mwalz9
mwalz9
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December 3rd, 2021 at 11:44:08 AM permalink
I don't care if there are 3 doors, 100 doors or 1,000,000 doors.

If the host opens all but 2, my odds were still the same if I had selected the one I selected or if I had selected the one he is trying to make me switch to!
unJon
unJon
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December 3rd, 2021 at 11:55:29 AM permalink
Quote: mwalz9

I don't care if there are 3 doors, 100 doors or 1,000,000 doors.

If the host opens all but 2, my odds were still the same if I had selected the one I selected or if I had selected the one he is trying to make me switch to!
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No. You are assuming the host is randomly opening doors. Heís not. He knows what door has the prize and is intentionally opening all the other doors.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
unJon
unJon
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December 3rd, 2021 at 11:59:37 AM permalink
Or think about it this way.

You pick door 1. The host asks if you want to keep door one and switch to both doors 2 and 3. Would you rather have just one door or two?

Thatís the Month Hall problem. The fact that the host opens 1 or 98 or 999,998 doors that he already knows weíre empty doesnít change the offer.

Switching letís you win EVERY time unless you picked the right door originally. (Reread the last sentence. Itís the key. Switching makes you win every single time, unless you got lucky and picked the winner originally.)

Since with three doors you pick right 1/3 of the time, switching letís you win 1 - 1/3 = 2/3.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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Wizard 
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December 3rd, 2021 at 12:22:14 PM permalink
Quote: mwalz9

If the host opens all but 2, my odds were still the same if I had selected the one I selected or if I had selected the one he is trying to make me switch to!
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You are wrong. As to why, I have an entire article on the Monty Hall Problem.
It's not whether you win or lose; it's whether or not you had a good bet.
billryan
billryan 
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December 3rd, 2021 at 12:26:32 PM permalink
I will accept I am wrong, but would love to have it explained.

I pick Door #1. Monty opens Door #2 and it is a goat. I can switch doors if I like. I introduce a new twist. I will flip a coin. There are two doors so each is a 50-50 prop.
heads means keep my door, tails means switch. The coin flip is 50-50. So if I let the coin decide, it is 50-50, but if switch on my own, it's 66%?
The difference between fiction and reality is that fiction is supposed to make sense.
ThatDonGuy
ThatDonGuy
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December 3rd, 2021 at 12:37:10 PM permalink
Quote: billryan

I will accept I am wrong, but would love to have it explained.

I pick Door #1. Monty opens Door #2 and it is a goat. I can switch doors if I like. I introduce a new twist. I will flip a coin. There are two doors so each is a 50-50 prop.
heads means keep my door, tails means switch. The coin flip is 50-50. So if I let the coin decide, it is 50-50, but if switch on my own, it's 66%?
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Yes. Look at it this way: suppose you are playing a game where somebody rolls a 6-sided die, keeps the number hidden, and you have to decide whether or not the number is at least 3. If you toss a coin, you have a 50% chance of being right; if you choose "yes, the number is at least 3," you have a 2/3 chance of being right.

Going back to the original problem:
There is a 1/3 chance you chose the winning door; Monty will always open a door with a goat behind it.
There is a 2/3 chance you chose a door with a goat; Monty will always open the other door with a goat behind it.
In every case, Monty opens a door and shows you a goat.
If you switch, then, if you originally chose the winning door, you now have a goat, and if you originally chose a goat, you now have a winning door. The probability of originally choosing a door with a goat was 2/3, so that is the probability of you winning if you always switch.
Wizard
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Wizard 
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December 3rd, 2021 at 4:09:42 PM permalink
Here is a Bayesian solution.

Assume the player picks his door randomly and if the player picks the car, the host will pick another door to open randomly.

Then, the player picks door B and the host opens door C. What is the probability the car is behind door A?

Pr(Car in A given Player picks B and host reveals goat in C) =
Pr(Car in A and Player picks B and host reveals goat in C) / Pr(Player picks B and host reveals goat in C) =
Pr(Player picks B and Car in A and host reveals goat in C) / [Pr(Player picks B and car in A and host reveals goat in C) + Pr(Player picks B and car in B and host reveals goat in C)]=
(1/3)*(1/3) / [(1/3)*(1/3) + (1/3)*(1/3)*(1/2)] =
(1/3) / [(1/3) + (1/3)*(1/2)] =
(1/3) / (3/6) =
(2/6) / (3/6) =
2/3

In, fact let's do it again but without the part about the player randomizing his choice.

Pr(Car in A given host reveals goat in C) =
Pr(Car in A and host reveals goat in C) / Pr(Host reveals goat in C) =
Pr(Car in A and host reveals goat in C) / [Pr(car in A and host reveals goat in C) + Pr(car in B and host reveals goat in C)]=
(1/3) / [(1/3) + (1/3)*(1/2)] =
(1/3) / (3/6) =
(2/6) / (3/6) =
2/3
It's not whether you win or lose; it's whether or not you had a good bet.
gordonm888
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gordonm888
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Thanks for this post from:
OnceDear
December 3rd, 2021 at 5:08:29 PM permalink
Let's try this as an explanation:

Before you pick a door you are told this: "After you pick a door, Monty Hall will knowingly choose one of the other two doors and reveal that the prize was not behind that door. So go ahead and choose a door."

When you go ahead and choose a door, will your odds be 1 in 3 of picking the correct door, or will they be 1 in 2?

I hope you said "1 in 3." Now when Monty Hall does exactly what you were told he would do, the odds that you originally picked the correct door are still 1 in 3. Which means that the odds of the only remaining door being the correct door must be whatever is left, namely 2 in 3.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
Wizard
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Wizard 
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December 3rd, 2021 at 5:35:55 PM permalink
Quote: gordonm888

Let's try this as an explanation:

Before you pick a door you are told this: "After you pick a door, Monty Hall will knowingly choose one of the other two doors and reveal that the prize was not behind that door. So go ahead and choose a door."

When you go ahead and choose a door, will your odds be 1 in 3 of picking the correct door, or will they be 1 in 2?

I hope you said "1 in 3." Now when Monty Hall does exactly what you were told he would do, the odds that you originally picked the correct door are still 1 in 3. Which means that the odds of the only remaining door being the correct door must be whatever is left, namely 2 in 3.
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I like that.
It's not whether you win or lose; it's whether or not you had a good bet.
Gialmere
Gialmere
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December 3rd, 2021 at 6:31:41 PM permalink
I'll give this a shot...

I think it helps to label the goats A and B. As mentioned above, all you know for sure is that after you select a door Monty will open a different door and show you a goat. There are three scenarios...

1) You pick goat A and get shown goat B. You switch and win the car.
2) You pick goat B and get shown goat A. You switch and win the car.
3) You pick the car and get shown either goat. You switch and lose the car.

So 2 of the 3 switch scenarios win. But what if you don't switch?

1) You pick goat A and get shown goat B. You stay and lose the car.
2) You pick goat B and get shown goat A. You stay and lose the car.
3) You pick the car and get shown either goat. You stay and win the car.

So 2 of the 3 stay scenarios lose.
Have you tried 22 tonight? I said 22.

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