Showing cards when all in (and before decided how many times to run it) do you show?
Bug bearer of mine is people not showing cards . I often shout "cards over!" as you would in a tournament but not eve
Further proof you don’t know what you don’t know.
Why? It doesn't alter the play before the decision of how many times. And if anyone takes the huff then the default is that it's run only once.
this run it more than once faf started with poker on TV. If you watch some of the early stuff even the dealers had to be told how to do it.
IYKYK
Didace is spot on.
Assume we are at the river. Player behind has 1 out. We know 8 cards, 2+2+4 on board, so 44 unknown. Run it once. Behind wins 1 and loses 43 times.RIT. Now first board behind wins 1 of 43 times but if he wins he only gets 50%. For second board, his chances are 1 win and 42 losses but only if he lost first board. So again 1 chance to win, 42 chance to lose, plus 1 chance he has
Apologies in advance if this has been questioned/answered further down the thread. My brain is such that I will not remember to raise the question should I decide to peruse much further.
Why in your equations have you not made account for the cards discarded by other players which could be as many as 12 in a table of 8?
It may be just as simple as you using basic maths to illustrate your point and taking licence from but you have framed this as a right or wrong answer.
The answer may or may not support your conclusion, but I am genuinely interested eitherwise. I'm quietly hoping a math fanatic can provide me with a concrete answer that I have seen long dismissed many times.
P.S. To stay on topic, I sometimes ask the other player if they would like to play face up before the cards because I prefer the cards to tell me if I took a bad beat, rather than a smug fish rolling his cards over after the fact thinking he played well.
Apologies in advance if this has been questioned/answered further down the thread. My brain is such that I will not remember to raise the question should I decide to peruse much further.Why in your equations have you not made account for the cards discarded by other players which could be as many as 12 in a table of 8?It may be just as simple as you using basic maths to illustr
Or discarded burn cards for that matter.
What are the odds of 0, 1, 2, 3, ... of your outs being in the muck and how does that change the odds of you winning? Add that up and how does it differ form using just the number of unknown cards? The answer is -- it's the same.
I’ll try it more mathematically, but Didsce is right - we ignore discards because they are irrelevant. Let’s start by making a concrete, but simplified problem. Player A has TT. Player B has 88. Flop is T83r and we want to know what the probability of the turn being an 8 is. Assume heads up but with normal burn cards.
Now, the turn being an 8 is the conjunction of three events - burn card 1 is not an 8, burn card 2 is not an 8, and turn card is an 8. Since we know 7 cards, there are 45 remaining. The probability that burn card 1 is not an 8 is 44/45. Given that burn 1 is not an 8, we have 44 cards left, so burn 2 is not an 8 with probability 43/44. Given both burns aren’t an 8, we have 43 cards left, and the turn is an 8 with probability 1/43.
Our probability of the turn being an 8 is therefore 44/45 x 43/44 x 1/43, but the common factors of 44 and 43 in the numerators and denominators cancel, leaving the value of 1/45 for our probability - exactly what we get by ignoring the burns.
Adding a player or players doesn’t change this; it just adds more fractions to our calculation. With one player getting two cards the calculation would be
44/45 x 43/44 x 42/43 x 41/42 x 1/41 which is still 1/45.
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What are the odds of 0, 1, 2, 3, ... of your outs being in the muck and how does that change the odds of you winning? Add that up and how does it differ form using just the number of unknown cards? The answer is -- it's the same.
You didn't answer the question but to put it in other terms.....
How does the Math that was ascribed work if the two outs you are depending on for a win have been mucked?
It's certainly not the same is it.
Because the outs are equally likely to be anywhere. What if all the outs are on the bottom of the deck? That's the same as them being in the muck. Or maybe on the river you have one out but it is the third card in the deck? The point is until you know where they are you don't know where they are.
Quite simply - You don't know what the cards are, so why would they factor into the odds?They are unknown, just like the remaining cards in the stub. In the end, it all evens out. If other players folded most of your outs, your odds are hitting are lower. But if very few to none of your outs are folded, your odds have now increased. It evens out and is therefore irrelevant.
Because the outs are equally likely to be anywhere. What if all the outs are on the bottom of the deck? That's the same as them being in the muck. Or maybe on the river you have one out but it is the third card in the deck? The point is until you know where they are you don't know where they are.
OK and Thank-You. Please disregard the question that I asked, I was hoping for a more sophisticated explanation from the guy I quoted, but there you go.
