Thread: Conditional probability involving deck of cards.

I had a discussion with a friend the other day:

Deck of cards (52 cards). 13 of them are hearts. Well shuffled, randomly dealt.

Situation 1
I'm sitting at a table by myself with 2 cards in my hands. Both are hearts.
The rest of the deck is on the table.
X = The probability that the top card of the remaining deck, when turned, is a heart.

Situation 2
I'm sitting at a table together with 4 other people we all have 2 cards in our hands. I do not know which cards are held by the other players. Both my cards are hearts.
The rest of the deck is on the table.
Y = The probability that the top card of the remaining deck, when turned, is a heart.

Question
X = Y ?

And why / why not?

Hello, drwmn!

I had a discussion with a friend the other day.

Deck of cards (52 cards). 13 of them are hearts. Well shuffled, randomly dealt.

Situation 1
I'm sitting at a table by myself with 2 cards in my hands; both are hearts.
The rest of the deck is on the table.
X = The probability that the top card of the remaining deck, when turned, is a heart.

Situation 2
I'm sitting at a table together with 4 other people; we all have 2 cards in our hands.
I do not know which cards are held by the other players.
Both my cards are hearts. .The rest of the deck is on the table.
Y = The probability that the top card of the remaining deck, when turned, is a heart.

Question
X = Y ?

And why / why not?

The answer is YES; the probabilities are equal.


Situtation 1

You have two Hearts.
The rest of the deck has 50 cards: 11 Hearts and 39 Others.
. .


Situtation 2

You have two Hearts.
The rest of the deck has 50 cards: 11 Hearts and 39 Others.

Randomly select six cards from the deck
. . and move them to the bottom of the deck.
(These are the cards dealt to the other three people.)

No matter what six cards were moved,
. . the probability that the top card is a Heart remains the same.