Combinations

Ok...

I am trying to figure out the odds of being dealt 7 cards with none of those cards being a face card.

Now, I understand how to get all the possible combinations of 7 cards out of 52. But how to get all the combinations that do/don't include the 12 face cards?

Calculate all the possible combinations of 7 cards out of 40 cards -40 is the number of cards in a deck that are not a face card.

Derp. To easy.

I then subtract one from the other for the number of combinations WITH face cards?

40/52 * 39/51 etc

Quote: Dalex64

40/52 * 39/51 etc

To expand on what Dalex wrote..... There are 12 picture cards and 40 non picture cards.
The odds on being dealt 1 card and it not being a picture card is thus 40/52.

If you were not dealt a picture card, there are 39 non picture cards left, and still 12 picture cards.

The odds NOW of the second card not being a picture card is 39/51

Third card 38/50.... fourth 37/49 .....seventh 34/46

So 40/52 x 39/51 x 38/50 x 37/49 x 36/48 x 35/47 x 34/46 = about 13.9%, or around 1 in 7.2 hands....

here is another method using combinations
from
http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

the number of 7 card combos I think you know
52C7 = 52! / (7! (52 - 7)!) = 133,784,560

the number of 7 card combos without a face
40C7 = 40! / (7! (40 - 7)!) = 18,643,560

so
18,643,560 / 133,784,560
should be the fraction you seek
(0.139355094 or 13.9355%)

I do hope you see why

to subtract
133,784,560 - 18,643,560 = 115,141,000
this being 7 cards with at least 1 face card

115,141,000 / 133,784,560
yes another fraction
(0.860644906 or 86.06449%)

calculators are fun
Sally

Quote: SOOPOO

To expand on what Dalex wrote..... There are 12 picture cards and 40 non picture cards.
The odds on being dealt 1 card and it not being a picture card is thus 40/52.

If you were not dealt a picture card, there are 39 non picture cards left, and still 12 picture cards.

The odds NOW of the second card not being a picture card is 39/51

Third card 38/50.... fourth 37/49 .....seventh 34/46

So 40/52 x 39/51 x 38/50 x 37/49 x 36/48 x 35/47 x 34/46 = about 13.9%, or around 1 in 7.2 hands....

Thank you. It's so damn simple once you see it...

Ok, now to expand on that concept, pai-gow deck. Let's say I want to know the chances of not getting the joker, a face card, AND/OR a face or joker.

I imagine you would get all the combinations for 40/53 descending, first, since that is the hands with only point cards. Then figure out how many with only faces, then with only the joker, and the remainder is the hands with jokers and faces?

It's the multi-variable ones that get confusing to me. I really want to wrap my head around the concept.

Quote: 21Flip

Thank you. It's so damn simple once you see it...

working with combinations is challenging for most and for most requires practice
I do agree

Quote: 21Flip

Ok, now to expand on that concept, pai-gow deck. Let's say I want to know the chances of not getting the joker <snip>

well you already know how many 7 card hands there are in a 52 card deck
52C7
try 53C7 and get your fraction for No Joker

or
you have 1 joker so 1C1 (you have 1 and want 1)
that leaves you with 52 cards and you want to choose 6
52C6
so
1C1 * 52C6 / 53C7 is another fraction for hands with a joker
and simple subtraction gets to the answer you seek for no joker
1C1 * 52C6 = 20,358,520 correct?

try it and see
and the rest is history (hertory on my end)

154,143,080 = 53C7
133,784,560 = 52C7

solve the others the same way
or use a different method that accounts for all the combinations

Sally

added:
many times this notation becomes useful, as in Excel or WolframAlpha
C(53,7)
C(52,7)

example
http://www.wolframalpha.com/input/?i=C%2852%2C7%29+%2F+C%2853%2C7%29