mega test question please help
manuel
what is the probability that a hand of 13 cards contains
a) the ace and king of some suit
b) all 4 of at least 1 of the 13 denominations
For a) there contains 4 possibilities the ace and king are hearts,
spades, clubs or diamonds
There is an exclusion-inclusion formula where one takes the union of the
probability of all the sets. Is the probability of getting an ace and
king of hearts equal to c(13,2)*c(4,1)/c(52,13)?
If so is the probability of getting an ace and king of hearts and
spades equal to c(13,2)*c(4,1)/c(52,13) + c(13,2)*c(4,1)/c(52,13)?
For b) I need help.
manuel
Horst Kraemer
wrote:
> what is the probability that a hand of 13 cards contains
> a) the ace and king of some suit
> b) all 4 of at least 1 of the 13 denominations
>
> For a) there contains 4 possibilities the ace and king are hearts,
> spades, clubs or diamonds
> There is an exclusion-inclusion formula where one takes the union of the
> probability of all the sets. Is the probability of getting an ace and
> king of hearts equal to c(13,2)*c(4,1)/c(52,13)?
No. The probability of getting AK of Hearts is
13*12 3
C(50,11)/C(52,13) = ----- = --
52*51 51
Let's forget the eternal 1/C(52,13) and just speak about the number of
possible distinct hands including AK of Hearts. A hand with AK of
Hearts may contain additionnally any combination of 11 cards out of
the remaining 50 cards. There are C(50,11) combinations of this kind.
Therefore there are C(50,10) hands containing two _specific_ cards.
In general there are C(52-k,13-k) hands containing k _specific_ cards.
In both of your problems you have several sets of "specific cards". In
a) the 4 AK pairs. In b) the 13 quadruples with the same rank. And the
problem is to find the number of hands which contain at least one of
this set (not "a specific set of these sets").
This is a typical situation where you would apply the
inclusion-exclusion. Assume that you have M "special" subset of your
basic set. In a) the 4 sets of {A,K}. In b) the 13 sets of cards of
the same rank. Now the problem is to find the number of hands which
would contain at least one of these combinations. It is usually easy
to find the number of cards which contain a specific one of these
special sets, or two specific ones out of the special sets. Its just
C(52-k,13-k) Where k is the number of cards which make up all the
combined cards of the specific sets. For instance the number of hands
containing AK of Hearts and AK of spades would be C(48,9). The problem
is that these counts do not exclude that such a hand would contain AK
of Diamonds, too.
In general the principle of inclusion-exclusion for M sets of equal
size which are mutually exclusive goes like this:
The number of hands containing _at_ _least_ one of the special sets
is:
M * Number of hands containing a selected set
- C(M,2) * Number of hands containing two selected sets
+ C(M,3) * Number of hands containing three selected sets
...
? C(M,M) * Number of hands containing all sets
Note that this always means ...containing _at_ least one/two/three
selected sets.
Now for the AK example this boils down to.
Number of hands containing at leanst one AK combination
= 4 * Number of hands containing AK of Hearts
- C(4,2) * Number of hands containing AK of Hearts and
AK of Spades
+ C(4,3) * Number of hands containing AK of Hearts and
Ak of Spades and AK of Diamonds
- C(4,4) * Number of hands containing all AK.
(The naming is arbitrary. It just denotes any selected combination
and all selected combinations of k sets have the same size)
= 4 * C(50,11)
- 6 * C(48,9)
+ 4 * C(46,7)
- 1 * C(44,5)
If you divide this by C(52,13) you get about 21%, i.e. about 21% of
the hands you will hold at least one AK of the same suit.
The same for the sets of quadruples of the same rank.
13 * C(48.9)
- C(13,2) * C(44.5)
+ C(13,3) * C(40,1)
Here it stops because a hand may contain no more of 3 quadruples of
the same rank.
Hope it helps
Horst