I'm asking the question of how the Math works if your outs are in the muck already and the obvious answer is they do not.
Math does not work when there an element of incomplete information (obviously) and that is what I was looking for an eloquent or scientific answer to. I.E. Math is great at calculating statistical probabilities which may help you rationalise long term profit, but it does not help in the slightest, in the moment, when your outs are simply not there and you are bedrock still counting on the numbers.
Quite simply - You don't know what the cards are, so why would they factor into the odds?They are unknown, just like the remaining cards in the stub. In the end, it all evens out. If other players folded most of your outs, your odds are hitting are lower. But if very few to none of your outs are folded, your odds have now increased. It evens out and is therefore irrelevant.
Quite simply - You don't know what the cards are , so they cannot be factored into the odds.
It may or may not even out in the long term, but that was not the question I asked.
You use the word irrelevant assuming that over the long term the statistics will ring true. Apologies, but that is simply not the case.
Starting to feel like the odds that this is just OP on an alt account is about 108%
Okay, here's another way to look at it. Let's say it's the turn card we're interested in.
Dealer takes the cards out of the shuffler and cuts the deck. At this point, the turn card has been determined.
Dealer deals two cards each to eight players. That's 16 cards.
Dealer burns a card and deals the flop. That's four more cards, 20 total.
Dealer burns a card, that's 21 so far. The next card will be the turn card.
All we're interested in is the likelihood of that 22nd card being what we want it to be. The only information we have is the cards we've seen. It doesn't matter how the cards are distributed in this particular deck; unseen cards are unseen cards. All we care about is that 22nd card.
Quite simply - You don't know what the cards are , so they cannot be factored into the odds.
It may or may not even out in the long term, but that was not the question I asked.
You use the word irrelevant assuming that over the long term the statistics will ring true. Apologies, but that is simply not the case.
You are essentially asking about conditional probability.
The probability of A is different than the probably of A given B.
In this case, we are all describing your outs as the probability of A, and you're asking why we don't incorporate B. The reason is that we don't have any event B to include. We don't know what cards are in the stub, or discards, or other hands. If we did, then (and only then, and always then) we would instead have to calculate the probability of A given B to determine the probability of winning.
Say, for example, if one player says "oh, I folded the last 8" or if a spade gets exposed by a player who is folding while you are on the spade draw. Now you would have a different probability to calculate, A given B. Absent that knowledge, though, all distributions of the unknown cards are equally probable, and all you can calculate is the probability of A.
Another related example: in a normal poker game, again, we are all just calculating the probability of A. So, say, every time you're on a flush draw you know you have 9 outs, and the probability of every player on every flush draw is exactly the same. But on televised poker, they know what all the players cards were, they show them as every player looks at their cards and folds. Now there is additional information available about how many of the suit you are drawing to got folded. When they show the probabilities on screen, they are now not always identical, the percentages will vary based on what cards are known to be folded and therefore cannot be in the stub still to possibly come out on turn or river. This is conditional probability.
You are essentially asking about conditional probability.The probability of A is different than the probably of A given B.In this case, we are all describing your outs as the probability of A, and you're asking why we don't incorporate B. The reason is that we don't have any event B to include. We don't know what cards are in the stub, or discards, or other hands. If we did,
Hallelujah. Thank you Dinesh.
Exactly what I was looking for, someone who is able to think a little outside their bounds.
I've now got something interesting to read up on.
BTW, I may be a newbie on the forum, but I'm certainly not an imposter.
OK and Thank-You. Please disregard the question that I asked, I was hoping for a more sophisticated explanation from the guy I quoted, but there you go.I'm asking the question of how the Math works if your outs are in the muck already and the obvious answer is they do not. Math does not work when there an element of incomplete information (obviously) and that is what I was look
Math is all we have when we have incomplete information. Based on what information we have, the probability of a turn card being our one out is exactly 1/45. That doesnÂ’t change, as my calculation in the prior post demonstrated, if we account for the probability of discards being our out or if we ignore that probability. Math certainly canÂ’t tell us whether we will hit our out or not on any given run out; we shouldnÂ’t expect it to. It canÂ’t tell you what number will come up on the next roulette wheel spin or what the dice will be on the next throw at the craps table either. It can only tell you how likely any particular outcome is.
