I expect the MHP will come up on the course. Imagine Richard’s response when the professor tells the class the single remaining door has the probability of 2/3 the car: “And you have the Gall – the bare-faced cheek to lecture me about probability ?- go to the back of the class and sit on the naughty seat until you get your mind synchronized with reality.”
'Seeing as I appear to be off Richard’s Xmas card list at the moment – ’
*Happy Holidays, Palmer. With very best wishes, Freddie./*
We’ve been here before. If you want to change the rules then you’re playing a different game. Obviously picking 2 doors is better than picking 1 door…problem with that is you don’t get to pick 2 doors.
Forget about the goats and Monty Hall revealing new information. When the game starts one door has a car behind it, the other two doors have nothing. You get to pick one door. After you’ve picked a door, the host changes the rules and says you can have whatever is behind the door you already picked, or you can switch and have whatever is behind the other TWO doors. Obviously choosing two doors is better than keeping one.
I’ve always explained the problem to people as follows:
Imagine, instead of choosing between doors, you roll a six-sided dice, and instead of numbers, this dice has a single colored dot on each surface - five of them white, one of them red. If you roll the red dot, Mr. Hall will give you a brand new car. Before rolling the dice, however, Mr. Hall covers each side with a piece of tape to conceal the color of the dots. Finally, you roll the dice.
At this point, you know that the probability of having rolled the red dot is only 1/6, so odds are that you’ve rolled a white dot (5/6). Mr. Hall then removes four pieces of tape to reveal four white dots (leaving your inital roll concealed), returns it to the position it was in after you rolled it, and asks you if you’d like to switch to the only other unrevealed dot. Because the position of the dice hasn’t changed, you know that the probability of having rolled a red dot hasn’t changed either. In other words, whatever dot you rolled in the first place, it’s still the same dot regardless of which dots have since been revealed. Those odds are still 1/6 red and 5/6 white. Knowing this, the odds are clearly in your favor if you switch.
The same reasoning applies to the Monty Hall problem, as the contents of the doors do not change when Mr. Hall reveals one of the losing doors after you make your initial selection. The probability that your original choice was a losing door will always be 2/3 because its contents never change (in the same way that the probability that you rolled a white dot will always be 5/6 because the color of the dot never changes), so the probability that the remaining unchosen door is the winner will always be equal to the probability that your original choice was a loser.
Hello folks back again!
Not for the MH talk, for a question to some of you.
I’m investigating such comment streams as this. Big ones like slashdot, BBC HYS and so on.
What I’m interested in is why some people keep going at it. I’m not having a go at anyone in particular, just interested in the social network stuff.
If you’re interested in getting into a discussion, tell me. I’m found by googling Chris Saltmarsh CERN (I used to work there, the CERN bit gets me, there are others of my name but none as ancient as I am.)
Or you could email me. You can find that on the net, or from this site.
Lovely to hear from you.
Cheers
The problem that both sides of the spectrum are having is consolidating two different factors into one problem.
Thus we have people on both sides using two completely different processes to explain something where the problem lies in the ability of the problem itself to align fractional mathematics and boolean logic.
To illustrate that both sets of people are both wrong and right in different ways:
The explanation of the issue is that the contestant is more likely to have chosen something “other than a car” originally. This is true. However this statement concludes that the real loss is in the initial choice. This is true.
What is lost in translation however is that you were never under any circumstances going to be allowed to pick “Monty’s” door. The problem itself has to follow its own rules which state that “Monty” is entitled to a door which is never going to be a door that you pick.
Mathematicians are using a choice which exists only under the assumption of the contestant assuming he can pick each door.
The contestant can pick (1) goat door - it does not matter which, because “Monty” is already assigned to the other (1) variable.
The contestant has thus the choice of “goat” or “car” because “Monty” will always be a placeholder for one of those goat doors.
You can never choose “Monty’s Door” and therein lies the problem with the fraction being represented as n/3 wherein n = 1 or 2.
To label this:
Monty’s Door = can only be assigned to him
Initial Choice = contestant choice
Switch Choice = contestant choice
How many choices do we really have available? It appears to be 3 if we include “Monty’s Door” however that is a constant not a variable. Monty will always own (1) goat, with the door being the “placeholder” for that goat. The goat he owns is independent of the door. He will always have a goat behind his door, rendering it impossible for you to “choose” that particular goat.
The problem here is simple:
With (3) doors and (1) prize you will always have a 1/3 chance of car and 2/3 chance of a goat.
Monty will always have a goat door so that door is not available for choice. Monty existing and owning a door in this entire problem has changed the denominator. The denominator has the illusion of being 3 so mathematicians are using the right operations and operands but the wrong denomination, due to the assumption that “Monty’s Door” was ever to be considered in the pool of options.
You only ever had two choices:
Stay or Switch
One will always be a goat and one will always be a car.
From those two axioms, the conclusion should be evident.
The choice was always:
Car on initial
Car on switch
Monty’s door always holds one goat. Its a choice you can never make.
