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Gialmere
Gialmere
Joined: Nov 26, 2018
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December 10th, 2021 at 7:48:24 AM permalink
Meanwhile, down the rabbit hole in Palindromeland, here is a classic Sam Loyd puzzle I should have posted on 12/02/2021



How many different ways can you read WAS IT A CAT I SAW, i.e., WASITACATISAW, moving in single steps up, down, left or right?
Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
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December 11th, 2021 at 11:05:59 PM permalink
BUZZZZZZZZZZZZZZZZZZZZZZZZ!!


---------------------------------------------------------

$0.00
Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
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December 15th, 2021 at 8:17:26 AM permalink
From Riddler comes yet another take on...



The Monty Hall problem is a classic case of conditional probability. In the original problem, there are three doors, two of which have goats behind them, while the third has a prize. You pick one of the doors, and then Monty (who knows in advance which door has the prize) will always open another door, revealing a goat behind it. Itís then up to you to choose whether to stay with your initial guess or to switch to the remaining door. Your best bet is to switch doors, in which case you will win the prize two-thirds of the time.

Now suppose Monty changes the rules. First, he will randomly pick a number of goats to put behind the doors: zero, one, two or three, each with a 25 percent chance. After the number of goats is chosen, they are assigned to the doors at random, and each door has at most one goat. Any doors that donít have a goat behind them have an identical prize behind them.

At this point, you choose a door. If Monty is able to open another door, revealing a goat, he will do so. But if no other doors have goats behind them, he will tell you that is the case.

It just so happens that when you play, Monty is able to open another door, revealing a goat behind it.

Should you stay with your original selection or switch? And what are your chances of winning the prize?
Have you tried 22 tonight? I said 22.
unJon
unJon 
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Thanks for this post from:
Gialmere
December 15th, 2021 at 9:08:38 AM permalink
Fun problem!


Letís assume you picked door #1 to start.

The fact that Monty could open a door means you arenít in a world with zero goats or a world with one goat that is behind the door #1. That eliminated 33% of the options (25% zero goat and 1/3*25% one goat in door #1).

If you stick with your door, you win 37.5% of the time (if thereís one goat (2/3*25%) or if thereís two goats but behind doors #2 and 3 (1/3*25%). So thatís 25% winners dividing by total of 66.67% of possible worlds = 37.5%.

If you switch you win 50% of the time. (Win all one goat options and two goat options where one goat is behind door #1. That adds to 4/3*25% = 33.33% of the total 66.67%).

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
charliepatrick
charliepatrick
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Thanks for this post from:
Gialmere
December 15th, 2021 at 9:24:12 AM permalink
There are three possible initial conditions about the number of goats (since 0 has been ruled out).
As will be seen their probabilities given the info now known aren't identical!
(a) There's only one goat, it doesn't matter whether you switch or not as the others contain a prize. Chance of winning 1 (or 3/3).
(b) There are two goats, as in regular Monty, you should switch so your chances are 2/3.
(c) There are three goats, therefore you have no chance as there's one behind every door, your chances are 0/3.

Thus regardless of the chances of each you are better off (or identical) by switching.

The initial chances of (a) was 1/3, but one of those has a goat in the box you chose, so the conditional chance is 2/8.
(b) and (c) have chances of 3/8.

Thus the total chance is (2/8*3/3+3/8*2/3+3/8*0/3) = (6+6+0)/24 = 50%.
(Sticking is (6+3+0)/24 = 3/8.)
ThatDonGuy
ThatDonGuy
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December 15th, 2021 at 11:02:49 AM permalink

Since a goat was revealed, then either 1, 2, or 3 doors have goats.
1/3 of the time, one door has a goat; before the reveal, you have a 2/3 chance of winning.
1/3 of the time, two doors have goats; before the reveal, you have a 1/3 chance of winning.
1/3 of the time, all three doors have goats; you have zero chance of winning.
The overall chance of winning if you do not switch = 1/3 x 2/3 + 1/3 x 1/3 + 1/3 x 0 = 1/3.
The chance of winning if you do switch = 1 - 1/3 = 2/3.

Therefore, you should switch; you have a 2/3 chance of winning the prize.

unJon
unJon 
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December 15th, 2021 at 12:19:04 PM permalink
Quote: ThatDonGuy


Since a goat was revealed, then either 1, 2, or 3 doors have goats.
1/3 of the time, one door has a goat; before the reveal, you have a 2/3 chance of winning.
1/3 of the time, two doors have goats; before the reveal, you have a 1/3 chance of winning.
1/3 of the time, all three doors have goats; you have zero chance of winning.
The overall chance of winning if you do not switch = 1/3 x 2/3 + 1/3 x 1/3 + 1/3 x 0 = 1/3.
The chance of winning if you do switch = 1 - 1/3 = 2/3.

