2
$\begingroup$

Update (14th Jun 18)

enter image description here

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?

$\endgroup$
  • 2
    $\begingroup$ An important sentence in the link is: "What makes CA easy for baccarat is the simple observation that every hand in baccarat is completely determined by a sequence of six cards. While it is true that the 5th and 6th cards may not be used, by always including them as a possibility, the baccarat universe can be fully written out: there are exactly 302500 unique baccarat hands." The focus in the linked blog is on the six cards that might be used, not just on the ones that are used. I'm not saying the blog is correct, but if you want to understand it, you need to start from the same point of view. $\endgroup$ – BruceET Jun 14 '18 at 2:03
  • $\begingroup$ Hmmmm, I think my "mathematical/statistical" background brought my eyes straight to the maths here and completely ignored this statement. I wonder what would happen if we just consider "removing the unused cards" now, as it would "significantly reduce" the number of combinations. $\endgroup$ – Chen Stats Yu Jun 14 '18 at 2:07
  • 1
    $\begingroup$ That issue is addressed in the Answer by @MikeEarnest (+1). See especially the last paragraph. $\endgroup$ – BruceET Jun 14 '18 at 2:13
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ But does it not creat lots of invalid combinations towards to total number of combinations here? See updated post. I am happy with the coin example, as they are valid outcomes. $\endgroup$ – Chen Stats Yu Jun 14 '18 at 9:05
  • $\begingroup$ It's not creating extra combinations, its just taking some of the valid combinations of five cards, and splitting them into several combinations of six cards. In your example, you noted (1,4,2,4,2) is a complete line of play with only five cards. Under our their scheme, this would be split into the 47 sequences of the form (1,4,2,4,2,x), where x ranges over the cards not equal to the first five. $\endgroup$ – Mike Earnest Jun 14 '18 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.