Getting tired of having Instants shoved down my throat

[Ashrem]

Right. I more correctly should have said that the odds may change rather than increase. Not mathematically, of course, but simply by chance. The odds may increase, decrease or remain the same.

that's a gambler's fallacy. The odds do not change. You have exactly the same chance after 100 failures as you did after 1 success. with a 25% chance, the odds are still 4:1 against you getting the daily building, no matter how many times you've already failed. Odds do not change based on statistics.

 

[DeletedUser7370]

Right. I more correctly should have said that the odds may change rather than increase. Not mathematically, of course, but simply by chance. The odds may increase, decrease or remain the same.

I read a book on statistics and probability in which the author even managed to succumb to this error.

In one chapter he correctly identified that if you have a pea placed in one of 3 boxes with the other 2 empty, then looking in one box and not finding the pea does not change the fact that you have 1 pea and 3 boxes. Assuming you are allowed to open another box without any further randomization step then you would know not to look in the box you checked the first time and that would change the probability to 50/50, but given a randomization you are back to 3 boxes and 1 pea with no other information known.

In a later chapter he went on to try to explain the Monty Hall problem from Let's Make a Deal. This problem like the 3 boxes and 1 pea has 3 doors and 1 prize. Now after you pick a door, Monty reveals that one of the doors that you did not pick does not contain the prize. At this point you have 2 doors and 1 prize, and Monty is asking if you want to stick with your pick or change to the other door. The author of the book incorrectly stated that changing doors changes the probability of getting the prize. When considered objectively it is exactly the same as the 3 boxes and 1 pea with one box known to not contain the pea; the probability is 50/50 for either door and that is regardless of which door you picked that was not revealed.

Another way to think of that problem is if your friend flips a coin and has it on the back of one hand. You then pick heads or tails. Your friend peeks without showing you and then asks if you want to change your pick. The coin is still only heads or tails and you know nothing knew so the probability of picking correctly is still 50/50.

The funny part is the author was a college professor of mathematics and in chapter 1 explained how poorly everyone does at understanding probabilities, odds, and gambling type things.

 

[Ashrem]

The author of the book incorrectly stated that changing doors changes the probability of getting the prize.

After a 10,000 people have chosen one with a 50% odds of having a prize, about 5,000 of them should have a prize. After 10,000 people have chosen the one out of three (and stayed with it), only about 3333 will have picked one with a prize. Their odds appear not to be different, but the probability that they will find themselves in the class of "people who got a good prize" actually is higher.

 

[DeletedUser7370]

@Ashrem reread the problem. It entirely hinges on Monty revealing that 1 of the 2 doors not picked does not contain the prize. That leaves the player with a 50/50 situation. Even if there was a 'good' prize and a 'not so good' prize the player will still have a 50/50 choice. It does not matter what they chose the first time and the probability does not change.

 

[Ashrem]

@Ashrem reread the problem. It entirely hinges on Monty revealing that 1 of the 2 doors not picked does not contain the prize. That leaves the player with a 50/50 situation. Even if there was a 'good' prize and a 'not so good' prize the player will still have a 50/50 choice. It does not matter what they chose the first time and the probability does not change.

I'm fully aware of the detail of the exercise. You are likely making the most common error of looking at the event as a singularity.

If 10,000 people pick from A, B, and C, where there is only a prize in C, ~3,333 are going to end up with the prize. At that point in time, If you eliminate either A or B (depending what the person selected), what you are actually doing is selecting ~3,333 people who have a prize, and ~6,666 people who don't have a prize. (it kind of sounds impossible until you consider it is being done one person at a time). So when he eliminates one wrong choice, at that moment 2/3 people have the other wrong choice. If everyone of the 10,000 picks the other door, 2/3 are picking right and 1/3 is picking wrong.

The individual's odds at that moment are the same, but the aggregate answer is that there's a 66% chance of improving your position.

 

[DeletedUser7370]

You are likely making the most common error of looking at the event as a singularity.

