1. ## Re: Do you know the correct answer?

Originally Posted by peb
The table in post 6 is NOT incomplete. The first two columns give all 9 possibilities. 3 spots where the car can be, 3 guesses. 3 times 3 equals 9. There are 9 rows, each row shows 8ne possible game scenario.

It us a complete mathematical proof.
Yes it is incomplete. It groups two possibilities as single. And it does this three times.

To not count them because the result is the same is like saying it doesn’t matter if you remove one or four empty chambers from the Russian roulette revolver.
Last edited by Decourcy; 06-12-2022 at 06:14 PM.

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## Re: Do you know the correct answer?

Originally Posted by Decourcy
Let’s play a different game. Russian roulette. One bullet in 6 chambers. 1 in 6 chance yes? One squeeze and a “click”. Don’t rotate the cylinder. Remaining 5 are now 1in 5 yes? Squeeze again “click”. Remaining are now one in 4. Same same.

The point being that of course probabilities change as options reduce

The loaded chamber was always the loaded chamber but the probability that it will be the next chamber increases until, if it’s the last chamber, it’s 6/6 or 100%

But also, if you intend to pull the trigger 6 times without spinning the cylinder, the the odd are 6 in 6 (100%) that you will shoot yourself by the time you are you reach the 6th pull if not sooner. Your example looks at the probability for a single pull while counting how many many pulls have already beet taken.

Going back to Monty. The contestant makes one choice and has a 33% chance Monty had two doors - 66% chance. The contestant does not open the door until Monty has opened both of his doors. That means there is a 66% chance that one of his doors is the car. If he does not trade either door with the contestant, then he still has 66% chance. His chances haven't changed, because he has not changed either of the two doors in his possession. But, his odds don't change after he has opened the first door assuming it is a goat because we are saying it is a 66% chance that one the doors has a car. It could be either one and ther ren=mains a 33% chance that neither door has the car: i.e. the contestant is the winner.

3. ## Re: Do you know the correct answer?

Originally Posted by Tom Montgomery
.

Are you saying that you DO accept his first, formal, explanation of the probabilities?
Nope, it has the same result as the Informal one, just less clarity in amongst the waffle. If you switch, you have one chance of being correct 1/3. You can only switch once, not twice. Remember, one choice, with its 1/3 chance, has been taken out of the equation.

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## Re: Do you know the correct answer?

Originally Posted by Peerie Maa
Nope, it has the same result as the Informal one, just less clarity in amongst the waffle. If you switch, you have one chance of being correct 1/3. You can only switch once, not twice. Remember, one choice, with its 1/3 chance, has been taken out of the equation.
I just don't see this being settled here. The answer is as clear to me as yours is to you. I think I'm going to call it a day. Have a good night all!

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## Re: Do you know the correct answer?

.
I just played the game 20 more times using the "always switch" strategy. The contestant scored 14 wins and 6 losses. A winning percentage of 70%.

I have now played the game 80 times with the contestant scoring 53 wins and 27 losses for a winning percentage of 66.25%.

6. ## Re: Do you know the correct answer?

you didnt have to play it out
you could use maths
Originally Posted by Tom Montgomery
.
I just played the game 20 more times using the "always switch" strategy. The contestant scored 14 wins and 6 losses. A winning percentage of 70%.

I have now played the game 80 times with the contestant scoring 53 wins and 27 losses for a winning percentage of 66.25%.

7. ## Re: Do you know the correct answer?

​...

8. peb
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## Re: Do you know the correct answer?

Originally Posted by Decourcy
Yes it is incomplete. It groups two possibilities as single. And it does this three times.

To not count them because the result is the same is like saying it doesn’t matter if you remove one or four empty chambers from the Russian roulette revolver.
I have ignored all Russian Roulette posts, sorry, that's my quirk. When I was in high school the boyfriend of a girl in my class killed himself by actually playing Russian roulette when drunk one night. I can't hardly stand to read the term in a small town, that caused a huge amount of trama.

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## Re: Do you know the correct answer?

Originally Posted by Paul Pless
you didnt have to play it out
you could use maths
Yeah. But I rather enjoyed playing.

