1. ## Card probability

In a deck of 40 cards, from 1 to 10, 4 suits, I want the probability to get a straight with 8-9-10, just those 3 cards, the suit doesnt matter, drawing 6 cards. Any idea?

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2. ## Re: Card probability

Ran a few simulations: getting ~75000 per million.

So 75000/1000000 = 3/40 : do you know the solution?

3. ## Re: Card probability Originally Posted by luca2019 In a deck of 40 cards, from 1 to 10, 4 suits, I want the probability to get a straight with 8-9-10, just those 3 cards, the suit doesnt matter, drawing 6 cards. Any idea?
It is not clear what you are asking.
Are you asking for the particular straight $8-9-10$ OR just any straight of three cards?
Which ever it is, do we understand that the straight will be contained in a draw of six cards?

4. ## Re: Card probability

Hi,
the problem is to find a particular straight 8 - 9 - 10, the straight will be contained in a draw of six cards.
The deck contains 40 cards and there are 4 suits for each rank: e.i. [8, clubs][8,diamonds][8,spades][8,hearts] and so on for every rank.

5. ## Re: Card probability

Hi Denis,
your solution is close to mine: I got with a simulation an 0,0878, 87800 per million, now I'd like to confirm this value with a math expression...

6. ## Re: Card probability Originally Posted by luca2019 Hi Denis,
your solution is close to mine: I got with a simulation an 0,0878, 87800 per million, now I'd like to confirm this value with a math expression...
Looks good...I goofed by picking from 40 each time: should be from 40,39,38,37,36,35
Ran the damn thing 5 times to get:
88181, 87814, 88140, 87735, 88081

Trying to "formulate" but keep getting a headache!

I'm sure our resident expert Plato will soon put us to shame EDIT: the 6 cards must include AT LEAST 1 eight, 1 nine, 1 ten.
So stuff like 8,8,8,8,9,10 and 2,7,8,9,9,10 is ok.

7. ## Re: Card probability

$p = \dfrac{\left(\dbinom{4}{2}^3+6 \dbinom{4}{1} \dbinom{4}{3} \dbinom{4}{2}+3 \dbinom{4}{1}^2 \dbinom{4}{4}\right) \dbinom{28}{0} + 3 \left(\dbinom{4}{3} \dbinom{4}{1}^2+\dbinom{4}{2}^2 \dbinom{4}{1}\right) \dbinom{28}{1} + 3 \dbinom{4}{1}^2 \dbinom{4}{2} \dbinom{28}{2} + \dbinom{4}{1}^3 \dbinom{28}{3}}{\dbinom{40}{6}} = \dfrac{802}{9139} \approx 0.0877558$

These are grouped by how many cards are dealt that are NOT 8, 9, or 10. You can see these by looking at the $\dbinom{28}{k}$ term, $k=0,1,2,3$

8. ## Re: Card probability Originally Posted by romsek ......
= approx 0.0877558
YA!! Quite close to my 4th run of 87735 9. ## Re: Card probability

I just had a quick look and I'm getting a different answer (I'm OK with it being incorrect but at the moment I don't see why):
There are 4 ways to get an 8. After that there are 4 ways to get a 9, and then 4 ways to get a 10. So $4^3$ ways to get a 3 card straight.
Then there are $\binom {37}{3}$ ways to get the remaining three cards and $\binom {40}{6}$ ways to draw 6 cards. So I get$$\frac {4^3\binom {37}{3}}{\binom {40}{6} }= \frac{32}{247}=.12955$$

10. ## Re: Card probability Originally Posted by Walagaster I just had a quick look and I'm getting a different answer (I'm OK with it being incorrect but at the moment I don't see why):
There are 4 ways to get an 8. After that there are 4 ways to get a 9, and then 4 ways to get a 10. So $4^3$ ways to get a 3 card straight.
Then there are $\binom {37}{3}$ ways to get the remaining three cards and $\binom {40}{6}$ ways to draw 6 cards. So I get$$\frac {4^3\binom {37}{3}}{\binom {40}{6} }= \frac{32}{247}=.12955$$
Prof. Walagaster, I agree and that was my first reaction. But then post #4 gave me pause that I my not understand the setup.
Then I saw Dennis' post. He is usually spot on with his simulations. But I still agree with you.

