Odds of getting the same exact hand 5 times in a row.

[Joshssj4]

Today on PokerStars i got 6 /images/graemlins/heart.gif 3 /images/graemlins/club.gif 5 times in a row, hightly unlikely obviously. What is the probability?

[Jordan Olsommer]

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Today on PokerStars i got 6 /images/graemlins/heart.gif 3 /images/graemlins/club.gif 5 times in a row, hightly unlikely obviously. What is the probability?

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((2/52)(1/51))^5 = 4,099,374,736,625,375-to-1

Youll win the lottery before that ever happens again.

[kyro]

(2/52)(1/51)(16/17)^4 would be more relevant

[MikeL05]

where's the 16/17 coming from?

Also... I'd bet the odds are at least 100-to-1 that you DIDN'T get the same hand 5 times in a row.

[kiddj]

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Today on PokerStars i got 6 /images/graemlins/heart.gif 3 /images/graemlins/club.gif 5 times in a row, hightly unlikely obviously. What is the probability?

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But the EV is ~0, which isn't too bad... /images/graemlins/wink.gif

[kyro]

I kinda pulled it out of air. It's really the odds of getting the same non-pocket pair 5 times in a row. Not sure why I did it that way.

[Jordan Olsommer]

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I kinda pulled it out of air. It's really the odds of getting the same non-pocket pair 5 times in a row. Not sure why I did it that way.

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You don't need to add that term.

You have 2 choices for the first card - either the 6/images/graemlins/heart.gif or 3/images/graemlins/club.gif out of 52 - 2/52

For the second card, you only have one choice - whichever of the two that you didn't get on the first go-round - out of the 51 remaining cards: 1/51.

So (2/52)(1/51) are your odds of getting dealt 6/images/graemlins/heart.gif3/images/graemlins/club.gif on one particular hand, so five in a row is just that term to the fifth power. No (16/17) needed, least of all for the reason that it gives you an incorrect answer.

Here's another way to look at it, simpler than the way I originally posted:

# of possible holdem starting hands = 52c2 = 1326
odds of getting one particular starting hand = 1/1326
odds of getting one particular hand 5x in a row = 1/1326^5 = 4,099,374,736,625,375-to-1.

[kyro]

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I kinda pulled it out of air. It's really the odds of getting the same non-pocket pair 5 times in a row. Not sure why I did it that way.

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You don't need to add that term.

You have 2 choices for the first card - either the 6/images/graemlins/heart.gif or 3/images/graemlins/club.gif out of 52 - 2/52

For the second card, you only have one choice - whichever of the two that you didn't get on the first go-round - out of the 51 remaining cards: 1/51.

So (2/52)(1/51) are your odds of getting dealt 6/images/graemlins/heart.gif3/images/graemlins/club.gif on one particular hand, so five in a row is just that term to the fifth power. No (16/17) needed, least of all for the reason that it gives you an incorrect answer.

Here's another way to look at it, simpler than the way I originally posted:

# of possible holdem starting hands = 52c2 = 1326
odds of getting one particular starting hand = 1/1326
odds of getting one particular hand 5x in a row = 1/1326^5 = 4,099,374,736,625,375-to-1.

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Right, but you're talking about a specific hand, 6h3c (or whatever). That's a pretty useless statistic to figure out unless you feel like telling me the odds of getting,
8/images/graemlins/spade.gif 4/images/graemlins/spade.gif,
6/images/graemlins/heart.gif 2/images/graemlins/club.gif,
K/images/graemlins/heart.gif K/images/graemlins/diamond.gif,
A/images/graemlins/club.gif 2/images/graemlins/heart.gif,
4/images/graemlins/spade.gif 3/images/graemlins/diamond.gif

in a row.

[Jordan Olsommer]

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Right, but you're talking about a specific hand, 6h3c (or whatever). That's a pretty useless statistic to figure out unless you feel like telling me the odds of getting,
8/images/graemlins/spade.gif 4/images/graemlins/spade.gif,
6/images/graemlins/heart.gif 2/images/graemlins/club.gif,
K/images/graemlins/heart.gif K/images/graemlins/diamond.gif,
A/images/graemlins/club.gif 2/images/graemlins/heart.gif,
4/images/graemlins/spade.gif 3/images/graemlins/diamond.gif

in a row.

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Or you feel like calculating the odds of receiving 6/images/graemlins/heart.gif3/images/graemlins/club.giffive times in a row, which is what the OP asked.

edit: or if you wanted to do the "what are my chances of getting 6-3 five hands in a row?" there's a much more intuitive way to do it than with the (16/17): You have 8 choices for the first card you get , and 4 choices for the second card = (8/52)(4/51)^5. Much easier to grok.

[shday]

Given any hand, (1/1326)^4 would be the odds of getting the same hand 4 more times... another way to answer the original question.

[AaronBrown]

That's one time in 3,091,534,492,176. If everyone in the world dealt five hands every day for a year from perfectly shuffled decks, the odds are against any one every dealing five identical hands.

I agree that the odds of getting any hand five times is a row is probably more relevant than asking about a specific hand. But if it was pocket Aces five times in a row, he wouldn't be alive to post, that's survivorship bias.

[RiverTheNuts]

I got jacks 4 hands in a row on party a few months back, thats my biggest "same hand streak"

[AaronBrown]

That's much more likely. There are six ways to get Jacks, so 6^4 = 1,296 more ways to get Jacks for times in a row than a specific hand like Js Jd. You only have to play 2,385,443,281 hands to expect to get Jacks four times in a row. If everyone in the world dealt four hands, about three of them would get Jacks every time.

