March 10th, 2020 at 6:50:17 AM
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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?

March 10th, 2020 at 7:18:40 AM
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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.

Show Spoiler

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.

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.

March 10th, 2020 at 7:33:45 AM
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Question -- Can each contestant choose the second door after seeing what is behind the first door?

Last edited by: Wizard on Mar 10, 2020

It's not whether you win or lose; it's whether or not you had a good bet.

March 10th, 2020 at 7:41:50 AM
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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.

March 10th, 2020 at 7:42:38 AM
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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?

I heart Crystal Math.

March 10th, 2020 at 7:44:40 AM
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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/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁

March 10th, 2020 at 7:58:44 AM
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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/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁

March 10th, 2020 at 7:59:03 AM
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Secret Santa logic:

Show Spoiler

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’s the car behind the door, he’s done. If it’s the keys (#2) he opens door 2 next. If it’s 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.

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

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

March 10th, 2020 at 7:59:28 AM
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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?

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?

It's not whether you win or lose; it's whether or not you had a good bet.

March 10th, 2020 at 8:00:37 AM
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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’s 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.

This gives a 2/3 success rate.

I heart Crystal Math.

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