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February 10th, 2019, 11:21 AM   #1
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Card computation with a deck of 40 cards (no pictures)

Using a deck of 40 cards, I turn 6 card face up. If the only straights allowed are the ones from 8 to 10, let’s say 8-9-10, what is the probability to get a straight? I’m trying to find the solution but I’m not sure of my findings. Thanks for your help.


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February 10th, 2019, 12:40 PM   #2
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Quote:
Originally Posted by lucadelvecchio View Post
Using a deck of 40 cards, .....
Does that mean 1 1 1 1, 2 2 2 2 .......... 9 9 9 9 , 10 10 10 10 ?

Is you question same as:
what is probability of getting 8, 9 and 10 in any order?
If so, looks like: 12/40 * 8/39 * 4/38 = 8/1235

Last edited by Denis; February 10th, 2019 at 12:50 PM.
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February 10th, 2019, 12:57 PM   #3
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Quote:
Originally Posted by Denis View Post
Does that mean 1 1 1 1, 2 2 2 2 .......... 9 9 9 9 , 10 10 10 10 ?

Is you question same as:
what is probability of getting 8, 9 and 10 in any order?
If so, looks like: 12/40 * 8/39 * 4/38 = 8/1235
Your calculation does not take into account choosing 6 cards, not just 3, so he needs 3 out of six to form his straight.
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February 10th, 2019, 01:03 PM   #4
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Your calculation does not take into account choosing 6 cards, not just 3, so he needs 3 out of six to form his straight.
Too bad, so sad
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February 10th, 2019, 01:13 PM   #5
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What I did till now is create a simulation and try to solve with math. The math suggests to count all the possible straight which in my opinion are 4^3 times Newton binomial (40 on 3) which means all the combination of the remaining 3 cards, divided Newton binomial (40 on 6)....


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February 10th, 2019, 02:46 PM   #6
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Do you need to get exactly one 8, one 9, one 10 (plus 3 others)
or are these allowed: 8 8 9 9 10 10 or 8 9 9 9 9 10 and other similars?
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February 10th, 2019, 03:55 PM   #7
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Card computation with a deck of 40 cards (no pictures)

At least one 8, one 9 and one 10, if there are two or more 8s or 9s or 10s it is considered anyway a straight. So 8 8 9 9 10 10 is one straight A’s 8 9 10 1 2 8.


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February 12th, 2019, 08:48 AM   #8
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$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, $k=0,1,2,3$
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