Do you think this a fair bet?

Question

A friend offers you the following wager. He takes 3 playing cards from a pack; 2 kings and an ace. Having mixed them up, he then puts them face down on a table and asks you to slide one card out still keeping it face down. The object of the game is to pick the ace. The card you have selected is purely random because you don't know in what order they were dealt BUT HE DOES and with this knowledge he proceeds to turn over one of the 2 remaining cards, showing it to be one of the kings. "Now there are only 2 cards left, he says, "one you selected and the last one remaining. If you have picked the ace I will give you £1 if not, you give me £1".

Is this a fair bet?

Answer 1

No they know what order they put the cards down and they are watching what order you are putting them back in so they know they have a much bigger chance of winning than you  because you are picking them at random.

Answer 2

Mathematically yes it is fair bet. If there is NO other trick. One is ace and one is king so chances of the selected card is an ace are 50-50.

But if he KNOWS the cards by making any markings on the back side or something then it is not worth.

Answer 3

No it isn't a fair bet.

The odds are still stacked in your opponent's favour. There are 3 cards to choose from, your chance of selecting the ace is 1 in 3. You choose a card. Your opponent has two cards, at least one of which must be a king. Him turning over a card to reveal a king does not change the odds from your point of view. It is still 1 in 3 for you and 2 in 3 for him.

However, if a king was revealed and you were given the option to exchange your card with the one yet to be revealed, then you have a better chance. Believe it or not, you should always exchange. It is hard to explain, but basically you are moving from a 1 in 3 situation to a 1 in 2 situation. (You don't know whether the unrevealed card is a king or ace, regardless of whether your opponent does or not). In other words, your new 1 in 2 situation is more likely to give you the ace than your original 1 in 3. So you should choose the option from 1 in 2, ie not the card you originally chose, but the card you are given the option to swap. This counter intuitive situation is commonly used in business/team building exercises to demonstrate that you should be flexible with your ideas. Very few people initially accept that their original choice might not be the best one despite the change of circumstances.

Answer 4

how does the dealer know where it is, as he mixed them up b4 putting them down.

if he was  watching where the ace was going, why couldnt you

Answer 5

It's not a fair bet.

Revealing one of the remaining cards to be king stacks the odds in the friend's favour, unless he gives me a second chance to either stay with my choice or pick the other remaining card. There are bigger odds that the last remaining card is the ace.

It's called The Monty Hall Problem and it's one of the most puzzling probability puzzles.

Answer 6

No. This is the Monty Hall problem. The probability that you have the ace is 1/3. The probability that you have a king is 2/3. If he offered you £2 for having the ace and took £1 from you for not having the ace it would be a fair bet...