The Game Show Problem

Wacky · 2003-07-15, 15:19

In this weekend's paper, there was comment about a previous puzzle:
(http://www.parade.com/current/columns/askmarilyn.html)

Say you’re on a game show and given a choice of three doors. Behind one door is a car; behind the other two doors are goats. You choose door #1. The show’s host, who knows what’s behind the doors, opens door #3. A goat appears. The host then asks, “Do you want to choose door #2, instead?” Should you switch? You said, “Yes. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance.”

Do you think her answer is correct? Why?

Orgasmic Troll · 2003-07-15, 15:25

Why you gotta be doing this here? This was a peaceful forum!

Orgasmic Troll · 2003-07-15, 15:33

That said, the probabilites stated are correct. The key piece of information is that the host of the show knows which door the prize is behind and will never open that door.

Before the host opens up the door, you have a 1/3rd chance at every door, but if you could choose between *A* and *B and C* you would take the B and C combo, 2 for 1 deal. When the host opens up the third door, that's what you're getting when you choose door #2, a 2 for 1 deal.

Wacky · 2003-07-15, 17:05

But aren't you choosing between *A and C* and *B and C* ?

Whether you change or not, you still get door *C* for free.

ET_ · 2003-07-15, 17:42

Quote:

That said, the probabilites stated are correct.

Same for me. I would change.

And find a goat racing car.

Luigi :(

smh · 2003-07-15, 18:03

Try to imagine that there are 100 doors. You pick one and the host opens 98 of the remaining doors. Would you switch?

Orgasmic Troll · 2003-07-15, 18:04

Quote:

Originally Posted by Wackerbath

But aren't you choosing between *A and C* and *B and C* ?

Whether you change or not, you still get door *C* for free.

No, because you're switching from A, not choosing A from 2 closed doors and 1 open door. You gave up B and C when you chose A initially, and now you are getting the chance to switch to the option you gave up (B and C) compacted into one door.

The possibilities are as follows

(a) The prize is in A, Host opens B, you don't switch (win)
(b) The prize is in A, Host opens B, switch to C (lose)

(c) The prize is in A, Host opens C, you don't switch (win)
(d) The prize is in A, Host opens C, switch to B (lose)

(e) The prize is in B, Host opens C, you don't switch (lose)
(f) The prize is in B, Host opens C, switch to B (win)

(g) The prize is in C, Host opens B, you don't switch (lose)
(h) The prize is in C, Host opens B, switch to C (win)

so you say "Ah-Ha! 2 out of 4 times, you switch and lose! so it's 1/2 probability!" .. but these aren't all equal probabilities!

The chances that the host opens door B or C are not independent of which door the prize is in, if the prize is in A, the probability B or C get opened is 50% each, but if the prize is in B, the probability that C gets opened is 100%, and if the prize is in C, the probability B gets opened is 100% .. so this translates to

(a) 1/3 * 1/2 * 1/2 = 1/12
(b) 1/3 * 1/2 * 1/2 = 1/12 (switch and lose)
(c) 1/3 * 1/2 * 1/2 = 1/12
(d) 1/3 * 1/2 * 1/2 = 1/12 (switch and lose)
(e) 1/3 * 1 * 1/2 = 1/6
(f) 1/3 * 1 * 1/2 = 1/6 (switch and win)
(g) 1/3 * 1 * 1/2 = 1/6
(h) 1/3 * 1 * 1/2 = 1/6 (switch and win)

so the probability of winning given that you switch (i.e. P(win|switch) )
is (f+h)/(b+d+f+h) = (1/6+1/6)/(1/12+1/12+1/6+1/6) = (1/3)/(1/2) = 2/3

edit: added quote

Wacky · 2003-07-15, 19:32

I still don't buy it. Initially, there are three equally likely situations: The car is behind A, B, or C. The only thing that the host's action has done is to remove the third possibility. It doesn't change the fact that the other two are still equally likely.

On what basis do you assign probabilities to the actions of the host. The host is not a random selector. How do you know that he isn't operating under the rule " Open door *C* if there is goat behind it because if makes for a better camera shot"?

ColdFury · 2003-07-15, 19:48

Isn't this the Monty Hall problem?

I don't remember the reasoning, but my Discrete Mathmatics professor was quite firm the best strategy was always to stick with your first choice.

Orgasmic Troll · 2003-07-15, 20:09

Yeah, this is the Monty Hall problem, the infamous never ending Monty Hall problem.

