Title: probability question Post by: Oxford_HRV on October 24, 2012, 07:04:52 AM 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....? Title: Re: probability question Post by: Tal on October 24, 2012, 08:17:28 AM 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... Title: Re: probability question Post by: Jon MW on October 24, 2012, 08:33:14 AM ... 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. Title: Re: probability question Post by: Doobs on October 24, 2012, 08:35:03 AM The answer to both your questions is it is irrelevant. The chance of a second ace is 1 in 17 wherever you are sat.
Title: Re: probability question Post by: bobAlike on October 24, 2012, 09:50:51 AM 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.
Title: Re: probability question Post by: aaron1867 on October 24, 2012, 09:59:25 AM Why does it matter?
You always seem to get that bloody deuce? :) :) :) Title: Re: probability question Post by: Leatherman on October 24, 2012, 12:23:33 PM Why does it matter? You always seem to get that bloody deuce? :) :) :) Or even worse the 4d which for second you think is the other Ace. :( Title: Re: probability question Post by: Tal on October 24, 2012, 01:10:18 PM Good point. Live poker is rigged. Everyone knows that.
Title: Re: probability question Post by: SuuPRlim on October 24, 2012, 01:10:30 PM 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. Title: Re: probability question Post by: George2Loose on October 24, 2012, 04:06:34 PM 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) Title: Re: probability question Post by: aaron1867 on October 24, 2012, 04:07:44 PM 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. Title: Re: probability question Post by: Cf on October 24, 2012, 05:30:09 PM 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. Title: Re: probability question Post by: George2Loose on October 24, 2012, 05:32:26 PM Ok so calculate the above pls
Title: Re: probability question Post by: Tal on October 24, 2012, 05:35:56 PM First Broadway card is 5/13. Second card being the same suit is 12/51.
60/663 is about 1/11 or 9% Title: Re: probability question Post by: Tal on October 24, 2012, 05:37:09 PM Have I read that correctly or are you asking for both cards to be Broadway cards?
Title: Re: probability question Post by: skolsuper on October 24, 2012, 05:37:20 PM Ok so calculate the above pls Suited broadways are 3% of all hands. The 6O question it is relevant whether you're asking for hands with at least a pair, or only hands with just 1 pair. Title: Re: probability question Post by: George2Loose on October 24, 2012, 05:43:33 PM At least one pair
Title: Re: probability question Post by: skolsuper on October 24, 2012, 06:07:03 PM card 1: 1
card 2: 48/51 (3 cards pair us) card 3: 44/50 (6 cards pair us) card 4: 40/49 etc card 5: 36/48 card 6: 32/47 multiply together to get the probability of being dealt an unpaired hand = 34.5% therefore the probability of at least 1 pair in 6 cards (or 2 pair or trips or a full house or quads, but ignoring straights and flushes) is 65.5% Title: Re: probability question Post by: Oxford_HRV on October 25, 2012, 03:31:57 AM if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace?
it's approx 20% of the deck with 3 aces in, it must be worked out! :) ty Title: Re: probability question Post by: AdamM on October 25, 2012, 08:24:13 AM The way I'm seeing the original question, if your first card is an ace, of the 51 remaining cards 3 are aces.
9 cards are dealt before it comes back to you and the question is what are the chances of one of them being an ace. each card has a 5.88% chance of being an ace (3/51*100) multiplied by 9 shots, that's 52.92% chance that someone elses first card is an ace. look right? Title: Re: probability question Post by: AdamM on October 25, 2012, 08:33:36 AM Ok so calculate the above pls Suited broadways are 3% of all hands. The 6O question it is relevant whether you're asking for hands with at least a pair, or only hands with just 1 pair. This chances of card 1 being a suited broadway card 20/52 chances of second card being the same suit and aslo 10+ 4/51 (20/52) x (4/51) = 80/2652 = 3.02% Title: Re: probability question Post by: AdamM on October 25, 2012, 08:39:50 AM if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace? that's a different question and the answer isn't 20%it's approx 20% of the deck with 3 aces in, it must be worked out! :) ty not sure if I'm awake enough to work out what it is though. the way you've phrased it, from a 51 card deck, containing 3 aces, what are the chances of 1of the next 10 being an ace (not 2 or 3) Title: Re: probability question Post by: Oxford_HRV on October 25, 2012, 08:44:40 AM I got to assume it surely can't be odds in favour another ace is dealt out.
