Klopp
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March 10th, 2020 at 6:50:17 AM permalink
In the final segment of a game show for couples, the last remaining couple is presented with three closed doors, behind which, in random order, they will find a car, the key to the car, and a goat. One of the two players is given the task of finding the car, the other must find the car key. If both are successful, they get to keep the car. Each of the two players may open two doors; the second player may not see what is behind the doors chosen by the first player. The couple may discuss a strategy before the game starts. What strategy will afford the best odds of winning the car, and what are the winning odds?
EdCollins
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March 10th, 2020 at 7:18:40 AM permalink
Just to be clear, if the first person is given the task of finding the car, and the second person is given the task of finding the keys, but the first person finds the keys and the second person finds the car, that's a miss, correct, since neither was successful on their own, despite being successful together.

If so, I offer this simple solution, which I would hope can easily be improved upon.
The couple agree beforehand to each pick Door #1 and Door #2.

That's it.

If the goat is behind Door #3, they lose. Otherwise, they will win. There's a 1/3 chance the goat is behind Door #3. Thus, they have a 2/3 chance of finding the car and the keys.
Wizard
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March 10th, 2020 at 7:33:45 AM permalink
Question -- Can each contestant choose the second door after seeing what is behind the first door?
Last edited by: Wizard on Mar 10, 2020
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
unJon
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March 10th, 2020 at 7:41:50 AM permalink
Seems like a variant of the secret Santa problem.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
CrystalMath
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March 10th, 2020 at 7:42:38 AM permalink
Even if the players cannot see anything, they can have player 1 choose doors 1&2 and player 2 choose doors 2&3 for a 50% success rate.


Can the second player see which doors player 1 opened?
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DJTeddyBear
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March 10th, 2020 at 7:44:40 AM permalink
I was going to agree with Ed, but then as I was typing it, it all fell apart in my head. ☹️
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
DJTeddyBear
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March 10th, 2020 at 7:58:44 AM permalink
After further noodling the problem, and gasp taking out a pen and paper, I agree with CrystalMath.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
unJon
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CrystalMath
March 10th, 2020 at 7:59:03 AM permalink
Secret Santa logic:

I assume that a person picks his second door after opening the first.

Label the car #1, keys #2 and goat #3.

Player #1 must find car (#1). He begins by opening door #1. If it痴 the car behind the door, he痴 done. If it痴 the keys (#2) he opens door 2 next. If it痴 the goat (#3), he opens door 3 next.

Player #2 does the same algorithm but opens door #2 first, looking for keys.

Of the 6 ways to sort the three objects behind three doors, this method produces a winner in the following:

123
132
213
321

Or 2/3 of the time.
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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March 10th, 2020 at 7:59:28 AM permalink
While we wait for clarification on the rules from Klopp, allow me to ask the following Wizard Variant:

In the final segment of a game show for couples, the last remaining couple is presented with three closed doors, behind which, in random order, they will find a car, the key to the car, and a goat. One of the two players is given the task of finding the car, the other must find the car key. If both are successful, they get to keep the car. Each of the two players may open two doors; the second player may not see what is behind the doors chosen by the first player. Both players may choose their second door after seeing what is behind their first door. Neither player shall know any results from the other player. The couple may discuss a strategy before the game starts. What strategy will afford the best odds of winning the car, and what are the winning odds?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
CrystalMath
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March 10th, 2020 at 8:00:37 AM permalink
Assuming player 2 can see which doors player 1 opened, but not their contents, then player 1 will choose door 1. If it contains player 2痴 item, then open door 3, otherwise door 2. Player 2 will open door 1if player 1 opened door 3, otherwise player 2 will open doors 2&3.

This gives a 2/3 success rate.
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unJon
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March 10th, 2020 at 8:01:06 AM permalink
Quote: Wizard

While we wait for clarification on the rules from Klopp, allow me to ask the following Wizard Variant:

In the final segment of a game show for couples, the last remaining couple is presented with three closed doors, behind which, in random order, they will find a car, the key to the car, and a goat. One of the two players is given the task of finding the car, the other must find the car key. If both are successful, they get to keep the car. Each of the two players may open two doors; the second player may not see what is behind the doors chosen by the first player. Both players may choose their second door after seeing what is behind their first door. Neither player shall know any results from the other player. The couple may discuss a strategy before the game starts. What strategy will afford the best odds of winning the car, and what are the winning odds?



See my post above. I get 2/3 chance of winning.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Klopp
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March 10th, 2020 at 8:08:59 AM permalink
Yes, each contestant is allowed to open up to two doors. It is not relevant whether or not a contestant opens a second door if he/she is successful at the first trial. Of course, the contestant will always open a second door if the first trial is not successful. It is essential that the second player cannot see which doors were opened by the first player: suppose the second player would see that the first player stops after having opened one door, the second player would have perfect oinformation what was behind that door.
unJon
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March 10th, 2020 at 8:16:01 AM permalink
Quote: CrystalMath

Assuming player 2 can see which doors player 1 opened, but not their contents, then player 1 will choose door 1. If it contains player 2痴 item, then open door 3, otherwise door 2. Player 2 will open door 1if player 1 opened door 3, otherwise player 2 will open doors 2&3.

