Card probabilities

[Guest]

I am calculating probabilities for cards in an 8 deck situation, you draw the first 4 cards from the deck without putting the cards back.

I am looking for the probability of hitting certain hands:

4 red or 4 black 9's
16/416 * 15/415 * 14/414 * 13/413 * 2 = 546/184489578

4 other nines (this one does not match up with the solution of 0.0000262779)
since there are 32 nines in the deck, then it should be 32/416 * 31/415 * 30/414 * 29/413 right?

3 suited nines
To me this one is trickier. You have 8 nines per suit, so is it a combination?
8 choose 3 * 4 since there are 4 suits?

4*(8 choose 3)* 1/416*1/415*1/414 Does that seem correct?

4 suited nines

3 unsuited nines

2 suited nines

2 unsuited nines

1 nine

any help is appreciated

 

[soroban]

Hello, astnfan!

When drawing cards without replacement, I prefer using Combinations.

With eight decks, there are 416 cards
. . and $\displaystyle {416\choose4} \:=\:1,229,930,520$ possible four-card hands.
Call this number $\displaystyle D.$

I am calculating probabilities for cards in an 8 deck situation.
You draw 4 cards from the deck without putting the cards back.

I am looking for the probability of hitting certain hands:

[1] 4 red or 4 black Nines

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: $\displaystyle P(\text{4 red or 4 black Nines}) \:=\:\frac{3640}{D}$

[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: $\displaystyle P(\text{any 4 Nines}) \:=\:\frac{35,960}{D}$

[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: $\displaystyle P(\text{3 suited Nines})\:=\:\frac{64,512}{D}$

[4] 4 suited nines

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: $\displaystyle P(\text{4 suited Nines}) \;=\;\frac{280}{D}$

[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: $\displaystyle \,P(\text{3 unsuited Nine})\;=\;\frac{589,824}{D}$

[6] 2 suited nines

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 . . .

 

[Guest]

Thanks a lot for the help Soroban. I have a few questions though...

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...

[4] 4 suited nines

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.

[6] 2 suited nines

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 . . .
[/quote]

Thanks for all the help, any further assistance is much appreciated!