I just completed 100 plays of the game employing the "always switch" strategy.

The first twenty plays resulted in 10 contestant wins and 10 losses for a winning percentage of 50%.
The second twenty plays resulted in 15 contestant wins and 5 losses for a winning percentage of 75%.
The third twenty plays resulted in 14 contestant wins and 6 losses for a winning percentage of 70%.
The fourth twenty plays resulted in 14 contestant wins and 6 losses for a winning percentage of 70%.
The fifth twenty plays resulted in 15 contestant wins and 5 losses for a winning percentage of 75%.

The final total was 100 plays resulting in 68 contestant wins and 32 losses for a winning percentage of 68%.

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## Re: Do you know the correct answer?

Originally Posted by Tom Montgomery
.
So I just played the game 20 more times following your requirement. I turned over the card to my left (the contestant card) every time. The I discarded a Joker from the two cards to my right. And then always making the switch.
I may have misunderstood your earlier post. If you find the ace in the two cards held by Monty, then I will agree that's a win for switch.

I just played it 45 times and it came out closer to 50:50, but in the process I realized that it's actually difficult to shuffle 3 cards well. I started seeing patterns in the outcome that I'm not sure are random.

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## Re: Do you know the correct answer?

Originally Posted by bamamick
I like goats.

Mickey Lake
Me too and I already have a car.

12. ## Re: Do you know the correct answer?

Suppose you had only two choices to begin with. You'd have a 50/50 chance. Now a third choice is added before you choose. Now you've a 1 in 3 chance, no?

Once the 3rd choice is eliminated, it's no longer a factor as to odds. The host may have some sort of established pattern. If the prize is behind the even numbered curtain, he eliminates one of the odd numbers, and if that's known by the contestant, then the contestant is more likely to choose the even numbered curtain. That would show up in long term results.

13. ## Re: Do you know the correct answer?

Originally Posted by John Smith
Suppose you had only two choices to begin with. You'd have a 50/50 chance. Now a third choice is added before you choose. Now you've a 1 in 3 chance, no?

Once the 3rd choice is eliminated, it's no longer a factor as to odds. The host may have some sort of established pattern. If the prize is behind the even numbered curtain, he eliminates one of the odd numbers, and if that's known by the contestant, then the contestant is more likely to choose the even numbered curtain. That would show up in long term results.
The host does have a pattern. He always eliminates one of the goats. The contestant knows that

14. J P
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## Re: Do you know the correct answer?

I’m seeing the ‘problem’ as a paradox of time, with both arguments being correct.

Looked at as a singular event/game sample, the odds at the second choice – to switch, or not – are 50-50. However, with increased frequency of the events/samples, over time, say, playing 100 games, it would seem to be logical to choose the switch, because statistically the first choice would have a 66% probability of being wrong. Therefore, the ‘correct answer’ is to switch if playing multiple games, but in a singular instance it’s 50-50.

But, there is randomness, intuitiveness, woo-woo, in the first selection – how do you put numbers to that?

15. ## Re: Do you know the correct answer?

Originally Posted by J P
I’m seeing the ‘problem’ as a paradox of time, with both arguments being correct.

Looked at as a singular event/game sample, the odds at the second choice – to switch, or not – are 50-50. However, with increased frequency of the events/samples, over time, say, playing 100 games, it would seem to be logical to choose the switch, because statistically the first choice would have a 66% probability of being wrong. Therefore, the ‘correct answer’ is to switch if playing multiple games, but in a singular instance it’s 50-50.

But, there is randomness, intuitiveness, woo-woo, in the first selection – how do you put numbers to that?
Prolly not.

When I shared the puzzle with the mob at dinner, Son1 said - quickly and with no hesitation - 'switch, every time'. He said he visualized the 2/3 chance being compressed into the 'switch' door in terms of quantum physics. He tried to explain it... but it was a leap too far, for me. I suspect he was wrong also. Or correct for the wrong reasons.

16. ## Re: Do you know the correct answer?