Roll 2 dice having 6 sides calling a number [2..12] for the roll. Now roll the same 2 dice a second time also calling a number [2..12] for the second the roll. Does your accuracy in the first roll affect the probability of your second roll? NO!

A person picks a number from [1..3]. They are then asked to pick from [1..2]. Does their first pick affect the probability of there second pick? Again NO!

where there is only a prize in C

This would create a non-random system in which probability no longer applies! The only way the probability is affected by Monty offering information and a chance to change the pick is if Monty has a bias to making that offer when a player has chosen a certain way. In order to understand that you might want to look at the history of WWII when bingo games were used to randomly select the encoding of messages and when the callers learned that they accidentally the patterning of the codes allowing them to be broken.

Do I have to post an entire probability matrix to demonstrate this or are you capable of building your own?

Another way:
You flip a coin once you have 50/50.
You flip a coin 1000 times you have many results, but each of them can only be 50/50. The 1000th flip does not care about anything from the previous 999. Each flip or question is a singularity and it is an error to not view each as such.

 

[Ashrem]

This would create a non-random system in which probability no longer applies!

That's what Monty does, by removing a known-bad choice from the mix. There is nothing random about his choice. He is deliberately changing the odds in your favor.

 

[DeletedUser7370]

That's what Monty does, by removing a known-bad choice from the mix. There is nothing random about his choice. He is deliberately changing the odds in your favor.

Facepalm!

And I edited my previous post.

 

[Ashrem]

Roll 2 dice having 6 sides calling a number [2..12] for the roll. Now roll the same 2 dice a second time also calling a number [2..12] for the second the roll. Does your accuracy in the first roll affect the probability of your second roll? NO!

A person picks a number from [1..3]. They are then asked to pick from [1..2]. Does their first pick affect the probability of there second pick? Again NO!

Except none of those things are what you are doing. Monty flips three coins. he says, "1 is heads, two are tails, without looking, tell me which of the three coins is heads." After you pick, he takes away one of the tails, and says "Pick which coin is heads." There is still a 66% chance that the coin you picked is tails, because your selection event hasn't been altered. You are still selecting from the first three, and there is now a 66% chance the coin you didn't select is heads, because you now get to pick two coins, the one that is unpicked, and the one he removed are both in the pool of winning choices for you.

He's not flipping again, and he's not rolling again. There is no new random event being generated. The result is no longer truly random.

 

[Ashrem]

@TedGrau I'll suggest a matrix for you, with the "player" trying to pick the highest number
generate a large number of sets of three, with your hypothetical participant always selecting the third option. (to represent x people who picked door#3)
Now remove whichever of 1 or 2 is lower

I can tell you right now, that at the end, 2/3 of the higher numbers will be in column 1 or column 2. If your user had given up "door" 3 for the one that was left, they'd win 2/3 times.

 

[DeletedUser7370]

tell me which of the three coins is heads.

Pick.

After you pick

Confirming my assessment that was a pick.

he takes away one of the tails

Altering selection event.

because your selection event hasn't been altered.

Begin separate and distinct pick based on new parameters.

says "Pick which coin is heads."

50/50 chance. First pick is completely irrelevant.

You are still selecting from the first three

And this is Vegas makes billions DAILY.

 

[Ashrem]

Please look at my matrix suggestion above.

In my simulation of 5000 people picking door three, 1616 people got the winning number by sticking with door three while 3384 got the highest number by switching to whichever of door 1 or 2 was left after the lowest was removed.

 

[SoggyShorts]

https://betterexplained.com/articles/understanding-the-monty-hall-problem/

I think the best part of the explanation is the 100 door problem.
Imagine you pick 1 door from 100
Then Monty examines the other 99 doors and eliminates 98 of them
Would you rather keep your door (1 in 100 chance)
Or the other door that is the best door from 99?