Now, for Wackerbath, I'm assuming you mean that if the prize is in A, he always opens door C, instead of picking between both. (because he only has a choice when the prize is behind A)

Well, that changes our sample space to the following ( options (a) and (b) are eliminated))

(c) The prize is in A, Host opens C, you don't switch (win)
(d) The prize is in A, Host opens C, switch to B (lose)

(e) The prize is in B, Host opens C, you don't switch (lose)
(f) The prize is in B, Host opens C, switch to B (win)

(g) The prize is in C, Host opens B, you don't switch (lose)
(h) The prize is in C, Host opens B, switch to C (win)

well, now, the probability of the second statement (Host opens X) is 100%, so now we have

(c) 1/3 * 1 * 1/2
(d) 1/3 * 1 * 1/2 (switch, lose)
(e) 1/3 * 1 * 1/2
(f) 1/3 * 1 * 1/2 (switch, win)
(g) 1/3 * 1 * 1/2
(h) 1/3 * 1 * 1/2 (switch, win)

in other words, every choice is equal probability (1/6) but now out of the three switch possibilities, 2 are wins, so the probability remains at 2/3 for switching

toferc · 2003-07-15, 20:29

Just what this forum needed, the 500th "Monty Hall" flamewar.

I personally have been involved in at least a half-dozen of these.

TravisT has it right, but I'll try to say it another way for those still not convinced.

key point

According to the statement of the problem, the host always opens one door. The door always satisfies the following requirements:
1) No prize
2) Not the one the contestant picked.

hand-waving argument

If you employ the switching strategy, the only way to lose is if you picked the correct door in the first place. To see that this is true, consider the case where your initial guess was wrong. Since we know that you picked the wrong door, the prize must be behind one of the two remaining doors. The host's game now looks like this:

- There are two doors, and behind one of them is a prize
- The host has to show you which one doesn't have a prize

If you pick the door he didn't open, you win. Switching always wins when you guess wrong the first time.

I think it's fairly obvious that switching always loses when you were right the first time.

So... since your first guess is wrong 2/3 of the time and right 1/3 of the time, switching wins 100% of 2/3 of the time and 0% of 1/3 of the time, for a total of 2/3 of the time.

more rigorous argument

Define events A1, A2, A3 as
A1 - prize behind door 1
A2 - prize behind door 2
A3 - prize behind door 3

Without loss of generality, assume the contestant initially picks door number 1. The host must open door 2 or 3. Define B2, B3 as
B2 - host opens door 2
B3 - host opens door 3

Since the host must not open the prize door, the probabilities of B2 and B3 conditioned upon A1, A2, and A3 are as follows:

P(B2|A1) = p
P(B3|A1) = 1-p
P(B2|A2) = 0
P(B3|A2) = 1
P(B2|A3) = 1
P(B3|A3) = 0

(p is introduced here to cover cases where the host has a door-opening preference)

Define events S2 and S3 as follows:
S2 - contestant switches to door 2
S3 - contestant switches to door 3

If the contestant always employs the switching strategy, the probabilities of S2 and S3 conditioned on A1, A2, and A3 are completely determined by the probabilities of B2 and B3:

P(S2|A1) = 1-p
P(S3|A1) = p
P(S2|A2) = 1
P(S3|A2) = 0
P(S2|A3) = 0
P(S3|A3) = 1

Now, the probability of the contestant winning is
P(win) = P(A2 and S2) + P(A3 and S3)
= P(A2)P(S2|A2) + P(A3)P(S3|A3)
= 1/3 * 1 + 1/3 * 1
= 2/3

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2003-07-15, 15:19	#1
Wacky Jun 2003 The Texas Hill Country 3²·11² Posts	The Game Show Problem In this weekend's paper, there was comment about a previous puzzle: (http://www.parade.com/current/columns/askmarilyn.html) Say you’re on a game show and given a choice of three doors. Behind one door is a car; behind the other two doors are goats. You choose door #1. The show’s host, who knows what’s behind the doors, opens door #3. A goat appears. The host then asks, “Do you want to choose door #2, instead?” Should you switch? You said, “Yes. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance.” Do you think her answer is correct? Why?

2003-07-15, 15:25	#2
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 1010000001₂ Posts	Why you gotta be doing this here? This was a peaceful forum!

2003-07-15, 15:33	#3
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 641₁₀ Posts	That said, the probabilites stated are correct. The key piece of information is that the host of the show knows which door the prize is behind and will never open that door. Before the host opens up the door, you have a 1/3rd chance at every door, but if you could choose between A and B and C you would take the B and C combo, 2 for 1 deal. When the host opens up the third door, that's what you're getting when you choose door #2, a 2 for 1 deal.

2003-07-15, 17:05	#4
Wacky Jun 2003 The Texas Hill Country 2101₈ Posts	But aren't you choosing between A and C and B and C ? Whether you change or not, you still get door C for free.

