This is a discussion on How to calculate odds on a multicard draw. within the online poker forums, in the Learning Poker section; I initially put this question to the administrator, and they answered back a few minutes ago recommending I post it in the forum... so here 

How to calculate odds on a multicard draw. 
#1




How to calculate odds on a multicard draw.
I initially put this question to the administrator, and they answered back a few minutes ago recommending I post it in the forum... so here goes :
My friends and I have been struggling with calculating the odds in various cardgame situations. I managed to find a person with a PHD in Statistics, but his answers/solutions did not hold up to testing, and he won't help anymore... so I'm asking you for help. We have several examples, but I'll stick with one to start with, and maybe ask the 2nd question on a followup email... assuming we get this 1st question solved. Imagine you have a deck of 52 cards, face down on the table. The odds of randomly pulling a spade on a single draw from the deck is 25% (13 spades, 52 cards, easy). What if I pull out four random cards from the deck. What are the odds that there is at least one spade among them ? Same problem, but if I pull seven random cards from the deck of 52, what are the odds that there is at least one spade among the seven ? I need to see the math... the method used to derive the odds of finding a spade in those cards. I hope to hear back from you with good clear examples of how to calculate it !
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Similar Threads for: How to calculate odds on a multicard draw.  
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How to calculate odds?  3  May 4th, 2020 11:02 PM  Learning Poker 
#2




OK, I'll take a stab. I like math.
The reason why there are several answers, all of which are defensible, is that there are several approaches. Let's play a game. Maybe you know this one? There's a purpose here, to demonstrate a principle. Let's say that you have three pairs of hole cards. One pair is aces. one pair is kings. One pair is ace + king. You randomly pick one of the six cards and it's an Ace. What are the odds that the other card of that pair is an ace?
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#3




I hate math, but this made me smile..even I got it
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#4




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#5




I hate math too, unless its poker math, so...
50/50
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#6




Wow! You take 4:20 seriously, eh? Don't let math drive you so crazy. You'll end up like Sheldon Cooper.
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#7




As I recall, it was Bertrand Russell who in the early 20th century refined the relationships between philosophy/mathematics/logic. He didn't invent it all, but Russell was remarkable both for being not insane and for being a gifted speaker, teacher, and writer. Russell used my example of 3 pairs of cards (he used coins I think) to demonstrate that often the events which we think are discrete, or mutually exclusive, are not so. According to Russell, you're hunting for the pair of aces so if you flip an ace you've flipped one of the three. For two of the three aces on the board, the other card of its pair is an ace, and one of the aces has a king with it. Russell would say that the flipping of the cards isn't mutually exclusive in the information you've gleaned. Rather than simply eliminating the pair of Kings by flipping an ace and thus making it a 50/50 that you picked the pair of aces, you're at 66.6% of flipping a second ace once you've flipped the first. Thinking about it in this way leads different approaches of solving a problem, to Heisenberg's uncertainty principle and Shroedinger's cat, essentially saying that the unflipped cards literally have no value until they are flipped and thus the probability function (which is a wave) collapses and a single result gels. In quantum physics, it's impossible to determine the precise location of an electron and trying to do so affects the path of it; all you can do is create probabilities of where the electron might be along its path depending on your knowledge of its properties. So back to OP's question. You have 52 cards laid out and you want to pick a spade on your first pick, so one in four. Does selecting four of the 52 cards into a smaller pile and then flipping one of those four change the odds? Does eliminating cards one at a time from the 52 until there's only one left, does that change the odds of the last remaining card being a spade? Does it matter if you look at each of the 51 cards you eliminate as you do so? Most people would say "of course not!" and they are probably right. But most people would also have said that flipping an Ace leads to the odds of a second ace flip being 50%. So it's no wonder that OP is having a hard time sorting this out, and even with the help of a PhD in statistics the answers aren't clear. (However, I suspect that the PhD is an exaggeration because someone that smart could answer the questions easily I'm sure). I hope that OP comes back so we can discuss it.
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#8




