How to calculate odds on a multi-card draw.

Odysseus101 · Dec 26, 2016

rhombus said:
oops in all the excitement I was thing of a card i.e. 4

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

Yes, that now makes perfect sense. It's what I suggested as an answer, to figure out the probability of getting no spades on the four card draw and then the remaining probability is accounted for by the other results of at least one spade.

I always like it when people agree with me!

Dwarf · Dec 26, 2016

Odysseus101 said:
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?

Odysseus101 said:
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.

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

Odysseus101 · Dec 26, 2016

Dwarf said:
I love Bertrand Russel.

And now I love you, Dwarf.

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.

Dwarf · Dec 26, 2016

Odysseus101 said:
And now I love you, Dwarf.

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.

Two in 5 leading to 40%

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

karl coakley · Dec 26, 2016

dj11 said:
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.

It is impossible for it to be a 100% chance you will get a spade, let alone 103%. Obviously your math is wrong. I agree that Odysseus got it right.

firstcrack · Dec 26, 2016

rhombus said:
oops in all the excitement I was thing of a card i.e. 4

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

The above is indeed the elegant solution, and extending for 7 cards it would be approximately 88.5%. [1-(39/52*38/51*37/50*36/49*35*48*34/47*33/46)] = .885 or 88.5%

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%

Fan_of_EP · Dec 26, 2016

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 100-30.38 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.

Fan_of_EP · Dec 26, 2016

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,N-1) / 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"

firstcrack · Dec 26, 2016

Fan_of_EP said:
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,N-1) / 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"

My Direct approach to the four draw outcome: (FWIW and, again, not so elegant of an approach)

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%

Fan_of_EP · Dec 26, 2016

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).

firstcrack · Dec 26, 2016

Fan_of_EP said:
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).

Sure. First there is this rough, verbal explanation: (then the numbers)

Explanation:

13C4 = 715 (four spades chosen from thirteen spades possible)

13C3 * 39C1 = 11154
(three spades chosen from thirteen possible spades, multiplied by 39 possible non-spade cards for remaining fourth card)

13C2 * 39C2 = 57798
(two spades chosen from thirteen spades possible multiplied by the result of two combinations of non-spade cards chosen from a possible 39 remaining cards)

13C1 * 39C3 = 118807
(1 spade chosen from 13 possible spades multiplied by three combinations of non-spade 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...

Fan_of_EP · Dec 26, 2016

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

firstcrack · Dec 26, 2016

I will try to use your notation:

Here's an example of one calculation.

C(n,r) = n! / [r!(n-r)!]
C(13,4) = 13! / [4!(13-4)!]

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

firstcrack · Dec 26, 2016

...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

firstcrack · Dec 26, 2016

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)

braveslice · Dec 27, 2016

@Odysseus101, Is Monty Hall problem related?

https://en.wikipedia.org/wiki/Monty_Hall_problem

MREM · Dec 27, 2016

With these problems its easier to look at the inverse question. Chance(At least one spade) = 1-Chance(No Spades)

For two cards drawn you have 1-(39/52*38/51)

For x cards
1-(39/52*38/51*....*(39-x+1)/(52-x+1))
or 1-((39!/(39-x)!)/(52!/(52-x)!))

Works up to 39 since after that there are only the 13 spades left on the table so 100%.

Fan_of_EP · Dec 27, 2016

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 "5-of-a-kind" hand falls in the poker hand rankings... higher than a Royal Flush, or between 4-of-a-kind and a straight flush.

https://www.cardschat.com/forum/general-poker-13/kings-peasants-311703/#post3386625

firstcrack · Dec 27, 2016

Fan_of_EP said:
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 "5-of-a-kind" hand falls in the poker hand rankings... higher than a Royal Flush, or between 4-of-a-kind and a straight flush.

https://www.cardschat.com/forum/general-poker-13/kings-peasants-311703/#post3386625

I will check it out when I have some time later this week. (If someone doesn't beat me to it, first.)

Vaguito1986 · Dec 28, 2016

Yes

Sent from my SM-J111M using CardsChat mobile app

Odysseus101 · Dec 30, 2016

braveslice said:
@Odysseus101, Is Monty Hall problem related?

https://en.wikipedia.org/wiki/Monty_Hall_problem

Hello, BraveSlice. Yes the Monty Hall problem is certainly related to Russell's poser. It's also related to the "three prisoners" paradox which I studied in math at university. The prisoners and Russell are mentioned in that wikipedia article.

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.

Odysseus101 · Dec 30, 2016

firstcrack said:
I will check it out when I have some time later this week. (If someone doesn't beat me to it, first.)

Hi Firstcrack, I'm back. I love how you've approached and explained these problems, given the limitations of writing math formulae here in the forum.

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

firstcrack · Dec 30, 2016

^^Thank you Odysseus^^

How to calculate odds on a multi-card draw.

Odysseus101

Rock Star

Dwarf

Rock Star

Odysseus101

Rock Star

Dwarf

Rock Star

karl coakley

Visionary

firstcrack

Rock Star

Fan_of_EP

Rising Star

Fan_of_EP

Rising Star

firstcrack

Rock Star

Fan_of_EP

Rising Star

firstcrack

Rock Star

Fan_of_EP

Rising Star

firstcrack

Rock Star

firstcrack

Rock Star

firstcrack

Rock Star

braveslice

Pull-ups!

MREM

Enthusiast

Fan_of_EP

Rising Star

firstcrack

Rock Star

Vaguito1986

Rising Star

Odysseus101

Rock Star

Odysseus101

Rock Star

firstcrack

Rock Star

About CardsChat

Our Community

Our Site