I"m not sure I entirely understand the question. What do you mean by an "occurrence of same kind in a hand"?

Are you talking about Texas Hold "em? Maybe you mean that YOU are dealt a pair of 8"s and I am also dealt a pair of 8"s. Is that what you mean? Assuming that is the question, here is my answer:

Let"s just deal one card at a time - one to you, then one to me, then one more to you, then one more to me.

The first card doesn"t matter at all - whatever it is, it is. Let"s say it happens to be a "9" dealt to you.

The next card dealt (to me) MUST be a "9". There are three 9"s left, but there are only 51 cards left in the deck. So far we are at a 3/51 chance.

Now the next card (to you) MUST be a "9". There are two 9"s left, but there are only 50 cards left in the deck. That is a 2/50 chance.

Now the last card (to me) MUST be a "9". There is only one 9 left, and there are only 49 cards left in the deck. That is a 1/49 chance.

So our formula is 3/51 * 2/50 * 1/49. You could simplify that to 1/17 * 1/25 * 1/49.

That is a 1 out of 20,825 chance.

That is your answer IF, and only IF it was just you and I playing. As soon as we start adding more people to the table, it will be EASIER to get 2 people with the same pocket pair. That"s because if YOU have a pair of 9"s, ANYONE at the table might also have a pair of 9"s. PLUS, someone else might also have a different pair at the same time that you have a pair of 9"s. If someone else has a pair of Queens, ANYONE at the table might also have a pair of 9"s like you, but they could have a pair of Queens.

Figuring out the chances with multiple people at the table is beyond my mathematical ability.

Heads up, the chance that 2 people share the same pocket pair is 1 out of 20,825. If this is NOT what your question was asking for, please clarify.

***EDIT***

AH HAH!!! NOW I understand!

You are talking about how many "4-of-a-kind" there are in poker! Now that I understand the question, this one is an easy answer. Allow me to explain:

What you need to remember about poker is that it is a 5-card game. You need 5 cards to have a complete poker hand.

So you are imagining there is only ONE hand that can be Queen-Queen-Queen-Queen. The truth is, there are 48 hands that can have Queen-Queen-Queen-Queen.

You could have Q-Q-Q-Q with the Ace of clubs.

You could have Q-Q-Q-Q with the Ace of diamonds.

You could have Q-Q-Q-Q with the Ace of hearts.

You could have Q-Q-Q-Q with the Ace of spades.

You could have Q-Q-Q-Q with the Two of clubs.

You could have Q-Q-Q-Q with the Two of diamonds.

You could have Q-Q-Q-Q with the Two of hearts.

You could have Q-Q-Q-Q with the Two of spades.

You could have Q-Q-Q-Q with the Three of clubs.

You could have Q-Q-Q-Q with the Three of diamonds.

You could have Q-Q-Q-Q with the Three of hearts.

You could have Q-Q-Q-Q with the Three of spades.

OK - you understand what I mean I hope. I only described 12 different ways to have a "4-of-a-kind" with Queens. If you keep going on with 4-queens with a "4", 4-queens with a "5", 4-queens with a "6", etc., you will see there are FORTY-EIGHT ways to make 4-queens!

THEN, you need to do that with EVERY "4-of-a-kind" that there is. You can have 2-2-2-2-3, 2-2-2-2-4, 2-2-2-2-5, 2-2-2-2-6, etc., etc., etc.

OK - to your FINAL statement in your original answer. You say, "Then answer should be 13 * 52 instead of 13 * 48".

Let"s look at the "4-of-a-kind" in Queens again. Without stating EVERY possible suit, you get four of each of the following:

QQQQ-Ace, QQQQ-Kind, QQQQ-Jack, QQQQ-10, QQQQ-9, QQQQ-8, QQQQ-7, QQQQ-6, QQQQ-5, QQQQ-4, QQQQ-3, and QQQQ-2. That"s 12 different ways. Multiply that by 4 different suits and you get 48.

Where you went wrong in your thinking is that you can"t have Q-Q-Q-Q- with another QUEEN! The 4 queens are already gone!