So, the odds of drawing one black card are 13/52, or 25%. The odds of drawing two black cards are 13/52 * 12/51, or 5.88%. No problem there. Similarly, the odds of drawing three black cards in a row are 1.29%, four in a row is 0.2641%, five in a row is 0.0495%, six in a row is 0.008429%, and seven in a row is 0.0010994%. Intuitively we know (at least if we play a lot of cards) that drawing seven spades in a row is extremely unlikely, but the odds are even lower than we'd probably guess. Even getting 4 in a row is only a quarter of one per cent. So in holdem if you have hole spades, what are the odds of drawing three spades on the flop? That depends on what cards are taken out of the deck by the other hole cards. Here's where our approach yields to different results, and ties into OP's question. Do we assume equal distribution among all hole cards including yours? Do we assume equal distribution among all hole cards other than yours? Or do we ignore all the other hole cards as if they don't exist? Do we assume that we're taking 5 cards out of the remains of the deck and flip from that, and if we do does that make a difference? These questions are the assumptions upon which you build your formula, and those answers are sometimes counterintuitive. Which is why different results come from different approaches; it all comes back to Bertrand Russell. In OP's question there is an answer and it's not super complicated, but a fundamental understanding of the question and of your assumptions is the key to getting that answer.
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#9




Poker is a game of math, but it is also a game of people.
If you don't understand the math you will make bad decisions, but discussions like this make me think of the Married with Children episode where the blond bimbo Kelly tried to remember a ton of information and when she learned something new she had to forget something else. Don't get distracted by the math too much. Learn it, understand it, but in the middle of a hand your focus should be on the hand, (the villain, position, betting line, ect...) not trying to get exact figures. Most players just use the Rule of 4 and 2. You count your outs and multiply them by 4 on the flop (with 2 cards to come) and by 2 on the turn (with 1 card to come) which gives pretty close to your odds of making your hand. Make sure you are getting a good count of your outs (don't count outs twice) and you will have a rough idea of where you are at. Always remember, it is a game of incomplete information. Unless heads up, you will never know what was mucked and you could think you have better odds to make a hand than you do.
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#10




Thanks Odysseus101, Let me try to refine the question. A deck of 52 cards, face down on the table. I select four of them, discarding the remaining 48. I scoop up the four cards in front of me and flip them over at the same time. How do I calculate the odds that there is at least one spade among them ? Not just ONE spade among the four... there could be two, or three, or all four (an unlikely event as you pointed out), but the odds of ANY spade showing up in the four card blind pull. I figure it's a simple math problem, but haven't found the answer yet.
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#11




Final Anecdote: The math portion of poker shouldn't be underestimated. It can help build confidence in both your approach and in your abilities. It can also help you deal with bad beats and bad runs, if you let it. For example, think of the free throw shooter in a basketball game who is making 75% of his free throws. As a fan, should you be frustrated if this shooter is missing a free throw every other time he is approaching the foul line for a two shot try? Not if you understand the math. Making 3 out of every 4 shots, on average, means you should expect to miss a free throw every other time you approach the foul line (assuming two shots, and many trials, in this example). Most people don't think of it this way, though. We tend to focus too much on the favored outcome. Similarly, if you are a twotoone favorite in a hand, this means you are going to lose one hand in three. Yet, some folks react to these situations on the table as if it is some form of highway robbery or something, when they lose. So really, when evaluating RNGs and so forth, we are talking confidence intervals. Bad beats and bad runs are part of these intervals. Having said all this, humans are flawed, and programmers are human, so I am an advocate for certification and a watchdog approach to online poker. It's just easier to say than it is to do, obviously.
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#12