2003-07-15, 18:03	#6
smh "Sander" Oct 2002 52.345322,5.52471 4A5₁₆ Posts	Try to imagine that there are 100 doors. You pick one and the host opens 98 of the remaining doors. Would you switch?

2003-07-15, 19:32	#8
Wacky Jun 2003 The Texas Hill Country 3²×11² Posts	I still don't buy it. Initially, there are three equally likely situations: The car is behind A, B, or C. The only thing that the host's action has done is to remove the third possibility. It doesn't change the fact that the other two are still equally likely. On what basis do you assign probabilities to the actions of the host. The host is not a random selector. How do you know that he isn't operating under the rule " Open door C if there is goat behind it because if makes for a better camera shot"?

2003-07-15, 19:48	#9
ColdFury Aug 2002 500₈ Posts	Isn't this the Monty Hall problem? I don't remember the reasoning, but my Discrete Mathmatics professor was quite firm the best strategy was always to stick with your first choice.

2003-07-15, 20:09	#10
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 1201₈ Posts	Yeah, this is the Monty Hall problem, the infamous never ending Monty Hall problem. Now, for Wackerbath, I'm assuming you mean that if the prize is in A, he always opens door C, instead of picking between both. (because he only has a choice when the prize is behind A) Well, that changes our sample space to the following ( options (a) and (b) are eliminated)) (c) The prize is in A, Host opens C, you don't switch (win) (d) The prize is in A, Host opens C, switch to B (lose) (e) The prize is in B, Host opens C, you don't switch (lose) (f) The prize is in B, Host opens C, switch to B (win) (g) The prize is in C, Host opens B, you don't switch (lose) (h) The prize is in C, Host opens B, switch to C (win) well, now, the probability of the second statement (Host opens X) is 100%, so now we have (c) 1/3 * 1 * 1/2 (d) 1/3 * 1 * 1/2 (switch, lose) (e) 1/3 * 1 * 1/2 (f) 1/3 * 1 * 1/2 (switch, win) (g) 1/3 * 1 * 1/2 (h) 1/3 * 1 * 1/2 (switch, win) in other words, every choice is equal probability (1/6) but now out of the three switch possibilities, 2 are wins, so the probability remains at 2/3 for switching

2003-07-15, 20:29	#11
toferc Aug 2002 2·3·5 Posts	Just what this forum needed, the 500th "Monty Hall" flamewar. I personally have been involved in at least a half-dozen of these. TravisT has it right, but I'll try to say it another way for those still not convinced. key point According to the statement of the problem, the host always opens one door. The door always satisfies the following requirements: 1) No prize 2) Not the one the contestant picked. hand-waving argument If you employ the switching strategy, the only way to lose is if you picked the correct door in the first place. To see that this is true, consider the case where your initial guess was wrong. Since we know that you picked the wrong door, the prize must be behind one of the two remaining doors. The host's game now looks like this: - There are two doors, and behind one of them is a prize - The host has to show you which one doesn't have a prize If you pick the door he didn't open, you win. Switching always wins when you guess wrong the first time. I think it's fairly obvious that switching always loses when you were right the first time. So... since your first guess is wrong 2/3 of the time and right 1/3 of the time, switching wins 100% of 2/3 of the time and 0% of 1/3 of the time, for a total of 2/3 of the time. more rigorous argument Define events A1, A2, A3 as A1 - prize behind door 1 A2 - prize behind door 2 A3 - prize behind door 3 Without loss of generality, assume the contestant initially picks door number 1. The host must open door 2 or 3. Define B2, B3 as B2 - host opens door 2 B3 - host opens door 3 Since the host must not open the prize door, the probabilities of B2 and B3 conditioned upon A1, A2, and A3 are as follows: P(B2\|A1) = p P(B3\|A1) = 1-p P(B2\|A2) = 0 P(B3\|A2) = 1 P(B2\|A3) = 1 P(B3\|A3) = 0 (p is introduced here to cover cases where the host has a door-opening preference) Define events S2 and S3 as follows: S2 - contestant switches to door 2 S3 - contestant switches to door 3 If the contestant always employs the switching strategy, the probabilities of S2 and S3 conditioned on A1, A2, and A3 are completely determined by the probabilities of B2 and B3: P(S2\|A1) = 1-p P(S3\|A1) = p P(S2\|A2) = 1 P(S3\|A2) = 0 P(S2\|A3) = 0 P(S3\|A3) = 1 Now, the probability of the contestant winning is P(win) = P(A2 and S2) + P(A3 and S3) = P(A2)P(S2\|A2) + P(A3)P(S3\|A3) = 1/3 * 1 + 1/3 * 1 = 2/3