 

[satchmo33]

I think this is part of the problem- we tend to only notice how our own luck is going when it sucks. Even though your personal data of 400 rolls getting you 12% instead of 25% is fairly comprehensive, the player who got 40% wins isn't going to come here to post their stats.

My own conspiracy theory is that the nuts we collect around our city has gone down. I recall(perhaps falsely) getting a 3x almost every day, and this time across 2 live and 1 Beta account I'm at 8 total.


https://www.gamersgemsofknowledge.com/eldrasils-ascending-daily-prizes

Thanks, Soggy.

 

[DeletedUser7370]

https://betterexplained.com/articles/understanding-the-monty-hall-problem/

I think the best part of the explanation is the 100 door problem.
Imagine you pick 1 door from 100
Then Monty examines the other 99 doors and eliminates 98 of them
Would you rather keep your door (1 in 100 chance)
Or the other door that is the best door from 99?

And again equally wrong.
Pick from 1 to 100 doors...I chose 7.
Doors 1..6, 9..82, and 84..100 eliminated by Monty as bad. Thank you Monty.
Pick from door 7 or 83.
What do I actually know about what is behind either of those doors? Only that one has a prize and the other does not.

Everything else is fallacy and assumption.

Anyhow here is a full matrix for the 3 door problem:
https://docs.google.com/spreadsheets/d/189EqRujPgVva1fOE-z6cmioOaslYUG8qAVfhd33hs7k/edit?usp=sharing

On the second sheet I have removed all the cases where Monty stole the winner from the player on the repicks. Note that the probability is 50%. Now you can all enjoy a Violent Femmes song.

 

[SoggyShorts]

door 1 (your pick) has 33% chance
door 2+3 have 66% chance

By eliminating a bad door monty is letting you pick BOTH doors 2&3 instead of door 1.
Did you try the simulator I linked?

 

[DeletedUser7370]

door 1 (your pick) has 33% chance
door 2+3 have 66% chance

By eliminating a bad door monty is letting you pick BOTH doors 2&3 instead of door 1.
Did you try the simulator I linked?

View attachment 2562

Monty is not allowed to eliminate winning results. Did you look at the full matrix and the second sheet showing what happens when we clean up for that GIANT silent restriction in the game?

Monty never showed that door 1 was the winner and then let them pick again. Monty can't eliminate the winner because he has to show what is behind the door. Check your graphic as line 1 requires eliminating the winner.

 

[SoggyShorts]

Monty never showed that door 1 was the winner and then let them pick again. Monty can't eliminate the winner because he has to show what is behind the door. Check your graphic as line 1 requires eliminating the winner.

No it doesn't.
In line 1 where the car is behind door 1 Monty eliminates 2 or 3 (it doesn't matter which)

Here's an even better sim:
http://www.mathwarehouse.com/monty-hall-simulation-online/

Spoiler: My results with 1,000 simulations of each

 

[DeletedUser7370]

No it doesn't.
In line 1 where the car is behind door 1 Monty eliminates 2 or 3 (it doesn't matter which)

Person picked 1. Car was behind 1. Person won.

See my full matrix. The problem is always that Monty can't eliminate a win. Eliminating or even questioning a win in any case, even a repick, would result in massive lawsuits. So in your line 1, person picked 1 and won; no repicks are possible.


You lose.

 

[SoggyShorts]

Eliminating or even questioning a win in any case, even a repick, would result in massive lawsuits.

For 23 years Monty hall did this, and questioned your potential win every single episode.

1. You select a door without opening it.
2. Monty selects a different door that has a goat and shows you the goat.
3. You decide if you want to open the door you picked originally or if you want to switch.

You pick door 1 (pick, not open)
Monty opens door #2(or 3) and shows you a goat, and asks if you want to change your mind.
That is line 1 of the matrix, and the only one where sticking wins.
In both of the other possibilities (car behind 2 or 3) switching makes you win. 1 vs 2

There's the whole matrix of all possible choices. I'm not sure where you went wrong with yours, but every single article I can find via google agrees with the above.