Further:
The mental confusion arises when assigning fixed identifiers to rotating objects. There is never a door 1, 2, or 3 or A, B, or C. We assigned them identifiers to fix their positions, however “Monty” will never be assigned to A, B, or C because he has to go where a goat goes, not with a specific door.
Using fixed identifiers changes the way we view the problem. A variable cannot be a constant and a constant cannot be a variable. The two are boolean opposites.
The problem lies in information transference, not in mathematics. In a scenario without Monty, you have n/3, however if Monty is assigned to one door fulltime, you can never pick his door, regardless of the identifier (A,B,C) which means your probability can only be expressed in n/2 because his door will never be an option in Stage 2.
The very construct of this problem points to how things change in each stage. The problem both groups are having is not realizing that the conditions of Monty being there have impacted the original idea of (3) choices. Without Monty you have (3) choices. At the event in which he exists you only have (2) choices. You cannot choose his goat door even though you cannot presently put a fixed identifier on it.
It makes no difference whether (1) goat is in China and the (2) is in France. Revealing geography is not the same as revealing existence. We already knew both goats existed, so Monty gives useless information. When he opens the door, he is merely displaying to observers that he has chosen a Chinese or French goat. You were always going to be left with the remaining goat or a Car. You never had the option to pick because he would have already chosen the specific goat before ever presenting the option to you. This is irrelevant though because even from Stage 1 we are assuming Monty will be assigned a goat. You cannot assign his goat (though so far he is unidentified in Stage 1) as an option for us to pick.
Based on the actual Monty Hall scenario, even in the beginning you have a choice of (3) doors, but a choice of only (2) prizes - A car and Monty’s least favorite goat (language, name or other identifier to be determined in Stage 2)
Also, he will always be gifting you the entirety of his door’s probability. As with any removal, the contents are distributed evenly to the “stay” and “switch” doors.
You didn’t need to see him do anything to know this. The key point is that he is actually removing both the denominator and the enumerator from the fraction. Your options are (1) car and (1) unidentified goat.
Monty states: "You cannot choose from my pool of options because then I’ll just switch to the other goat in the door you didn’t pick. Regardless, I win because I get a goat either way, which is my function. You can have the leftover doors, but you would have always had a goat and a car option regardless of which language your goat ends up speaking.
Further, if another individual was playing and would receive Monty’s leftovers and yours upon your final decision, then he wins wherever you lose. If you always choose to switch, his prize swaps also, even though its not by choice. If he was originally going to win, now you’ve swapped and made him a loser. If you swap every time, his prize swaps every time and he has also been impacted by variable change…that is to say he should make the same assumption you are making. One person loses from another person’s gain. The leftover door is the other person playing with you. You switch every time so it switches every time to align with your rejected prize.
The door and yourself cannot both have a 66.6 percent chance of a car. But you never had a 66.6 chance, in fact Monty had 100% of a goat which means you could never get “his” goat, “Monty’s Goat”, so you never had (3) options for prizes, only (3) options for doors in Stage 1. Stage 2 shifts the focus by reducing the number of doors to the number of “types” of prizes (car or goat). And that is where the math gets confused. Performing numerical operations alone without separating objects from their assigned placeholders.
Another incredibly long post - full of inaccuracies and untruths.
The problem you’re havi
ng is you don’t understand probability and how it applies to the MHP. If you’d bothered to read any of the previous posts there are numerous explanations as to why switching doors increases your chances of winning to 2/3 (you can disregard Richard’s explanation since it’s nonsense)
These statements you made are incorrect:
“You have two objects you’re dealing with.”. You have THREE objects at the start.
“In fractions you don’t just change the enumerator, you change the denominator. The denominator represents possible choices and the enumerator represents quantity of choices or guesses.” As I said, you don’t understand probability.
“The problem is actually simple if you understand”. Yes it is, and no you don’t.
“Use that formula in each stage (as information is revealed)” Gives the wrong answer. “Then you will see that Stage 1 and Stage 2 are entirely separate fractions.” No they’re not, they’re the same, that’'s why the CORRECT answer is 2/3
“That number is evenly distributed to the 2 remaining mystery doors. (16.67 percent to each)”. An unsubstantiated assertion that is completely incorrect.
I could go on but you get the drift. Read some of Jonathon’s earlier posts, he provides some very clear and concide explanations as to why switching doors is a 2/3 chance.
I am not sure why people think this is an assault on their math skills, but rather examining what constitutes a “decision”.
You can change your mind as many times as you want in your head. Your first choice was based on the information given at the time which was 1 guess /3 doors, your second choice is based on the new information which is 1 guess /2 doors. The third door can no longer be selected, though it was available before he opened it.
In effect, all Monty did is change your problem from 1/3 to 1/2.
As has already been illustrated by both groups, it makes no difference how many goat doors are in front of you if they are all going to be removed before you make a “final”, “actual” choice.
All of those doors are useless information because they will not even exist when you actually make your final choice. They are one by one removed from being possible choices.
By the time you are down to the last 2 doors, you went from 1/1000 to 1/2… where 1000 is the amount of total doors we have used to start with.
