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  Poker - What are the Odds
 
  #1  
17-07-2006, 9:06 AM
withawedge
CardsChat Regular
 
Location: Chorley, UK
Plays at: Poker Stars
Likes: Holdem
Posts: 534
What are the Odds

All,

Not being a mathematical genius I am asking for help here to work out the odds of the following hands being dealt.

Playing in a $5+$0.50 SnG. There is still 9 of us left, hands 5 , 6 , 7 read as follows:

Hand 5 = Qs Ad
Hand 6 = As Qc
Hand 7 = Qh Ad

Playing in a $12+$1 SnG. Only 5 of us left. Three hands consecutively that showed as:

Qh 7s
7d Qs
Qc 7d

Can anyone out there workout the odds please for both tables. If any one thinks I might be seeing things I can attach the Instant Hnad Histories to show the hands
 

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  #2  
17-07-2006, 9:58 AM
ChuckTs
whitebread
 
Location: lopping off my C-game
Posts: 11,485
Here's my attempt:


16 (4 aces * 4 queens) possible AQ comb.'s and 2652 (52 any card*51 any card except already chosen one) possible 2-card holdings preflop.

So % of getting AQ(suited or not) dealt = 16/2652 = ~0.6%

Then to get the probability of it happening 3 times in a row, just take 0.6^3 = ~0.2%

For offsuit only, it's 12 (take away the 4 suited possibilities) possible AQs combos and the same total possible holdings, so 12/2652 = ~0.45%

For % of getting it 3 times in a row, it's 0.45^3 = ~0.09%

You should get the same answers for Q7 because it is the same type of hand - two random cards (and therefore the same type of calculations).

Correct me if I'm wrong - I think I did this right
  #3  
17-07-2006, 11:19 AM
starfall
CardsChat Regular
 
Location: London, England
Plays at: Stars
Likes: Omaha Hi/Lo
Posts: 576
The chance of the first card dealt being an Ace or a Queen is 8/52 (it can be either an Ace or a Queen), and the chance of the 2nd card being the other is 4/51 (it has to be one of the 4 of the other card).
This gives a chance of it happening one hand of 0.012066... or 1.2% (Chuck, you got half that because you didn't factor in the higher likelihood of getting a matching card on the first one dealt).
For suited, the probability is 3/4 the probability of either suited or unsuited, because the first card probability remains the same, but you have to eliminate the matching suit card from the 2nd deal possibilities.

The 2nd mistake was on the probability calculation - you can't just multiply the % value, you need to multiply the value where 1= every time, so it's 0.012 x 0.012 x 0.012 = 0.000001728, which is 1 in every 578,704 times.
 


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