# Odds

#### Craig234

DP Veteran
I realize I've forgotten a good amount about statistics. Does anyone know a good formula for calculating things like cards odds, for example 'the odds of at least three hearts in the next four cards when two have been played so far'?

#### Crovax

DP Veteran
I realize I've forgotten a good amount about statistics. Does anyone know a good formula for calculating things like cards odds, for example 'the odds of at least three hearts in the next four cards when two have been played so far'?

If you are talking about for the purposes of playing cards, then knowing approximate odds is a much better option unless you are quick at doing lots of calculations in your head. The base formula is quite simple it's just the time it takes to calculate it

#### Gateman_Wen

##### Official disruptive influence
Supporting Member
DP Veteran
I realize I've forgotten a good amount about statistics. Does anyone know a good formula for calculating things like cards odds, for example 'the odds of at least three hearts in the next four cards when two have been played so far'?
Careful, you don't want the pit boss to think you're counting. That is if you wanna keep both kneecaps.

#### Craig234

DP Veteran
Careful, you don't want the pit boss to think you're counting. That is if you wanna keep both kneecaps.
Counting is blackjack and not what I'm talking about.

##### The tolerant left? I'm the intolerant left.
DP Veteran
I realize I've forgotten a good amount about statistics. Does anyone know a good formula for calculating things like cards odds, for example 'the odds of at least three hearts in the next four cards when two have been played so far'?
Most of statistics is just trying to logically move through the steps.

For example, the odds of drawing any given card in a 52 card deck is 1/52. There are 13 hearts so the odds begin as 13/52.

Assuming the first two cards played were hearts, the odds of an individual draw being a heart would be 11/50.

You now compare the number of scenarios where you “win” to the number where you lose. There are two scenarios where you “win” since you said at least 3. The odds of drawing exactly 3 and the odds of drawing exactly 4. You calculate those two odds and add them together. First let’s calculate drawing exactly 3.
*note; adding all probabilities together should = 1. By that I mean the odds of drawing 0/4, 1/4, 2/4, 3/4, and 4/4 being hearts. At those represent all possible outcomes it reasons that adding all those outcomes together should be 1 (or 100% probability)

There are 11 hearts left, and you need to draw 3. So the number of possible combinations is: 11!/((11-3)! * 3!)

The last card can be anything, so the number of possible combinations out of the remaining cards would be: 39!/((39 - 1)! * 1!)

You multiply those two numbers and divide it by the total number of possible 4 card hands out of the remaining cards: 50!/((50 - 4)! * 4!)

Doing that math gets me 165 * 39/230300 = ~0.0279 = ~2.79%

Repeating that math for drawing 4 hearts gets me: 330/230300 = ~0.00143 = ~0.143%

So that’s approximately ~2.933%

For good measure, I added up the other probabilities and they did indeed = 100%. If you are interested in calculating card draw probabilities permutations is what you are looking for. I included some instructional links.

Hopefully this was helpful. I tried to explain it in such a way that you could replicate it yourself with different numbers. It’s been a bit since I took statistics so it was fun to do some again.

• Yakshi

#### Craig234

DP Veteran
Most of statistics is just trying to logically move through the steps.

For example, the odds of drawing any given card in a 52 card deck is 1/52. There are 13 hearts so the odds begin as 13/52.

Assuming the first two cards played were hearts, the odds of an individual draw being a heart would be 11/50.

You now compare the number of scenarios where you “win” to the number where you lose. There are two scenarios where you “win” since you said at least 3. The odds of drawing exactly 3 and the odds of drawing exactly 4. You calculate those two odds and add them together. First let’s calculate drawing exactly 3.
*note; adding all probabilities together should = 1. By that I mean the odds of drawing 0/4, 1/4, 2/4, 3/4, and 4/4 being hearts. At those represent all possible outcomes it reasons that adding all those outcomes together should be 1 (or 100% probability)

There are 11 hearts left, and you need to draw 3. So the number of possible combinations is: 11!/((11-3)! * 3!)

The last card can be anything, so the number of possible combinations out of the remaining cards would be: 39!/((39 - 1)! * 1!)

You multiply those two numbers and divide it by the total number of possible 4 card hands out of the remaining cards: 50!/((50 - 4)! * 4!)

Doing that math gets me 165 * 39/230300 = ~0.0279 = ~2.79%

Repeating that math for drawing 4 hearts gets me: 330/230300 = ~0.00143 = ~0.143%

So that’s approximately ~2.933%

For good measure, I added up the other probabilities and they did indeed = 100%. If you are interested in calculating card draw probabilities permutations is what you are looking for. I included some instructional links.

