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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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.
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.
ready?
$p = \dfrac{\left(\dbinom{4}{2}^3+6 \dbinom{4}{1} \dbinom{4}{3} \dbinom{4}{2}+3 \dbinom{4}{1}^2 \dbinom{4}{4}\right) \dbinom{28}{0} +
3 \left(\dbinom{4}{3} \dbinom{4}{1}^2+\dbinom{4}{2}^2 \dbinom{4}{1}\right) \dbinom{28}{1} +
3 \dbinom{4}{1}^2 \dbinom{4}{2} \dbinom{28}{2} +
\dbinom{4}{1}^3 \dbinom{28}{3}}{\dbinom{40}{6}} = \dfrac{802}{9139} \approx 0.0877558$
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, $k=0,1,2,3$
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$$
\frac {4^3\binom {37}{3}}{\binom {40}{6} }= \frac{32}{247}=.12955
$$
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.
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.
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.