1 coin, 3 cups ;)

[Snakedoc]

Take 1 coin, place it under a cup without being seen. Ask a friend to choose a cup but do not reveal what's under it. Take one cup away which does not have a coin under it, but not the cup they selected.

Ask them if they want to change their mind.

If they do, there is a much greater chance of winning. Try it, it works. Simple maths really. Who knows how it works? I do!

[Pullen]

Because then they have a 50/50 chance

[Snakedoc]

Nope! It's better than 50/50

[Pullen]

o yea its 2/3s isnt it, I read about it a couple of times before but never understood it

[Mr. Grapes]

probability increases from 33% to 50%

edit: DOH!

[Col]

Yeah, when you take the cup away it increases, but the odds don't change if they change their mind as far as I can work out.

[M4T VW]

If you only have 2 cups left then is it not 50%, 50/50?

What about if you had 2 girls and only one cup?

[Snakedoc]

Ok, to start with you have a 1 in 3 chance or 33%. By being told one of the cups but not the one you choose is empty, you increase the chances (if you swap) to 66% getting it right. Something like that lol.

[Smifis]

What about if you had 2 girls and only one cup?
Then you're a sick :censored:.

3 cup, coin under one, they choose a cup and you lift up a cup without the coin under, the other guy gets a bit :censored:y and punches you anbd gets the coin

[Lorem-Ipsum]

What about if you had 2 girls and only one cup?

Now I see why you use that avatar.............. lol.

[Terbinator]

I'd say both 33% and 50% are reasonable answers, 50% being the pedants take.

[coiler]

If you only have 2 cups left then is it not 50%, 50/50?

What about if you had 2 girls and only one cup?


Uh oh! Don't let Tystar read that, he's a right sick puppy!!! :lol:

[Rampageturke]

I totally misread this title

[Snakedoc]

That kinda was the idea lol.

[Snakedoc]

This may help:


The Monty Hall problem is a probability puzzle based on the American television game show Let's Make a Deal. The name comes from the show's host, Monty Hall. The problem is also called the Monty Hall paradox, as it is a veridical paradox in that the result appears absurd but is demonstrably true.

The problem can be unambiguously stated as follows:

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice? (Krauss and Wang 2003:10)

As there is no way for the player to know which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In fact, the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

[David84]

what i really want to know snake if you take that cup and coin trick to Vegas what's the probability you will ever come back?

[Snakedoc]

LOL, never would! Count some cards too!

[David84]

just make sure there's enough room in your suit case for me :)