think possibly you will have to add all fractions up 3/51----- 3/43 and multiply by chances you not receiving an ace, in a tree diagram possibly? I was saying 10 cards is approx 20% of 51 cards, not really written well :s Title: Re: probability question Post by: Oxford_HRV on October 25, 2012, 08:51:43 AM if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace? that's a different question and the answer isn't 20%it's approx 20% of the deck with 3 aces in, it must be worked out! :) ty not sure if I'm awake enough to work out what it is though. the way you've phrased it, from a 51 card deck, containing 3 aces, what are the chances of 1of the next 10 being an ace (not 2 or 3) that's the question I would like answered, I said 10 cards instead of 9 because I'm including the second card that would be dealt to me Title: Re: probability question Post by: doubleup on October 25, 2012, 08:59:08 AM if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace? it's approx 20% of the deck with 3 aces in, it must be worked out! :) ty There are 12777711870 combunations of 10 cards in 51 and 6540715896 combinations of 10 cards in 48 (the 3 aces removed). So the probability of no aces in the 10 cards dealt is 51% (6540715896/12777711870) and therefore of an ace being dealt 49%. Title: Re: probability question Post by: AdamM on October 25, 2012, 08:59:27 AM I got to assume it surely can't be odds in favour another ace is dealt out. think possibly you will have to add all fractions up 3/51----- 3/43 and multiply by chances you not receiving an ace, in a tree diagram possibly? I was saying 10 cards is approx 20% of 51 cards, not really written well :s your questions are all over the place. In your OP you're asking what are the chances of one of your opponents getting an ace as their first card, knowing that one of them is in your hand. Each of them has a 3/51 chance of getting an ace, and there's 9 shots. 3/51*100=5.88% * 9 shots = 52.94% try it by taking an ace out of the dec and dealing 9 cards in a row doesn't seem unreasonable that there'd be an ace around half the time Title: Re: probability question Post by: AdamM on October 25, 2012, 09:01:01 AM if you took out any ace, then deal out 10 cards from a shuffled deck afterwards what is the probability one is an ace? that's a different question and the answer isn't 20%it's approx 20% of the deck with 3 aces in, it must be worked out! :) ty not sure if I'm awake enough to work out what it is though. the way you've phrased it, from a 51 card deck, containing 3 aces, what are the chances of 1of the next 10 being an ace (not 2 or 3) that's the question I would like answered, I said 10 cards instead of 9 because I'm including the second card that would be dealt to me your original question specifically says what are the chances of another ace being dealy BEFORE it comes back to you. that's 9 cards Title: Re: probability question Post by: AdamM on October 25, 2012, 09:02:42 AM Do you want to know if you have an ace as either of your cards, what are the chances of 1 or mor other aces being held by one of the other 9 players?
Title: Re: probability question Post by: AdamM on October 25, 2012, 09:06:25 AM doubting my method now :)
Title: Re: probability question Post by: kinboshi on October 25, 2012, 09:06:38 AM Do you want to know if you have an ace as either of your cards, what are the chances of 1 or mor other aces being held by one of the other 9 players? I'm not sure he knows, as that question has been answered by doubleup (well he included all ten players as requested, but the method is there). Title: Re: probability question Post by: AdamM on October 25, 2012, 09:11:52 AM not quite, my suggested question includes any of the 18 cards we haven't seen.
