Secondly, what is the probability of this happening 3 times in a ten draw set, assuming the decks are shuffled each time?

What is your attempt?

The probability of two persons drawing the same card from two standard decks is when both people draw either 2 Aces of Spade, 2 Two of Spades,.

Let's take one. The probability that 2 Aces of Spade are drawn.

P(2 Aces of Spade) = 2/(52+52) x 1/(52+51) = 1/5356

Now this could be for any of the 52 different cards present in the deck. So... can you complete it now?

Now, you can use a binomial distribution for that.

Let X be the event 2 similar cards are picked,

X ~ B(10, a)

Where a is the probability you obtained in the first part.

Find P(X = 3)

That's great, but is the initial probability you've figured an attempt to draw a specific or named card? Or any card similar to the "first" card drawn?

Isn't the probability for success calculated by the formula of 52 favorable outcomes / 2704 possible outcomes?

The first one is for a specific card.

Since there are 52 different cards, there are 52 times this probability to get the probability of any other cards.

I don't understand what you mean by the other question :confused:

And where did you get 2704 from?

Maybe my logic is flawed... I'm certainly no probability expert. Two decks. I draw a card from one, you draw a card from the other. I have 52 possible outcomes. For each possible outcome of mine, you have 52 possible outcomes. Isn't that a simple 52 x 52 for possible outcomes? With 52 possible matches?

Not quite...

In the first you have 104 possible outcomes (probability of taking any particular card = 2/104 = 1/52).

When you took it, I can take any of the (51+52) cards left, which changes the probability.

The way I interpret the question is: You have ONE deck of cards and so do I. Let's both pick a card. What is the probability they match?
Answer: 1/52 Why? I pick a card, we don't care what it is, but you have to match it.. Your chance of picking THAT particular card is 1/52.

What is the probability of two people drawing the same card from two 52-card decks.

:)

TWO 52-card decks... not ONE deck with 104 cards... the decks are indeed separate.

Correct... simplified. Which is why my math works as well. Person One, Deck One... 52 possible outcomes. Person Two, Deck Two... 52 possible outcomes. Therefore... 52 Favorable outcomes out of 2704 possibilities. Same probability. 1.923%

Now... who can compute the binomial distribution of that 1.923% probability happening in 3 out of 10 trials (assuming cards are replaced and shuffled after each)? I am lost on that formula... anyone care to do the math?

Ah, well, that changes some things.

And yes, then that becomes as you described.

For the second part... are the ten cards drawn one after the other or are the cards replaced each time until it is done 10 times?

Replaced and reshuffled...

Ok, so, you can use the binomial distribution.

Great.. so what is the answer? :P

Uh... you don't know how to use the binomial distribution?