Getting tired of having Instants shoved down my throat

[SoggyShorts]

I thought of another way to look at it:

Imagine you pick a door, but before you open it Monty asks you
"Are you sure you want that door, or do you want BOTH of the other doors?"

 

[DeletedUser7370]

I'm not sure where you went wrong

I am sure where you went wrong.

That is the first layer of the matrix and does not consider Monty eliminating any of the selections. Again see my full matrix.

And really, I expected more from both of you.

 

[DeletedUser7370]

or do you want BOTH door #2 and door #3?

Yes thank you for the win.

GIANT silent restriction in the game
EDIT: Monty has to show what is behind the door. If he shows the prize as being eliminated then why play?

Monty can't eliminate the winner

When people finally get this restriction then can get the math right.

 

[SoggyShorts]

And really, I expected more from both of you.

Let me take one more shot at this:

We agree there are only 9 possibilities, right?
3 doors with you can pick any one of those 3.

Spoiler: 9 possible combinations of door and pick

Now what happens if Monty opens a loser and you ignore him, choosing to open the door you picked?
You win 3 out of 9 times, right?

 

[DeletedUser7370]

Let me take one more shot at this:

Refer to the full matrix I posted. I covered all those possibilities plus all repicks, and then filters the repicks based on the simple rule that Monty is not allowed to steal a win from a player.

 

[SoggyShorts]

the simple rule that Monty is not allowed to steal a win from a player.

There are 3 rules:
(1) Monty never opens the door you chose initially; (2) Monty always opens a door concealing a goat; (3) When the first two rules leave Monty with a choice of doors to open (which happens in those cases where your initial choice was correct) he makes his choice at random.

I've looked at your matrix, and there are too many possibilities listed. There are only 9.

For example you pick door #1
1. Door #1 has the car, monty shows you a goat in either #2 or #3.
Stay = win
Switch =lose
2. Door #2 has the car, monty shows you a goat in #3
Stay = lose
Switch = win
3. Door #3 has the car, Monty shows you a goat in #2
Stay = lose
Switch = win

The same pattern happens if you pick #2, or #3 as your initial choice.

Or you can set aside the matrix for a moment and look at it this way:

• At the beginning of the game you have a 1/3rd chance of picking the car and a 2/3rds chance of picking a goat.
• Switching doors is bad only if you initially chose the car, which happens only 1/3rd of the time.
• Switching doors is good if you initially chose a goat, which happens 2/3rds of the time.
• Thus, the probability of winning by switching is 2/3rds, or double the odds of not switching.

Hopefully this helps. If you're still not convinced, then anything I say at this point would just be repeating myself and others that have solved this puzzle over the last 50 years. Try one of the sims I posted, make your own sim, read the wikipedia page or anything else you find on google there's loads of information on it, and all of it proves that switching yields better results.

 

[Ashrem]

Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy. Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating the predicted result (Vazsonyi 1999). @TedGrau is in good company. I still implore him to run the numbers for himself.
Without the incorrect assumptions that Monty does not clear a door and Monty can not take the prize away. The whole point of the game is that you can lose the prize if you pick wrong. Not try to make a matrix that contains invalid possibilities.

 

[DeletedUser5800]

So there are 3 pieces of pie, two pieces have a dead cat in it and one has a live cat in it... Schroedinger throws one piece in the trash and you had a 50/50 shot the whole time.

 

[DeletedUser7370]

Now that it is morning and I have had some coffee instead of evening beverages.

This is what I meant by Monty can't take a win away.

(2) Monty always opens a door concealing a goat

I mixed that rule up a bit in some of my posts. My apologies for that confusion.

For example you pick door #1
1. Door #1 has the car, monty shows you a goat in either #2 or #3.
Stay = win (outcome 1,2)
Switch =lose (outcome 3,4)
2. Door #2 has the car, monty shows you a goat in #3
Stay = lose (outcome 5)
Switch = win (outcome 6)
3. Door #3 has the car, Monty shows you a goat in #2
Stay = lose (outcome 7)
Switch = win (outcome 8)

You list 6 outcomes. If we separate your 1. into "monty shows you a goat in #2" and "monty shows you a goat in #3", then we would have 8 outcomes, 4 winners and 4 losers evenly split on stay and switch. You can't account for all possible outcomes by ignoring some.

Here is another representation you gave which again falsely combines the 2 possibilities and that is why it does not come away with the right result.

Assuming pick of door 1. Line 1 in that image: car, goat, goat. Monty eliminates door 2:goat, stay wins, change loses. Monty eliminates door 3:goat, stay wins, change loses. Line 2, Monty can only eliminate door 3, no changes to this line, stay loses and change wins. Line 3, Monty can only eliminate door 2 no changes to this line, stay loses and change wins. Now count: winners on stay 2, losers on stay 2, winners on change 2, losers on change 2.

I've looked at your matrix, and there are too many possibilities listed.

The first sheet leaves out your rule 2 (above) so that it can clearly be seen to cover all possible combinations. The second sheet applies that rule by removing lines that would violate the rule.

If you would like I can write out the math to show that 3=4, and I will tell you it is wrong and it uses the same method of falsely ignoring some outcomes to reach that conclusion.

 

[Ashrem]

There is nothing random involving the player after the first pick.

After the initial selection is made, if you picked a loser, changing makes you winner, and if you picked a winner, changing makes you a loser. Every single time. There is no random there, there is no other possible outcome. It is a fixed result.

