In a 4 deck blackjack game, what are the probabilities of these events occurring?

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September 13th, 2013 at 9:40:39 AM permalink

Suited	probability	one in:
15	0.000000000167	6005148842
14	0.000000000850	1176266268
13	0.000000004251	235253253.6
12	0.000000020829	48010868.08
11	0.000000100079	9992109.601

The formula that I used was
=HYPGEOMDIST(15,15,52,208)

and yes I rounded of the probabilities to 12 positions.

Woops - I see I made and error. I calculated for the first x cards all to be suited. one quick change and I will re-post.

beachbumbabs

beachbumbabs

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September 13th, 2013 at 9:41:48 AM permalink

Rats! Back to the books....

If the House lost every hand, they wouldn't deal the game.

odd1

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September 13th, 2013 at 9:52:47 AM permalink

	probability	one in:
15	0.000115366040	8668.062078
14	0.000550265653	1817.30405
13	0.002188038574	457.0303339
12	0.007263506623	137.6745492
11	0.020125228107	49.6888778

=HYPGEOMDIST(15,26,52,208)
=HYPGEOMDIST(14,26,52,208)
=HYPGEOMDIST(13,26,52,208)
=HYPGEOMDIST(12,26,52,208)
=HYPGEOMDIST(11,26,52,208)

and so on

miplet

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September 13th, 2013 at 10:01:36 AM permalink

Quote: odd1
probability one in:
15 0.000115366040 8668.062078
14 0.000550265653 1817.30405
13 0.002188038574 457.0303339
12 0.007263506623 137.6745492
11 0.020125228107 49.6888778

=HYPGEOMDIST(15,26,52,208)
=HYPGEOMDIST(14,26,52,208)
=HYPGEOMDIST(13,26,52,208)
=HYPGEOMDIST(12,26,52,208)
=HYPGEOMDIST(11,26,52,208)

and so on

Nope. Hint: there are 4 different suits. And with 13 or less, you can have 2 different sets of different suits.

�Man Babes� #AxelFabulous

7craps

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September 13th, 2013 at 10:10:35 AM permalink

Quote: beachbumbabs
Rats! Back to the books....

It is the (1)hypergeometric distribution and (2)Multivariate hypergeometric distribution
lots of info about this online as well as online calcs

15 and 14 are the first and
13 and less is the second.

for 15 of same suit
N=208
K_1=52
k=15
draw_n=26

Excel
=COMBIN(4,1) * COMBIN(K_1,k) * COMBIN(N-K_1,draw_n-k) / COMBIN(N,draw_n)

added: other formula
=COMBIN(4,1) * COMBIN(draw_n,k)*COMBIN(N-draw_n,K_1-k)/COMBIN(N,K_1)

For the OP

Quote: highroller
1.In a 4 deck blackjack game

Why does this have to be a Blackjack game?

winsome johnny (not Win some johnny)

beachbumbabs

beachbumbabs

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September 13th, 2013 at 10:16:28 AM permalink

Quote: 7craps
It is the (1)hypergeometric distribution and (2)Multivariate hypergeometric distribution
lots of info about this online as well as online calcs

15 and 14 are the first and
13 and less is the second.

for 15 of same suit
N=208
K_1=52
k=15
draw_n=26

Excel =COMBIN(4,1) * (COMBIN(K_1,k) * COMBIN(N-K_1,draw_n-k) / COMBIN(N,draw_n))

For the OP
Why does this have to be a Blackjack game?

OOH, OOH, OOH (Horshack)! I DID have the right concept, or at least one workable option! Yay me! Thanks, 7craps!

I heart learning.

If the House lost every hand, they wouldn't deal the game.

odd1

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September 13th, 2013 at 10:56:08 AM permalink

OK,

So I did not take into account all the suits.

The following statements come out to the same results - correct?

=HYPGEOMDIST(15,26,52,208)*4

equals

=COMBIN(4,1) * COMBIN(52,15) * COMBIN(208-52,26-15) / COMBIN(208,26)

highroller

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September 13th, 2013 at 11:42:56 AM permalink

u guys are incredible!

I greatly appreciate all the help!

