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Oxford_HRV
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« on: October 24, 2012, 07:04:52 AM » |
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If you are dealt in SB on a 10 handed table and recieve an ace
what is the chance another ace has been dealt out before it comes back to you?
I'm struggling to work it out I can't seem to think of the right equation
the first ace is always 13/1 but as cards are dealt the fraction changes to the position you're sat in through card elimination
for example if cut off is dealt an ace aswell that is the 9th card with 3 aces left which gives you 3/44
so in the 10 cards dealt out before you receive your second card, chances an ace has already been dealt out is....?
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To win at poker is to not have to play
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Tal
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« Reply #1 on: October 24, 2012, 08:17:28 AM » |
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Unless you know what those cards are, I wouldn't think it affects the probability of you being dealt a second ace.
In theoretical physics terms, if seat 5's card isn't known by someone, it doesn't have a value.
If memory serves, that is...
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"You must take your opponent into a deep, dark forest, where 2+2=5, and the path leading out is only wide enough for one"
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Jon MW
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« Reply #2 on: October 24, 2012, 08:33:14 AM » |
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...
the first ace is always 13/1 but as cards are dealt the fraction changes to the position you're sat in through card elimination
...
As Tal suggests - if you don't know what the cards are then they don't affect the probability. Drawing a red card from a pack is 26/52 =1/2 If beforehand you had already taken 40 cards away then the odds of drawing a a red would still be 26/52 = 1/2 (if you don't know which 40 cards you'd removed). The fact that there's only 12 cards left in the pack you're drawing from is irrelevant. |
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Jon "the British cowboy" Woodfield
2011 blonde MTT League August Champion
2011 UK Team Championships: Black Belt Poker Team Captain - - runners up - -
5 Star HORSE Classic - 2007 Razz Champion
2007 WSOP Razz - 13/341
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Doobs
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« Reply #3 on: October 24, 2012, 08:35:03 AM » |
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The answer to both your questions is it is irrelevant. The chance of a second ace is 1 in 17 wherever you are sat.
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Most of the bets placed so far seem more like hopeful punts rather than value spots
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bobAlike
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« Reply #4 on: October 24, 2012, 09:50:51 AM » |
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Probabilty shmobabilty! Can't remember the last time I had 2 aces at the same time. Must be storing them up for the next deep stack.
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Ah! The element of surprise
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Leatherman
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« Reply #6 on: October 24, 2012, 12:23:33 PM » |
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Why does it matter? You always seem to get that bloody deuce?  Or even worse the  which for second you think is the other Ace.  |
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Satellite King
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Tal
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« Reply #7 on: October 24, 2012, 01:10:18 PM » |
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Good point. Live poker is rigged. Everyone knows that.
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"You must take your opponent into a deep, dark forest, where 2+2=5, and the path leading out is only wide enough for one"
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SuuPRlim
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« Reply #8 on: October 24, 2012, 01:10:30 PM » |
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Probabilty shmobabilty! Can't remember the last time I had 2 aces at the same time. Must be storing them up for the next deep stack.
Moaning this one in like a champ! Rest. |
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George2Loose
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« Reply #9 on: October 24, 2012, 04:06:34 PM » |
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Odds on getting any suited broadway in hold em (A-T) assuming 9 handed
Odds on getting a pair in 6 card Omaha (assuming 6 handed)
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Ole Ole Ole Ole!
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aaron1867
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« Reply #10 on: October 24, 2012, 04:07:44 PM » |
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Odds on getting any suited broadway in hold em (A-T) assuming 9 handed
Odds on getting a pair in 6 card Omaha (assuming 6 handed)
Odds are probably smaller than beating Alex in a race. |
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Cf
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« Reply #11 on: October 24, 2012, 05:30:09 PM » |
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Odds on getting any suited broadway in hold em (A-T) assuming 9 handed
Odds on getting a pair in 6 card Omaha (assuming 6 handed)
The amount of players at the table makes no difference to the odds of being dealt a specific hand. |
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Blue text
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George2Loose
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« Reply #12 on: October 24, 2012, 05:32:26 PM » |
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Ok so calculate the above pls
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Ole Ole Ole Ole!
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Tal
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« Reply #13 on: October 24, 2012, 05:35:56 PM » |
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First Broadway card is 5/13. Second card being the same suit is 12/51.
60/663 is about 1/11 or 9%
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"You must take your opponent into a deep, dark forest, where 2+2=5, and the path leading out is only wide enough for one"
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Tal
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« Reply #14 on: October 24, 2012, 05:37:09 PM » |
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Have I read that correctly or are you asking for both cards to be Broadway cards?
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"You must take your opponent into a deep, dark forest, where 2+2=5, and the path leading out is only wide enough for one"
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