Leaving out the jokers, leaving only 52 cards. I tried two ways and came with different results. Method 1: Some player holding 4 aces would have a probability of NCR(48,9)/NCR(52,13), NCR is a function meaning n choose r. And for four players, each player has the same probability, so total probability of any player holding 4 aces is 4*NCR(48,9)/NCR(52,13)=0.015642.
Method 2: The probability of ace of spade belonging to some player is 1/4, and for ace of heart, the probability is also 1/4, so the probability of 4 aces to belong to the same player is 1/4 * 1/4 * 1/4 * 1/4=1/256, and by same reasoning as above, the probability should be 4 * 1/256=1/64=0.015625.
so, which one is correct?
Thank you...
Being told that the first answer is the right one isn’t going to help you, so let’s look more closely.
If you think about the second answer, you know it’s wrong. Instinctively it’s wrong because it seems as if dealing (say) the ace of spades doesn’t change the state of play. You can formalise the wrongness by taking an extreme example. What is the probability of a player getting all the red cards in the pack? You and I know that the probability is zero, but the method of the second answer gives $(\frac 1 2)^{26}$, which is not zero.
The key to seeing what the answer should be is to consider what happens when someone has the ace of spades. There are then 51 cards left in the pack, and the player has 12 cards to go - so the probability of getting the ace of hearts is $\frac{12}{51}$, not $\frac{13}{52}$ as your second answer would suggest.
The aces of diamonds and clubs would get you factors of $\frac{11}{50}$ and $\frac{10}{49}$. Multiplying it all out, the result equals your first answer.
You can if you like reduce all this to a formula, as you did in your description. But the key to problems of this kind is not to “guess which formula to use” but go down to first principles and work from there.