7 craps: u are correct..it doesn't need to be a blackjack game...I should have asked "with 4 decks of 52 cards"...

odd1 or 7 craps:are u guys able to tell me what the answers are to my 7 Qs (for 9 cards, 10 cards, etc, etc, 15 cards)...?

highroller

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September 13th, 2013 at 11:46:28 AM permalink

i almost forgot about miplet...

miplet, odd1, or 7 craps:are u guys able to tell me what the answers are to my 7 Qs (for 9 cards, 10 cards, etc, etc, 15 cards)...?

odd1

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September 13th, 2013 at 12:24:41 PM permalink

I would use HYPGEOMDIST again.

=HYPGEOMDIST(target number of queens, number of cards drawn, number of queens in the deck, total number of cards)

7 queens with 9 cards drawn from a 4 deck boot.
=HYPGEOMDIST(7,9,16,208)

highroller

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September 13th, 2013 at 12:58:01 PM permalink

odd1:

not 7 queens...by "Q", I meant "questions"....sorry about my poor abbreviation...I should have written "my 7 questions...", rather than "my 7 Q..."...

have u been able to determine, the answers to my 7 questions...?

using 4 standard decks, what are the probabilities of the following events occurring>>>

a.after turning over 26 cards from the 208 cards in a 4 deck game, 15 of the cards being the same suit?

b.after turning over 26 cards from the 208 cards in a 4 deck game, 14 of the cards being the same suit?

c.after turning over 26 cards from the 208 cards in a 4 deck game, 13 of the cards being the same suit?

d.after turning over 26 cards from the 208 cards in a 4 deck game, 12 of the cards being the same suit?

e.after turning over 26 cards from the 208 cards in a 4 deck game, 11 of the cards being the same suit?

f.after turning over 26 cards from the 208 cards in a 4 deck game, 10 of the cards being the same suit?

g.after turning over 26 cards from the 208 cards in a 4 deck game, 9 of the cards being the same suit?

odd1

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September 13th, 2013 at 1:18:58 PM permalink

Quote:
not 7 queens...by "Q", I meant "questions"....sorry about my poor abbreviation...I should have written "my 7 questions...", rather than "my 7 Q..."...

That is too funny - 7 queens - woops

highroller

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September 13th, 2013 at 1:27:06 PM permalink

odd1:

I apologize for my bad abbreviation...

have u been able to determine, the answers to my 7 questions...?

using 4 standard decks, what are the probabilities of the following events occurring>>>

a.after turning over 26 cards from the 208 cards in a 4 deck game, 15 of the cards being the same suit?

b.after turning over 26 cards from the 208 cards in a 4 deck game, 14 of the cards being the same suit?

c.after turning over 26 cards from the 208 cards in a 4 deck game, 13 of the cards being the same suit?

d.after turning over 26 cards from the 208 cards in a 4 deck game, 12 of the cards being the same suit?

e.after turning over 26 cards from the 208 cards in a 4 deck game, 11 of the cards being the same suit?

f.after turning over 26 cards from the 208 cards in a 4 deck game, 10 of the cards being the same suit?

g.after turning over 26 cards from the 208 cards in a 4 deck game, 9 of the cards being the same suit?

odd1

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September 13th, 2013 at 1:49:21 PM permalink

I am somewhat of a hack and can do basic calculation.

As pointed out by miplet

Quote:

Nope. Hint: there are 4 different suits. And with 13 or less, you can have 2 different sets of different suits.

my corrected formula should work for 15 and 14, but it does not take into account two suites when doing 13 or less.

I would listen to 7craps

Quote:

It is the (1)hypergeometric distribution and (2)Multivariate hypergeometric distribution
lots of info about this online as well as online calcs

15 and 14 are the first and
13 and less is the second.

for 15 of same suit
N=208
K_1=52
k=15
draw_n=26

Excel
=COMBIN(4,1) * COMBIN(K_1,k) * COMBIN(N-K_1,draw_n-k) / COMBIN(N,draw_n)

added: other formula
=COMBIN(4,1) * COMBIN(draw_n,k)*COMBIN(N-draw_n,K_1-k)/COMBIN(N,K_1)

highroller

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September 13th, 2013 at 3:31:15 PM permalink

odd1:

i greatly appreciate your help...thank you...

7 craps & millet:

u guys & odd1 are all incredible...far above my math level...

odd1 answered 2 of my questions(on 14 cards & on 15 cards).....

do one of u guys know the answers to my 5 questions which odd1 has not been able to answer(in 9 , 10, 11, 12, & 13 cards)...?

thank you

odd1

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September 13th, 2013 at 4:10:01 PM permalink

7craps has two formulas right in his post. He says the first is for 14 and 15 while the second is for 13 and less.

