How Many Possible Poker Hands Are In A 52 Card Deck?

Suicane75 · 08-25-2004, 02:19 AM

Ok, i'm not ashamed to admit that trying to figure this out makes my head hurt. How many 2 card starting hands are possible in a game of poker using a regulation 52 card deck? It isn't 104 is it? That seems low.

thesloppy · 08-25-2004, 02:23 AM

My probability is rusty at best, but I'd go with 51 * 52 = 2652.

On second thought, that's sure to be wrong, but I'm sure some of the gentlemen around here will be able to help you out.

Logically I would say you take one beginning card, and there are 51 possible hands involving that card and any other card. Throw your original card out, since it's already been accounted for in all possibilities, and move onto the next card, leaving you with 50 possible hands incolving your second card and any other possible card except for the first. Continuing this trend with every possible card would give you 51+50+49+48.....etc etc until you get to one possible hand remaining. Excel tells me that would be 1326 But I'm certainly not 100% sure on that answer either..

Brillig · 08-25-2004, 02:36 AM

Well, there isn't a simple answer for this.

52 * 51 is correct (2652) in one way.

If you don't care about the order the hands are dealt, it's half that (1326)

If you want to know how many different hands are possible, doing the typical Hold'em separation between suited and off-suit hands, there are 169 possible hands.

Suicane75 · 08-25-2004, 02:44 AM

But, say my first card is a 2, there are 51 possible hands with that.
So wouldn't it be 13 * 51= 663?

Pumpy Tudors · 08-25-2004, 02:49 AM

Quote:

Originally Posted by Suicane75

But, say my first card is a 2, there are 51 possible hands with that.
So wouldn't it be 13 * 51= 663?

No, if your first card is a specific 2, there are 51 possible hands with that. Since there are 4 2's in a deck, then the total is 4 * 13 * 51 == 2652, unless you don't care in which order the cards are dealt, as stated earlier.

sooner333 · 08-25-2004, 02:50 AM

Quote:

Originally Posted by Suicane75

But, say my first card is a 2, there are 51 possible hands with that.
So wouldn't it be 13 * 51= 663?

No, because there are four types of 2's...you can start out with any of 52 cards. Then get dealt any of the remaining 51.

Suicane75 · 08-25-2004, 02:59 AM

I don't understand, and I doubt I will, lol. It just seems that no matter which i'm 2 dealt, i have 51 possible hands that can come out of that, the same with 3,4,5 etc.

TLK · 08-25-2004, 03:02 AM

Quote:

Originally Posted by Suicane75

I don't understand, and I doubt I will, lol. It just seems that no matter which i'm 2 dealt, i have 51 possible hands that can come out of that, the same with 3,4,5 etc.

Quote:

Listen, here's the thing. If you can't spot the sucker in the first half hour at the table, then you ARE the sucker.

Just playin

sooner333 · 08-25-2004, 03:06 AM

Quote:

Originally Posted by Suicane75

I don't understand, and I doubt I will, lol. It just seems that no matter which i'm 2 dealt, i have 51 possible hands that can come out of that, the same with 3,4,5 etc.

But you're now assuming you WILL get a two. However, if you get a 2, then there are only 51 possibly hands.

I guess the only way I know how to say it is that you start out with zero cards, you get any one of 52 cards in the deck. So now you have your one card (it could be any card in the deck). Then your next card can be any other card in the deck, but since you already have one, it cannot be that. So you have 52 possibilities times 51 possibilities.

Suicane75 · 08-25-2004, 03:09 AM

Ok, i understand the theory, i just dont comprehend it, but thats nothing new, lol.

Suicane75 · 08-25-2004, 03:10 AM

Quote:

Originally Posted by TheLionKing

Just playin

HEY!!! 2nd place baby.

Suicane75 · 08-25-2004, 03:14 AM

Quote:

Originally Posted by sooner333

But you're now assuming you WILL get a two. However, if you get a 2, then there are only 51 possibly hands.

