The scams of Statistics...

[Scott Mayers]

I totally agree with your diagram, thanks to make it, it is very similar to the cars example i give to you, but it is well presented. (and i am happy you finally come to use this "12")

I only choose this because you seem to think it requires it but it is not necessary as each Option is necessarily a repeat of each other. That is, whatever odds is in Case 1 = the odds in Case 2 = the odds in Case 3. As I see it confuses Leo more, this was the reason I only used one Option in the beginning. I still hold to it. Do you agree that each Option, representing wherever the host places the car in is superfluously unnecessary?

Please, go back to my example with the car, and particularly the example i make with a dice roll, if you read it carefully and honestly, i think you will agree with me.

I think we need to eliminate the excess complexity first and why I'm asking whether you agree or not that using all 12 cases are doing that here? That is, what is true of all 12 events = what is true of one specific Option as I presented it above. For example, do you agree to the following very simplified diagram as a beginning to the problem?:

Simple Diagram.jpg (38.25 KiB) Viewed 3167 times

Do you agree that whatever is true about Option 1, Option 2, or Option 3 as equivalent to each other, as the same probability?
These options are the doors to which the host knows are placed in them and represents how the initial option acts as 1/3. Correct?

[Scott Mayers]

No, i was really speaking about P(option1 ligne 1) = P(option1 line 3) not P(option1 ligne 1) = P(option1 line 2)
I mean by it, do you think the probability to be in the first case, equal the probability to be in the 3th case in you diagram.

I will let you going further before giving you more to read, i just wanted to confirm it wasn’t a mistake. Good night

I am definitely not equating them as they are illustrated as being completely different events. Why you would think that is odd to me.

Maybe you are not reading this correctly. See my last post so that we may have a chance to reconstruct the problem step by step as given in the puzzle.

[HexHammer]

I don't see this response here but am guessing it is from another thread? Anyways, the 100 door example as I heard it before to aid in following the original construction of this Monty Hall problem is as follows.

Originally pick a door. You have 1/100 chance to get the car. Then the host reveals 98 of the doors as duds. He always knows where the car is and so if he knows you picked it, he can arbitrarily pick from any of the 98/99 doors remaining. If you didn't pick the car initially, He must remove ALL doors except for the one with the car for the other option. Since in all cases from the start you only have a 1/100 chance, 99/100 of the time, the car will be in the door the host MUST reveal thus assuring without a doubt that it increases your odds to Switch as 99/100 to Win by switching.

This is perfectly rational for 100 doors. But note that the host gets to pick from 98/100 doors to eliminate. This is

100/100 - 1/100[Guest] - 1/100[Host Leaving door unveiled] for either case = 98/100.

But, given only three doors, the host only removes 1 door which is also equal to the quantity remaining.

3/3 - 1/3[Guest] - 1/3[Host Leaving door unveiled] = 1/3.

This 1/3 is the same chance that the original Guest had at the beginning and why in this special case using only 3 doors balances out the apparent logic using more than 3 doors.

In the 100 door sample, the Guest has 100/100 - 1/100[Host Leaving door unveiled] = 99/100[Chance to Win by Switching]

In the 3 door sample, the Guest has 3/3 - 1/3[Host Leaving door unveiled] = 2/3[Chance to Win by Switching]

But,

In the 100 door sample, the Guest has 100/100 - 99/100[Switch Odds to Win] = 1/100 chance to Lose by Staying

and
In the 3 door sample, the Guest has 3/3 - 2/3[Switch Odds to Win] = 1/3 chance to Lose by Staying.

Now Given you have 99/100[Chance to Win by Switching] and 1/100[chance to Win by Staying (= not Switching)] for the 100 doors, the ratio of Stays/Switches for Winning is:

(1/100)/(99/100) = 1/99 odds to Stay over Switching for a Win [So Switching does improve your odds here].

In contrast, you have 2/3[Chance to Win by Switching] and 1/3[Chance to Win by Staying] given the 3 doors. The ratio of Stays/Switches for Winning is:

(1/3)/(2/3) = 1/2 odds to Stay over Switching for a Win [So Switching does not improve your odds here] .

This definitively proves how and why given only three doors, that staying or switching is always only 1/2. The Stays to Switch odds are equal with three doors only. That is, it is also 1/2 odds to Switch over Staying.