If this site allowed attachments, I would append my excel file.

highroller

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September 13th, 2013 at 6:07:46 PM permalink

odd1:

do u know how to use the 2nd formula to calculate the 5 probabilities for these 5 remaining questions on 9,10,11,12, & 13 cards...?

thank you...

odd1

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September 13th, 2013 at 7:19:18 PM permalink

Set k to your target number of suited cards.

highroller

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September 13th, 2013 at 7:52:21 PM permalink

odd1:

odd1:

i have another set of questions for you>>>

using 4 standard decks of 52 cards per deck, what are the probabilities of the following events occurring>>>

A.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 15 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

B.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 14 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

C.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 13 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

D..after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 12 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

E.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 11 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

F.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 150cards more cards than the number of cards of "the suit with the lowest number of cards"...?

G.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 9 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

highroller

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September 13th, 2013 at 7:55:10 PM permalink

odd1:

correction>>>>

in the question labeled "F"i ment to write "10", rather than "150"...

i have another set of questions for you>>>

using 4 standard decks of 52 cards per deck, what are the probabilities of the following events occurring>>>

A.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 15 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

B.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 14 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

C.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 13 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

D..after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 12 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

E.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 11 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

F.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 10 more cards than the number of cards of "the suit with the lowest number of cards"...?

G.after turning over 52 cards from the 208 cards in a 4 deck game, the number of cards of the "suit with the highest number of cards", is 9 cards more cards than the number of cards of "the suit with the lowest number of cards"...?

miplet

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September 17th, 2013 at 6:34:16 AM permalink

Why do want these numbers? I could write a quick program, but it won't give any formulas.

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7craps

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September 17th, 2013 at 8:50:35 AM permalink

Quote: odd1
7craps has two formulas right in his post. He says the first is for 14 and 15 while the second is for 13 and less.

If this site allowed attachments, I would append my excel file.

That is not what I said.
The two formulas I showed in my post

Excel
=COMBIN(4,1) * COMBIN(K_1,k) * COMBIN(N-K_1,draw_n-k) / COMBIN(N,draw_n)
added: other formula
=COMBIN(4,1) * COMBIN(draw_n,k)*COMBIN(N-draw_n,K_1-k)/COMBIN(N,K_1)
and in Excel we do lose precision too with very large values

both solve the same problem looking at it in a different way and arrive at the same answer.
The second one is used many times in Keno and Lottery
as it looks at the question from the players perspective. (the math produces smaller values)

***Below is for the first set of questions in the OP***
The OP is not clear on really what he wants
(would help to state why he wants these since he has not asked how to do the math)
because 13,13 could be a CD, CH, CS, DH, DS or HS or each reverse sequence (12 other ways)

So for 13
the hypergeometric function in Excel or using COMBIN() or just doing this by hand
returns a value that includes 13,13
(2 suits could have exactly 13, not probable but possible)

Not much of a difference between 13 and 13,13 as I can see
0.008752154
0.008752149 for 13,13
one must subtract out (exclude) the outcomes where 2 suits have exactly 13.
the multivarate hypergeometric distribution has more terms in it
(I do not see excel having a function for it (maybe an add-in),
so vba solution comes to mind also or use an R package)

=12 * COMBIN(K_1,k) * COMBIN(K_2,k_2a) * COMBIN(N-K_1-K_2,draw_n-k-k_2a) / COMBIN(N,draw_n)

N	208
draw_n	26
K	52
k	13
K_2	52
k_2a	13

now is the OP looking for X or more?
for 15 that would include 16-26 (small values)
That would seem reasonable.

For 12
one must also exclude 12,12 because it is already counted in just 12
12,13 and 12,14 also excluded or included?

