Mixing
Alternative analysis of the Monty Hall problem
The widely accepted answer for the Monty Hall problem is that it is better for a contestant to switch doors because there is a $frac23$ probability he picked the door with a goat behind it the first time and only a $frac13$ probability the he picked the door with the car behind it.
We can look at this problem in a different way. There are three ways of arranging one car and two goats behind three doors, assuming that we do not distinguish between the two goats:
Door1 Door2 Door3
1. CAR goat goat
2. goat CAR goat
3. goat goat CAR
Consider arrangement $1$. If the contestant first picks door $1$ and then the host opens say, door $2$ to reveal a goat and the contestant then switches to door $3$, the contestant ends up with a goat. Thus we have a combination of contestant's first pick, the door the host opens and whether the contestant then switches doors or not. There are eight such combinations for each arrangement.
The eight combinations and their outcomes for arrangement $1$ are listed here.
The contestant gets the car in four of these eight combinations, in two of them by switching doors and in two by not switching doors. The other arrangements will be similar. So considering all three arrangements, the contestant will get the car in $12$ of the $24$ total combinations, in $6$ of them by switching doors and $6$ by not switching.
Analyzing the Monty Hall problem this way we see that switching doors or not switching doors is all the same. What is wrong with this analysis ?
probability monty-hall
edited Jul 29 '17 at 4:37
Siong Thye Goh
99.9k1465117
asked Jul 29 '17 at 4:30
CrankyShaftCrankyShaft
1091
add a comment |
The widely accepted answer for the Monty Hall problem is that it is better for a contestant to switch doors because there is a $frac23$ probability he picked the door with a goat behind it the first time and only a $frac13$ probability the he picked the door with the car behind it.
We can look at this problem in a different way. There are three ways of arranging one car and two goats behind three doors, assuming that we do not distinguish between the two goats:
Door1 Door2 Door3
1. CAR goat goat
2. goat CAR goat
3. goat goat CAR
Consider arrangement $1$. If the contestant first picks door $1$ and then the host opens say, door $2$ to reveal a goat and the contestant then switches to door $3$, the contestant ends up with a goat. Thus we have a combination of contestant's first pick, the door the host opens and whether the contestant then switches doors or not. There are eight such combinations for each arrangement.
The eight combinations and their outcomes for arrangement $1$ are listed here.
The contestant gets the car in four of these eight combinations, in two of them by switching doors and in two by not switching doors. The other arrangements will be similar. So considering all three arrangements, the contestant will get the car in $12$ of the $24$ total combinations, in $6$ of them by switching doors and $6$ by not switching.
Analyzing the Monty Hall problem this way we see that switching doors or not switching doors is all the same. What is wrong with this analysis ?
probability monty-hall
edited Jul 29 '17 at 4:37
Siong Thye Goh
99.9k1465117
asked Jul 29 '17 at 4:30
CrankyShaftCrankyShaft
1091
5
"widely accepted" is wrong - it's the correct answer. It's a mathematical problem, not an empirical one.
– MathematicsStudent1122
Jul 29 '17 at 4:32
4
The eight outcomes you list are not all equally probable.
– Graham Kemp
Jul 29 '17 at 4:39
add a comment |
The widely accepted answer for the Monty Hall problem is that it is better for a contestant to switch doors because there is a $frac23$ probability he picked the door with a goat behind it the first time and only a $frac13$ probability the he picked the door with the car behind it.
We can look at this problem in a different way. There are three ways of arranging one car and two goats behind three doors, assuming that we do not distinguish between the two goats:
Door1 Door2 Door3
1. CAR goat goat
2. goat CAR goat
3. goat goat CAR
Consider arrangement $1$. If the contestant first picks door $1$ and then the host opens say, door $2$ to reveal a goat and the contestant then switches to door $3$, the contestant ends up with a goat. Thus we have a combination of contestant's first pick, the door the host opens and whether the contestant then switches doors or not. There are eight such combinations for each arrangement.
The eight combinations and their outcomes for arrangement $1$ are listed here.
The contestant gets the car in four of these eight combinations, in two of them by switching doors and in two by not switching doors. The other arrangements will be similar. So considering all three arrangements, the contestant will get the car in $12$ of the $24$ total combinations, in $6$ of them by switching doors and $6$ by not switching.
Analyzing the Monty Hall problem this way we see that switching doors or not switching doors is all the same. What is wrong with this analysis ?
probability monty-hall
edited Jul 29 '17 at 4:37
Siong Thye Goh
99.9k1465117
asked Jul 29 '17 at 4:30
CrankyShaftCrankyShaft
1091
The widely accepted answer for the Monty Hall problem is that it is better for a contestant to switch doors because there is a $frac23$ probability he picked the door with a goat behind it the first time and only a $frac13$ probability the he picked the door with the car behind it.
We can look at this problem in a different way. There are three ways of arranging one car and two goats behind three doors, assuming that we do not distinguish between the two goats:
Door1 Door2 Door3
1. CAR goat goat
2. goat CAR goat
3. goat goat CAR
Consider arrangement $1$. If the contestant first picks door $1$ and then the host opens say, door $2$ to reveal a goat and the contestant then switches to door $3$, the contestant ends up with a goat. Thus we have a combination of contestant's first pick, the door the host opens and whether the contestant then switches doors or not. There are eight such combinations for each arrangement.
The eight combinations and their outcomes for arrangement $1$ are listed here.
The contestant gets the car in four of these eight combinations, in two of them by switching doors and in two by not switching doors. The other arrangements will be similar. So considering all three arrangements, the contestant will get the car in $12$ of the $24$ total combinations, in $6$ of them by switching doors and $6$ by not switching.
Analyzing the Monty Hall problem this way we see that switching doors or not switching doors is all the same. What is wrong with this analysis ?
probability monty-hall
probability monty-hall
edited Jul 29 '17 at 4:37
Siong Thye Goh
99.9k1465117
asked Jul 29 '17 at 4:30
CrankyShaftCrankyShaft
1091
edited Jul 29 '17 at 4:37
Siong Thye Goh
99.9k1465117
asked Jul 29 '17 at 4:30
CrankyShaftCrankyShaft
1091
edited Jul 29 '17 at 4:37
Siong Thye Goh
99.9k1465117
edited Jul 29 '17 at 4:37
Siong Thye Goh
99.9k1465117
edited Jul 29 '17 at 4:37
Siong Thye Goh
99.9k1465117
99.9k1465117
asked Jul 29 '17 at 4:30
CrankyShaftCrankyShaft
1091
asked Jul 29 '17 at 4:30
CrankyShaftCrankyShaft
1091
asked Jul 29 '17 at 4:30
CrankyShaftCrankyShaft
1091
1091
5
"widely accepted" is wrong - it's the correct answer. It's a mathematical problem, not an empirical one.
– MathematicsStudent1122
Jul 29 '17 at 4:32
4
The eight outcomes you list are not all equally probable.
– Graham Kemp
Jul 29 '17 at 4:39
add a comment |
5
"widely accepted" is wrong - it's the correct answer. It's a mathematical problem, not an empirical one.
– MathematicsStudent1122
Jul 29 '17 at 4:32
4
The eight outcomes you list are not all equally probable.
– Graham Kemp
Jul 29 '17 at 4:39
5
5
"widely accepted" is wrong - it's the correct answer. It's a mathematical problem, not an empirical one.
– MathematicsStudent1122
Jul 29 '17 at 4:32
"widely accepted" is wrong - it's the correct answer. It's a mathematical problem, not an empirical one.
– MathematicsStudent1122
Jul 29 '17 at 4:32
4
4
The eight outcomes you list are not all equally probable.
– Graham Kemp
Jul 29 '17 at 4:39
The eight outcomes you list are not all equally probable.
– Graham Kemp
Jul 29 '17 at 4:39
add a comment |
2 Answers
2
active
oldest
votes
The second choice you list, the door the host opens, is not independent of the door the contestant chooses. If the contestant chooses a door with a goat, Monty has only one door to open to show a goat.
Now that your spreadsheet has been posted, the first four lines are half as probable as the last four lines.
answered Jul 29 '17 at 4:40
Ross MillikanRoss Millikan
292k23197371
Hey, be a little nicer. The OP has presented a question very nicely.
– zhw.
Jul 29 '17 at 6:54
the first four lines are half as probable as the last four lines.
– CrankyShaft
Jul 31 '17 at 23:58
add a comment |
There are only two distinct outcomes in a million dollar lottery: win or lose. Therefore #favoured/#total tells us that the probability for winning is $1/2$ ... no?
You cannot just count the outcomes, you have to weight their probability too.
Assuming no bias when offered equally viable choices.
There is a probability of $1/3$ for the contestant to pick any one from the three doors.
If the contestant happens to pick the door hiding the car, then the host has a probability of $1/2$ for selecting either of the remaining doors.
However, if the contestant picks either of the doors hiding the goat, then the host has only one viable choice -- she certainly reveals the other door hiding a goat.
$begin{array}{|l:l:l|l|} hline
text{Door }1 & text{Door }2 & text{Door } 3 & text{Probability} & text{Win by}\ hdashline
text{(Car)} & text{(Goat)} & text{(Goat)} \hline
text{Pick} & text{Reveal} & & tfrac 13cdottfrac 12 & text{Stay} \
text{Pick} & & text{Reveal} & tfrac 13cdottfrac 12 & text{Stay} \
&text{Pick} & text{Reveal} & tfrac 13cdottfrac 11 & text{Switch} \
& text{Reveal} &text{Pick} & tfrac 13cdottfrac 11 & text{Switch} \ hline
end{array}$
Thus the probability that "Stay" is the winning strategy is $1/3$ while the probability that "Switch" is the winning strategy is $2/3$.
Of course, if you completely ignore this advice and just randomly decide to stay or switch (say by a coin toss), then the probability that you win is $1/2$.
answered Jul 29 '17 at 4:59
Graham KempGraham Kemp
84.7k43378
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The second choice you list, the door the host opens, is not independent of the door the contestant chooses. If the contestant chooses a door with a goat, Monty has only one door to open to show a goat.
Now that your spreadsheet has been posted, the first four lines are half as probable as the last four lines.
answered Jul 29 '17 at 4:40
Ross MillikanRoss Millikan
292k23197371
Hey, be a little nicer. The OP has presented a question very nicely.
– zhw.
Jul 29 '17 at 6:54
the first four lines are half as probable as the last four lines.
– CrankyShaft
Jul 31 '17 at 23:58
add a comment |
The second choice you list, the door the host opens, is not independent of the door the contestant chooses. If the contestant chooses a door with a goat, Monty has only one door to open to show a goat.
Now that your spreadsheet has been posted, the first four lines are half as probable as the last four lines.
answered Jul 29 '17 at 4:40
Ross MillikanRoss Millikan
292k23197371
Hey, be a little nicer. The OP has presented a question very nicely.
– zhw.
Jul 29 '17 at 6:54
the first four lines are half as probable as the last four lines.
– CrankyShaft
Jul 31 '17 at 23:58
add a comment |
The second choice you list, the door the host opens, is not independent of the door the contestant chooses. If the contestant chooses a door with a goat, Monty has only one door to open to show a goat.
Now that your spreadsheet has been posted, the first four lines are half as probable as the last four lines.
answered Jul 29 '17 at 4:40
Ross MillikanRoss Millikan
292k23197371
The second choice you list, the door the host opens, is not independent of the door the contestant chooses. If the contestant chooses a door with a goat, Monty has only one door to open to show a goat.
Now that your spreadsheet has been posted, the first four lines are half as probable as the last four lines.
answered Jul 29 '17 at 4:40
Ross MillikanRoss Millikan
292k23197371
edited Jul 29 '17 at 14:46
answered Jul 29 '17 at 4:40
Ross MillikanRoss Millikan
292k23197371
answered Jul 29 '17 at 4:40
Ross MillikanRoss Millikan
292k23197371
answered Jul 29 '17 at 4:40
Ross MillikanRoss Millikan
292k23197371
292k23197371
Hey, be a little nicer. The OP has presented a question very nicely.