We know that 1 of our cards is and ace and the other is a random non-ace that means 3/50 cards are an ace what are the chances that if we deal 18 out, 1, 2 or 3 of the remaining aces are dealt Using my shitty shorthand, I get a 100%+ answer, which is why I think I might be doing it wrong. I'm just calculating the way I would count outs before pot odds based call on the turn Title: Re: probability question Post by: doubleup on October 25, 2012, 09:58:22 AM not quite, my suggested question includes any of the 18 cards we haven't seen. We know that 1 of our cards is and ace and the other is a random non-ace that means 3/50 cards are an ace what are the chances that if we deal 18 out, 1, 2 or 3 of the remaining aces are dealt Using my shitty shorthand, I get a 100%+ answer, which is why I think I might be doing it wrong. I'm just calculating the way I would count outs before pot odds based call on the turn You have to use combinations to work out these kind of problems. Your method is ok for a quick approximation of a simple scenario but the errors get magnified in complicated calculations. http://en.wikipedia.org/wiki/Combination Title: Re: probability question Post by: david3103 on October 25, 2012, 10:25:18 AM Loads of help and some answers here
http://en.wikipedia.org/wiki/Poker_probability_%28Texas_hold_%27em%29 confirms the probability of suited broadway cards as 0.0302 and although this isn't directly the question Oxford asked, this table gives some clues (http://i1236.photobucket.com/albums/ff442/pokerpops/ProbabilityofholdingdominatedAce.jpg) Title: Re: probability question Post by: atdc21 on October 25, 2012, 12:54:46 PM I have lost track of what the OP was asking or why, but that is a well handy table David, ty.
Title: Re: probability question Post by: youthnkzR on October 25, 2012, 01:36:07 PM Why does it matter? You always seem to get that bloody deuce? :) :) :) haha too true! never the same suit ever :( Title: Re: probability question Post by: Oxford_HRV on October 25, 2012, 05:16:07 PM your questions are all over the place. Each of them has a 3/51 chance of getting an ace, try it by taking an ace out of the dec and dealing 9 cards in a row doesn't seem unreasonable that there'd be an ace around half the time sorry, I have too many questions to ask! the way I came to this scenario was by thinking, if I get a specific value card and the person next to me receives the same, what are the chances I pair up when it gets back to me? simply 1/13 1/17 1/21 multiplied I am always SB being dealt first in these situ's then I wanted to take into account the variables, some one at random may actually receive the same card as me, which lead me to the OP, after my workings out failed to satisfy my belief I was correct. the first ace is GTD and you get 1/13 but say you are on BTN 10 handed, (no aces known to be out) you technically have better odds of receiving an ace as you cannot account for what has been dealt out you can only assume your chances of an ace is 4/43 it's this which makes it unable for everyone to have a 3/51 chance as the deck gets shorter, so to take into account the variables here I wanted to know the probability of a 3 ace 51 card deck having ten cards dealt off top (effectively any 10 removed) one being an ace. Title: Re: probability question Post by: Oxford_HRV on October 25, 2012, 05:16:40 PM Loads of help and some answers here http://en.wikipedia.org/wiki/Poker_probability_%28Texas_hold_%27em%29 confirms the probability of suited broadway cards as 0.0302 and although this isn't directly the question Oxford asked, this table gives some clues (http://i1236.photobucket.com/albums/ff442/pokerpops/ProbabilityofholdingdominatedAce.jpg) good link TY :) Title: Re: probability question Post by: Cf on October 25, 2012, 05:43:15 PM your questions are all over the place. Each of them has a 3/51 chance of getting an ace, try it by taking an ace out of the dec and dealing 9 cards in a row doesn't seem unreasonable that there'd be an ace around half the time sorry, I have too many questions to ask! the way I came to this scenario was by thinking, if I get a specific value card and the person next to me receives the same, what are the chances I pair up when it gets back to me? simply 1/13 1/17 1/21 multiplied I am always SB being dealt first in these situ's then I wanted to take into account the variables, some one at random