I don't think it matters if you pick four cards out of the 52 and then flip them all at once. The answer is the same as whether or not you pick four random cards, and flip them one at a time or altogether. Can we agree on that? If not, why do you think that picking four cards and putting them in your hand and then flipping (from) them makes a difference? OK, let's look at that specific situation. Out of four cards, you'll get five potential results: zero spades, one spade, two spades, three spades, or four spades. The odds of each of those five results must add up to 100%. The easiest approach is to eliminate the one result of zero spades, which we don't want. To get that we multiply together 39/52 * 38/51 * 37/50 * 36/49 to get 30.38175. That means on a four card draw, the odds of getting no spades is 30.38%. Which means that the odds of getting any other result (at least one spade) is 10030.38 or 69.62%. Does that make sense?
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#13




Same approach on a seven card draw. The odds of getting no spades at all is 39/52 * 38/51 * 37/50 * 36/49 * 35/48 * 34/47 * 33/46 = .114967953 if I hit my calculator buttons correctly. So, on a seven card draw the odds of at least one spade are 88.5%
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#14




This formula might work;
(13/52) + (13/51) + (13/50) + (13/49) = 1.0343261 which is 103% chance one will be a spade. But we all know for a certainty, that occasionally we will whiff in the draw, and get no spades. Which shows that you have a problem with no absolute answer.
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#15




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#16




re: Poker & How to calculate odds on a multicard draw.
It's gonna be 25%. Over the long haul, it will work out to 25%. Perhaps a better way to wonder about this would be to get into the statistics of doing this, perhaps with some fixed sample size, and achieving the expected long term results. And yes, the Heisenberg thing is valid, and adds the uncertainty. Also it is useless here, except that it shows that we can probably expect a random card if the deck is properly shuffled.
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#17




Let's say you're flipping a coin. It's 1/2 for heads. So if you are doing more than one flip, you multiply rather than add. Odds of a head on one flip = 1/2 or 50% Odds of two heads on two flips = 1/2 * 1/2 = 1/4 or 25% Odds of three heads on three flips = 1/2 * 1/2 * 1/2 =1/8 or 12.5% Or use a six sided dice. Odds of rolling a one are 1/6 or 16.667%. Roll two die, and the odds of getting two 1s are 1/6 * 1/6 or 1/36. This makes sense intuitively, because you know that there are 36 possible outcomes when rolling two die, and double ones is 1 of those 36 outcomes. Make sense? Right?
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#18




And you multiply them if you are looking for the odds of getting two spades in a row from start: 13/52 * 12/51 = 5.88%. These are fairly simple calculations because there's really only one way to approach it. My point about Heisenberg and Schrodinger is that when the problem gets more complicated we have to look closely at the assumptions upon which we base the problem; the wrong assumptions will lead to a wrong answer, or a different but perhaps equally right answer for different circumstances/assumptions.
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#19




If you multiply fractions less than 1, the results shrink. So in the card game we're playing (find a spade) according to your method here, the results of multiplying those fractions is something way less than 5% of the time we find a spade in 4 cards. My addition goes; I get a chance, I get another (slightly better) chance, I get another (even better) chance I get a last (still better) chance. We got 4 opportunities (chances) at 25% or better each, to find the spade. I think the ultimate solution will involve properly framing the question.
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#20




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#21




So, back the the original problem;
In this case the reducing odds are additive, and the math is (13/52) + (13/51) + (13/50) + (13/49) = 1.0343261 which is 103% chance one will be a spade. That formula can be open ended, but at some point it might be more useful to consider the odds against a spade showing, which will not be certain until the 40th card is pulled. If this alone were a betting game, the house would only pay 97%
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#22