[Joshssj4]

For the people who do not believe me because of the sheer numbers keep in mind I am not a person to come onto 2+2 and make things up to create sheer conversation. It did in fact happen.

[2+2 wannabe]

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That's much more likely. There are six ways to get Jacks, so 6^4 = 1,296 more ways to get Jacks for times in a row than a specific hand like Js Jd. You only have to play 2,385,443,281 hands to expect to get Jacks four times in a row. If everyone in the world dealt four hands, about three of them would get Jacks every time.

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plus the poster probably would have made his post with any pair, so you could probably multiply it by 13

[Onaflag]

I was bored and goofing around in one of the FPP tournies today and a guy piped in that he got the same hand 3 times in a row. Don't know if it's true. He didn't show them.

Onaflag..........

[AaronBrown]

With something this unlikely, it's fair to rule out that it happened by random chance. If it happened, it must have been a bug in PokerStars software. In a home game, these things happen because people don't shuffle well.

But with all due respect, this is a hand you would fold immediately and not focus on the details. Suppose you got one offsuit 6 3, then another. By the time you got the third you really noticed it and focused on the suits. When the fourth and fifth came in with identical suits; you remembered it as five exactly identical in a row? That's still unlikely, but it's 144 times more likely than five exactly identical in a row.

Or do you play more than one game at a time? Could you have looked at the hand, thought you folded it and went on to another game? When you came back to the first game and saw the same cards, you would have thought you got the same hand again. Four times in a row is 1,326 times more likely than five.

I'm not saying these things happened, just that it's easy to misremember a detail or two that is insignificant at the time, but dramatically changes the odds after the fact.

[Paul2432]

Everyone in this thread has the wrong answer.

The event happened in the past. The question was not "what is the probability I get 6h 3c five times in a row again?"*

The probability that this event happened is a function of two probabilities. If M=the probability that the OP making up this story for fun, and B=the probability the OP is a bad observer and believes he had the hand five times in a row but really did not, then the probability he actually received 6h 3c fives times in a row is (1-B)(1-M). I'll leave it to others to assign values to B and M.

Paul

*The question of the event repeating has not really been answered either. It is anywhere from 0-100% depending on how much the OP plays. If he never plays poker again, then it is 0%, if he plays an infinite number of hands, then it is 100%. The only question that has been answered is the probability that the next five hands (or some other earmarked set of 5 hands) will be 6h 3c.

[Jordan Olsommer]

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Everyone in this thread has the wrong answer.

The event happened in the past. The question was not "what is the probability I get 6h 3c five times in a row again?"*

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I think it was more than obvious that that was what he meant. (here's a quick puzzler: what do you think the likelihood is that he meant "what are the odds against getting 6/images/graemlins/heart.gif3/images/graemlins/club.gif 5x in a row?" versus the likelihood that he meant "hey guys, evaluate the probability that I'm full of sh*t!")

You remind me of a friend of mine who used to try to pay for sodas in college with only a penny because the sign said "Coke, .99 cents"

[AaronBrown]

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You remind me of a friend of mine who used to try to pay for sodas in college with only a penny because the sign said "Coke, .99 cents"

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I do that. I also argue when it says "Buy one, get one free," because if I have to pay for the first one, neither of the items are free. When it says "we will beat any price in town," I say "I offer a penny, beat that."

Yeah, we're annoying, but the world needs more of us.

[Joshssj4]

I come here to ask a simple factual question and you all assume I'm lying. As if I have nothing better to do with my life than spread lies on an internet forum. You should have better faith in those that seriously post here. This did happen and I would like to thank the posters that rather then asked pointless actually gave me the answer I had come here to find.

[shday]

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I come here to ask a simple factual question and you all assume I'm lying. As if I have nothing better to do with my life than spread lies on an internet forum. You should have better faith in those that seriously post here. This did happen and I would like to thank the posters that rather then asked pointless actually gave me the answer I had come here to find.

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Dude, no one was rude or flamed you... just expressing some scepticism here and there. You really shouldn't expect otherwise. Personally, I think a software glitch was more likely the cause of what you saw (compared to it happening by chance alone).

[ajmargarine]

I was playing in a tournament last week, and the first 2 flops of the tourney were exactly the same, suits and values (for real)(no, i'm serious)(lol). So to figure the probability of that happening you would go (and this is a question):

((3/52)(2/51)(1/50))^2= the probability?!???

(despite a year of college calculus, and a flirtation with majoring in math, I barely remember anything I learned back then--this forum looks like a nice place to relearn and learn, especially as it applies to poker)

[Jordan Olsommer]

The probability of any one specific flop if you don't know any hole cards is 1/(52c3). So two in a row would be (1/(52c3))^2 = .00000000247

But I'm assuming you knew what your hole cards were, so that means the number of possible flops is 50c3, not 52c3, so the probability would then be (1/(50c3))^2 = .000000002603, or 384,159,999-to-1.

If you want really rare, the probability that you'll get the same board twice in a row (that's same flop, same turn, same river) is .00000000000000047055, or 2,125,136,240,639,999-to-1

[SumZero]

I just got 5d 4h, 3h 2s, Ah Tc, 4c 6h, and 6s 2h in a row. What are the odds I ever get that same 5 hand in a row sequence? What are the odds that I'm making that up?

[AaronBrown]

Jordan Olsommer computed what you asked mathematically, ((3/52)(2/51)(1/50))^2, but not what you asked in words. Your formula gives the probability that the first two flops will be some specific hand, say 5c 9d Kh.

The chance that the first two flops will be the same as each other, without specifying what they are, is the square root of that (in other words, remove the square from your formula). That's 1/22,100.