Once the player picks a loser, Monty has no choices, he can only eliminate the other loser. Anyone who initially picked the wrong door is guaranteed a win by switching. Once the player picks a winner, nothing Monty does can change the result. Either door he eliminates leaves another loser. Anyone who picked the right door initially is guaranteed a loss by switching.

66% of initial picks are losers, changing has to result in a winner for every one of those people. It's not possible to switch from one loser to another, because Monty is forced to eliminate the other loser.

 

[DeletedUser7370]

Either door he eliminates leaves another loser. Anyone who picked the right door initially is guaranteed a loss by switching.

Either door means 2 doors. Lets assume the player always picks door 1 first. and the prize is behind door 1.

Monty has two possible choices. Display door 2 or display door 3.
After which the player has 2 choices change or stay.
Put them in a matrix and you have 4 possible outcomes.

Monty must select randomly in that situation or the game would become predictable. The matrices that Soggy showed leave that out. I am picking through the simulator's code now.

 

[Ashrem]

Monty must select randomly in that situation or the game would become predictable. The matrices that Soggy showed leave that out. I am picking through the simulator's code now.

But it doesn't change the options for the player. They aren't choosing between three doors, they only ever get two choices at that stage. Which of those two choices is random doesn't add to the number of possibilities for them.

The game as set has exactly four outcomes available.
You picked wrong and keep it and lose
You picked wrong and change it and win
You picked right and keep it and win
You picked right and change it and lose

Those are the only four possibilities for the entire game. There are no others.

If the players always change their pick at the second stage, then there are only two possible outcomes:

They picked wrong and change and win (66%)
They pick right and change and lose. (33%)

If everyone changes at stage two, those are the only possible outcomes. It is not possible to change from a loser to a loser or from a winner to a winner. At the point where the choice is offered, there is one winner and one loser, there is nothing else.

 

[DeletedUser7370]

I finally see it using a different matrix of possibilities. The key for me was to ignore Monty's action as completely irrelevant.

The initial selection is made.
If you picked a loser (2/3), changing makes you winner (2/3), staying makes you a loser (2/3).
If you picked a loser (2/3), changing makes you winner (2/3), staying makes you a loser (2/3).
If you picked a winner (1/3), changing makes you a loser (1/3), staying makes you a winner (1/3).

All possibilities covered.

 

[DeletedUser5800]

Now that it is morning and I have had some coffee instead of evening beverages.

Oh well if we are doing that, I actually agreed with @Ashrem the whole time, it all hinges on the odds being higher that your original choice was wrong and it's a fun probability exercise which I was curious and amused to see play out (again).

Schroedinger

was just a weak attempt at being humorous and had no bearing on my actual position.

 

[Ashrem]

It occurred to me while snoozing this morning why it's rough. We all think of the game as starting with the initial choice. The game doesn't actually start until Monty has eliminated one box.

You could conceptualize the game as:
Monty gives the contestant a box that has a 33% chance of a prize in it but they can't look. Then he offers to trade them another box that is always the exact opposite prize status of the contestant's initial box.

Only that wouldn't be as good television.....

 

[DeletedUser7370]

@Ashrem I think you got that backwards. If the game started after Monty eliminated one box then the result is even odds and the question would be very different.

The game has to start with the initial choice of 1 out X, where X>2. Because the question is, "does changing from the initial choice after X-2 non-win boxes are eliminated affect the probability of getting the win item?" The initial choice is critical because it affects how Monty behaves in order to eliminate the box(es), and without some form of initial choice the question becomes something else which is what I think trips so many people up. Losing sight of the correct question certainly tripped me up in some of my replies and analyses.

And as you show in your alternate conceptualization Monty can be the one that makes the initial choice, but he has to peek to find out what his choice resulted in so he can bring out the opposite; meaning Monty's initial choice directs his behavior to supply a second box upon which the player can make the choice of switching or keeping.

 

[Ashrem]

@Ashrem I think you got that backwards. If the game started after Monty eliminated one box then the result is even odds and the question would be very different.

Key to my statement is that he starts by giving you a 66% chance of having a losing box. (All the player's initial choice does is remove that responsibility from him, it doesn't change the game)

Monty always peeks.

 

[SoggyShorts]

The key for me was to ignore Monty's action as completely irrelevant.

Yep. That's why Line 1 only has 1 outcome, not 2. Regardless of which door Monty picks switching still gets you 2 doors. You either get 2&3 or 3&2. It looks like 2 outcomes, but it isn't.

To look at it in the simplest way possible:
You stand in front of a door, and monty asks if you want to open that door or both other doors.

 

[Ashrem]

Burned up 14 five minute instants this morning, extending a scouting tendril to reach a player who regularly posts nice trades. Would have been near $10 worth of diamonds to do it that way, which I would never have done.

 

[DeletedUser5800]

Would have been near $10 worth of diamonds to do it that way, which I would never have done.

I think that's the real point of instants, there is probably a chunk of people whom are happy never spending to speed things up but will develop a new habit of doing so with instants so when they find themselves with none will be more likely to hit the diamond button than they otherwise would have been. I have to admit I'm already guilty of this to an extent. I used all of my time instants (the few I had) and still needed to speed up one 2 day production a bit more so I could run through about 60 quest in the event I finally started yesterday so I threw a handful of diamonds at it I'm sure I wouldn't have otherwise. Having pushed that one button a few times made it easier to push the other.

Granted they were diamonds left over from before I quit buying waaaaay back and I have no intention to restart... but had I started playing six months to a year later than I did and had yet to develop a habit (and broken it) or was yet to have a strong opinion this could have well been the trigger. They have made my hurry up finger a little more twitchy!