0.029050996 for 12,12
this still includes 12,13 and 12,14

instead of this for 12
0.029054026

the excluded probabilities for 9,9 and 10,10 and 11,11 are much higher
(because there are more of them)
but requires more work (not fun for me)

Hey, I see if miplet really wants to code this up and the other idea(s) the OP has
(maybe he has an infinite number of them)
excellent

winsome johnny (not Win some johnny)

miplet

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September 17th, 2013 at 10:59:00 AM permalink

Highest Suit	Combinations	Probability
7	69402819291824671709113774080000	0.07682463280415907648
8	279763785574064061996605607000000	0.30968122502713248462
9	280678314434880522238839961200000	0.31069355197059004769
10	164312367826222050132120338916000	0.18188363891030561313
11	72593439788370329330684080953600	0.08035645255681021461
12	26245661336927768831987537973600	0.02905232547446819544
13	7906630659378142556244515520000	0.00875215161751193690
14	1988424092741842351384157460000	0.00220106261305482155
15	416883395995531086083299219200	0.00046146416163356258
16	72633365655043301385232355400	0.00008040064802434074
17	10463390010330367546492056000	0.00001158232624600346
18	1237225170140415081510885000	0.00000136953181962878
19	118871510518612823091120000	0.00000013158341749877
20	9153106309933187378016240	0.00000001013192314740
21	554208896722539254676480	0.00000000061347500608
22	25688510464112914017600	0.00000000002843559388
23	875993536713142848000	0.00000000000096967072
24	20619977731072356000	0.00000000000002282504
25	297991936242594048	0.00000000000000032985
26	1983674131792416	0.00000000000000000219
Total	903392788986651583539793735039584	1

�Man Babes� #AxelFabulous

miplet

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September 17th, 2013 at 11:48:27 AM permalink

Highest Suit - Lowest Suit	Combinations	Probability
0	162604288696898081959751951021289584665600000000	0.00406392559878018819
1	0	0.00000000000000000000
2	2566272112861522230797790059808683881477464000000	0.06413815414385568871
3	2894830756444505554307878947223708857371481600000	0.07234973265176532410
4	4835937824762186445136501713560039205989546420000	0.12086330365365902826
5	5469659865192870180947073114005416333176145920000	0.13670174951050928855
6	5925973119650074062305858471940775021294176540000	0.14810626491851343511
7	5144648026025466698291306542043898777304871520000	0.12857881533894437153
8	4313756539491916791576452669912890337949232670000	0.10781256612748287403
9	3244387606006735722741019035638653589159390400000	0.08108611371864429346
10	2257417099401459186754087797757419083531245500000	0.05641902320597721009
11	1430903790865067159548251851755346085389220000000	0.03576219662894469922
12	850790746454294954448152750127258287903597067500	0.02126358610482868404
13	469702609480132107633848306552988736491490560000	0.01173915198533491487
14	241595931326551693733372179823987268531080460000	0.00603814264523666062
15	115825218020514250901858452239950176588557552000	0.00289478876768914902
16	52058457747065224653579415755071824668817349000	0.00130108314341988830
17	21892810942336405005235763009909106564720720000	0.00054716118209933278
18	8614443440844183227219651944741846916265318000	0.00021529848627638688
19	3170660009791220134217344053313191939268128000	0.00007924345957958111
20	1092260576904971921046694986322200352029583500	0.00002729857714452281
21	351389548299541875323106450557131640986464000	0.00000878218521739123
22	105388444177070036317367025509686955289960000	0.00000263394526392330
23	29411498087380027232844094650428106027782400	0.00000073507372366162
24	7623772441346032575457613323179417978861776	0.00000019053891033227
25	1830001804332271707607440788015605830678528	0.00000004573674678595
26	405621809852948382835922439984647335083840	0.00000001013759765929
27	82761612500931987180826167552994282626048	0.00000000206843889748
28	15492813664540594923702984324456870097000	0.00000000038720775788
29	2650212871503009625392616646177062185600	0.00000000006623606312
30	412529485795407846361502299390780410000	0.00000000001031023936
31	58163828849045755085914963094634384000	0.00000000000145367305
32	7390864536585702677770868834606984700	0.00000000000018471790
33	841629328928309118592375766715384000	0.00000000000002103461
34	85363889993089480233295562034106800	0.00000000000000213347
35	7660113429749097414824418738446400	0.00000000000000019144
36	603663273602174948143616042435300	0.00000000000000001508
37	41434532177750875049022591436800	0.00000000000000000103
38	2454600609294151543806298536000	0.00000000000000000006
39	124229142472001143133292960000	0.00000000000000000000
40	5310165158514379132625983600	0.00000000000000000000
41	189194590915821629630419200	0.00000000000000000000
42	5534063479008209227336800	0.00000000000000000000
43	130550608274068592064000	0.00000000000000000000
44	2431411064511996173880	0.00000000000000000000
45	34812988208921395200	0.00000000000000000000
46	370441289116429440	0.00000000000000000000
47	2798676041616000	0.00000000000000000000
48	14070857375900	0.00000000000000000000
49	42433060800	0.00000000000000000000
50	64125360	0.00000000000000000000
51	32448	0.00000000000000000000
52	4	0.00000000000000000000
Total	40011630317667415166387164671974219643751106143824	1

�Man Babes� #AxelFabulous

odd1

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September 17th, 2013 at 12:35:11 PM permalink

7craps,

I am easily confused.

