Thread: probabiliy of equal cards

If we have two decks of 52 cards each that are shuffled randomly.And if the positions of cards for each deck are numbered from 1 to 52.
What is the probability that these two decks contain the same card at the same position.

Originally Posted by billobillo

If we have two decks of 52 cards each that are shuffled randomly.And if the positions of cards for each deck are numbered from 1 to 52. What is the probability that these two decks contain the same card at the same position.

If a standard decks is well shuffled what is the probability that no card is in its original position?

Originally Posted by Plato

If a standard decks is well shuffled what is the probability that no card is in its original position?

Well I guess your question is the same of mine, because having a non shuffled and a shuffled deck is like having two different decks.

Originally Posted by billobillo

Well I guess your question is the same of mine, because having a non shuffled and a shuffled deck is like having two different decks.

That is correct. So if is the probability of not having any card in its correct place then is the probability of at least one card in its correct place.

Do you know about derangements ?

Originally Posted by Plato

That is correct. So if is the probability of not having any card in its correct place then is the probability of at least one card in its correct place.

Do you know about derangements ?

Actually I'm not a math specialist, I tried to solve it in a easy way and I got p=1.
A card in position 1 in deck 1 has probability 1/52 to be equal to a card in position 1 in deck 2.
And since we have 52 cards, then the probability to get two equal cards at same position is 52 * 1/52 = 1
Could it be p=1?

Originally Posted by billobillo

Actually I'm not a math specialist, I tried to solve it in a easy way and I got p=1.
A card in position 1 in deck 1 has probability 1/52 to be equal to a card in position 1 in deck 2.
And since we have 52 cards, then the probability to get two equal cards at same position is 52 * 1/52 = 1
Could it be p=1?

You cannot work this question with out derangements. As far as I know there is no other way.
Did you read the link I gave you?
Why were you given this problem to do?

Two decks are shuffled. The first is the reference deck and I interpret that the question is what is the P that the second deck will have the exact same order as the first. I would answer 52 factorial is the number of possible events.P= 1/8*10^67

Originally Posted by bjhopper

Two decks are shuffled. The first is the reference deck and I interpret that the question is what is the P that the second deck will have the exact same order as the first. I would answer 52 factorial is the number of possible events.P= 1/8*10^67

You have got to be kidding!

Originally Posted by Plato

You cannot work this question with out derangements. As far as I know there is no other way.
Did you read the link I gave you?
Why were you given this problem to do?

Yes I read the link, it contains a similar question about persons and hats but with no solution,
I am a computer specialist and I am taking a course about data security and random attacks and it contains a probability part.

Originally Posted by bjhopper

Two decks are shuffled. The first is the reference deck and I interpret that the question is what is the P that the second deck will have the exact same order as the first. I would answer 52 factorial is the number of possible events.P= 1/8*10^67

The question is to find the probability for one common card at the same position,and not all common cards