Update (14th Jun 18)
My argument here is that if we assume all hands are 6-card hands, we have created a lot of extra invalid combinations to the "total". For example in this game, an extra of 37446746112/31=1207959552 combinations were created. Am I correct?
OP
Baccarat is a popular CASINO game played by a lot of people. One of my friend recently asked me a probability question and further brought my interest in this as I have a PhD in Statistics.
I read on this page where Combinatorial Analysis is used. Looking at the example here,
- Player’s two cards: 1, 4
- Banker’s two cards: 2, 4
- Card 5: 2 (dealt to Player)
- Card 6: 7 (not used)
I understand the method of working out the number 4030726144. But my question is
Why would we consider the combinations of the SIX cards? As we only use FIVE card for this hand (according to the rules of this game). The answer for this hand would be 4030726144/32 = 125960192.
It seems to me that the total 4,998,398,275,503,360 is an agreed number where it's been referred to on many sites, for example, the calculator seem to be using the same method. But none of which seem to explain why.
As far as I am concerned, we should only consider VALID hands. So the example above should only be considered as a FIVE-card-hand
Can someone help me understand this better?
There is nothing wrong with including "extra" information when computing a probability. For example, when you flip two fair independent coins, there are two ways to calculate the probability that the first flip is heads.
There are two possibilities for the first coin, heads or tails, each equally likely. Therefore, $P(\text{first flip is heads})=\frac12$.
There are four possibilities for the outcome of the two flips, $HH,HT,TH,TT$, all equally likely. In two of those, the first flip is heads. Therefore, $P(\text{first flip is heads})=\frac24$.
In the second method, the extra information about the second flip was unnecessary, but it did not affect the calculation.
In the Baccarat scenario, it is true that sometimes you will only need four or five cards to determine the outcome, and sometimes you need six. Like the coin example, even though the sixth card is sometimes unnecessary, it is OK to include in in the calculations.
In fact, including the sixth card for all cases makes the computations simpler. If you were considering deals of four, five and six cards, then these would each have different probabilities you would have to keep track of, whereas when you all possibilities for the top six cards of the deck, all possibilities are equally likely.