That's what Russell's problem I mentioned earlier (where you have three pairs of cards and you're looking for the odds of hitting a second ace) demonstrates. Sometimes the information gleaned as the events proceed affects the ultimate probability, so the events in the chain aren't as mutually exclusive as we sometimes assume. You wrote "Difference here is that we are trying to figure out the odds of getting 1 specific result (1 or more spades)." and that's incorrect. Getting more than one spade is the sum of several results: getting one spade, getting two spades, getting three spades, and getting four spades. We could calculate each of those four results and add them up, but it's easier to calculate the odds of getting no spades (30.38%) and recognize that the other four results (at least one spade) added up will account for the other 69.62%. Now, that is an easy calculation. It doesn't tells us the odds of hitting one, two, three, or four spades. But we don't need to know that since OP wants the odds of at least one spade. Lastly, how can the odds of something be over 100%? It can't. Your approach of adding it all up will yield a result of much higher than 103% if you take it to seven draws. And let's take it to 13 draws; obviously the probability is utterly tiny of drawing all 13 spades in 13 draws, but if you add it all up it's much higher than 100%. That should tell you that your understanding of the problem is flawed. Not to be condescending, or mansplaining, or whatever the term is these days. Don't worry, you're not alone by a long shot. And I won't calculate the odds of that long shot.
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#23




=1  48/52 * 47/51 * 46/50 * 45/49 ~ 28.13%
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#24




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#25




so maybe =1  39/52 * 38/51 * 37/50 * 36/49 ~ 69.62 similar to when you want to calculate a flush draw with 2 remaining cards. Calculate not hitting and subtract from 100% =1  38/47 * 37/46 0.349676226
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#26




I always like it when people agree with me!
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#27




I love Bertrand Russel. In this hypothetical, based on how we are picking I believe that 66.6% is an INCORRECT answer. We know we have 3 sets AA AK KK. When we flip 1 card we find an Ace. We can therefore deduct the KK from our potential hand. We are left with two possible hands AK or AA. When we flip the second card we are flipping from one of two potential sets, either the set AK or the set AA. So 50% If the information we gleaned isn't mutually exclusive. We are not choosing one card out of a pool of 3. We are choosing one card out of a pool of 5. There would still be 3 kings left over, and only two aces, leaving us with 40%? Correct me where you can
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#28




I'm reading your post, and I admit to being puzzled as to how you figure it's a one in five, leading to 40%. Can you please explain in some more detail? I'm not saying you're wrong. Please go back to the basics for those of us in the cheap seats.
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#29




I'm going back to the idea of the information being gleaned basing it on the statement not being mutually exclusive. When you flip over 1 ace, you know you either have AK or AA. The obvious way to look at it is, of the two pairs AK or AA I have A? 50% of the time I will flip over the second ace, 50% of the time I am flipping over the third king. That is because I know of AA AK KK, I'm deciding between A? A? and KK. By flipping over an Ace, I know one Ace is in my pair, one pair has 2 kings, and one pair has 1 ace. If you look at the information being gleaned as not mutually exclusive, (while also assuming you can't draw one of any of the other 5 cards) two assume I'm drawing from a pool of 2 aces and 1 king, is incorrect. I know that of the 3 sets, there still exist 3 kings out there and only 2 aces. Therefore I'm at 40% to draw an ace
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#30




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#31




Now, if you are a glutton for punishment, you may solve this directly as follows: First, the number of desired outcomes: 13C7 = 1,716 13C6 * 39C1 = 66,924 13C5 * 39C2 = 953,667 13C4 * 39C3 = 6,534,385 13C3 * 39C4 = 23,523,786 13C2 * 39C5 = 44,909,046 13C1 * 39C6 = 42,414,099 Total Number of Desired Outcomes = 118,403,623 Next, the total number of possible outcomes: 52C7 = 133,784,560 Probability = Desired Outcomes divided by Total Possible Outcomes 118,403,623 / 133,784,560 = 0.88503204704638562177877626536276 or Approximately 88.5%
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#32




