www.damninteresting.com/lets-make-a-deal/#comment-25585 Alucin Veritas Sat, 26 Dec 2009 08:23:27 +0000 http://www.damninteresting.com/?p=211#comment-25585 An inversion of thought might provide clarity. Instead of having the probability of being right, look at the probability of being wrong. Initially, there is a two thirds probability of being wrong. Then, in the switch there is a one in two probability of being wrong. Since the two probabilities are connected, they are multiplied. The result is that if a switch has occurred, then there is a one in three probability of being incorrect. An inversion of thought might provide clarity. Instead of having the probability of being right, look at the probability of being wrong. Initially, there is a two thirds probability of being wrong. Then, in the switch there is a one in two probability of being wrong. Since the two probabilities are connected, they are multiplied. The result is that if a switch has occurred, then there is a one in three probability of being incorrect.

]]> By: Bob Nesbo http://www.damninteresting.com/lets-make-a-deal/#comment-25471 Bob Nesbo Mon, 09 Nov 2009 09:48:30 +0000 http://www.damninteresting.com/?p=211#comment-25471 I believe Edgar Allen Poe had an entire dissertation regarding something similar to this. I read it years ago, and I will have to search it... I believe Edgar Allen Poe had an entire dissertation regarding something similar to this. I read it years ago, and I will have to search it…

]]> By: sulkykid http://www.damninteresting.com/lets-make-a-deal/#comment-24339 sulkykid Sun, 19 Apr 2009 00:53:05 +0000 http://www.damninteresting.com/?p=211#comment-24339 Previous odds and outcomes ARE relevant as any horseplayer knows. Previous odds and outcomes ARE relevant as any horseplayer knows.

]]> By: tactilejones http://www.damninteresting.com/lets-make-a-deal/#comment-24338 tactilejones Sat, 18 Apr 2009 23:05:46 +0000 http://www.damninteresting.com/?p=211#comment-24338 Okay, head hurts, nausea setting in, and yet I'm compelled to exacertbate the situation by tediously registering, logging in, and commenting via T9 text from my cellphone!! So, I think I get the concept, initially... switching to the 2/3 odds from the 1/3 odds and all that. But then, consider this scenario: No previous gameplay. Two doors, one contains a prize, one contains nada. You "own" one door. You're given the choice to switch doors, or not, before the prize-containing door is revealed. Whether you switch or not, your chances are 1/2. Other than the "no previous gameplay", this is the scenario you're presented with in Monte's game, no? So if previous odds and outcomes are not relevant, as according to the Gambler's Dillemma (as presented here), shouldn't "pick one of two doors randomly containing a prize" remain at 1/2 odds, regardless of whatever silly game you were playing a moment before? HELP!! I will not eat or sleep until I understand this! Okay, head hurts, nausea setting in, and yet I’m compelled to exacertbate the situation by tediously registering, logging in, and commenting via T9 text from my cellphone!! So, I think I get the concept, initially… switching to the 2/3 odds from the 1/3 odds and all that. But then, consider this scenario: No previous gameplay. Two doors, one contains a prize, one contains nada. You “own” one door. You’re given the choice to switch doors, or not, before the prize-containing door is revealed. Whether you switch or not, your chances are 1/2. Other than the “no previous gameplay”, this is the scenario you’re presented with in Monte’s game, no? So if previous odds and outcomes are not relevant, as according to the Gambler’s Dillemma (as presented here), shouldn’t “pick one of two doors randomly containing a prize” remain at 1/2 odds, regardless of whatever silly game you were playing a moment before? HELP!! I will not eat or sleep until I understand this!

]]> By: howardfrankfort http://www.damninteresting.com/lets-make-a-deal/#comment-23986 howardfrankfort Sat, 14 Feb 2009 17:42:19 +0000 http://www.damninteresting.com/?p=211#comment-23986 The least likely senario oddswise is you have chosen the winner @ 1 of 3 choices. If the odds favor you have chosen a goat in step 1, accept that you probably have. What doors would this senario leave Monty? A winner and a looser. He cannot choose the winner, so if odds are your only indicator, and they favor you have chosen a goat, HE had no choice to make, you forced him to eliminate the other goat. So if odds are you chose a goat to begin with, and you know what that would have forced Monty to do, what door would you pick? The switch takes into account your choice and Montys choice. Instead of using your odds of 33% getting the prize in step 1, use your odds of 66% of picking a goat in step 1 plus if that happend, 100% odds of Monty picking the other goat for you. The least likely senario oddswise is you have chosen the winner @ 1 of 3 choices. If the odds favor you have chosen a goat in step 1, accept that you probably have. What doors would this senario leave Monty? A winner and a looser. He cannot choose the winner, so if odds are your only indicator, and they favor you have chosen a goat, HE had no choice to make, you forced him to eliminate the other goat. So if odds are you chose a goat to begin with, and you know what that would have forced Monty to do, what door would you pick? The switch takes into account your choice and Montys choice. Instead of using your odds of 33% getting the prize in step 1, use your odds of 66% of picking a goat in step 1 plus if that happend, 100% odds of Monty picking the other goat for you.

]]> By: tikigod4000 http://www.damninteresting.com/lets-make-a-deal/#comment-23658 tikigod4000 Mon, 05 Jan 2009 17:57:14 +0000 http://www.damninteresting.com/?p=211#comment-23658 MY BRAIN!!! MY BRAIN!! MY BRAIN!!! MY BRAIN!!

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