Thanks a lot for the help Soroban. I have a few questions though...
soroban said:
With eight decks, there are 416 cards
. . and \(\displaystyle {416\choose4} \:=\:1,229,930,520\) possible four-card hands.
Call this number \(\displaystyle D.\)
[1] 4 red or 4 black Nines[/size]
There are 16 red Nines and \(\displaystyle {16\choose 4} \:=\:1820\) ways to draw four red Nines.
Similarly, there are \(\displaystyle 1820\) ways to draw four black Nines.
Therefore: \(\displaystyle P(\text{4 red or 4 black Nines}) \:=\:\frac{3640}{D}\)
I understand this part now, thank you!
[2] 4 other nines . Does this mean any 4 Nines?
There are 32 Nines and \(\displaystyle {32\choose4} \,=\,35,960\) to draw 4 Nines.
Therefore: \(\displaystyle P(\text{any 4 Nines}) \:=\:\frac{35,960}{D}\)
This means the probability of getting 4 nines that are not all one color, so I took
COMBIN(32,4)/COMBIN(416,4) - (Prob of above solution)
[3] 3 suited nines . This means "of the same suit"?
There is a choice of \(\displaystyle 4\) suits.
Then we draw 3 of the 8 Nines in that suit: \(\displaystyle {8\choose3}\,=\,56\) ways.
And the fourth card must not be a Nine . . . There are \(\displaystyle 288\) choices.
Hence, there are: \(\displaystyle \,4\,\times\,56\,\times\,288\:=\:64,512\) ways to get exactly 3 suited Nines.
Therefore: \(\displaystyle P(\text{3 suited Nines})\:=\:\frac{64,512}{D}\)
This part I still do not understand, 3 suited nines means the probability that you will get no more than 3 nines, and all 3 of those nines will be of the same suit.
I do not understand how you get 288 for the 4th card to not be a nine. If there are a total of 416 cards, and 32 of them are nines, you already chose 3 of them, so there are 29 nines left in the deck, then wouldnt it be 416-3 choose 1? I am still not sure...
There are 4 choices for the suit.
Then we draw 4 of the 8 Nines of that suit: \(\displaystyle {8\choose4}\,=\,70\) ways.
Hence, there are: \(\displaystyle \,4\,\times\,70 \:=\:280\) to draw 4 suited Nines.
Therefore: \(\displaystyle P(\text{4 suited Nines}) \;=\;\frac{280}{D}\)
This part I understand fully =p
[5] 3 unsuited nines . 3 different suits?
We choose three of the four suits: \(\displaystyle \,{4\choose3} \,=\,4\) choices.
From the first suit, there are 8 choices of Nines.
For the second suit, there are 8 choices of Nines.
For the third suit, there are 8 choices of Nines.
. . There are: \(\displaystyle \,8^3\,=\,512\) ways to get 3 Nines.
And the fourth card must not be a Nine . . . \(\displaystyle 288\) choices.
Hence, there are: \(\displaystyle \,4\,\times\,512,\,\times\,288\:=\:589,824\) ways.
Therefore: \(\displaystyle \,P(\text{3 unsuited Nine})\;=\;\frac{589,824}{D}\)
This one I need to calculate 2 different ways. The first one is for 3 suited nines, meaning 3 completely different suits
The other way I need to calculate getting 3 nines, 2 of the same suit and one of another suit.
There are 4 choices for the suit.
We draw 2 of the 8 Nines from that suit: \(\displaystyle \,{8\choose2}\,=\,28\) ways.
Then we draw 2 of the other 288 (non-Nine) cards: \(\displaystyle {288\choose2}\,=\,41,328\) ways.
Hence, there are: \(\displaystyle \,4\,\times\,28\,\times\,41,328\:=\:4,628,736\) ways.
Therefore: \(\displaystyle \,P(\text{2 suited Nines})\;=\;\frac{4,628,736}{D}\)
I'll let you work out the last two . . .
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Thanks for all the help, any further assistance is much appreciated!
