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

[Joshssj4]

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

[Jordan Olsommer]

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

[/ QUOTE ]

((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]

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

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

[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]

[ QUOTE ]
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.

[/ QUOTE ]

You don't need to add that term.

You have 2 choices for the first card - either the 6[img]/images/graemlins/heart.gif[/img] or 3[img]/images/graemlins/club.gif[/img] 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[img]/images/graemlins/heart.gif[/img]3[img]/images/graemlins/club.gif[/img] 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]

[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]

You don't need to add that term.

You have 2 choices for the first card - either the 6[img]/images/graemlins/heart.gif[/img] or 3[img]/images/graemlins/club.gif[/img] 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[img]/images/graemlins/heart.gif[/img]3[img]/images/graemlins/club.gif[/img] 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.

[/ QUOTE ]

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[img]/images/graemlins/spade.gif[/img] 4[img]/images/graemlins/spade.gif[/img],
6[img]/images/graemlins/heart.gif[/img] 2[img]/images/graemlins/club.gif[/img],
K[img]/images/graemlins/heart.gif[/img] K[img]/images/graemlins/diamond.gif[/img],
A[img]/images/graemlins/club.gif[/img] 2[img]/images/graemlins/heart.gif[/img],
4[img]/images/graemlins/spade.gif[/img] 3[img]/images/graemlins/diamond.gif[/img]

in a row.

[Jordan Olsommer]

[ QUOTE ]

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[img]/images/graemlins/spade.gif[/img] 4[img]/images/graemlins/spade.gif[/img],
6[img]/images/graemlins/heart.gif[/img] 2[img]/images/graemlins/club.gif[/img],
K[img]/images/graemlins/heart.gif[/img] K[img]/images/graemlins/diamond.gif[/img],
A[img]/images/graemlins/club.gif[/img] 2[img]/images/graemlins/heart.gif[/img],
4[img]/images/graemlins/spade.gif[/img] 3[img]/images/graemlins/diamond.gif[/img]

in a row.

[/ QUOTE ]

Or you feel like calculating the odds of receiving 6[img]/images/graemlins/heart.gif[/img]3[img]/images/graemlins/club.gif[/img]five 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.