I guess the only way I know how to say it is that you start out with zero cards, you get any one of 52 cards in the deck. So now you have your one card (it could be any card in the deck). Then your next card can be any other card in the deck, but since you already have one, it cannot be that. So you have 52 possibilities times 51 possibilities.

Ok, and i don't wanna belabour (sp?) a point that I admit i have no clue about, but I WILL get 1 of 13 cards right? So once i get that 1 card then I have 51 other possibilities, which means...13 * 51? Arghhh, i'm arguing and I know I have to be wrong, whats the matter with me? Thanks for your replies guys, I do apreciate it, even if i sound like a simpleton.

TLK · 08-25-2004, 03:21 AM

Quote:

Originally Posted by Suicane75

HEY!!! 2nd place baby.

see you this weekend

..... say you prayers, eat your vitamins.... brother

Pumpy Tudors · 08-25-2004, 03:37 AM

Quote:

Originally Posted by Suicane75

Ok, and i don't wanna belabour (sp?) a point that I admit i have no clue about, but I WILL get 1 of 13 cards right? So once i get that 1 card then I have 51 other possibilities, which means...13 * 51? Arghhh, i'm arguing and I know I have to be wrong, whats the matter with me? Thanks for your replies guys, I do apreciate it, even if i sound like a simpleton.

No, you'll get 1 of 52 cards. If you only want to talk ranks, then you'll get 1 of 13 ranks. Your next card will also be one of 13 ranks. So it's 13 * 13, which means 169.

korme · 08-25-2004, 04:10 AM

suicane has never sounded so dumb

QuikSand · 08-25-2004, 08:19 AM

Good heavens, someone lock this thread.

Joe · 08-25-2004, 08:53 AM

yeah, stop being so dumb

Bee · 08-25-2004, 08:56 AM

From my experience the odds of getting a 2,7 offsuit is about 1 in 5 so I'll go with that.

Butter · 08-25-2004, 09:22 AM

http://binomial.csuhayward.edu/poker.html

Butter · 08-25-2004, 09:24 AM

Dola, I just realized this doesn't really answer your question at all. Oh well. I'm a poor reader.

Samdari · 08-25-2004, 09:40 AM

Quote:

Originally Posted by Suicane75

I don't understand, and I doubt I will, lol. It just seems that no matter which i'm 2 dealt, i have 51 possible hands that can come out of that, the same with 3,4,5 etc.

Why are you allowing the second card to be distinguished by suit, but not the first?

GabeRivers · 08-25-2004, 11:30 AM

I'm wondering the same as Samdari. I haven't played poker in years, and even when I did it wasn't much, but doesn't poker rank suits similar to bridge: spades, hearts, diamonds, clubs? In other words, if the two best hands were Ace (spades)/ten versus Ace (clubs)/ten, doesn't the Ace (spade) win the hand?

If so, it would seem to my tired old mind that the answer to the question would be 52 times 51.

Samdari · 08-25-2004, 11:32 AM

Quote:

Originally Posted by GabeRivers

I'm wondering the same as Samdari. I haven't played poker in years, and even when I did it wasn't much, but doesn't poker rank suits similar to bridge:

Actually, no.

Like any card game, some people may play differently than the "according to Hoyle" rules, but the official rules of poker view all suits equally.

GabeRivers · 08-25-2004, 11:38 AM

Too much bridge, I guess, and not enough poker.

Whar · 08-25-2004, 11:41 AM

Sloppy covered all this. If I only read the whole post.

1326 is the correct answer.

Suicane75 · 05-10-2013, 08:59 AM

I think this is right around the time I started smoking pot.

sovereignstar v2 · 05-11-2013, 02:50 PM

Quote:

Originally Posted by Suicane75

I think this is right around the time I started smoking pot.

This is right around the time that I was the hottest driver on the circuit. I'm talking Sliders here.

Suicane75 · 05-14-2013, 10:48 AM

Listen, I know you aren't Finnish.

