Chances of getting the exact same hand?

soulsey · Jun 28, 2012

Just thought i would log onto titan poker have a quick game, after sitting down at two tables the first hand was exactly the same. Just wondering what people think the chances of these sort of things are seems odd to me

WVHillbilly · Jun 28, 2012

I'd be more suspicious if it never happened.

stg1969 · Jun 28, 2012

I'd be more suspicious if the flops were the same too, lol.

xUnrated · Jun 28, 2012

about the same odds as me winning a hand..

BlueNowhere · Jun 28, 2012

Odds of getting it once x odds of getting it once. So here there are 6 combos of 66 and 1326 total combos. 6/1326 = 1//221 (1/221)*(1/221) = (1/221)^2 = 1/48841 = 0.002% chance.

Big_Rudy · Jun 28, 2012

BlueNowhere said:
Odds of getting it once x odds of getting it once. So here there are 6 combos of 66 and 1326 total combos. 6/1326 = 1//221 (1/221)*(1/221) = (1/221)^2 = 1/48841 = 0.002% chance.

Well, as you know, I'm definately (and, defiantly

) NOT a math guy, so my thinking could be way off, but this seems wrong. OP isn't asking about odds of getting this, or any, specific hand back-to-back, but just the odds of getting the exact same hand twice in a row. It seems to me (and my potentially flawed logic) that your first part (odds of getting it once) doesn't apply.

I guess what I'm trying to say is I THINK what the OP is asking is the odds of getting ANY hand twice in a row. I would think, then, that your odds of getting it the first time is 100% as we are just interested in matching our first hand with the second one and you have this chance literally every time you are dealt a hand.

Having said all that, I have no idea what the correct formula is (and you may even be right for all I know), but I don't think it's anywhere near as rare as 0.002%. I will concede it's a fairly rare event; just not THAT rare. Now, if OP IS asking the odds of getting EXACTLY this specific hand twice in a row, then I believe your formula is correct.

WVHillbilly · Jun 28, 2012

Not twice in a row. Same hand on 2 different tables at same time.

BlueNowhere · Jun 28, 2012

Big_Rudy said:
Well, as you know, I'm definately (and, defiantly ) NOT a math guy, so my thinking could be way off, but this seems wrong. OP isn't asking about odds of getting this, or any, specific hand back-to-back, but just the odds of getting the exact same hand twice in a row. It seems to me (and my potentially flawed logic) that your first part (odds of getting it once) doesn't apply.

I guess what I'm trying to say is I THINK what the OP is asking is the odds of getting ANY hand twice in a row. I would think, then, that your odds of getting it the first time is 100% as we are just interested in matching our first hand with the second one and you have this chance literally every time you are dealt a hand.

Having said all that, I have no idea what the correct formula is (and you may even be right for all I know), but I don't think it's anywhere near as rare as 0.002%. I will concede it's a fairly rare event; just not THAT rare. Now, if OP IS asking the odds of getting EXACTLY this specific hand twice in a row, then I believe your formula is correct.

I see what you're saying. Well presuming suits don't factor into the same hand thing, there is 78 starting hands with 16 combos each and 13 starting hands with 6 combos.
So chance of 2nd hand being a PP (preusming 1st hand is already a PP) = 6/1326 chance of 2nd hand bieng a non PP (presuming 1st hand is non PP) 16/1326

[(13/91)*(6/1326)] + [(78/91)*16/1326)] = 1/91

1/91*100 = 1.0989...% chance of getting the same hand on one table presuming the hand has already been dealt on the other.

If you wanted to include suitedness it's the same formula but a longer version.

Big_Rudy · Jun 28, 2012

BlueNowhere said:
I see what you're saying. Well presuming suits don't factor into the same hand thing, there is 78 starting hands with 16 combos each and 13 starting hands with 6 combos.
So chance of 2nd hand being a PP (preusming 1st hand is already a PP) = 6/1326 chance of 2nd hand bieng a non PP (presuming 1st hand is non PP) 16/1326

[(13/91)*(6/1326)] + [(78/91)*16/1326)] = 1/91

1/91*100 = 1.0989...% chance of getting the same hand on one table presuming the hand has already been dealt on the other.

If you wanted to include suitedness it's the same formula but a longer version.

Ok, NOT going to comment on the math/formula thing, but this seems to be more in line with what I was thinking and your numbers come out pretty-much where I intuitively feel they should be. Like I said, pretty rare, but not so rare as in the first calculation.

moemtg · Jun 28, 2012

BlueNowhere said:
Odds of getting it once x odds of getting it once. So here there are 6 combos of 66 and 1326 total combos. 6/1326 = 1//221 (1/221)*(1/221) = (1/221)^2 = 1/48841 = 0.002% chance.

