How dumb are you? (Riddle)

[Bostaevski]

[ QUOTE ]
You are playing a game in which you and two other players sit at a round table facing each other. At the start of each round, a light behind you will flash either red or white. The color is completely random for each player. You can not see your own light, but can see each of the other two players' lights.

After the lights flash, each player must guess which color their own light was. They may either guess red or white, or pass.

If all three players pass, the round is a push. If at least one player guesses and all players who guess are correct, the players win $1,000. But if at least one player guesses and any players are wrong, the players lose $1,000.

You can not communicate with the other players in any way during the game. No speaking, no hand signals, etc. However, you may meet with the players beforehand to come up with a strategy.

Obviously, if all players pass each round the players will break even. If all players guess each round, all three will be right only one time out of eight (although random chance may swing that either way) and would expect to lose money. If two players pass each round and one player guesses each time, the players can expect to break even.

Is there a valid strategy for this game that allows the players to have a positive expectation?

[/ QUOTE ]

I dunno if this counts
I would say "ok if you see a red light above the guy's head who is to your left, then you say 'pass'". Then the guy to the left says "Red". If nobody says "pass" then you can say "white".

Or did you forget to mention that they have to guess in order?

[Maple Leafs]

[ QUOTE ]
Or did you forget to mention that they have to guess in order?

[/ QUOTE ]
Let's say they guess at the same time -- they have buttons in front of them labelled "red" "white" and "pass" and they all lock in at the same time.

The answer does not involve the players signalling to each other in any way.

[dotbum]

Only guess when the two colours in front of you match, and then guess the same colour.

If my Maths is right, 1/4 of the time the lights will be monotone, and they will win 3k. 3/8 of the time one person will guess wrong and they will lose 1k.

EV: +$375?

[Maple Leafs]

[ QUOTE ]
If my Maths is right, 1/4 of the time the lights will be monotone, and they will win 3k. 3/8 of the time one person will guess wrong and they will lose 1k.


[/ QUOTE ]
They win $1K each time they're right as a group. It's not $1K per person who's right.

Same deal if they're wrong, they lose $1K whether it was one person or all of them.

[FishSticks]

lol wtf is this? If one guy guesses, it's 50/50. Seeing the other people's lights doesn't matter. If two guys guess, it's not like they hit a parlay and double their win for both hitting, so they just screw themselves by both guessing.

What the hell kind of a ridiculous game is this anyways? It doesn't sound fun, probably won't catch on in the casinos.

[T-God]

OP: I solved the riddle of which banned poster you are.

[emKay]

[ QUOTE ]
[ QUOTE ]
Or did you forget to mention that they have to guess in order?

[/ QUOTE ]
Let's say they guess at the same time -- they have buttons in front of them labelled "red" "white" and "pass" and they all lock in at the same time.

The answer does not involve the players signalling to each other in any way.

[/ QUOTE ]

I made up a +500 ev situation when they are allowed to lock in at a different time. Before they start the game, they pick a leader. The leader acts before the other 2 guys. He passes each time the other 2 players have the same color. Other 2 guys see the color of each other, guess it right and win 1k. Thats the case 50% of the time.
The other 50% of the time they show different lights. Then the leader guesses any of the 2 colors. The other 2 guys now know they have to pass. In this case they are breakeven, so overall they make 0 half of the time and and 1k half of the time.

I don't believe theres a +ev strategy when they have to log in at the same time.

[Maple Leafs]

[ QUOTE ]
I don't believe theres a +ev strategy when they have to log in at the same time.

[/ QUOTE ]
There is.

I'll post the answer later today, no sense having it drag on.

[doormat]

[ QUOTE ]
You are playing a game in which you and two other players sit at a round table facing each other. At the start of each round, a light behind you will flash either red or white. The color is completely random for each player. You can not see your own light, but can see each of the other two players' lights.

After the lights flash, each player must guess which color their own light was. They may either guess red or white, or pass.

If all three players pass, the round is a push. If at least one player guesses and all players who guess are correct, the players win $1,000. But if at least one player guesses and any players are wrong, the players lose $1,000.

You can not communicate with the other players in any way during the game. No speaking, no hand signals, etc. However, you may meet with the players beforehand to come up with a strategy.

Obviously, if all players pass each round the players will break even. If all players guess each round, all three will be right only one time out of eight (although random chance may swing that either way) and would expect to lose money. If two players pass each round and one player guesses each time, the players can expect to break even.

Is there a valid strategy for this game that allows the players to have a positive expectation?

[/ QUOTE ]

I believe they can win 6 times out of 8 with the following strategy: Pass whenever you see two different colors and bet the opposite color when you see two of the same color.

For example, all the combinations with two red lights are:

1) RRR
2) RRW
3) RWR
4) WRR

In case #1 all three guess white and the group loses. The group wins in cases 2, 3, and 4 because only one of them sees two reds and correctly guesses white. So the group wins three times and loses once. They also win three times out of four when at least two whites appear.

doormat

[PBAR]

Interview question:
You have 9 jars containing an infinite number of marbles of identical size. All the jars contain marbles weighing 1 gram, except one jar has marbles weighing 1.1 grams. You have a scale and are allowed ONE measurement. How can you figure out which jar contains the marbles weighing 1.1 grams?