If so, I offer this simple solution, which I would hope can easily be improved upon.

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.

Can the second player see which doors player 1 opened?

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.

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?

This gives a 2/3 success rate.

Quote:WizardWhile 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.

Quote:CrystalMathAssuming 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.

You don’t need this sort of signal sending to get to 2/3.

Quote:unJonQuote:CrystalMathAssuming 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.

You don’t need this sort of signal sending to get to 2/3.

Ahh, yes. Very good.

Apparently, if you want to know if your answer is correct or not, ask me if I agree with it. 😡

Quote:unJonI get 2/3 chance of winning.

So do I.

Quote:GialmereIs it an immediate loss if either player picks the goat?

Not in my variant.

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.

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.

Quote:gordonm888Wizard'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.