Playing card probability and combinations

[Hrodis]

Hi All.

I'm attempting to develop an early report for a baccarat casino game and encountering some difficulty determining the proper way to calculate some odds.

I’m trying to calculate the likelihood of hands that would end with 0-Value cards. In total, there are 416 cards in the deck and 128 of them are 0-Value.
Additionally, the number of cards you draw vary based on this chart.

Total Cards	Probability
4 Cards	0.378868%
5 Cards	0.303444%
6 Cards	0.317687%

I need solutions or formulas to calculate the following

• Hands where 6 cards are drawn and 6 of them have 0-Value
• Hands where 6 cards are drawn and 5 of them have 0-Value
• Hands where 6 cards are drawn and 4 of them have 0-Value
• Hands where 6 cards are drawn and 3 of them have 0-Value
• Hands where 6 cards are drawn and 2 of them have 0-Value
• Hands where 6 cards are drawn and 1 of them have 0-Value
• Hands where 6 cards are drawn and 0 of them have 0-Value
• Hands where 5 cards are drawn and 4 of them have 0-Value
• Hands where 5 cards are drawn and 3 of them have 0-Value
• Hands where 5 cards are drawn and 2 of them have 0-Value
• Hands where 5 cards are drawn and 1 of them have 0-Value
• Hands where 5 cards are drawn and 0 of them have 0-Value
• Hands where 4 cards are drawn and 3 of them have 0-Value
• Hands where 4 cards are drawn and 2 of them have 0-Value
• Hands where 4 cards are drawn and 1 of them have 0-Value
• Hands where 4 cards are drawn and 0 of them have 0-Value

The Following chart is provided.

	4 Cards Drawn​	5 Cards Drawn​	6 Cards Drawn​
Six 0-Value Cards	Not Possible	Not Possible
Five 0-Value Cards	Not Possible	Not Possible
Four 0-Value Cards	Not Possible
Three 0-Value Card
Two 0-Value Cards
One 0-Value Card
No 0-Value Cards

 

[MarkFL]

Suppose we are drawing \(n\) cards from this deck, and we wish to calculate the probability that \(m\) cards have zero value, where \(0\le m\le n\). I would state:

[MATH]P(X)={n \choose m}\frac{\dfrac{128!}{(128-m)!}\cdot\dfrac{288!}{(288-(n-m))!}}{\dfrac{416!}{(416-n)!}}[/MATH]
Here is a page where you can calculate the probabilities to complete your table:

nCr(6,6)(nPr(128,m)*nPr(288,n-m))/(nPr(416,n)) where n=6,m=0 - Wolfram|Alpha

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