Hopefully this was helpful. I tried to explain it in such a way that you could replicate it yourself with different numbers. It’s been a bit since I took statistics so it was fun to do some again.

First, thanks for putting in the effort to try to answer.

I was thinking more in terms of a more general formula around various types of events with various numbers of 'rolls'.

I think you're on the right track, but I realized I was having to use 'hard numbers' more than I'd like to do the calculation. It's not hard to do the match by playing out each possibility - just tedious - but I think there's a more general formula for it I've forgotten.

I'm realizing it's not even easy to check your math after the first part, which is pretty simply and clearly correct about the odds for each card being a heart.

For example, to change it a little (I should have said this in the OP), imagine the question, 'what are the odds at least three cards in the next five cards are hearts when two have been drawn'. For the first of five cards, it's 11/50, as you said. IF a heart is drawn, the odds for the second card are 10/49, and so on. But of those five cards, the three hearts could be 'HHH--', 'HH--H', 'HH-H-', 'H--HH', and so on, many possibilities.

And each of them needs to be calculated and added. And there's where I meant there should be an easier formula than all that by hand. If my quick count is correct, there are none possible orders for the three hearts in the next five cards, PLUS four more possibilities for four hearts, plus the one for five hearts. each needing to be calculated and then added.

• #### Crovax

DP Veteran
Most of statistics is just trying to logically move through the steps.

For example, the odds of drawing any given card in a 52 card deck is 1/52. There are 13 hearts so the odds begin as 13/52.

Assuming the first two cards played were hearts, the odds of an individual draw being a heart would be 11/50.

You now compare the number of scenarios where you “win” to the number where you lose. There are two scenarios where you “win” since you said at least 3. The odds of drawing exactly 3 and the odds of drawing exactly 4. You calculate those two odds and add them together. First let’s calculate drawing exactly 3.
*note; adding all probabilities together should = 1. By that I mean the odds of drawing 0/4, 1/4, 2/4, 3/4, and 4/4 being hearts. At those represent all possible outcomes it reasons that adding all those outcomes together should be 1 (or 100% probability)

There are 11 hearts left, and you need to draw 3. So the number of possible combinations is: 11!/((11-3)! * 3!)

The last card can be anything, so the number of possible combinations out of the remaining cards would be: 39!/((39 - 1)! * 1!)

You multiply those two numbers and divide it by the total number of possible 4 card hands out of the remaining cards: 50!/((50 - 4)! * 4!)

Doing that math gets me 165 * 39/230300 = ~0.0279 = ~2.79%

Repeating that math for drawing 4 hearts gets me: 330/230300 = ~0.00143 = ~0.143%

So that’s approximately ~2.933%

For good measure, I added up the other probabilities and they did indeed = 100%. If you are interested in calculating card draw probabilities permutations is what you are looking for. I included some instructional links.

Hopefully this was helpful. I tried to explain it in such a way that you could replicate it yourself with different numbers. It’s been a bit since I took statistics so it was fun to do some again.

That's not quite right drawing three of any suit out 4 cards is 1.648%

##### The tolerant left? I'm the intolerant left.
DP Veteran
That's not quite right drawing three of any suit out 4 cards is 1.648%
You are forgetting I started off with the assumption that 2 hearts had already been played. That drastically changes the odds.

#### Craig234

DP Veteran
I think the gist of the difficulty is that jump from 'what are the odds of the next card being a heart' - which is pretty trivial like '11/50', and multiple evens like 'the odds of one of the three next cards being a heart', which would be 1 - (39/50)*(38/50(*(37/50), to when it's 'three of the next five cards, or 'x of the next y cards'. No idea of a formula for that I realized, just tedious lists by hand, but I think there are formulas.

#### Crovax

DP Veteran
First, thanks for putting in the effort to try to answer.

I was thinking more in terms of a more general formula around various types of events with various numbers of 'rolls'.

I think you're on the right track, but I realized I was having to use 'hard numbers' more than I'd like to do the calculation. It's not hard to do the match by playing out each possibility - just tedious - but I think there's a more general formula for it I've forgotten.

I'm realizing it's not even easy to check your math after the first part, which is pretty simply and clearly correct about the odds for each card being a heart.

For example, to change it a little (I should have said this in the OP), imagine the question, 'what are the odds at least three cards in the next five cards are hearts when two have been drawn'. For the first of five cards, it's 11/50, as you said. IF a heart is drawn, the odds for the second card are 10/49, and so on. But of those five cards, the three hearts could be 'HHH--', 'HH--H', 'HH-H-', 'H--HH', and so on, many possibilities.