This gives a 2/3 success rate.



You don稚 need this sort of signal sending to get to 2/3.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
CrystalMath
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March 10th, 2020 at 8:41:24 AM permalink
Quote: unJon

Quote: CrystalMath

Assuming player 2 can see which doors player 1 opened, but not their contents, then player 1 will choose door 1. If it contains player 2痴 item, then open door 3, otherwise door 2. Player 2 will open door 1if player 1 opened door 3, otherwise player 2 will open doors 2&3.

This gives a 2/3 success rate.



You don稚 need this sort of signal sending to get to 2/3.



Ahh, yes. Very good.
I heart Crystal Math.
DJTeddyBear
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March 10th, 2020 at 8:41:39 AM permalink
Sigh...

Apparently, if you want to know if your answer is correct or not, ask me if I agree with it. 😡
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Wizard
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March 10th, 2020 at 9:50:41 AM permalink
Quote: unJon

I get 2/3 chance of winning.



So do I.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
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March 10th, 2020 at 10:35:15 AM permalink
Is it an immediate loss if either player picks the goat?
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Wizard
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March 10th, 2020 at 1:14:32 PM permalink
Quote: Gialmere

Is it an immediate loss if either player picks the goat?



Not in my variant.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
charliepatrick
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March 11th, 2020 at 4:32:09 AM permalink
Suppose Player 1 is trying to find the Car and Player 2 trying to find the Key.
This is similar to solution already given but in more detail...

Player 1 opens box #1.
(i) If it is a car then no further box is needed - also in boxes 2 and 3 there are a Goat and a Key (in either order) - thus successes are C G K or C K G.
(ii) If it is a Goat then Player 1 opens box #2 - thus if success boxes are G C K.
(iii) If it is a Key then Player 1 opens box #3 - thus if success boxes are K G C.
Therefore 4 of the 6 possibilities result in successfully finding the Car.

Player 2 opens box #3.
(i) If it is a Key then success (C G K or G C K).
(ii) If it is a Car, then Player 1 found the Car so must have seen the Key in Box #1 (K G C) - so now open Box #1.
(iii) If it is a Goat, then the Car was in Box #1 (C K G, with K C G player 1 would have opened Box 3 and failed) - so now open Box #2.
Thus if Player 1 found the Car, Player 2 can always find the Key.

So pr(success) - pr(Player 1 finds Car) = 2/3.
Wizard
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March 11th, 2020 at 7:31:19 AM permalink
The solution to this problem I found to be similar to that of Secret Santa Problem #2.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
gordonm888
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March 11th, 2020 at 12:30:42 PM permalink
Wizard's problem


I assume that when the second player picks, that there is no visual evidence as to which doors have been picked by the 1st player.

They agree that the first player will pick door #1, and then door#2 if Player#2's item was behind door#1, otherwise Player#1 will pick Door#3.

Player#2 then picks Door#2. If door#2 has the goat he picks door#3. If door#2 has player#1's item then he picks door#1.

When the items 1, 2 and G(for Goat) are arranged in doors #1,2,3 as shown the outcome of this strategy is as shown.

12G WIN
1G2 WIN
21G WIN
2G1 LOSE
G12 LOSE
G21 LOSE

So the odds are 50%, which is slightly better than 4/9 which is the probability if they both pick doors randomly.

Basically, this strategy guarantees a WIN whenever the items are arranged as 21G, but is no better than random for the 5 out of 6 arrangements. Other similar strategies can also enable a win for any one of the six arrangements, but will be no better than random for the 5 arrangements.
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unJon
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March 11th, 2020 at 12:47:46 PM permalink
Quote: gordonm888

Wizard's problem


I assume that when the second player picks, that there is no visual evidence as to which doors have been picked by the 1st player.

They agree that the first player will pick door #1, and then door#2 if Player#2's item was behind door#1, otherwise Player#1 will pick Door#3.

Player#2 then picks Door#2. If door#2 has the goat he picks door#3. If door#2 has player#1's item then he picks door#1.

When the items 1, 2 and G(for Goat) are arranged in doors #1,2,3 as shown the outcome of this strategy is as shown.

12G WIN
1G2 WIN
21G WIN
2G1 LOSE
G12 LOSE
G21 LOSE

So the odds are 50%, which is slightly better than 4/9 which is the probability if they both pick doors randomly.

Basically, this strategy guarantees a WIN whenever the items are arranged as 21G, but is no better than random for the 5 out of 6 arrangements. Other similar strategies can also enable a win for any one of the six arrangements, but will be no better than random for the 5 arrangements.



G21 is a winner not a loser under your strategy.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
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