Therefore, you should switch; you have a 2/3 chance of winning the prize.


link to original post



You neglected a piece of information.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
ThatDonGuy
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December 15th, 2021 at 12:47:53 PM permalink
Quote: unJon

Quote: ThatDonGuy


Since a goat was revealed, then either 1, 2, or 3 doors have goats.
1/3 of the time, one door has a goat; before the reveal, you have a 2/3 chance of winning.
1/3 of the time, two doors have goats; before the reveal, you have a 1/3 chance of winning.
1/3 of the time, all three doors have goats; you have zero chance of winning.
The overall chance of winning if you do not switch = 1/3 x 2/3 + 1/3 x 1/3 + 1/3 x 0 = 1/3.
The chance of winning if you do switch = 1 - 1/3 = 2/3.

Therefore, you should switch; you have a 2/3 chance of winning the prize.


link to original post



You neglected a piece of information.
link to original post



I assume you are thinking that I "forgot" that if only one door has a goat behind it, then the probability of winning is 100% if I keep my door. Doesn't this cause the same problem as the original Monty Hall problem involving, "Well, there's a 50/50 chance whether or not I switch"?

unJon
unJon 
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December 15th, 2021 at 12:50:30 PM permalink
Quote: ThatDonGuy

Quote: unJon

Quote: ThatDonGuy


Since a goat was revealed, then either 1, 2, or 3 doors have goats.
1/3 of the time, one door has a goat; before the reveal, you have a 2/3 chance of winning.
1/3 of the time, two doors have goats; before the reveal, you have a 1/3 chance of winning.
1/3 of the time, all three doors have goats; you have zero chance of winning.
The overall chance of winning if you do not switch = 1/3 x 2/3 + 1/3 x 1/3 + 1/3 x 0 = 1/3.
The chance of winning if you do switch = 1 - 1/3 = 2/3.

Therefore, you should switch; you have a 2/3 chance of winning the prize.


link to original post



You neglected a piece of information.
link to original post



I assume you are thinking that I "forgot" that if only one door has a goat behind it, then the probability of winning is 100% if I keep my door. Doesn't this cause the same problem as the original Monty Hall problem involving, "Well, there's a 50/50 chance whether or not I switch"?


link to original post



Not that.

There are one goat scenarios that are impossible given our data set. So the conditional probability of 1, 2, and 3 goats is not 1/3, 1/3 and 1/3.
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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GialmereTwelveOr21
December 15th, 2021 at 1:56:22 PM permalink

Let's just say the other doors have a car.

Pr(original door has a car given goat revealed) = Pr(original door has car and goat revealed)/Pr(goat revealed)

Pr(original door has car and goal revealed) = Pr(0 goats)*Pr(goat revealed)*pr(chosen door has car) + Pr(one goat)*Pr(goat revealed)*pr(chosen door has car) + Pr(two goats)*Pr(goat revealed)*pr(chosen door has car) + Pr(3 goats)*Pr(goat revealed)*pr(chosen door has car) =
(1/4)*0*1 + (1/4)*(2/3)*1 + (1/4)*1*(1/3) + (1/4)*1*0 = 2/12 + 1/12 = 3/12 = 1/4.

Pr(goat revealed) = Pr(0 goats)*Pr(goat revealed) + Pr(one goat)*Pr(goat revealed) + Pr(two goats)*Pr(goat revealed)+ Pr(3 goats)*Pr(goat revealed) =
(1/4)*0 + (1/4)*(2/3) + (1/4)*1 + (1/4)*1 = 2/12 + 3/12 + 3/12 = 6/12 = 8/12 = 2/3.

So, Pr(original door has a car given goat revealed)= (1/4) / (2/3) = (1/4)*(3/2) = 3/8.

----

Pr(other door has a car given goat revealed) = Pr(other door has car and goat revealed)/Pr(goat revealed)

Pr(other door has car and goal revealed) = Pr(0 goats)*Pr(goat revealed)*pr(other door has car) + Pr(one goat)*Pr(goat revealed)*pr(other door has car) + Pr(two goats)*Pr(goat revealed)*pr(other door has car) + Pr(3 goats)*Pr(goat revealed)*pr(other door has car) =
(1/4)*0*1 + (1/4)*(2/3)*1 + (1/4)*1*(2/3) + (1/4)*1*0 = 2/12 + 2/12 = 4/12 = 1/3.

The probability of a goat revealed is still 2/3.

Thus, the probability the other door has a car, given a goal is revealed is (1/3)/(2/3) = 1/2.

Thus, by staying the player has a 3/8 chance at a car. By switching it is 1/2. Thus, he should switch.
It's not whether you win or lose; it's whether or not you had a good bet.

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