Simple probability problem

[bonanova]

Four cards are taken from a standard deck, one from each suit.

They are shuffled and dealt face down on a table.

Coins are then placed at random on two of the cards.

What is the probability that the two chosen cards are the same color?

[Morningstar]

1/3?

By simple counting, there are six ways the coins can be placed on the cards, because the coins are not differentiated. There is one way per color to choose two cards of that color, so two out of six possibilities satisfy the conditions, or 1 in 3.

Edited May 8, 2012 by Morningstar

[Barcallica]

1/6?

[WitchOfSecrets]

Two answers.

1/3. imagine putting down one coin on any card. Three remain. There is only a 1/3 chance the second coin will be on the other red/black card.

100%. If they're face-down, all are the same color.

Edited May 8, 2012 by WitchOfSecrets

[Peekay]

1/3 There are 2 black and 2 red cards on the table. The first coin can be put on any card (p=1). Once that is done, there are 3 cards left out of which one is the same color as the one with the first coin. So probability = 1/3.

[Morningstar]

Something makes me feel like this is too easy. Maybe it's bonanova's mysterious smiley face avatar. Then again, I may be over-thinking.

[Molly Mae]

1/5

My reasoning is thus: For the first coin, the colour chosen is arbitrary. Of the remaining three cards, one of them is the colour you need. So even though my psychiatrist disagrees, a fifth is the answer to all problems.

Edited May 10, 2012 by Molly Mae

[mmiguel]

Four cards are taken from a standard deck, one from each suit.

They are shuffled and dealt face down on a table.

Coins are then placed at random on two of the cards.

What is the probability that the two chosen cards are the same color?

4 cards from each suit.

2 are red, 2 are black.

Placed on table at random.

Coins placed on 2 of them.

There are 4 choose 2 = 6 ways of placing the coins.

Of these combinations, there are only two in which cards under coins are same color.

Each combination is equally likely (P = 1/6).

Two combinations are mutually exclusive and satisfy event of interest.

Hence probability is 2*1/6 = 1/3