Gretzky...what are the odds?

So I'm playing a S'n'G and catch 99 in the hole, 1 guy all-in for minimal raise, call, and i call. i catch a 9 on the flop and raise that pushes out the other guy and the all-in dude catches straight w/e...next hand i get 99 in the hole again, catch a 9 on the flop and have the same guy against me and get him all-in on the river but beat his straight with a full boat this time
What are the odds of catching the same two hole cards (even the suits were the same), two hands in a row, hitting the set on the flop with the same card (suit wise) ?

Had a similar thing happen to me with 66 a few weeks ago, all same suit and everything but I caught it twice in between 5 hands and had 66 three times between 15 hands

1 billion to 1

(I rounded up; by how much, is for you to decide)

haha 1,000,000,000 : 1 is that for real? that's pretty nutz, but 2 hands in a row is quite a stretch

um no, not for real.

haha i was going to say woooow where u get this from? but neeeevermind

I put 5 cents up to any takers that zach is the first to reply to this with the answer

well, aside from suits it would be 220 1 of getting dealt the same hand twice in a row. but specific two hands would be 48400 to 1. hitting a set would be 7.5 to 1. so i believe its 12375 to 1 of getting dealt the same hand twice in a row and hitting a set with each. im not exactly sure, it could be 1650 to 1 but i think you multiply by 7.5 twice. i could be off on this, but this is how i remember it. you never really have to do these calculations while playing poker. at least, it's not going to effect how well you play.

^^^ I dont think thats right if im honest, once in every 12k hands seems to often in my opinion. I dont think ive ever been dealt the same pp twice in a row, never mind hitting sets with both, and ive played more than 12k hands

i would take that 5 cent bet but we all know Zach is gonna come rushin in with some crazy math that we cant comprehend anytime now

it's going to be alot higher if your talking about 2 specific hands. like if you say whats the odds of me getting pocket 9's twice in a row on the next two hands then its going to be alot higher. the odds of being dealt 2 specific hands in a row are 1 in 48400. but the odds of getting dealt the same hand twice in a row is 220 to 1. as for the exact same suits, im not sure, but it doesn't matter either.

After utilizing my abacus and probability text from high school I came up with this. PRETTY DAMN HIGH!!!

odds of this happening: "who cares?"

Seriously, waste your energy on something productive...

.......that was my point alien, thanks for your support.

killjoy

lol it's not that crazy.

Odds of pocket pair 1st hand = 3/51 = 0.0588 (no matter what 1st card is, 3 out of the remaining 51 will make a pair)

Odds of set = 1-(48/50)(47/49)(46/48) = 0.117 (the opposite of the odds of missing the set on all 3 cards)

Odds of getting the same cards again same suits (without order mattering) = (2/52)(1/51) = 0.075% = 0.00075


odds of setting with the same card = 1-(49/50)(48/49)(47/48) = 0.06

Odds of them all happening = 0.0588*0.117*0.00075*0.06 = .000031% = 1 in 3,230,162.

But then again this makes no difference. Sure it's unlikely, but so is me being dealt A5o, JJ, 38h, 98o, KK, A8o, 22, 23o, 8Td all in a row. In fact I'd bet that sequence is less likely than your hand.

Anyway the only thing remotely complicated in the math I did was the fact that 1-any probability is the probability of it not happening. Everything else is pretty basic probability, so I don't know what y'all are talking about me doing insane complicated math.

Now I'm currently taking an upper level college course in probability. That DOES have insane math including integrals, infinite sums, etc. in probability. All I'm using is arithmetic.

haha sorry everyone i didn't mean to figure it out literally when i first posted, but when people actually started doing the math i thought it would be very interesting to see. if you dont care to see then dont waste your time reading the post.

o and thanks zachvac for coming up with an answer! thats some good stuff you know how to do it

when u say getting dealt a pocket pair is 3/51, isn't it 220 to 1 of being dealt a specific pocket pair? like they say the odds of getting dealt aces are 220 to 1

The math is different for a specific pocket pair. You'd have to calculate:
4/52 x 3/51 = 0.0769 x 0.0588 = 0.0045 = 0.45% or 222:1

dammit zach, you were too quick for anyone to take up my bet

But as mentioned, the weird part wasn't that it was 9s specifically. Like I said it's extremely strange to see any sequence of cards that you'll see (just list your last 10 hands. Odds of those happening in a row are probably worse than the 9s setting twice with the same card). So the first one was the odds of any pocket pair. First card can be any of 52, but second has to be one of the 3 left to match the first out of 51 cards left. The SECOND time however, I did restrict it to being 9s, in fact the same 2 9s. The main point though is that if this had happened with 5s he'd have been just as surprised.

Thanks, now I have a headache.

Where were you when I needed my x of 30 question array odds question answered?

Must have missed that one