may actually receive the same card as me, which lead me to the OP, after my workings out failed to satisfy my belief I was correct. the first ace is GTD and you get 1/13 but say you are on BTN 10 handed, (no aces known to be out) you technically have better odds of receiving an ace as you cannot account for what has been dealt out you can only assume your chances of an ace is 4/43 it's this which makes it unable for everyone to have a 3/51 chance as the deck gets shorter, so to take into account the variables here I wanted to know the probability of a 3 ace 51 card deck having ten cards dealt off top (effectively any 10 removed) one being an ace. You simply have to ignore what other people have. Let's use the 2 aces example. I'll show why even taking dealing to others into account you still have a 3/51 chance of receiving it. However, i'm going to simplify the problem a bit to illustrate the point: You are being dealt to first. The first card is always the ace. (this is just to make the numbers easier to work with, it'd also work if we assumed you had a 4/52 chance of being dealt an ace in the first place). There are only two aces in the deck so only one remains. (again, makes the numbers easier to work with without changing the logic behind the problem) We are dealing to 9 people before you. So now you have a 1/51 (1.96%) chance of receiving the second ace. And we're going to deal the cards face up so they're no longer unknown. The chances of no-one else being dealt the ace are: 50/51 * 49/50 * 48/49 * 47/48 * 46/47 * 45/46 * 44/45 * 43/44 * 42/43 = 14/17 (82.35%) So the odds have changed for us now we know if it's been dealt or not. 17.65% of the time we will have a 0% chance of receiving it. 82.35% of the time we will have a 1/42 (2.38%) chance of receiving it. Note that this is higher than the 1.96% of the time we have if the cards are unknown. So sometimes we now have no chance at all, but sometimes our chance has increased. Let's find the average of these chances... 17.65% * 0 = 0 82.35% * 2.38% = 1.96% Add 'em up and we get 1.96% (1/51) which is what the chance would be if we consider the cards unknown. This works for any number of players. So it is completely irrelevent how many players sit at a table for the purposes of determining odds of which cards will be dealt to you, nor does it matter where you are sat at the table. Title: Re: probability question Post by: skolsuper on October 25, 2012, 06:06:59 PM your questions are all over the place. Each of them has a 3/51 chance of getting an ace, try it by taking an ace out of the dec and dealing 9 cards in a row doesn't seem unreasonable that there'd be an ace around half the time sorry, I have too many questions to ask! the way I came to this scenario was by thinking, if I get a specific value card and the person next to me receives the same, what are the chances I pair up when it gets back to me? simply 1/13 1/17 1/21 multiplied I am always SB being dealt first in these situ's then I wanted to take into account the variables, some one at random may actually receive the same card as me, which lead me to the OP, after my workings out failed to satisfy my belief I was correct. the first ace is GTD and you get 1/13 but say you are on BTN 10 handed, (no aces known to be out) you technically have better odds of receiving an ace as you cannot account for what has been dealt out you can only assume your chances of an ace is 4/43 it's this which makes it unable for everyone to have a 3/51 chance as the deck gets shorter, so to take into account the variables here I wanted to know the probability of a 3 ace 51 card deck having ten cards dealt off top (effectively any 10 removed) one being an ace. You simply have to ignore what other people have. Let's use the 2 aces example. I'll show why even taking dealing to others into account you still have a 3/51 chance of receiving it. However, i'm going to simplify the problem a bit to illustrate the point: You are being dealt to first. The first card is always the ace. (this is just to make the numbers easier to work with, it'd also work if we assumed you had a 4/52 chance of being dealt an ace in the first place). There are only two aces in the deck so only one remains. (again, makes the numbers easier to work with without changing the logic behind the problem) We are dealing to 9 people before you. So now you have a 1/51 (1.96%) chance of receiving the second ace. And we're going to deal the cards face up so they're no longer unknown. The chances of no-one else being dealt the ace are: 50/51 * 49/50 * 48/49 * 47/48 * 46/47 * 45/46 * 44/45 * 43/44 * 42/43 = 14/17 (82.35%) So the odds have changed for us now we know if it's been dealt or not. 17.65% of the time we will have a 0% chance of receiving it. 82.35% of the time we will have a 1/42 (2.38%) chance of receiving it. Note that this is higher than the 1.96% of the time we have if the cards are unknown. So sometimes we now have no chance at all, but sometimes our chance has increased. Let's find the average of these chances... 