And each of them needs to be calculated and added. And there's where I meant there should be an easier formula than all that by hand. If my quick count is correct, there are none possible orders for the three hearts in the next five cards, PLUS four more possibilities for four hearts, plus the one for five hearts. each needing to be calculated and then added.

A standard formula for drawing x amount of suits in x amount of cards

Here is 3 of the same suit in 4 cards

4 x 4 x 13/52 x 12/51 x 11/50 x 39/49

You can modify based on the variables you want

#### Crovax

DP Veteran
You are forgetting I started off with the assumption that 2 hearts had already been played. That drastically changes the odds.

Yes it would lessen the odds not increase them

Edit: 3 hearts out of 4 cards with the first and only 2 cards played being hearts would be 0.0838%

Last edited:

##### The tolerant left? I'm the intolerant left.
DP Veteran
Yes it would lessen the odds not increase them

Edit: 3 hearts out of 4 cards with the first and only 2 cards played being hearts would be 0.0838%
I probably typo’d something in excel. Let me check.

##### The tolerant left? I'm the intolerant left.
DP Veteran
Yes it would lessen the odds not increase them

Edit: 3 hearts out of 4 cards with the first and only 2 cards played being hearts would be 0.0838%
I went to a card draw probability calculator. It matches my numbers. ##### The tolerant left? I'm the intolerant left.
DP Veteran
For example, to change it a little (I should have said this in the OP), imagine the question, 'what are the odds at least three cards in the next five cards are hearts when two have been drawn'. For the first of five cards, it's 11/50, as you said. IF a heart is drawn, the odds for the second card are 10/49, and so on. But of those five cards, the three hearts could be 'HHH--', 'HH--H', 'HH-H-', 'H--HH', and so on, many possibilities.

And each of them needs to be calculated and added. And there's where I meant there should be an easier formula than all that by hand.
My math accounts for this. The term is a combination. The formula is n!/((n-r)! * r!), where 'n' is the number of cards you are starting off picking from and 'r' is how many you want. n choose r. The factorial (!) accounts for the repeating odds that you noticed (i.e, 11/50, 10/49, 9/48...etc). So for for drawing 5 cards all you would change in my math is the number of cards drawn. So the first half would be (11!/((11-3)! * 3!) * (39!/((39-2)! * 2!) / (50!/((50 - 5)! * 5!)

#### Crovax

DP Veteran
I went to a card draw probability calculator. It matches my numbers.
View attachment 67385688

That calculator is for a TCG and is based of those rules.

This shows 3 separate ways to calculate 3 out of 4 cards being the same suit all with the same 1.648% much different than the 4.3% the deck builder is claiming

##### The tolerant left? I'm the intolerant left.
DP Veteran
Here is 3 of the same suit in 4 cards

4 x 4 x 13/52 x 12/51 x 11/50 x 39/49
This is wrong. You can easily tell because this assumes every card you draw was a heart. This math is right for the odds of drawing 3 hearts in a row, but he didn't say you draw three in a row. Just that you draw AT LEAST 3 cards in a hand of 4. So for example it could be 13/52 x 11/51 x 11/50 (you draw a non heart) x 10/39. They way you deal with that is factorials, which is what I did in my formula.

#### Crovax

DP Veteran
This is wrong. You can easily tell because this assumes every card you draw was a heart. This math is right for the odds of drawing 3 hearts in a row, but he didn't say you draw three in a row. Just that you draw AT LEAST 3 cards in a hand of 4. So for example it could be 13/52 x 11/51 x 11/50 (you draw a non heart) x 10/39. They way you deal with that is factorials, which is what I did in my formula.

There are 16 ways to draw 3 suits of 1 and 1 of another. Which is the first 4 x 4. The next 4 fractions is one possibility of drawing 3 suits and 1 other. The math is correct. I was simply showing it in the easiest way to understand.

You can check the math 3 different ways here

##### The tolerant left? I'm the intolerant left.
DP Veteran
There are 16 ways to draw 3 suits of 1 and 1 of another. Which is the first 4 x 4. The next 4 fractions is one possibility of drawing 3 suits and 1 other. The math is correct. I was simply showing it in the easiest way to understand.

You can check the math 3 different ways here

I don't know what to tell you. I've put my numbers into like 5 different online calculators and they all agree with my math. #### Crovax

DP Veteran
I don't know what to tell you. I've put my numbers into like 5 different online calculators and they all agree with my math.
View attachment 67385695

Your 2 calculators don't even agree one with 2.9% and the other 2.7%