1 Name: Anonymous 2021-04-14 15:26

Quoted by: >>2,4,8

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## The solution?

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1 Name: Anonymous 2021-04-14 15:26

Quoted by: >>2,4,8

2 Name: Anonymous 2021-04-14 15:31

>>1

Step 1:Open random door

Step 2:If money, Congrats! If Goat Continue with instructions

Step 3:Sacrifice goat to Baphomet

Step 4:???

Step 5:Profit

Step 1:Open random door

Step 2:If money, Congrats! If Goat Continue with instructions

Step 3:Sacrifice goat to Baphomet

Step 4:???

Step 5:Profit

3 Name: Anonymous 2021-04-14 15:35

came to /prog/ to see if its really full of retards

rumours are true

rumours are true

4 Name: Anonymous 2021-04-14 15:37

Quoted by: >>10

>>1

Monty Hall problem only works when there are 3 or more doors.

Monty Hall problem only works when there are 3 or more doors.

5 Name: Anonymous 2021-04-14 15:41

Quoted by: >>6

x=prize being behind door 1 y=door 2 picked to be revealed after your pick z = pick of a door

P(X&Y) = P(xgY) p(Y)

P(Y&X) = P(YgX) *P(x)

P(xgY) p(z) = P(YgX) *P(z)

___?____1/10 = 1/9 * 1/10

P(xgy) = P(ygx)

z = x

the real formula

P(X&Y) = P(xgY) p(Y)

P(Y&X) = P(YgX) *P(x)

P(xgY) p(z) = P(YgX) *P(z)

___?____1/10 = 1/9 * 1/10

P(xgy) = P(ygx)

z = x

the real formula

6 Name: Anonymous 2021-04-14 15:44

>>5

slight mistake its actually

P(X&Y) = P(xgY) *p(z)

P(Y&X) = P(YgX) *P(z)

P(xgY) p(z) = P(YgX) *P(z)

___?____1/10 = 1/9 * 1/10

P(xgy) = P(ygx)

z = x

the real formula

slight mistake its actually

P(X&Y) = P(xgY) *p(z)

P(Y&X) = P(YgX) *P(z)

P(xgY) p(z) = P(YgX) *P(z)

___?____1/10 = 1/9 * 1/10

P(xgy) = P(ygx)

z = x

the real formula

7 Name: Anonymous 2021-04-14 15:50

I'm thinking of a random country.

Pick 1 and I will eliminate193 countries from the possibility and give you another country that you can switch to or stay with your picked country.

Pick 1 and I will eliminate193 countries from the possibility and give you another country that you can switch to or stay with your picked country.

8 Name: Anonymous 2021-04-14 15:54

Quoted by: >>9,12

>>1

Think of it this way, since the host will always pick a goat door whether you picked a goat door or not, the three possible scenarios after he reveals the goat door are:

1. You picked a goat door, there is a winning door unrevealed, and the host reveals the second goat door

2. You picked a goat door, there is a winning door unrevealed, and the host reveals the second goat door

3. You picked the winning door, there is a goat door unrevealed, and the host reveals the second goat door

In the first two scenarios, switching your pick will cause you to win and in the last scenario switching will cause you to lose. Since this is an exhaustive list of all possibilities, the probability of switching leading to a win goes to 0.66 repeating, up from 0.33 repeating to pick the correct door originally. If there are only two doors, the original probability is 0.50. Idk how you would apply the host opening the goat door here - if he opens it and gives you an opportunity to switch then there is no need to apply probabilities - just pick the other door

Think of it this way, since the host will always pick a goat door whether you picked a goat door or not, the three possible scenarios after he reveals the goat door are:

1. You picked a goat door, there is a winning door unrevealed, and the host reveals the second goat door

2. You picked a goat door, there is a winning door unrevealed, and the host reveals the second goat door

3. You picked the winning door, there is a goat door unrevealed, and the host reveals the second goat door

In the first two scenarios, switching your pick will cause you to win and in the last scenario switching will cause you to lose. Since this is an exhaustive list of all possibilities, the probability of switching leading to a win goes to 0.66 repeating, up from 0.33 repeating to pick the correct door originally. If there are only two doors, the original probability is 0.50. Idk how you would apply the host opening the goat door here - if he opens it and gives you an opportunity to switch then there is no need to apply probabilities - just pick the other door

9 Name: Anonymous 2021-04-14 15:58

Quoted by: >>12,16

>>8

BUT, just a thought, now bear with me. You could label the goat doors goat door A and goat door B giving you 4 possibilities making it 50-50

1. You picked goat door A, there is a winning door unrevealed, and the host reveals goat door B

2. You picked goat door B, there is a winning door unrevealed, and the host reveals goat door A

3. You picked the winning door, there is goat door A unrevealed, and the host reveals goat door B

4. You picked the winning door, there is goat door B unrevealed, and the host reveals goat door A

50-50

BUT, just a thought, now bear with me. You could label the goat doors goat door A and goat door B giving you 4 possibilities making it 50-50

1. You picked goat door A, there is a winning door unrevealed, and the host reveals goat door B

2. You picked goat door B, there is a winning door unrevealed, and the host reveals goat door A

3. You picked the winning door, there is goat door A unrevealed, and the host reveals goat door B

4. You picked the winning door, there is goat door B unrevealed, and the host reveals goat door A

50-50

10 Name: Anonymous 2021-04-14 16:08

Quoted by: >>12

Generalized Monty Hall choice:

1. You pick a door.

2. One of the doors you didn't pick is opened.

3. You are given the option of keeping your original door, or switching to ALL of the other doors.

When there are only two doors, the problem breaks down because you will be able to figure out 100% what's behind each door at step two, so it's no longer a matter of playing probabilities, you can simply choose the door that has the prize.