11. ## Re: Card probability Originally Posted by romsek $p = \dfrac{\left(\dbinom{4}{2}^3+6 \dbinom{4}{1} \dbinom{4}{3} \dbinom{4}{2}+3 \dbinom{4}{1}^2 \dbinom{4}{4}\right) \dbinom{28}{0} + 3 \left(\dbinom{4}{3} \dbinom{4}{1}^2+\dbinom{4}{2}^2 \dbinom{4}{1}\right) \dbinom{28}{1} + 3 \dbinom{4}{1}^2 \dbinom{4}{2} \dbinom{28}{2} + \dbinom{4}{1}^3 \dbinom{28}{3}}{\dbinom{40}{6}} = \dfrac{802}{9139} \approx 0.0877558$

These are grouped by how many cards are dealt that are NOT 8, 9, or 10. You can see these by looking at the $\dbinom{28}{k}$ term, $k=0,1,2,3$

We can go through it term by term

3 cards must be 8 9 10 to define the straight. This gets us a starting point of (1,1,1)

The other 3 cards may be anything but we must count carefully

There are 4 possibilities for these other 3 cards, they can contain 0-3 cards that are 8,9, or 10.

If they contain 0, then the 3 cards are all 8s, 9s, or 10s.

These can be arranged as (1,1,1) 1 way, as (0,1,2) 6 ways, and as (0, 0, 3) 3 ways.
This corresponds to the factor of $\dbinom{28}{0}$

If they contain 1, the 2 of the 3 cards are 8s, 9s, or 10s

There are 3 ways to do both (0, 0, 2) and (0, 1, 1).
This corresponds to the factor of $\dbinom{28}{1}$

Similarly if they contain 1 8,9, or 10, then there are 3 ways to do (0,0,1), thus the factor of $\dbinom{28}{2}$

If they contain 3 cards then we have the one way to do (0,0,0) and this corresponds to the factor of $\dbinom{28}{3}$

All of it normalized by $\dbinom{40}{6}$ i.e. choose 6 cards from 40.

This number matches up with my own sim. I'm pretty confident it's correct.

12. ## Re: Card probability Originally Posted by luca2019 In a deck of 40 cards, from 1 to 10, 4 suits, I want the probability to get a straight with 8-9-10, just those 3 cards, the suit doesnt matter, drawing 6 cards. Any idea?
That is badly worded; should be:
The jacks, queens, kings are removed from a regular deck of cards:
so you have a deck of 40 cards, from 1 to 10, 4 suits.

6 cards are picked at random.

What is the probability of picking AT LEAST one 8, one 9 and one 10?

13. ## Re: Card probability Originally Posted by romsek We can go through it term by term
3 cards must be 8 9 10 to define the straight. This gets us a starting point of (1,1,1)
The other 3 cards may be anything but we must count carefully
It is that point that I find most troubling.
Can we at least agree that there are $4^3$ ways to choose a "straight" $8-9-10$ ?

Now once that three-sum is selected there are $\dbinom{37}{3}$ ways to select three others to add to that fixed triple.
Note that the composition of those three makes no difference because the so-called straight has been fixed.

14. ## Re: Card probability

This is the right solution and confirms the simulation. I would never have understood it by myself!

15. ## Re: Card probability

Luca, you DO realise that the problem could have been presented
in a much clearer manner this way:
A deck of 40 cards is made up of 4 1's, 4 2's. ....... 4 9's and 4 10's.
6 cards are picked at random, with no replacement.
What is the probability that 3 given numbers are picked AT LEAST once?

The 3 given numbers can be any numbers from 1 to 10 of course:
no need to bring in stuff like "a straight" and the likes.