Hello, sapientiam!
Bridge-hand problems have HUGE numbders.
Do they expect you to crank out the exact answers? . . .
ack!
Calculate the probability that a bridge hand contains:
1. At least one black card.
2. More black cards than red cards
3. a 3-3-3-4 distribution (that is, at least 3 cards of each suit)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
First, there are \(\displaystyle C(52,13)\,=\,\frac{52!}{(13!)(39!)}\) possible bridge hands.
. . That is the denominator \(\displaystyle (D)\) for all answers.
(1) The opposite of "at least one Black" is "no Blacks" or "all Reds".
To get all Reds, there are: \(\displaystyle \,C(26,13)\) ways.
. . Hence: \(\displaystyle P(\text{all Reds})\:=\:\frac{C(26,13)}{D}\)
. . Therefore: \(\displaystyle \,P(\text{at least 1 Black})\;=\;1 \,-\,\frac{C(26,13)}{D}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
(2) "More Blacks than Reds" . . . a long tedious problem!
We must work out these numbers:
. . \(\displaystyle \;7B,\,6R\;=\;C(26,7)\cdot C(26,6)\)
. . \(\displaystyle \;8B,\,5R\;=\;C(26,8)\cdot C(26,5)\)
. . \(\displaystyle \;9B,\,4R\;=\;C(26,9)\cdot C(26,4)\)
. . \(\displaystyle 10B,\,3R\;=\;C(26,10)\cdot C(26,3)\)
. . \(\displaystyle 11B,\,2R\;=\;C(26,11,\cdot C(26,2)\)
. . \(\displaystyle 12B\,1R\;=\;C(26,12)\cdot C(26,1)\)
. . \(\displaystyle 13B,\,0R\;=\;C(26,13)\cdot C(26,0)\)
. . add them up and divide by \(\displaystyle D\).
But, of course, it's a trick question, isn't it?
Exactly HALF the hands will have more Blacks than Reds,
. . the other half will have more Reds than Blacks.
Ha! . . . \(\displaystyle P(\text{more Blacks})\,=\,\frac{1}{2}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
(3) A 3-3-3-4 distribution of suits.
There are
4 choices for the suit of the set-of-four.
There are \(\displaystyle C(13,4)\) ways to get the set-of-four.
There are \(\displaystyle C(13,3)\) ways to get the first set-of-three.
There are \(\displaystyle C(13,3)\) ways to get the next set-of-three.
There are \(\displaystyle C(13,3)\) ways to get the last set-of-three.
There are: \(\displaystyle \,4\,\times\,C(13,4)\,\times\,[C(13,3)]^3\) ways to get the distribution.
. . then divide by \(\displaystyle D\).
:shock: