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Topic: MONTY HALL SHOW (Read 14421 times) |
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sojibby
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Re: MONTY HALL SHOW
« Reply #50 on: Jun 1st, 2006, 4:23pm » |
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There are three doors labled 1, 2, and 3. For argument's sake, let's say that the car is behind door one and there are goats behind door 2 and door 3. We'll call them goat A and goat B (it makes a difference).
There are 8 scenarios here:
SCENARIO ONE
Contenstant choice: 1 (car)
Host opens: 2 (goat A)
Contestant choice: switch (lose car)
SCENARIO TWO
Contestant choice: 1 (car)
Host opens: 2 (goat A)
Contestant choice: no switch (win car)
SCENARIO THREE
Contestant choice: 1 (car)
Host opens: 3 (goat B)
Contestant choice: switch (lose car)
SCENARIO FOUR
Contestant choice: 1 (car)
Host opens: 3 (goat B)
Contestant choice: no switch (win car)
SCENARIO FIVE
Contestant choice: 2 (goat A)
Host opens: 3 (goat B)
Contestant choice: switch (win car)
SCENARIO SIX
Contestant choice: 2 (goat A)
Host opens: 3 (goat B)
Contestant choice: no switch (lose car)
SCENARIO SEVEN
Contestant choice: 3 (goat B)
Host opens: 2 (goat A)
Contestant choice: switch (win car)
SCENARIO EIGHT
contestant choice: 3 (goat C)
Host opens: 2 (goat A)
Contestant choice: no switch (lose car)
These are all the possible scenarios, 8 in total.
In half of them she switches and in half of them she wins.
Therefore the probability is a sensible 1/2.
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rmsgrey
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Re: MONTY HALL SHOW
« Reply #51 on: Jun 1st, 2006, 6:17pm » |
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on Jun 1st, 2006, 4:23pm, sojibby wrote:In half of them she switches and in half of them she wins.
Therefore the probability is a sensible 1/2. |
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Here's a rather less interesting game. I give the contestant a blank piece of paper and then ask her to choose whether to switch it for the pile of 10 $10 bills on the table or keep the piece of paper, nominal value rather less than a cent. There are two scenarios:
SCENARIO ONE
Contestant choice: switch (get $100)
SCENARIO TWO
Contestant choice: no switch (get nothing)
In half of them she switches and in half of them she wins, so the probability of her leaving with $100 is 50% if she tosses a coin to make her decision.
If, instead, she makes her choice based on analysing the situation, she will choose to switch every time and leave with $100 100% or the time...
Since the analysis is supposed to determine whether the contestant should switch or not, then you should keep it at four scenarios and jsut decide for each scenario whether switching is good or bad:
Scenario one:
Contestant choice: 1 (car)
Host choice: 2 (goat A)
Shouldn't switch
Scenario two:
Contestant choice: 1 (car)
Host choice: 3 (goat B)
Shouldn't switch
Scenario three:
Contestant choice: 2 (goat A)
Host has no choice: 3 (goat B)
Should switch
Scenario four:
Contestant choice: 3 (goat B)
Host has no choice: 2 (goat A)
Should switch
The catch here is that the four scenarios are reached by different processes - in each of the first two scenarios, there are two choices made; in the last two, only one choice each. As an example, if you roll an ordinary six-sided die, and reroll it if it comes up a six the first time, you get 11 possible outcomes: 1, 2, 3, 4, 5, 6&1, 6&2, 6&3, 6&4, 6&5, 6&6, with a 6 coming up on the first roll in 6 of the cases. If you try it about 35 times over, if the 11 cases were equally probable, you'd expect to see each one about 3 times. I would expect it to actually come up more like each of 1, 2, 3, 4, 5 coming up about 6 times each, and the other 6 possibilities coming up about once each - getting a 6 on the first roll is as likely as getting any other value, and that probability is then shared among the 6 scenarios that share that initial 6...
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Icarus
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Re: MONTY HALL SHOW
« Reply #52 on: Jun 1st, 2006, 7:35pm » |
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The point of rmsgrey's post, sojibby, is that you are assuming that your 8 scenarios are all equally likely. This is not true.
Like you, I will label the door with the car as "1", and the doors with the goats "2", and "3".
There is no reason for the contestant to favor one door over the other when making her initial choice, so we have 3 equally likely cases:
Case 1: (probability = 1/3) She picks door 1.
Case 2: (probability = 1/3) She picks door 2.
Case 3: (probability = 1/3) She picks door 3.
In Case 2, the host will always pick door 3.
In Case 3, the host will always pick door 2.
Both situations have probability 1/3 of happening.
In Case 1, the host is free to pick either of door 2 or door 3, and has no reason to favor one door over the other, so we assume that he picks randomly, and each would have equal chance of being picked (actually, it does not matter how he makes his decision, but for simplicity's sake, assume he does so at random). So we have 2 subcases:
Case 1a: She picks 1, he picks 2. Probability = (1/3)(1/2) = 1/6
Case 1b: She picks 1, he picks 3. Probability = (1/3)(1/2) = 1/6.
Your 8 scenarios come from adding her decision on switching to these 4 cases. But note that two of the cases, 2 and 3, are twice as likely to occur as the other two, 1a and 1b.
Let's look at the "switches" scenarios:
Scenario 1: (Case 1a - probability 1/6)
She chooses 1; He chooses 2; She switches - Loses car.
Scenario 3: (Case 1b - probability 1/6)
She chooses 1; He chooses 3; She switches - Loses car.
Scenario 5: (Case 2 - probability 1/3)
She chooses 2; He chooses 3; She switches - Wins car.
Scenario 7: (Case 3 - probability 1/3)
She chooses 3; He chooses 2; She switches - Wins car.
If she switches, she loses the car 1/6 + 1/6 = 1/3 of the time, but she wins the car 1/3 + 1/3 = 2/3 of the time.
Consider now the scenarios where she does not switch:
Scenario 2: (Case 1a - probability 1/6)
She chooses 1; He chooses 2; She stays - Wins car.
Scenario 4: (Case 1b - probability 1/6)
She chooses 1; He chooses 3; She stays - Wins car.
Scenario 6: (Case 2 - probability 1/3)
She chooses 2; He chooses 3; She stays - Loses car.
Scenario 8: (Case 3 - probability 1/3)
She chooses 3; He chooses 2; She stays - Loses car.
Here the probabilities are just reversed (as they have to be): 1/6 + 1/6 = 1/3 of the time she wins, 1/3 + 1/3 = 2/3 of the time she loses.
Better to switch.
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sojibby
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Posts: 2
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Re: MONTY HALL SHOW
« Reply #53 on: Jun 4th, 2006, 5:29am » |
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It's really common sense when you think about it isn't it.
She has a 2/3 chance of picking a goat - from which a switch would result in a win.
Can't believe I was so blind to that.
Thanks for making me think!
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rmsgrey
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Re: MONTY HALL SHOW
« Reply #54 on: Jun 4th, 2006, 3:56pm » |
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on Jun 4th, 2006, 5:29am, sojibby wrote:It's really common sense when you think about it isn't it.
She has a 2/3 chance of picking a goat - from which a switch would result in a win.
Can't believe I was so blind to that.
Thanks for making me think! |
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Thenk-you for actually stopping to think - a lot of people who start out with a particular answer in mind are so convinced by their initial reasoning that they refuse to accept any argument that contradicts it.
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