What are the odds of these hands?

twizzybop

twizzybop

Legend
Silver Level
Joined
Apr 24, 2005
Total posts
2,380
Chips
0
3 Straight hands I got the same 2 cards but they were also suited all 3 times.

1st BB Qh 8h (which won me a big pot, hit the straight)
2nd SB Qd 8d Folded
3rd Button Qh 8h also folded

After the 3rd straight one came. I sat there going wtf is this? How in the world does someone get the same starting 2 cards that are suited 3 times straight? Not to mention 2 out of the 3 times they were also of the same suit.(hearts)
 
t1riel

t1riel

Legend
Silver Level
Joined
May 20, 2005
Total posts
6,964
Awards
1
Chips
130
Yesterday, I got three straight A, Q. Not that I'm complaining or anything.
 
F Paulsson

F Paulsson

euro love
Silver Level
Joined
Aug 24, 2005
Total posts
5,799
Awards
1
Chips
1
The probability of getting the first [Qh][8h] is 1.
The probability of getting another one (any suited Q8) is 4/1326 ~ 1/332.
Getting the third one is therefore (1/332)*(1/332) = 1/110224.

Of course, you might not have been looking for an actual answer to the rhetorical question of what the odds are :p
 
C

colin_147

Visionary
Silver Level
Joined
May 16, 2005
Total posts
708
Chips
0
Where's Diablo when you want an argument on odds of recurring events......
 
robwhufc

robwhufc

Cardschat Elite
Silver Level
Joined
May 25, 2005
Total posts
5,587
Chips
0
colin_147 said:
Where's Diablo when you want an argument on odds of recurring events......

He never got his head around it did he!

Here's another one then - I was dealt EIGHT consecutive suited hold cards about a month ago at Paradise. What are the odds of that?
 
F Paulsson

F Paulsson

euro love
Silver Level
Joined
Aug 24, 2005
Total posts
5,799
Awards
1
Chips
1
There are two ways to look at that, but let's go with the one yielding a lower probability (and thus a larger "wow"-factor ;)):

Probability of getting two suited cards:
Card 1: Probality of 1.
Card 2: Probability of 12/51 = 0.235, which is then the probability for two suited hole cards.

0.234^8 = 9.39*10^-6 gives a probability of 1/106442.

EDIT: 1/106442, not 1/106441. But who's counting!
 
Schatzdog

Schatzdog

Visionary
Silver Level
Joined
Jun 29, 2005
Total posts
693
Chips
0
Isn't each hand statistically independent from the previous? Shouldn't the probability of getting any hand dealt remain the same?
 
Four Dogs

Four Dogs

Legend
Silver Level
Joined
Apr 13, 2005
Total posts
4,343
Awards
1
Chips
108
colin_147 said:
Where's Diablo when you want an argument on odds of recurring events......
colin I think you and some of the newer members should go back and review the diablo posts. It completely escapes me how he got a reputation as a statistician other than he always had an opinion. On everything. I think it came from is tendency to go postal whenever anyone used some super extreme event as anectodal evedence of the corruption of online poker without actually having any grasp of statistics or why it would benefit a poker room to juice the cards. While we always saw eye to eye in that respect, he seemed incapable of understanding the math behind the odds.
 
Poker Odds - Pot & Implied Odds - Odds Calculator Starting Hands - Poker Hand Nicknames Rankings - Poker Hands
Top