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 February 10th, 2019, 11:21 AM #1 Newbie   Joined: Feb 2019 Posts: 3 Thanks: 0 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, letâ€™s say 8-9-10, what is the probability to get a straight? Iâ€™m trying to find the solution but Iâ€™m not sure of my findings. Thanks for your help. Inviato dal mio iPhone utilizzando Tapatalk
February 10th, 2019, 12:40 PM   #2
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Quote:
 Originally Posted by lucadelvecchio 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.

February 10th, 2019, 12:57 PM   #3
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Quote:
 Originally Posted by Denis 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.

February 10th, 2019, 01:03 PM   #4
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Quote:
 Originally Posted by mathman 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

 February 10th, 2019, 01:13 PM #5 Newbie   Joined: Feb 2019 Posts: 3 Thanks: 0 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).... Inviato dal mio iPhone utilizzando Tapatalk
 February 10th, 2019, 02:46 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 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?
 February 10th, 2019, 03:55 PM #7 Newbie   Joined: Feb 2019 Posts: 3 Thanks: 0 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 Aâ€™s 8 9 10 1 2 8. Inviato dal mio iPhone utilizzando Tapatalk
 February 12th, 2019, 08:48 AM #8 Senior Member     Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 $p = \dfrac{\left(\dbinom{4}{2}^3+6 \dbinom{4}{1} \dbinom{4}{3} \dbinom{4}{2}+3 \dbinom{4}{1}^2 \dbinom{4}{4}\right) \dbinom{28}{0} + 3 \left(\dbinom{4}{3} \dbinom{4}{1}^2+\dbinom{4}{2}^2 \dbinom{4}{1}\right) \dbinom{28}{1} + 3 \dbinom{4}{1}^2 \dbinom{4}{2} \dbinom{28}{2} + \dbinom{4}{1}^3 \dbinom{28}{3}}{\dbinom{40}{6}} = \dfrac{802}{9139} \approx 0.0877558$ 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, $k=0,1,2,3$ Thanks from Denis and jks

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