Question
From a deck of 52 playing cards, 7 cards are dealt. What are the odds of the following event occuring
--4 From one suit, 3 from another
The probability of the event occurring is also acceptable. But, the answer is 99/16722971 (as far as the odds go
--4 From one suit, 3 from another
The probability of the event occurring is also acceptable. But, the answer is 99/16722971 (as far as the odds go
Answers
Reiny
Could be:
HHHHCCC
HHHHSSS
HHHHDDD
Now, each of those can be arranged in 7!/(4!3!) or 35 ways
let' calculate the prob of <b>one</b> of these, HHHHCCC
Prob(HHHHCCC)
= (13/52)(12/51)(11/50)(10/49)(13/48)(12/47)(11/46)
But the same prob is true for all of the 35 cases,
so multiply that by 35 to get
prob(4from one suit, 3 from another) = <b>1573/1029112</b>
so the prob of NOT any of those = 1 - 1573/1029112
= 1027539/1029112
So the odds in favour of 4 from one suit, 3 from another
= 1573 : 1027539
check my arithmetic, I did not get the same answer you gave.
HHHHCCC
HHHHSSS
HHHHDDD
Now, each of those can be arranged in 7!/(4!3!) or 35 ways
let' calculate the prob of <b>one</b> of these, HHHHCCC
Prob(HHHHCCC)
= (13/52)(12/51)(11/50)(10/49)(13/48)(12/47)(11/46)
But the same prob is true for all of the 35 cases,
so multiply that by 35 to get
prob(4from one suit, 3 from another) = <b>1573/1029112</b>
so the prob of NOT any of those = 1 - 1573/1029112
= 1027539/1029112
So the odds in favour of 4 from one suit, 3 from another
= 1573 : 1027539
check my arithmetic, I did not get the same answer you gave.
Reiny
oops, for some reason I read it as
4 Hearts, 3 from another
so it could be
HHHHCCC
HHHHSSS
HHHHDDD
DDDDHHH
DDDDCCC
DDDDSSS
SSSSHHH
SSSSCCC
SSSSDDD
CCCCHHH
CCCCSSS
CCCCDDD
So there are 12 of these and each can be arranged in 35 ways
so to get the prob of one of these 420 cases, multiply
(13/52)(12/51)(11/50)(10/49)(13/48)(12/47)(11/46) by 420
(my calculator is overheating)
4 Hearts, 3 from another
so it could be
HHHHCCC
HHHHSSS
HHHHDDD
DDDDHHH
DDDDCCC
DDDDSSS
SSSSHHH
SSSSCCC
SSSSDDD
CCCCHHH
CCCCSSS
CCCCDDD
So there are 12 of these and each can be arranged in 35 ways
so to get the prob of one of these 420 cases, multiply
(13/52)(12/51)(11/50)(10/49)(13/48)(12/47)(11/46) by 420
(my calculator is overheating)