Amazing to know... On Merge Network I have had Qc5h 4 times in a row... I have also had some other crazy things like pocket aces (different suits) 3 times in a row as well... Granted I have played well over half a million hands online... but when things like that happen it makes me wonder... ohhhhh it makes me wonder... I guess I am buying a stairway to heaven

Moe

P.S. These hands were on the same table... NOT multi-tabling.

moots · Jun 28, 2012

BlueNowhere said:
Odds of getting it once x odds of getting it once. So here there are 6 combos of 66 and 1326 total combos. 6/1326 = 1//221 (1/221)*(1/221) = (1/221)^2 = 1/48841 = 0.002% chance.

Incorrect. Either you're going to get the same hand or you're not going to get the same hand...so the odds are 50/50. :smokin:

kidkvno1 · Jun 28, 2012

try getting the same hand well on Carbon and on stars at the same time...
It's happened to me a few times.

Aldito · Jun 28, 2012

Isn't it just (1/52)*(1/51) = 1/(52*51) = 0.000377073906 or 0.0377%

Been a while since I took statistics lol

WVHillbilly · Jun 28, 2012

Playing in a home game a few years back (2 decks, pass the deal) I was dealt JJ 3 hands in a row. I never thought it might be rigged.

Big_Rudy · Jun 28, 2012

WVHillbilly said:
Playing in a home game a few years back (2 decks, pass the deal) I was dealt JJ 3 hands in a row. I never thought it might be rigged.

Everyone knows home games are rigged. Ldo.

Big_Rudy · Jun 28, 2012

Aldito said:
Isn't it just (1/52)*(1/51) = 1/(52*51) = 0.000377073906 or 0.0377%

Been a while since I took statistics lol

Man, WAY too much math itt. My head hurts

. This formula doesn't seem right, though. This seems to be the answer if you wanted to know the odds of drawing one specific hand from a deck, not that you'd have back-to-back identical hands.

sam1chips · Jun 28, 2012

haha well actually, the probably that in two consecutive hands you would be dealt the six of hearts as the left card and the six of diamond as the right card is 0.00001422% [(1/52)*(1/51)] ^ 2

I personally would prefer aces or kings, but still very interesting lol

lenstra · Jun 28, 2012

I think the 0.0377% answer is correct... there are 51*52=2652 different starting hands (including suits and order) and all have the same chance of appearing, 1/2652 = 0.0377%.

At the first table you're simply going to get one, it doesn't matter which. At the second table there's 0.0377% chance to get a specific hand, e.g. the same one as the first table.

Big_Rudy · Jun 28, 2012

sam1chips said:
haha well actually, the probably that in two consecutive hands you would be dealt the six of hearts as the left card and the six of diamond as the right card is 0.00001422% [(1/52)*(1/51)] ^ 2

I personally would prefer aces or kings, but still very interesting lol

Yeah, this seems right. It's possible to manipulate the numbers a few different ways depending upon if order and/or suitedness matters or not. In Blue's earlier example he ignored them, hence his % came out higher. Just depends on specifically what you're looking for.

sam1chips · Jun 28, 2012

lenstra said:
I think the 0.0377% answer is correct... there are 51*52=2652 different starting hands (including suits and order) and all have the same chance of appearing, 1/2652 = 0.0377%.

At the first table you're simply going to get one, it doesn't matter which. At the second table there's 0.0377% chance to get a specific hand, e.g. the same one as the first table.

Haha i guess thats true, if in the middle of the first hand, you're thinking, "what's the probability of getting the exact same hand in the same order?" it would be 0.0377%

I guess my calculation would be as you were logging on to the computer you ask "What's the probability that I will get a 6 of hearts and 6 of diamonds in the same order in the first 2 hands?"

fozziethebear · Jun 29, 2012

It's not that crazy. 1/52 * 1/51. The first hand is irrelevant unless you are talking about odds of getting a specific matching hand. But for your 2nd hand to match your first is just 1 in 2652.

BlueNowhere · Jun 29, 2012

Big_Rudy said:
Yeah, this seems right. It's possible to manipulate the numbers a few different ways depending upon if order and/or suitedness matters or not. In Blue's earlier example he ignored them, hence his % came out higher. Just depends on specifically what you're looking for.

Yea this. I just presumed suits didn't matter.

stg1969 · Jul 2, 2012

mootso said:
Incorrect. Either you're going to get the same hand or you're not going to get the same hand...so the odds are 50/50. :smokin:

Hoping this is a joke.... if not, can I have some of what you're smoking, and do you wanna play heads up?

Punter4444 · Jul 2, 2012

its quite rare to get the same hand on two diff tables at the same time on the same client dnt think thats happened to me tho it has happend that iv got the same pocket cards on carbon and pokerstars at the same time

gazrosenau · Jul 2, 2012

if you concider how many hands are delt every minute it is likely to happen more regular than you think, we just dont notice it often

Chances of getting the exact same hand?

soulsey

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