re: Poker & How to calculate odds on a multicard draw.
Hey Odysseus101, I think you answered my question perfectly by calculating the odds of NOT hitting any spade, and subtracting from one. I had not considered that approach.
"The easiest approach is to eliminate the one result of zero spades, which we don't want. To get that we multiply together 39/52 * 38/51 * 37/50 * 36/49 to get 30.38175. That means on a four card draw, the odds of getting no spades is 30.38%. Which means that the odds of getting any other result (at least one spade) is 10030.38 or 69.62%. Does that make sense?" I read DJ11's response, and of course you're not going to have odds greater than 100%... ie: a flawed solution. I'm going to run a few math tests in Excel to make sure they look good. By the way, I only mentioned that all four cards would be flipped at the same time because there seemed to be some concern that the first card flip would be affecting the odds on the next card flip. I was just trying to be as clear as possible as to what I was hunting for, and I think you found it for me. Let me do a quick test and I'll post again shortly.
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#33




That works ! I reached 99% on the 14th draw, and slowly worked to 100% by the 40th draw, as expected.
That fellow I mentioned in my initial post (the one with the PHD in Statistics) gave me this info to use : Prob (1 spade in N cards selected) = C(9,1) x C(38,N1) / C(47,N) Prob (1 spade in 1 card selected) = C(9,1) x C(38,0) / C(47,1) = 9 x 1 / 47 = 9/47 = 19.1% Prob (1 spade in 2 cards selected) = C(9,1) x C(38,1) / C(47,2) = 9 x (38x1) / (47x46/2x1) = 31.6% Prob (1 spade in 3 cards selected) = C(9,1) x C(38,2) / C(47,3) = 9 x (38x37/2x1) / (47x46x45/3x2x1) = 9 x 703 / 16215 = 39% Prob (1 spade in 7 cards selected) = C(9,1) x C(38,6) / C(47,7) C(9,1) = 9 C(38,6) = (38x37x36x35x34x33)/(6x5x4x3x2x1) = 2,760,681 C(47,7) = (47x46x45x44x43x42x41)/(7x6x5x4x3x2x1) = 62,891,499 = 9 x 2,760,681 / 62,891,499 = 39.5% I told him this was giving me invalid results, and suggested that maybe what he gave me was for finding JUST ONE spade in the blind draw, which is not what I was hunting for, but he dismissed me and insisted his examples were correct. Oh, the reason why his numbers look slightly different is because I told him that five cards had been dealt to me, four were spades, one was not. This is why he is showing nine 'outs' from a deck of 47. I do have another problem I would love to run by you.... it's another "odds" problem, but a completely different example. I'll start a new thread for it called "Kings and Peasants"
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#34




First, the number of desired outcomes: (Note: 13C4 reads 'From 13 choose 4') 13C4 = 715 13C3 * 39C1 = 11154 13C2 * 39C2 = 57798 13C1 * 39C3 = 118807 Total Number of Desired Outcomes = 188,474 Next, the total number of possible outcomes: 52C4 = 270,725 Probability = Desired Outcomes divided by Total Possible Outcomes 188,474 / 270,725 = 0.69618247298919567827130852340936 or 69.62%
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#35




Hey FirstCrack. Can you expand a bit on your solution... can you show me the numbers you are multiplying together to get your five results (4S, 3S1N, 2S1N, 1S3N) and the total possible number of outcomes too ? (3S1N = 3 Spades, 1 Not).
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#36




Explanation: 13C4 = 715 (four spades chosen from thirteen spades possible) 13C3 * 39C1 = 11154 (three spades chosen from thirteen possible spades, multiplied by 39 possible nonspade cards for remaining fourth card) 13C2 * 39C2 = 57798 (two spades chosen from thirteen spades possible multiplied by the result of two combinations of nonspade cards chosen from a possible 39 remaining cards) 13C1 * 39C3 = 118807 (1 spade chosen from 13 possible spades multiplied by three combinations of nonspade cards chosen from 39 possible remaining cards) Notice that the 'Cs' have to add up to four in each row... hence, four cards chosen in each row. In this way, we are accounting for all outcomes where there are exactly one spade, all outcomes when exactly two spades, all outcomes when exactly three spades and all outcomes when exactly four spades. Let me know if this does not make sense. Next, the numbers...
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#37