– zhw.
Jul 29 '17 at 6:54
the first four lines are half as probable as the last four lines.
– CrankyShaft
Jul 31 '17 at 23:58
add a comment |
Hey, be a little nicer. The OP has presented a question very nicely.
– zhw.
Jul 29 '17 at 6:54
the first four lines are half as probable as the last four lines.
– CrankyShaft
Jul 31 '17 at 23:58
Hey, be a little nicer. The OP has presented a question very nicely.
– zhw.
Jul 29 '17 at 6:54
Hey, be a little nicer. The OP has presented a question very nicely.
– zhw.
Jul 29 '17 at 6:54
the first four lines are half as probable as the last four lines.
– CrankyShaft
Jul 31 '17 at 23:58
the first four lines are half as probable as the last four lines.
– CrankyShaft
Jul 31 '17 at 23:58
add a comment |
There are only two distinct outcomes in a million dollar lottery: win or lose. Therefore #favoured/#total tells us that the probability for winning is $1/2$ ... no?
You cannot just count the outcomes, you have to weight their probability too.
Assuming no bias when offered equally viable choices.
There is a probability of $1/3$ for the contestant to pick any one from the three doors.
If the contestant happens to pick the door hiding the car, then the host has a probability of $1/2$ for selecting either of the remaining doors.
However, if the contestant picks either of the doors hiding the goat, then the host has only one viable choice -- she certainly reveals the other door hiding a goat.
$begin{array}{|l:l:l|l|} hline
text{Door }1 & text{Door }2 & text{Door } 3 & text{Probability} & text{Win by}\ hdashline
text{(Car)} & text{(Goat)} & text{(Goat)} \hline
text{Pick} & text{Reveal} & & tfrac 13cdottfrac 12 & text{Stay} \
text{Pick} & & text{Reveal} & tfrac 13cdottfrac 12 & text{Stay} \
&text{Pick} & text{Reveal} & tfrac 13cdottfrac 11 & text{Switch} \
& text{Reveal} &text{Pick} & tfrac 13cdottfrac 11 & text{Switch} \ hline
end{array}$
Thus the probability that "Stay" is the winning strategy is $1/3$ while the probability that "Switch" is the winning strategy is $2/3$.
Of course, if you completely ignore this advice and just randomly decide to stay or switch (say by a coin toss), then the probability that you win is $1/2$.
answered Jul 29 '17 at 4:59
Graham KempGraham Kemp
84.7k43378
add a comment |
There are only two distinct outcomes in a million dollar lottery: win or lose. Therefore #favoured/#total tells us that the probability for winning is $1/2$ ... no?
You cannot just count the outcomes, you have to weight their probability too.
Assuming no bias when offered equally viable choices.
There is a probability of $1/3$ for the contestant to pick any one from the three doors.
If the contestant happens to pick the door hiding the car, then the host has a probability of $1/2$ for selecting either of the remaining doors.
However, if the contestant picks either of the doors hiding the goat, then the host has only one viable choice -- she certainly reveals the other door hiding a goat.
$begin{array}{|l:l:l|l|} hline
text{Door }1 & text{Door }2 & text{Door } 3 & text{Probability} & text{Win by}\ hdashline
text{(Car)} & text{(Goat)} & text{(Goat)} \hline
text{Pick} & text{Reveal} & & tfrac 13cdottfrac 12 & text{Stay} \
text{Pick} & & text{Reveal} & tfrac 13cdottfrac 12 & text{Stay} \
&text{Pick} & text{Reveal} & tfrac 13cdottfrac 11 & text{Switch} \
& text{Reveal} &text{Pick} & tfrac 13cdottfrac 11 & text{Switch} \ hline
end{array}$
Thus the probability that "Stay" is the winning strategy is $1/3$ while the probability that "Switch" is the winning strategy is $2/3$.