08-25-2004, 02:23 AM	#2
thesloppy Pro Starter Join Date: Feb 2003 Location: PDX	My probability is rusty at best, but I'd go with 51 * 52 = 2652. On second thought, that's sure to be wrong, but I'm sure some of the gentlemen around here will be able to help you out. Logically I would say you take one beginning card, and there are 51 possible hands involving that card and any other card. Throw your original card out, since it's already been accounted for in all possibilities, and move onto the next card, leaving you with 50 possible hands incolving your second card and any other possible card except for the first. Continuing this trend with every possible card would give you 51+50+49+48.....etc etc until you get to one possible hand remaining. Excel tells me that would be 1326 But I'm certainly not 100% sure on that answer either.. Last edited by thesloppy : 08-25-2004 at 02:36 AM.

08-25-2004, 09:22 AM	#19
Butter Coordinator Join Date: Mar 2002 Location: Dayton, OH	http://binomial.csuhayward.edu/poker.html __________________ My listening habits

08-25-2004, 09:24 AM	#20
Butter Coordinator Join Date: Mar 2002 Location: Dayton, OH	Dola, I just realized this doesn't really answer your question at all. Oh well. I'm a poor reader. __________________ My listening habits

08-25-2004, 11:41 AM	#25
Whar Mascot Join Date: Jun 2004	Sloppy covered all this. If I only read the whole post. 1326 is the correct answer. __________________ Whar If you knew how much time and energy this took you would laugh at me. Last edited by Whar : 08-25-2004 at 11:47 AM.

08-25-2004, 02:36 AM	#3
Brillig College Prospect Join Date: Oct 2000 Location: Mountain View, California	Well, there isn't a simple answer for this. 52 * 51 is correct (2652) in one way. If you don't care about the order the hands are dealt, it's half that (1326) If you want to know how many different hands are possible, doing the typical Hold'em separation between suited and off-suit hands, there are 169 possible hands.

08-25-2004, 02:44 AM	#4
Suicane75 Coordinator Join Date: Sep 2003 Location: NJ	But, say my first card is a 2, there are 51 possible hands with that. So wouldn't it be 13 * 51= 663?

08-25-2004, 02:59 AM	#7
Suicane75 Coordinator Join Date: Sep 2003 Location: NJ	I don't understand, and I doubt I will, lol. It just seems that no matter which i'm 2 dealt, i have 51 possible hands that can come out of that, the same with 3,4,5 etc.

08-25-2004, 03:09 AM	#10
Suicane75 Coordinator Join Date: Sep 2003 Location: NJ	Ok, i understand the theory, i just dont comprehend it, but thats nothing new, lol.

08-25-2004, 04:10 AM	#15
korme Go Reds Join Date: May 2001 Location: Bloodbuzz Ohio	suicane has never sounded so dumb

08-25-2004, 08:19 AM	#16
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	Good heavens, someone lock this thread.

08-25-2004, 08:53 AM	#17
Joe Pro Starter Join Date: Jun 2004 Location: Minneapolis	yeah, stop being so dumb

08-25-2004, 08:56 AM	#18
Bee Pro Starter Join Date: Oct 2000 Location: Fairfax, VA	From my experience the odds of getting a 2,7 offsuit is about 1 in 5 so I'll go with that.

08-25-2004, 11:30 AM	#22
GabeRivers n00b Join Date: Mar 2001 Location: Texas, USA	I'm wondering the same as Samdari. I haven't played poker in years, and even when I did it wasn't much, but doesn't poker rank suits similar to bridge: spades, hearts, diamonds, clubs? In other words, if the two best hands were Ace (spades)/ten versus Ace (clubs)/ten, doesn't the Ace (spade) win the hand? If so, it would seem to my tired old mind that the answer to the question would be 52 times 51.

08-25-2004, 11:38 AM	#24
GabeRivers n00b Join Date: Mar 2001 Location: Texas, USA	Too much bridge, I guess, and not enough poker.

05-10-2013, 08:59 AM	#26
Suicane75 Coordinator Join Date: Sep 2003 Location: NJ	I think this is right around the time I started smoking pot.

05-14-2013, 10:48 AM	#28
Suicane75 Coordinator Join Date: Sep 2003 Location: NJ	Listen, I know you aren't Finnish.

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