Wikipedia has a good page on the problem. https://en.wikipedia.org/wiki/Monty_Hall_problem

When you first choose a door, you have a 2 in 3 chance of being wrong. Even after seeing in the second door, you still have the 2 in 3 chance of being wrong. Opening the door doesn't improve your chances if you stick with your first choice..

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## Re: Do you know the correct answer?

.
Twelve pages over the course of 32 hours!

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## Re: Do you know the correct answer?

Originally Posted by peb
If you play my the rule to always switch
You only lose if you picked the right door. That's a 33% chance of losing. It is very, very simple.
I thought this was a very compelling argument made by peb way back at post #123. It seems to have gone unnoticed. I would sure like to see one of the 50/50 guys comment on it.

19. ## Re: Do you know the correct answer?

About 10 or 12 pages in the middle, i didn't bother to read.

20. ## Re: Do you know the correct answer?

But some seem to still not believe that.

21. ## Re: Do you know the correct answer?

Originally Posted by Tom Montgomery
.
Twelve pages over the course of 32 hours!

“I returned, and saw under the sun, that the race is not to the swift, not the battle to the strong, neither yet bread to the wise, nor yet riches to men of understanding, nor yet favour to men of skill;

^^^^^^^ Maybe not, but it’s still the way to bet…

Jeff C

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## Re: Do you know the correct answer?

Originally Posted by Jimmy W
Wikipedia has a good page on the problem. https://en.wikipedia.org/wiki/Monty_Hall_problem

When you first choose a door, you have a 2 in 3 chance of being wrong. Even after seeing in the second door, you still have the 2 in 3 chance of being wrong. Opening the door doesn't improve your chances if you stick with your first choice..
It is the simplist explanation.

23. ## Re: Do you know the correct answer?

This is amazing to me, that people don't understand this.

Here's a simulator I wrote in the Go programming language: https://goplay.tools/snippet/l-0dNVMAhlq

It's in the [a] Go Playground, so you can compile the code and run it. Currently set up to do runs of 1,000,000 plays of the game.

The rules of the game are

- There are 3 doors. One of them a valuable prize behind it (a car!). The other two has worthless prizes (goats!).
- The contestant selects a door. It remains closed.
- Monty Hall, the preceptor of the game show, opens one of the two remaining doors. It cannot be the door containing the valuable prize, as that would ruin/end the game. It's always a worthless prize.
- The contestant is offered the chance to switch to the remaining door, and can either switch, or hold.

The question is, which strategy is more effective? Switch to the last door?, or holding on to your initial choice?

When the contestant first makes their selection, they have a 1/3 chance (1:2 odds) of selecting the door with the car behind it, and a 2/3 chance (2:1 odds) of selecting a door with a booby prize behind it.

After Monty Hall has opened the door, and revealed the booby prize behind it, he has taken 1 booby prize out of contention, because his pick was not random.

That makes the remaining door coalesce into a 1/2 chance (even odds) that it is the valuable prize, because it's either behind that door, or it's behind the door selected by the contestant. And the door selected by the contestant is still sitting there with a a 1/3 chance of it being the valuable prize.

It's not so much a probability & stats problem as it is

Here's output from a sample run:

Code:
```Strategy: Switching
+---------+-----------+---------+
| OUTCOME | FREQUENCY | PERCENT |
+---------+-----------+---------+
| Win     |    666164 |  66.616 |
| Lose    |    333836 |  33.384 |
+---------+-----------+---------+
|             1000000 | 100.000 |
+---------+-----------+---------+

Strategy: Holding
+---------+-----------+---------+
| OUTCOME | FREQUENCY | PERCENT |
+---------+-----------+---------+
| Win     |    332717 |  33.272 |
| Lose    |    667283 |  66.728 |
+---------+-----------+---------+
|             1000000 | 100.000 |
+---------+-----------+---------+```
This is the guts of the thing:
Code:
```func main() {

// Our sample size is 1,000,000 plays of the game
const sample = 1_000_000

// for each possible strategy (switching or holding)
for _, strategy := range []string{SWITCHING, HOLDING} {
// zero the counters
win, lose, total := 0, 0, 0

// play the game, over and over...
for n := 0; n < sample; n++ {

// play the game
didWin := play(strategy)

