How many ways can a 5-card poker hand be dealt so that 4 of

Question

How many ways can a 5-card poker hand be dealt so that 4 of the 5 cards have the same face value (e.g. Four of a Kind)?

shishirgupta77 · Answer

There are 13 ways you can do that. and then remaining one card can have 48 different value.

Answer = 13 * 48

Rupstar · Answer

13c1*3c3*48C1

13 sets choose 1 card
picking the next three cards of that set
and ways of picking one card from the remaining 48

is this right? or do we not even consider what the last cards outcome is or something?

HongHu · Answer

Looks right to me. 

What would be the possibility of it?

Rupstar · Answer

A correction to my previous post i should have written
13c1*4c4*48c1
(mere technicality)
but my question was, why consider what the other card is since we only want to know if we can get the set of 4, so why multiply it by the various different cards that can be pulled...does this make sense?
eg: if i get A4AAA i only want to know what the probability of getting AAAA in 5 chances would be
why account for possibility of AA5A; AAA4A....and so on

Antmavel · Answer

I would consider 52*3*2*1*48 What's the Oa for this problem ? If my answer is wrong, can you please explain why...thanks

HongHu · Answer

That's why I asked "what is the probability" in my last post. This question asks "how many ways". Since "AAAA5" and "AAAA4" is counted as two ways, you definitely need to multiply that 48 to get the total number of outcomes. However, if you were to calculate the probability, then you don't need to count it. Just remember you can't count it either when you do the denominater.

HongHu · Answer

Your problem is that your solution is ordered, as if the first A is somehow different from the second A. For example, if you pick spade A in your first pick, and heart A in your second pick, you'd count it as a pick that is different from a heart A first and then spade A. To get to the correct answer from your approach you'd have to divide your answer by P(4,4).

Rupstar · Answer

Awesome!
 Honghu thanks for clarifying that !

lastochka · Answer

To your point, why shouldn't you count the example you provided as two distinct ways? I still get confused when looking at computation vs. permutation... I do understand that with computation the order does matter, while with permutation it doesn't.

Antmavel · Answer

I really want to understand this problem

I understood your first post : 13c1*3c3*48C1 1 card out of 13 potential set + the 3 same + any other card However I don't get your second post with correction : 13c1*4c4*48c1 ??? what is the logic here, why 4C4 ? for me it seems your meaning is 4 identical cards + any other one but you multiply the four cards by 13 sets because you can not be sure of whcih set what is the difference between both expression ? except on the result ? what;s the logic behind it....moreover I stillhave difficulty to understand why my previous answer is considered as ordered : 52*3*2*1*48 the first pick there are 52 choices and then 3 choices then ....

HongHu · Answer

Combination is when order doesn't matter, while permutation is when order does matter.

This question asks how many ways a five card hand can be dealt. So as long as it's the same five card hand, it is one way. The order of the five cards in the same hand doesn't matter.

halahpeno · Answer

Not to be a pain, but what about jokers? Or am I thinking too much?

HongHu · Answer

We are assuming no jokers around. :-D

How many ways can a 5-card poker hand be dealt so that 4 of

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