Intuitive ideas surrounding probability are often misleading. That’s because we must be extremely careful in how we define our problem. In this particular case, do we know anything about the mucked cards or not? Typically, no we do not. Hence the 1/45 probability is correct, even though the “real” probability (if we did know all the information) might be zero. But we can also calculate how likely it is that one of the mucked cards is our out and factor that into our decision making - again there’s no math that can tell us whether our out was mucked or not; it only can give us a probability. That’s exactly what I did in my prior post. I calculated the probability that our card was NOT mucked. Given that our card was not mucked, there is a different probability that we hit our out on the turn - we know we are more likely to hit the turn if our out was not mucked. To get the unconditional probability that we hit our out, we multiply this conditional one by the probability that our card was not mucked. My post demonstrated that we don’t actually have to go through all this — our unconditional probability is exactly what it would be if we only knew the cards we see and none were mucked.
ThatÂ’s why I said the mucked cards are irrelevant. If we had more information (for example someone exposed our out when he folded his hand), that would certainly change our thinking. If we knew someone mucked our card, then we would know we have zero probability of hitting our out on the turn. Nobody is disputing that. But typically we donÂ’t have that information.
Information is critical in probability calculations. That’s why the “Monty Hall” problem still trips people up. If you aren’t familiar, the show Lets Make a Deal would allow contestants to choose one of three doors. There was a big prize (like a car) behind one and worthless items (goats) behind the other two. After picking a door, Monty would open one of the doors the contestant didn’t pick and show a goat. He then asked if the contestant wanted to change doors. Should the contestant switch?
Intuitively, it seems like showing a goat hasn’t changed anything - you had a 1/3 chance of winning, you knew he’d show a goat, so you still have a 1/3 chance of winning. But that is not actually true. Monty has given you information that changes your probability calculations. Had he asked “Do you want to switch?” BEFORE showing a goat, intuition is right; nothing has changed. He eliminated one wrong door for you though. That means you should switch. The probability that you have the car behind your door is 1/3, but the probability that it is behind the other unopened door is now 2/3 - you are twice as likely to win if you switch. There is actually an easy way to see this, but people don’t usually think about it this way. If my original choice was right, and I switch then I lose. If my original choice was wrong, and I switch then I win. We should be able to agree that the original choice was right with probability 1/3, so the original choice must be wrong with probability 2/3.
Anyway, TL;DR version - trust the math, not your intuition.
Math is all we have when we have incomplete information. Based on what information we have, the probability of a turn card being our one out is exactly 1/45. That doesnÂ’t change, as my calculation in the prior post demonstrated, if we account for the probability of discards being our out or if we ignore that probability. Math certainly canÂ’t tell us whether we will hit our ou
I'm aware of the problem and seen the film 21 but Thank-you.
I'm going to apologise again and say that your prior post just come across as very black and white and frankly dismissive and condescending. You doubled down on the word Irrelevant and now seem to be tripling down on it just to further re-enforce your position. You could not bring yourself to say, simply, the Math does not cover what we don't know which is understandable.
"Anyway, TL;DR version - trust the math, not your intuition"
This proves my point like yours is the only and absolute truth. I Get it, you are a Math guy, I am more of a Psychology guy.
There is a world of grey that joins black to white.
I trust the Math as far as it is factual and relevant to the moment I am facing. Other than that, it's intuition for me all the way and I have very good intuition which has served me well in the corporate and poker world.
I'm aware of the problem and seen the film 21 but Thank-you.I'm going to apologise again and say that your prior post just come across as very black and white and frankly dismissive and condescending. You doubled down on the word Irrelevant and now seem to be tripling down on it just to further re-enforce your position. You could not bring yourself to say, simply, the Math does
Intuition has its place, but what we are discussing IS a math problem. There is no room for intuition and there is ONE correct answer. Sorry if that seems dismissive, but there is a single correct answer to “What is the probability that the turn card will be an 8 given that we know three 8s have already been dealt?” The correct answer is 1/45. If intuition leads to another answer, r then that is like intuition telling you that 2+2 is actually 5. This is a well-defined math problem with a correct answer. In other situations I don’t doubt that your intuition might serve you well, but solving well-defined math problems calls for mathematics, not intuition
Intuition has its place, but what we are discussing IS a math problem. There is no room for intuition and there is ONE correct answer. Sorry if that seems dismissive, but there is a single correct answer to “What is the probability that the turn card will be an 8 given that we know three 8s have already been dealt?” The correct answer is 1/45. If intuition leads to another answ
Fair enough. I'm guessing that you don't set-mine that often unless you get in cheap against a big stack?