// record the outcome as either a win or a loss
if didWin {
win++
} else {
lose++
}

// record the total number of games played (wins+losses)
total++

}

// report the outcome. This just formats it as a nice table.

tbl := tablewriter.NewWriter(os.Stdout)
tbl.SetAutoMergeCells(true)
tbl.SetColumnAlignment([]int{tablewriter.ALIGN_LEFT, tablewriter.ALIGN_RIGHT, tablewriter.ALIGN_RIGHT})
tbl.SetFooterAlignment(tablewriter.ALIGN_RIGHT)
tbl.Append(row("Win", win, total))
tbl.Append(row("Lose", lose, total))
tbl.SetFooter(row("", total, total))

// render the table with a nice header to tell us which strategy was in use.
fmt.Println()
fmt.Printf("Strategy: %s\n", strategy)
tbl.Render()

}

return
}

func play(strategy string) (didWin bool) {

// populate the doors. The car is placed behind door at random and
// every other door has a goat behind it.

doors := []string{GOAT, GOAT, GOAT}
doors[randomDoor()] = CAR

// The contestant selects a door at random from the 3 available doors.

contestantSelection := randomDoor()

// Monty Hall opens a door that is neither the winning door
// or the door selected by the contestant.
//
// We don't really care what door he opened, as that is guaranteed
// not to be of any interest (he's not going to open the winning door),
// but we do care about the single door that is neither selected by the
// contestant or opened by Monty Hall.

_, remaining := montyHallOpensDoor(doors, contestantSelection)

// if our strategy is switching, we switch to the remaining door,
// otherwise we stand pat.

if strategy == SWITCHING {
contestantSelection = remaining
}

// The contestant wins if his selected door has the car behind it.

didWin = doors[contestantSelection] == CAR

// return the win/lose indicator to the caller
return didWin
}```

24. This diagram show why this is true.

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## Re: Do you know the correct answer?

From a link to a Reddit page in the link that Tom first posted - it sort of flips the problem on it head, and made more sense of it to me than all the charts and tables .

Quote:
To me, the clearest explanation is that the only way to get it wrong by switching is to have picked the correct door in the first place. The odds of picking the correct door first are 1 in 3.

Pete

26. ## Re: Do you know the correct answer?

I have laid it out in the form of a white board, as in Toms video above.
The error that the statistician with his sharpie made is that he did not model all the facts in his working. He ignored that we know that Monty's door ALWAYS has O probability of revealing the car. The probability that the car is behind the other two doors is equal, we simply don't know which door.
This is how it should be worked, without the waffle.

White board.jpg

I hope that is clear enough.
Last edited by Peerie Maa; 06-13-2022 at 04:21 AM.

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## Re: Do you know the correct answer?

I will change my answer to 'switch'.
Contestant's chances to win are 1 in 3
Monty's chances are 2 in 3.
If you switch,you get Monty's chances ,minus the door that he opened.
Not unlike Monty allowing you to pick 2 doors.
R

28. ## Re: Do you know the correct answer?

Originally Posted by Ron Williamson
I will change my answer to 'switch'.
Contestant's chances to win are 1 in 3
Monty's chances are 2 in 3.
If you switch,you get Monty's chances ,minus the door that he opened.
R
Nope that suggests that Monty's decision affects the location of the car, which is impossible.
Take out Monty's 1:3 and set it to zero, and you are left with 1:3 and 1:3, no more, no less.

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## Re: Do you know the correct answer?

Nick. How does Monty's removal of on of the doors that you did not pick retroactively change the chance of you choosing the right door first, from 1/3 to 1/2 ?

30. ## Re: Do you know the correct answer?