Got all that... its the results I'm curious about... the 715, 11154, etc.
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#38




I will try to use your notation:
Here's an example of one calculation. C(n,r) = n! / [r!(nr)!] C(13,4) = 13! / [4!(134)!] 13!/[4!9!] 13*12*11*10 / 4*3*2*1 715 or long handed... 13! = 13*12*11*10*9*8*7*6*5*4*3*2*1 9! = 9*8*7*6*5*4*3*2*1 4! = 4*3*2*1 so with 13! / 9! everything cancels out in the denominator leaving 13*12*11*10 in the numerator. Hence, 13*12*10*11 / 4*3*2*1 or 715
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#39




...more
13C4 = 715 13C3 * 39C1 = 11154 286 * 39 = 11154 13C2 * 39C2 = 57798 78 * 741 = 57798 13C1 * 39C3 = 118807 13 * 9139 = 118807 Total Number of Desired Outcomes = 188,474 Next, the total number of possible outcomes: 52C4 = 270,725 Probability = Desired Outcomes divided by Total Possible Outcomes 188,474 / 270,725 = 0.69618247298919567827130852340936
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#40




The following link will do the combinatory math calcs for you:
https://www.mathway.com/Statistics Look for the .C. button to the left of the Pi button on the calculator key pad. (except the '.'s are boxes)
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#41




@Odysseus101, Is Monty Hall problem related?
https://en.wikipedia.org/wiki/Monty_Hall_problem 
#42




With these problems its easier to look at the inverse question. Chance(At least one spade) = 1Chance(No Spades)
For two cards drawn you have 1(39/52*38/51) For x cards 1(39/52*38/51*....*(39x+1)/(52x+1)) or 1((39!/(39x)!)/(52!/(52x)!)) Works up to 39 since after that there are only the 13 spades left on the table so 100%.
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#43




Thank you firstcrack. I'll review the numbers.
Might you and Odysseus101 take a look at the "Kings and Peasants" thread I started... it's another odds question... trying to figure out where a "5ofakind" hand falls in the poker hand rankings... higher than a Royal Flush, or between 4ofakind and a straight flush. https://www.cardschat.com/forum/gene...3/#post3386625
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#44




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#45




Yes
Sent from my SMJ111M using http://r.tapatalk.com/byo?rid=272
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#46




It's a decent article, all told. A good overview of how results seem counter intuitive when information is spread across events which we want to view as discrete or mutually exclusive. In that article, I like how it was reported that pigeons quickly learn to "switch" while humans due to psychological factors (most likely) refuse to acknowledge the advantage in that switch even after its demonstration. The article goes on to give a few different approaches to the problem, and stresses how the initial assumptions can yield a different result. This doesn't have much to do with OP's original problem which is pretty simple. I mentioned it because if OP added extra conditions (like taking four cards from the 52 and then flipping those) and assumed that they make a difference, then that is likely what was frustrating the PhD advisor to the issues. OP was making the issue much more complicated than it needed to be.
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#47




It's always a solid approach, OP, to look at the number of possible ways you can get the result you're looking for and then divide that by the total number of possible results. (Or if it's easier, to do 1  #results you don't want / total possible results). The basic tools used in these calculations to make it go much faster than charting it all out and counting on your fingers are the "factorial" and then "choosing". These are taught early in introductory probability/combinatorics math classes. If OP is good at math, pick up a textbook and start reading it. It starts off very easy, and big numbers don't change the problem or make it harder. It's only when information starts bleeding across events that things start getting really strange. I'll look at the Peasants and Kings problem now
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#48




re: Poker & How to calculate odds on a multicard draw.
^^Thank you Odysseus^^
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Similar Threads for: How to calculate odds on a multicard draw. > Texas Hold'em Poker  
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How to calculate odds?  3  May 4th, 2020 11:02 PM  Learning Poker 