>>4 #

If there's a condition that the door that's opened at Step 2 has to contain a goat, then the Monty Hall choice cannot be presented in the 50% of games where the contestant's first choice is incorrect, as opening the door the contestant didn't pick necessarily reveals the prize.

1. You pick a door.

2. One of the doors you didn't pick is opened.

3. You are given the option of keeping your original door, or switching to ALL of the other doors.

When there are only two doors, the problem breaks down because you will be able to figure out 100% what's behind each door at step two, so it's no longer a matter of playing probabilities, you can simply choose the door that has the prize.

>>4 #

If there's a condition that the door that's opened at Step 2 has to contain a goat, then the Monty Hall choice cannot be presented in the 50% of games where the contestant's first choice is incorrect, as opening the door the contestant didn't pick necessarily reveals the prize.

11 Name: Anonymous 2021-04-14 16:10

Tu stii ca unele iubiri nu se uita niciodataa, de ale tale amintiri nimeni inima mea nu vrea sa se despartaa

12 Name: Anonymous 2021-04-14 16:12

Quoted by: >>13

13 Name: Anonymous 2021-04-14 16:19

Quoted by: >>14,15

>>12

These are both about the three door monty hall problem, and thus not relevant to OP.

I'm not sure what the point of labeling the goats is supposed to be, but lets roll. Goat A, Goat B, Car C.

At your initial pick, you can choose A, B, or C.

If you pick A, the host opens door B and gives you the option to keep A or switch to B&C.

If you pick B, the host opens door A and gives you the option to keep B or switch to A&C.

If you pick C, the host opens A or B and gives you the option of keeping C or switching to A&B.

The fact that in the case where you initially pick C, the host may open either of A or B doesn't change that the Monty Hall choice is still either to stick with C or switch to A&B.

These are both about the three door monty hall problem, and thus not relevant to OP.

I'm not sure what the point of labeling the goats is supposed to be, but lets roll. Goat A, Goat B, Car C.

At your initial pick, you can choose A, B, or C.

If you pick A, the host opens door B and gives you the option to keep A or switch to B&C.

If you pick B, the host opens door A and gives you the option to keep B or switch to A&C.

If you pick C, the host opens A or B and gives you the option of keeping C or switching to A&B.

The fact that in the case where you initially pick C, the host may open either of A or B doesn't change that the Monty Hall choice is still either to stick with C or switch to A&B.

14 Name: Anonymous 2021-04-14 16:21

>>13

You could actually skip the "host opens a door revealing a goat" step entirely if it helps you understand the choice being presented.

And in that case, the game still works with two doors, since the contestant still don't know for sure which door has a goat and which has the car.

You could actually skip the "host opens a door revealing a goat" step entirely if it helps you understand the choice being presented.

And in that case, the game still works with two doors, since the contestant still don't know for sure which door has a goat and which has the car.

15 Name: Anonymous 2021-04-14 16:30

>>13

Nope this is the real way to do it, stop grouping them together, do it individually.

1. You picked goat door A, there is a winning door unrevealed, and the host reveals goat door B

2. You picked goat door B, there is a winning door unrevealed, and the host reveals goat door A

3. You picked the winning door, there is goat door A unrevealed, and the host reveals goat door B

4. You picked the winning door, there is goat door B unrevealed, and the host reveals goat door A

50-50

Nope this is the real way to do it, stop grouping them together, do it individually.

1. You picked goat door A, there is a winning door unrevealed, and the host reveals goat door B

2. You picked goat door B, there is a winning door unrevealed, and the host reveals goat door A

3. You picked the winning door, there is goat door A unrevealed, and the host reveals goat door B

4. You picked the winning door, there is goat door B unrevealed, and the host reveals goat door A

50-50

16 Name: Anonymous 2021-04-14 16:35

>>9

The reason that isnâ€™t the way it works is that it is twice as likely for option 1 or 2 to happen than it is for option 3 or 4. Think of it like a river forking into 3 streams, and the third stream forks in half later on. Options 1 and 2 have a probability of 0.33 repeating to occur each, and options 3 and 4 split the remaining 0.33 repeating between them

The reason that isnâ€™t the way it works is that it is twice as likely for option 1 or 2 to happen than it is for option 3 or 4. Think of it like a river forking into 3 streams, and the third stream forks in half later on. Options 1 and 2 have a probability of 0.33 repeating to occur each, and options 3 and 4 split the remaining 0.33 repeating between them

Return

https://i.imgur.com/h7357aB.png

If so can you explain to me why the formula doesn't work with 2 doors?

you have 2 doors

you pick door 1

x= chance of the prize behind door 1 y = the chance of door 2 being opened before you picked door 1 g= given

P(X&Y) = P(xgY) p(Y)

P(y&x) = P(ygx) P(x)

p(YgX) p(X) = P(XgY) p(Y)

1/1 1/2 = 1/1 1/2

IF we were to do it the same way that she does it in this video making y = the chance of any door being opened after you have picked door 1 we would have y = the chance of door 2 being opened after you picked door 1 = 1/1

p(YgX) p(X) = P(XgY) p(Y)

1/1 1/2 = 1/2 1/1

which doesn't make sense because if there are only two doors and door 2 is revealed to be incorrect then logically you would have a 100% chance of initially picking the correct door.

https://www.youtube.com/watch?v=ugbWqWCcxrg&start=182

music thread too

https://www.youtube.com/watch?v=qAMxr7JwxN4