I'll admit not having read through all 14 pages, so sorry if I'm just repeating an already stated point (which I probably do)

Originally Posted by CWSmith
Okay, Nick, I'll see if this old dog can learn a new trick:

 Contestant Picks this Door Prize is Behind This Door Monty Shows This Door Stays Switches 1 1 2 Win Lose 1 1 3 Win Lose 1 2 3 Lose Win 1 3 2 Lose Win cut for brevity Win % 0.5 0.5

Okay, Tom, I told you what I thought was wrong with your chart. Now I'd like you to tell me what is wrong with mine.
What is wrong is the assumption that all options are equally plausible. Corrected version:

 Contestant Picks this Door Prize is Behind This Door Monty Shows This Door Scenario probability Stays Switches Weight 1 1 2 1/6th Win Lose .5 1 1 3 1/6th Win Lose .5 1 2 3 1/3rd Lose Win 1 1 3 2 1/3rd Lose Win 1 cut for brevity other scenarios analytically redundant Weighted win ratio 2/3rds 1/3rd 3

31. ## Re: Do you know the correct answer?

There are lots of ways to fiddle the math to fit the desired outcome.
But you need to justify why you chose those numbers. Why is the probability of any pick a variable. The probability of picking 1 from 3 is always the same.

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## Re: Do you know the correct answer?

So if your original pick remains at 1/3, the probability of the prize being one of the other two choices must be 2/3 and you now know not to pick one of those.

If, as you assert, that it was not an advantage to change now the probability of either staying or changing must be 1/2. Which means that the probability of you originally making the correct choice must have changed, which is obviously nonsense

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## Re: Do you know the correct answer?

I was in the 50/50 camp when I first saw the problem but now would switch.

Putting aside all the trees and statistics, the nub of the issue is what do you know after Monty opens the door that you did not know before. Answer nothing, Monty will always have the option to open a door with a goat therefore nothing has changed.

There are 2 sets, my first choice 33% the other 2 doors 66%. Monty selecting a goat tells us nothing about what is in the second set but there is still a 66% chance that the second set had the car.

Were Monty to open an other door at random and picked a goat then it would be 50/50.

Apologies to all posters who have already made the point, I do not have time to read all 470 replies.

34. ## Re: Do you know the correct answer?

Originally Posted by Peerie Maa
There are lots of ways to fiddle the math to fit the desired outcome.
But you need to justify why you chose those numbers. Why is the probability of any pick a variable. The probability of picking 1 from 3 is always the same.
Good point, I should've explicitly written that:
Only when a contestant picks correct gate, which will happen in 1 of every 3 games, Monty has to choose a goat to be revealed. Simply put, the scenarios 1-1-2 and 1-1-3 have to "share" the same 1/3rd chance of happening. They are, naturally, equivalent, meaning 50% chance of happening *within* the initial 1/3rd thus being half as likely as the other two scenarios where Monty has no choice in which goat to reveal.
If we're listing out every scenario in the table, the 1-1-2 and 1-1-3 have to be valued as input at half of what 1-2-3 and 1-3-2 are. Doing otherwise introduces bias, which coincidentally matches the false '50-50 on switching' solution.

Originally Posted by oldcodger
Were Monty to open an other door at random
This is the key: he's not. There is 0% chance he'll show you the car. This is the introduced bias.

35. ## Re: Do you know the correct answer?

Originally Posted by WszystekPoTrochu
Good point, I should've explicitly written that:
Only when a contestant picks correct gate, which will happen in 1 of every 3 games, Monty has to choose a goat to be revealed. Simply put, the scenarios 1-1-2 and 1-1-3 have to "share" the same 1/3rd chance of happening. They are, naturally, equivalent, meaning 50% chance of happening *within* the initial 1/3rd thus being half as likely as the other two scenarios where Monty has no choice in which goat to reveal.
If we're listing out every scenario in the table, the 1-1-2 and 1-1-3 have to be valued as input at half of what 1-2-3 and 1-3-2 are. Doing otherwise introduces bias, which coincidentally matches the false '50-50 on switching' solution.

This is the key: he's not. There is 0% chance he'll show you the car. This is the introduced bias.
But but but. . . Monty always has to open a door with a goat, whatever the contestant picks. So the probabilities are the same whichever the contestant picked, car or goat.

If we're listing out every scenario in the table, the 1-1-2 and 1-1-3 have to be valued as input at half of what 1-2-3 and 1-3-2 are. Doing otherwise introduces bias, which coincidentally matches the false '50-50 on switching' solution.
Which is what I meant, choosing numbers to agree with a desired outcome.

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