Of course, if you completely ignore this advice and just randomly decide to stay or switch (say by a coin toss), then the probability that you win is $1/2$.
answered Jul 29 '17 at 4:59
Graham KempGraham Kemp
84.7k43378
add a comment |
There are only two distinct outcomes in a million dollar lottery: win or lose. Therefore #favoured/#total tells us that the probability for winning is $1/2$ ... no?
You cannot just count the outcomes, you have to weight their probability too.
Assuming no bias when offered equally viable choices.
There is a probability of $1/3$ for the contestant to pick any one from the three doors.
If the contestant happens to pick the door hiding the car, then the host has a probability of $1/2$ for selecting either of the remaining doors.
However, if the contestant picks either of the doors hiding the goat, then the host has only one viable choice -- she certainly reveals the other door hiding a goat.
$begin{array}{|l:l:l|l|} hline
text{Door }1 & text{Door }2 & text{Door } 3 & text{Probability} & text{Win by}\ hdashline
text{(Car)} & text{(Goat)} & text{(Goat)} \hline
text{Pick} & text{Reveal} & & tfrac 13cdottfrac 12 & text{Stay} \
text{Pick} & & text{Reveal} & tfrac 13cdottfrac 12 & text{Stay} \
&text{Pick} & text{Reveal} & tfrac 13cdottfrac 11 & text{Switch} \
& text{Reveal} &text{Pick} & tfrac 13cdottfrac 11 & text{Switch} \ hline
end{array}$
Thus the probability that "Stay" is the winning strategy is $1/3$ while the probability that "Switch" is the winning strategy is $2/3$.
Of course, if you completely ignore this advice and just randomly decide to stay or switch (say by a coin toss), then the probability that you win is $1/2$.
answered Jul 29 '17 at 4:59
Graham KempGraham Kemp
84.7k43378
There are only two distinct outcomes in a million dollar lottery: win or lose. Therefore #favoured/#total tells us that the probability for winning is $1/2$ ... no?
You cannot just count the outcomes, you have to weight their probability too.
Assuming no bias when offered equally viable choices.
There is a probability of $1/3$ for the contestant to pick any one from the three doors.
If the contestant happens to pick the door hiding the car, then the host has a probability of $1/2$ for selecting either of the remaining doors.
However, if the contestant picks either of the doors hiding the goat, then the host has only one viable choice -- she certainly reveals the other door hiding a goat.
$begin{array}{|l:l:l|l|} hline
text{Door }1 & text{Door }2 & text{Door } 3 & text{Probability} & text{Win by}\ hdashline
text{(Car)} & text{(Goat)} & text{(Goat)} \hline
text{Pick} & text{Reveal} & & tfrac 13cdottfrac 12 & text{Stay} \
text{Pick} & & text{Reveal} & tfrac 13cdottfrac 12 & text{Stay} \
&text{Pick} & text{Reveal} & tfrac 13cdottfrac 11 & text{Switch} \
& text{Reveal} &text{Pick} & tfrac 13cdottfrac 11 & text{Switch} \ hline
end{array}$
Thus the probability that "Stay" is the winning strategy is $1/3$ while the probability that "Switch" is the winning strategy is $2/3$.
Of course, if you completely ignore this advice and just randomly decide to stay or switch (say by a coin toss), then the probability that you win is $1/2$.
answered Jul 29 '17 at 4:59
Graham KempGraham Kemp
84.7k43378
answered Jul 29 '17 at 4:59
Graham KempGraham Kemp
84.7k43378
answered Jul 29 '17 at 4:59
Graham KempGraham Kemp
84.7k43378
answered Jul 29 '17 at 4:59
Graham KempGraham Kemp
84.7k43378
84.7k43378
add a comment |
add a comment |
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5
"widely accepted" is wrong - it's the correct answer. It's a mathematical problem, not an empirical one.
– MathematicsStudent1122
Jul 29 '17 at 4:32
4
The eight outcomes you list are not all equally probable.
– Graham Kemp
Jul 29 '17 at 4:39