Your situation changed with him keeping you from having to choose between so many doors. But the doors have always been the bottom part of the fraction.
The above is the only thing that actually happens here.
You never had to choose between 3 doors because he was going to come along and remove one from being a mystery. You only ever had to choose from 2.
To show that the problem lies with the time things are evaluated…answer how many doors you can choose from both before and after he reveals a goat?
Are the choice possibilities the same? Or has he actually changed the rules which governed your quantity of possible selections.
We already knew one door was a goat. What does it matter where on the stage its positioned?
The only thing happening here is the changing of the quantity of possible selections from (3) to (2).
The questions asked in the MHP are "Do you stick with door A (original guess) or switch to the other unopened door? Does it matter?"
And the simple answers are : yes it does matter, switch to the unopened door because that gives a 2/3 chance of winning whereas staying with your original guess is a 1/3 chance.
You see, the answer isn’t 1/2, and if you’d bothered to read any of the earlier posts you would have understood WHY the answer isn’t 1/2.
Your entire argument is false.
You seem to think I don’t understand how people arrive at their conclusion. I understand exactly what people are doing because if I didn’t then how could I explain why their order of operations is invalid. This is similar to a Mensa problem from years ago.
There are no longer 3 doors to choose from how do you get 2/3?
Any mathematics book will tell you to divide by total parts.
You keep saying 2/3 so my question is how do you have n/3 when there are not 3 choices any longer?
If I had 3 lottery tickets and 1 was a winner…the lottery spokesperson tells me hes going to rip a losing ticket up. Its the act of ripping the ticket (removing one false, losing ticket) that increases my odds of winning.
Monty removed a door and a goat both, you now have 1 goat, 2 doors.
Before you had 2 goats, 3 doors.
Take the above numbers and order them according to the timeline. You originally had a 2/3 chance of losing. He removes 1 goat, and 1 door.
Subtract 1 from the enumerator and 1 from the denominator:
2-1 = 1
3-1 = 2
The goat and the door that Monty opens are no longer factors in your decision. You can’t use the numbers prior to the door open because he just changed the both factors, he removed a goat and a door.
Put simply: Stare at the remaining doors and explain how there is a factor of 3 of anything. There were only 3 choices originally, the third choice doesn’t carry over because it’s no longer a choice.
You need to go and learn some basic probability theory - you don’t know what you’re talking about.
If you played 300 games, always picked Door1 and always stayed with Door1 how many games would you win? (Hint: it’s not 150)
If I’m wrong 2/3 times, and my initial guess is most likely wrong therefore, why not change it immediately after I make it. What does opening a goat door reveal that we didn’t already know?
Nemo Nobody,
Try this experiment. 3 blocks (representing car, goat, car) of equal weight.
There is a see-saw. To the right of it are the three blocks. You choose ONE. you put your choice (Y) on the left side of the balance:
^ * *
n=total # blocks, in this case 3. You have taken 1. There are now n-1 (2) left on the ground.
Monty reveals one block to be a Goat, then puts the other on the balance.
^ G
Either Y or * must be the car. The odds would be closer to 50-50 if you flip a coin to choose. But there are 3 choices to consider. The odds of Y being the goat are still 1/n ; the odds of * being a goat are (n-1)/n
This becomes more apparent with larger numbers (stay open-minded; I dismissed the larger numbers as irrelevant obfuscation). In the case of 20 blocks (doors):
^ G G G G G G G G G G G G G G G G G G
the likelyhood of * being a car is n-1/n in other words 19/20 while the chance of Y is 1/20.
That diagram didnt work out properly. and I should have said ‘car goat goat’ .
Intuitively the revealed goat may make you feel better about your choice (dodged a goat! Yeah!)
I agree, the goat tells you nothing about the nature of your initial choice OR the remaining door.
By staying you are choosing one door.
By switching you are electing to choose ALL doors except your initial choice. That was the crux of my misunderstanding. For relatively low values (3 doors) it isn’t as apparent.
ZenasDad,
That illustration actually was suitable enough to demonstrate what I was looking for. I was trying to work out exactly “why”. Alot of people only seem care about the answer, but never about how we arrive at the answer, or the definition of the laws that make these principles true. After illustration and visualization, I now see that its making a choice between two separately instanced events, one with really bad odds and one with the inverse of those odds. The other door will always equal the remainder of the bad odds and vice versa. I was trying to figure out the method of transference and I see that the two ratios balance each other out. Thank you!
Nemo, perhaps you were trying to figure out the wrong question. The MHP is not “What are the chances of the contestant choosing that door?”, it’s “What are the chances that door had a car put behind it before the contestant made their first choice?”.
zenasDad, before we get Buxtonned again, can I take us one step further than “ALL doors” by restating the claim that the chance of car of the ‘third door’, the one neither chosen first by the contestant nor revealed by Monty, increases from 1/3 to 2/3.
Freddie Orell,
I guess so but I will leave that to mathematicians, Palmer Eldrich, etc. It makes a simpler explanation (and may be correct); However many won’t accept it, especially those who weren’t taught probability or are trying to work it out for themselves.