Originally Posted by Warpe

the dots were put on on in a dark room...

Originally Posted by Warpe

But, if there is a correct solution, and each logic master has an equal opportunity to solve it, then each logic master needs to see two red dots to face the same conundrum.

Originally Posted by thenonsequitur

Heh, we arrived at the same conclusion at the same time...You stated it better than me though (and besides I was in the middle of revising my post as you posted, so technically you got it right first).

Originally Posted by Warpe

In the meantime, let's start a thought experiment:

You're at a poker table where the players only have the option of four actions: check, fold, call or push. We also assume that there will be no action on any streets beyond the flop.

You are dealt AA on the Button. Villain, a thinking player, is UTG and raises 6xBB preflop, you reraise to 18xBB, Villain calls, while also correctly surmising what your hold cards are in the process.

Flop is dealt...KKK.

Villain pushes.

Call or fold?

Give reasons for your decision.

Originally Posted by Warpe

Three red dots get the same reaction as two red dots and one blue (three hands in the air), therefore one cannot know by eveyone's reaction what colour one's dot is. It could be blue or it could be red. But, if there is a correct solution, and each logic master has an equal opportunity to solve it, then each logic master needs to see two red dots to face the same conundrum. Anyone seeing one red dot and one blue dot and three hands in the air would know their dot is red, so that can't be correct. They all have to be red.

Edit: Damn, somone beat me to it. I didn't read before posting.

Originally Posted by givememyleg

?

Did you mean postflop errr what ?

Originally Posted by ark4748

So I'm trying to wrap my head around this... Initially, the puzzle is unsolveable because the reactions will be the same for either two or three red dots. But if we assume that the grandmaster has devised a fair test for the masters, then there has to be three red dots, otherwise one student would have an easy solution to the problem..... Can we assume that the grandmaster is fair? I guess we need to otherwise the puzzle isnt solveable.

Originally Posted by Warpe

let's start a thought experiment

Originally Posted by thenonsequitur

For this, are we to assume that the stacks are both very deep?

Originally Posted by Warpe

Villain calls, while also correctly surmising what your hole cards are in the process.

Originally Posted by thenonsequitur

For this, are we to assume that the stacks are both very deep?

Originally Posted by Warpe

That sounds like fun. 200BB should do it.

Originally Posted by givememyleg

Also, this experiment is kind of silly. Just the fact that the only options are "check, call, push, fold" sort of skews this. Is this a built in rule at this game? If that was the case, I would be more inclined to think we were ahead. Villian only has 2 optins here, check or push.

If this was a normal game, with normal betting and villian open pushed here how would you think your reaction should change?

Originally Posted by givememyleg

I don't see how Hero can fold here.... would yoooooou?

Originally Posted by yorib

The counter-argument is this: If there were a severe punishment for getting the answer wrong, then if you saw a red dot and a blue dot you would know the answer, but if you were to "slow play" you could convince the person with a blue dot, that his dot were, in reality red. The problem is, that two people have to agree to slowplay (kind of like checking down a small stack on the bubble).

Originally Posted by Warpe

On the face of it, quads are normally/often slowplayed, right? So it follows that if Villain has a K, pushing is the last thing we should expect him to do. But if our thinking villain knows that that is our expectation...

Originally Posted by pgil

the guy at the front says, i have a red hat. here goes:
there must be at least 1 red hat on one of the two front people, otherwise the last guy can simply say he has a red hat as there are only 2 blues. the middle guy knows this, and so if he sees that the guy in front of him has a blue hat, then his must be red (unless the guy in back has already spoken). therefore, the guy in front must be wearing a red hat.

Originally Posted by IowaSkinsFan

Here's another puzzle:

A man has 12 metal balls in front of him. 11 are the same weight however one is heavier or lighter (you are not sure which)than the other balls. All you have with you is a balance to weigh the balls against each other, the balance does not give the weight of the balls. In 3 tries, how can you figure out which ball does not weigh the same as the others. Note there are multiple answers and you have to answer many possible happening when coming to the answer.

Originally Posted by tom

1) split the balls into 2 groups of 6 and weigh them against each other
2) whichever group of 6 is heaver, split into 2 groups of 3 and weigh them against each other
3) take the group of 3 that is the heaviest and weigh just 2 of the balls. If they are the same weight, the ball you didn't weigh is the heaviest. If they are not the same weight, whichever ball is heavier of the 2 is your answer.

Originally Posted by IowaSkinsFan

Originally Posted by JKDS

rest them all on the balance and let them roll off at the same time onto the same surface. Do this three times. The one that bounces higher or lower than the rest weighs different since it would have a greater or weaker gravitational force and thus a greater or weaker normal force once hitting the ground.

ya i cheated but ship it.

Originally Posted by tom

1) split the balls into 2 groups of 6 and weigh them against each other
2) whichever group of 6 is heaver, split into 2 groups of 3 and weigh them against each other
3) take the group of 3 that is the heaviest and weigh just 2 of the balls. If they are the same weight, the ball you didn't weigh is the heaviest. If they are not the same weight, whichever ball is heavier of the 2 is your answer.