17.65% * 0 = 0 82.35% * 2.38% = 1.96% Add 'em up and we get 1.96% (1/51) which is what the chance would be if we consider the cards unknown. This works for any number of players. So it is completely irrelevent how many players sit at a table for the purposes of determining odds of which cards will be dealt to you, nor does it matter where you are sat at the table. Well put. Title: Re: probability question Post by: Oxford_HRV on October 25, 2012, 06:19:11 PM your questions are all over the place. Each of them has a 3/51 chance of getting an ace, try it by taking an ace out of the dec and dealing 9 cards in a row doesn't seem unreasonable that there'd be an ace around half the time sorry, I have too many questions to ask! the way I came to this scenario was by thinking, if I get a specific value card and the person next to me receives the same, what are the chances I pair up when it gets back to me? simply 1/13 1/17 1/21 multiplied I am always SB being dealt first in these situ's then I wanted to take into account the variables, some one at random may actually receive the same card as me, which lead me to the OP, after my workings out failed to satisfy my belief I was correct. the first ace is GTD and you get 1/13 but say you are on BTN 10 handed, (no aces known to be out) you technically have better odds of receiving an ace as you cannot account for what has been dealt out you can only assume your chances of an ace is 4/43 it's this which makes it unable for everyone to have a 3/51 chance as the deck gets shorter, so to take into account the variables here I wanted to know the probability of a 3 ace 51 card deck having ten cards dealt off top (effectively any 10 removed) one being an ace. You simply have to ignore what other people have. Let's use the 2 aces example. I'll show why even taking dealing to others into account you still have a 3/51 chance of receiving it. However, i'm going to simplify the problem a bit to illustrate the point: You are being dealt to first. The first card is always the ace. (this is just to make the numbers easier to work with, it'd also work if we assumed you had a 4/52 chance of being dealt an ace in the first place). There are only two aces in the deck so only one remains. (again, makes the numbers easier to work with without changing the logic behind the problem) We are dealing to 9 people before you. So now you have a 1/51 (1.96%) chance of receiving the second ace. And we're going to deal the cards face up so they're no longer unknown. The chances of no-one else being dealt the ace are: 50/51 * 49/50 * 48/49 * 47/48 * 46/47 * 45/46 * 44/45 * 43/44 * 42/43 = 14/17 (82.35%) So the odds have changed for us now we know if it's been dealt or not. 17.65% of the time we will have a 0% chance of receiving it. 82.35% of the time we will have a 1/42 (2.38%) chance of receiving it. Note that this is higher than the 1.96% of the time we have if the cards are unknown. So sometimes we now have no chance at all, but sometimes our chance has increased. Let's find the average of these chances... 17.65% * 0 = 0 82.35% * 2.38% = 1.96% Add 'em up and we get 1.96% (1/51) which is what the chance would be if we consider the cards unknown. This works for any number of players. So it is completely irrelevent how many players sit at a table for the purposes of determining odds of which cards will be dealt to you, nor does it matter where you are sat at the table. thanks bud, that's awesome! well that's sorted that one out for me! ;) Title: Re: probability question Post by: youthnkzR on October 25, 2012, 07:45:15 PM i get a first Ahrt around 50% of the time, a second Ad follows a further 50% of the time ;sark; ;sark;
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