Quote:

15 and 14 are the first and
13 and less is the second.

I was wondering why when I put your two and mine into a spreadsheet I got the same answer for all three at every level.

Thanks for the lesson.

7craps

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September 17th, 2013 at 12:52:31 PM permalink

Quote: odd1
7craps,

I am easily confused.

me too
"It is the (1)hypergeometric distribution and (2)Multivariate hypergeometric distribution
lots of info about this online as well as online calcs

15 and 14 are the first and
13 and less is the second."
I guess I should have made that more clear.

But I did keep it in order as I added later that second formula
because for Keno and Lottery calcs, as you might know, Excel can do too much rounding, the second
formula is easily handled in Excel without thinking. Not that the OP has access to Excel

too much rounding is relative
COMBIN(80,20) in Excel returns
3,535,316,142,212,180,000 (it is close)
3,535,316,142,212,174,320 (exact)

This thread about choose(208,26) and choose(208,52)
for exact precision BigInteger is really needed, as you know,
to show results as those by miplet. (nice work)
can not get his exact results in Excel (at least v2007)

Let us see if the OP returns to say why he wants (really needs) this info and
possibly what he is up to

Quote: odd1
I was wondering why when I put your two and mine into a spreadsheet I got the same answer for all three at every level.

Thanks for the lesson.

That was my plan.
Thanks, Good to see someone else tested them.

winsome johnny (not Win some johnny)

miplet

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September 17th, 2013 at 1:52:58 PM permalink

Quote: 7craps

***Below is for the first set of questions in the OP***
The OP is not clear on really what he wants
(would help to state why he wants these since he has not asked how to do the math)
because 13,13 could be a CD, CH, CS, DH, DS or HS or each reverse sequence (12 other ways)

So for 13
the hypergeometric function in Excel or using COMBIN() or just doing this by hand
returns a value that includes 13,13
(2 suits could have exactly 13, not probable but possible)

Not much of a difference between 13 and 13,13 as I can see
0.008752154
0.008752149 for 13,13
one must subtract out (exclude) the outcomes where 2 suits have exactly 13.
the multivarate hypergeometric distribution has more terms in it
(I do not see excel having a function for it (maybe an add-in),
so vba solution comes to mind also or use an R package)

=12 * COMBIN(K_1,k) * COMBIN(K_2,k_2a) * COMBIN(N-K_1-K_2,draw_n-k-k_2a) / COMBIN(N,draw_n)
N	208
draw_n	26
K	52
k	13
K_2	52
k_2a	13
now is the OP looking for X or more?
for 15 that would include 16-26 (small values)
That would seem reasonable.

For 12
one must also exclude 12,12 because it is already counted in just 12
12,13 and 12,14 also excluded or included?

0.029050996 for 12,12
this still includes 12,13 and 12,14

instead of this for 12
0.029054026

the excluded probabilities for 9,9 and 10,10 and 11,11 are much higher
(because there are more of them)
but requires more work (not fun for me)

Hey, I see if miplet really wants to code this up and the other idea(s) the OP has
(maybe he has an infinite number of them)
excellent

Small correction for 13,13 and 12,12: because they are the same use 6 instead of 12.

�Man Babes� #AxelFabulous

7craps

Threads: 18
Posts: 1977

Joined: Jan 23, 2010

September 17th, 2013 at 2:52:30 PM permalink

Quote: miplet
Why do want these numbers?

I throw out (in) a guess or two
1) An online casino offers a 4 deck blackjack game (say William Hill)
with this side bet
2) highroller has a new casino game idea

thanks for the correction earlier
(yes, I knew I had copied the wrong data, had no time to correct.
4C2 = 6
I knew you would create the correct data table)

winsome johnny (not Win some johnny)

highroller