Help with probability (specific case)
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Help with probability (specific case)
A while ago I had a gutshot straight draw in a home game. I figure I have 4 outs right. And clearly it is not worth calling whatever bet was made. In addition to that I can see the bottom card on the deck is one of the three cards i need (my dealer friend not knowing any better). would this make the chances of me hitting that card less? or would it stay the same?
I mean does it change from a 4 in 47 chance to a 3 in 47 chance? Just because i know one of the four wont hit?
I mean does it change from a 4 in 47 chance to a 3 in 47 chance? Just because i know one of the four wont hit?
- Mad Max
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MasterShake - Posts: 1745
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becuase the only reason why i thouhgt it might stay as 4 outs is becuase with 4 outs, 3 of the 4 definetly wont hit. so my thinking was would it matter if i could see one of those 3 that definetly wont hit. in my mind i was arguing that of those 3 that definetly wouldnt hit, what would it matter if it were on the bottom of the deck or mixed somewhere in between? you know what i mean?
- Mad Max
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- Joined: Thu Apr 01, 2004 7:07 pm GMT
Pedant Alert
Technically, the probability is still the same. as there are 4 cards left to hit your straight, there's a 4 in 47 chance.
(Unless you call the bottom card a seen and not unseen card)
That is just a statistical anomaly as what has changed is not the probability but the knowledge.
I would be surprised if this is the end of the debate though...
We always have a cutting card at the bottom of our deck, but you still get a slight advantage when one of our players, having watched WSOP, wears his shades, lifts his cards up to look at them, and everyone else gets to see the reflection in his shades!!! Not every hand, but he can't understand that we aren't psychic when we fold to his pocket hig hpairs or AK.
The worst thing about poker, is when we know he is on a semi bluff, he can still catch a miracle card on the turn/river and unwittingly get his own back!
Back to probability... If you flip a coin 49 times and it comes down heads 49 times, what is the chance of it being a tail next time??? Is it different if you allow it to land on the floor and not in your hand???
(Unless you call the bottom card a seen and not unseen card)
That is just a statistical anomaly as what has changed is not the probability but the knowledge.
I would be surprised if this is the end of the debate though...
We always have a cutting card at the bottom of our deck, but you still get a slight advantage when one of our players, having watched WSOP, wears his shades, lifts his cards up to look at them, and everyone else gets to see the reflection in his shades!!! Not every hand, but he can't understand that we aren't psychic when we fold to his pocket hig hpairs or AK.
The worst thing about poker, is when we know he is on a semi bluff, he can still catch a miracle card on the turn/river and unwittingly get his own back!
Back to probability... If you flip a coin 49 times and it comes down heads 49 times, what is the chance of it being a tail next time??? Is it different if you allow it to land on the floor and not in your hand???
- ZIPPY
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- Location: Edinburgh, Scotland
Re: Help with probability (specific case)
Mad Max wrote:I mean does it change from a 4 in 47 chance to a 3 in 47 chance? Just because i know one of the four wont hit?
It's 3 in 46 now. Just as you've eliminated it as an out, you also eliminate it as an unseen card.
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Sean_in_NJ - Posts: 3340
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"Technically, the probability is still the same. as there are 4 cards left to hit your straight, there's a 4 in 47 chance. "
But there are only 3 possible cards to hit that, and it's out of 46.
But there are only 3 possible cards to hit that, and it's out of 46.
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snoogins47 - Posts: 2358
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On the subject;
Does any of you take into consideration that there are, on a 10 player table, 9 other hands in action? That's 18 cards, and when you're on a flush draw after the flop, what are the chances of them holding/folding none of the desired suit? 25% of the cards dealt would statistically be from that suit, in a 10 player scenario that would mean an average of 4,5 cards from each suit. So your actual number of outs would not be 9 but 4,5.
I guess I'm way off thinking like this, but it would be nice to get some other people's thoughts on this.
Does any of you take into consideration that there are, on a 10 player table, 9 other hands in action? That's 18 cards, and when you're on a flush draw after the flop, what are the chances of them holding/folding none of the desired suit? 25% of the cards dealt would statistically be from that suit, in a 10 player scenario that would mean an average of 4,5 cards from each suit. So your actual number of outs would not be 9 but 4,5.
I guess I'm way off thinking like this, but it would be nice to get some other people's thoughts on this.
- Biff Malibu
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Biff Malibu wrote:On the subject;
Does any of you take into consideration that there are, on a 10 player table, 9 other hands in action? That's 18 cards, and when you're on a flush draw after the flop, what are the chances of them holding/folding none of the desired suit? 25% of the cards dealt would statistically be from that suit, in a 10 player scenario that would mean an average of 4,5 cards from each suit. So your actual number of outs would not be 9 but 4,5.
I guess I'm way off thinking like this, but it would be nice to get some other people's thoughts on this.
Here's the problem with that. If you're going to make the leap that you only have 4.5 outs, then you have also eliminated 13.5 non-outs: the remaining 13.5 off-suited cards dealt to the other 9 players. So your odds aren't 4.5 in 47...they're 4.5 in 29 (47 unseen - 18 player cards). It could be better, it could be worse, but it's not too far from the original 9 in 47. It's almost a meaningless exercise.
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Sean_in_NJ - Posts: 3340
- Joined: Fri Jan 02, 2004 5:57 pm GMT
- Location: New jersey
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