Thus whether you stay or switch given only three doors are equivalent odds and so it doesn't matter what you choose.

You see what you wants to see.

[Obvious Leo]

As I see it confuses Leo more,

It is a puzzle of such exquisite simplicity that it plays right into the hands of a philosopher of the bloody obvious, Scott. I am not in the least confused about it.

I presume you're aware of the fact that this test has been performed many many times and that switching guesses doubles one's winning chances. Since you haven't addressed this exact point I'll ask you now. Are you aware of the fact that you have been PROVEN WRONG?

[Philosophy Explorer]

The Monty Hall Problem: Before a door is selected, the car could be behind any of three doors so the odds are 1 out of three that the car is behind any of three doors. After door three is selected revealing a goat, then takes that door out of play. This leaves doors 1 and 2 and the car could be behind either one with the remaining goat behind the other so the contestant could stick with his initial choice or switch to the other door as the odds that the car is behind either door is now one out of two so the contestant can stand pat or switch with either choice being equally like for success.

The Secretary Problem: I have experience in this area. My cutoff was to go through all the candidates and make a selection at the end unless a particular candidate had strong skills upon which I made the selection.

PhilX

[Obvious Leo]

The Monty Hall Problem: Before a door is selected, the car could be behind any of three doors so the odds are 1 out of three that the car is behind any of three doors. After door three is selected revealing a goat, then takes that door out of play. This leaves doors 1 and 2 and the car could be behind either one with the remaining goat behind the other so the contestant could stick with his initial choice or switch to the other door as the odds that the car is behind either door is now one out of two so the contestant can stand pat or switch with either choice being equally like for success.

False.

[Obvious Leo]

This is why your interpretation ifs false, Phil.

Before a door is selected, the car could be behind any of three doors so the odds are 1 out of three that the car is behind any of three doors.

The car cannot be behind the door that the game host chooses to open.

[Philosophy Explorer]

This is why your interpretation ifs false, Phil.

Before a door is selected, the car could be behind any of three doors so the odds are 1 out of three that the car is behind any of three doors.

The car cannot be behind the door that the game host chooses to open.

But the two out of three doors is based on before any of the three doors are opened.

PhilX

[Obvious Leo]

It's too easy for somebody with some fluency in maths to outsmart himself with this one, Phil, but the problem is in fact childishly simple. So simple in fact that children and animals get it right a higher proportion of the time than adults do and maths geeks will be wrong more often than most. Many famous mathematicians have made complete fools of themselves with this puzzle because it's not a maths problem but a logic problem. The best winning strategy it to ALWAYS switch your choice because the game host has filtered out one of the losing options for you. You odds when you change are NOT 1/2 but 2/3. The host of the game CAN'T reveal the car so think about it.

[Scott Mayers]

The following lays out ALL possible cases, fully enumerated.

Each possible game begins first with the host (or staff of the show) placing a car in either Door 1, Door 2, or Door 3.

Starting with placing the car in Door 1, the Guest chooses either Door 1, Door 2, or Door 3, as a first guess.

Let's begin by numbering the incident in the left most column. The second column will list where the actual car is. The third column will represent the Guest's First Selection for each possible game. The fourth column will represent the door which the Host Reveals.

The fifth column will represent what remains after the Guest's First guess and the Host's Reveal. This left-over is what the Guest may opt to Switch to. Then I will enumerate the results according to WINs where the Guest chooses to switch and where he chooses to stay.

#.......Car is in this Door....Guest's First Selection....Host Reveal.....Alternative to Switch.....Switch WINS.....Stay WINS
1).............D1.......................D1.......................D2...................D3........................0.................1.......
2).............D1.......................D1.......................D3...................D2........................0.................1.......
3).............D1.......................D2.......................D3...................D1........................1.................0.......
4).............D1.......................D3.......................D2...................D1........................1.................0.......

5).............D2.......................D1........................D3...................D2........................1.................0.......
6).............D2.......................D2........................D1...................D3........................0.................1.......
7).............D2.......................D2........................D3...................D1........................0.................1.......
.............D2.......................D3........................D1...................D2........................1..................0.......

9).............D3.......................D1........................D2....................D3........................1.................0.......
10)............D3.......................D2........................D1....................D3........................1.................0.......
11)............D3.......................D3........................D1....................D2........................0.................1.......
12)............D3.......................D3........................D2....................D1........................0.................1.......

Count up all the Switch WINS. Divide by all possibilities. You get 6/12 = 1/2 odds to WIN when you Switch
Count up all the Stay WINS. Divide by all possibilities. You get 6/12 = 1/2 odds to WIN when you Stay

If those of you won't bother to carefully read through this to see that all possibilities are appropriately covered and is as above, this proves absolutely that the above conclusion on the odds are perfect. Thus no matter what the math one attempts to use to try to defeat it, it has to conform to this. There is appropriate math to this, but for those falling for the problem as a weird magical gift to gain by switching, you have to be using it in error.

[Obvious Leo]

Or alternatively simply ignore all of the above confusing bullshit and consider this. When you make your initial choice there is 2/3 chance you will be wrong and that the car will be behind one of the other doors. The host then takes one of those doors out of the game for you.

[Scott Mayers]

The Monty Hall Problem: Before a door is selected, the car could be behind any of three doors so the odds are 1 out of three that the car is behind any of three doors. After door three is selected revealing a goat, then takes that door out of play. This leaves doors 1 and 2 and the car could be behind either one with the remaining goat behind the other so the contestant could stick with his initial choice or switch to the other door as the odds that the car is behind either door is now one out of two so the contestant can stand pat or switch with either choice being equally like for success.

The Secretary Problem: I have experience in this area. My cutoff was to go through all the candidates and make a selection at the end unless a particular candidate had strong skills upon which I made the selection.

PhilX

Thank you for participating. I had these puzzles in mind with you as you like math and logic AND can competently argue on what I witnessed of you so far.

We obviously agree on the Monty Hall Problem.

On the Secretary Problem, without initiating that yet, I first learned of this by interviewing an employer who used this method too. But he didn't particularly use or care about any math. He was presently going to hire new hair stylists and showed me a whole stack a foot deep of resumes. Then he picked up about a third of it and threw it in the garbage in front of me when I asked how important a resume was to getting a job there.

[Scott Mayers]

Or alternatively simply ignore all of the above confusing bullshit and consider this. When you make your initial choice there is 2/3 chance you will be wrong and that the car will be behind one of the other doors. The host then takes one of those doors out of the game for you.

Leo, there are three mathematical ways that can actually appear to justify three different conclusions but the errors in them are due to mistaking perception and being incompletely consistent, even if well intended. In the last few posts I re-initiated this without the need for math to list all possible events with detail to which you simply opt to ignore. I did this first to at least show by pure OBSERVATION what the solution actually is. This is intended to first establish what is TRUE regardless of opinions on using any math before we investigate it.

I already mentioned that I recognize the problem similar to our difference of perspective on whether we begin with assuming a nothing, a something, or, like you, an infinity, of things. The error partially relates to this. You have to agree that if we cannot actually agree to trying what I laid out above, we won't get any resolution or try to find out what our errors in math or logic is about.

I hope that you at least take some time to eventually try to read each line above. Sorry if the diagram isn't the most friendly to view but I intended to make it easy for anyone here to work on in the HTML without getting too confused.

[Obvious Leo]

Scott. You are simply bloody wrong and what you're peddling here is mischievous pseudoscience which I can't just ignore. There is a wealth of literature on this subject, NONE of which supports what you claim, but why don't you just do the fucking experiment? It's been done millions of times before and EVERY SINGLE TIME the strategy of switching choices DOUBLES THE WINNING CHANCE.

How do you account for this empirical FACT???

[Scott Mayers]

Scott. You are simply bloody wrong and what you're peddling here is mischievous pseudoscience which I can't just ignore. There is a wealth of literature on this subject, NONE of which supports what you claim, but why don't you just do the fucking experiment? It's been done millions of times before and EVERY SINGLE TIME the strategy of switching choices DOUBLES THE WINNING CHANCE.

How do you account for this empirical FACT???

Stop it Leo. The experiment is in my previous few posts by enumerating for them. Accusing me of pseudoscience is trolling. Instead of just stating your disgust, prove or disprove me here by participating with fairness. I granted you as much yet you seem to be acting with clear disrespect for me here. I am asking you to please try and do so with respect.