BACCARAT: Probability on Pairs!
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I have a newbie question which I hope the more experienced of you can help me with (I've forgotten most of my high school math!)
Some baccarat tables offer a player and/or banker pair bet. The payouts vary. (ranging from 1 to 6, to 1 to 12?)
My question is as such: How do I calculate the probability that the very first hand of an 8-deck baccarat game will consist of a pair? And subsequently, as more and more cards are used, how do we calculate the changing probabilities?
My thoughts will be to simply assume the first card is, lets say a nine. Then the probability that the next card will be a nine will be (32-1)/(416-1)= 0.74699. Since one nine and one card out of the 416 cards have come out. This is not what the wizard says. Where am I going wrong? And how do I calculate the probabilities that a pair will appear once a number of cards (with their values recorded) have been used?
Thank you very very much in advance to anyone who replies.
Quote: MagicalI've forgotten most of my high school math!
....
(32-1)/(416-1)= 0.74699
Proved your point. ;-)
Sorry. Couldn't pass up the opportunity. I'll let someone else do the math correctly.
Quote: MagicalHello all,
I have a newbie question which I hope the more experienced of you can help me with (I've forgotten most of my high school math!)
Some baccarat tables offer a player and/or banker pair bet. The payouts vary. (ranging from 1 to 6, to 1 to 12?)
My question is as such: How do I calculate the probability that the very first hand of an 8-deck baccarat game will consist of a pair? And subsequently, as more and more cards are used, how do we calculate the changing probabilities?
My thoughts will be to simply assume the first card is, lets say a nine. Then the probability that the next card will be a nine will be (32-1)/(416-1)= 0.74699. Since one nine and one card out of the 416 cards have come out. This is not what the wizard (https://wizardofodds.com/baccarat/baccaratapx5.html) says. Where am I going wrong? And how do I calculate the probabilities that a pair will appear once a number of cards (with their values recorded) have been used?
Thank you very very much in advance to anyone who replies.
Your decimal is in the wrong spot. 31/415 = 0.074699 . That's the the probability that the players first 2 cards are a pair. Sometimes the player gets a third card, which may result in three of a kind, or may pair up one of the first 2 cards. (I think that the side bet here only counts pairs in the first 2 cards, and three of a kind, but not pairs made by the third card.)
Quote: DocI'll let someone else do the math correctly.
(32-1)/(416-1) = 31/415 = 0.074699
However the question is still the same, the answer i come up with is 0.074699. However the numbers on the WOO site shows 0.071663 and 0.071864 for player and banker respectively. And yes, I am talking about the first 2 cards of the player and banker, the third card is not involved in determining if the hand has a Pair. I realize my calculations ignore the fact that the cards are dealt player, banker, player, banker. How do I include the situation where the cards are dealt this way? And how are the numbers 0.071663 and 0.071864 calculated?
Thank you to those who replied so far :)
Quote: MagicalYes, that was a mistake on my part, i put the decimal at the wrong place. My bad.
However the question is still the same, the answer i come up with is 0.074699. However the numbers on the WOO site shows 0.071663 and 0.071864 for player and banker respectively. And yes, I am talking about the first 2 cards of the player and banker, the third card is not involved in determining if the hand has a Pair. I realize my calculations ignore the fact that the cards are dealt player, banker, player, banker. How do I include the situation where the cards are dealt this way? And how are the numbers 0.071663 and 0.071864 calculated?
Thank you to those who replied so far :)
You did the calculations just fine. The 0.071663 and 0.071864 aren't counting pairs the become 3 of a kind when a third card is delt.
Edit for clarification (hopefully): When the first two cards are a pair, you can't know whether they will remain a "pair" until you determine whether there will be a third card and what the chance is that it will change the "pair" into trips.
Quote: MagicalHmm... I'm not counting the pairs that become 3 of a kind either. I'm not including the third card into the calculations. What am I missing? And how do i calculate the probability of a pair after lets say all the aces and 8 kings are used up already?
If you are only calculating if the 1st 2 cards make a pair use (d*4-1)/(d*52-1) where d is the number of decks. This is what you already did. The side bet in the baccarat appendex pays more when the pair becomes a three of a kind. I think that is why you are confused. Here are the numbers for player pair:
pairs that become trips (0.003036) + pairs that don't become trips (0.071663) = original pairs (.074699)
The main baccarat page also lists pairs just on the 1st 2 cards.
I'll answere your other question when I get my brain into thinking mode. I know how, its putting it into words.
Add them up. that sum = p .
t = total number of cards. = t*(t-1)/2 or combin(t,2) in excel. call this x.
p/x is the number you want.
Arrrg that looks confusing. I'll make a google docs spread sheet later tonight.
Your second post is so over my head tho. It might be simple math, but don't understand ANY of it. Thank you very much for the replies and I look forward to the google doc spread sheet miplet!
You can ignore the colunms for colored,suited,non colored as I will be adding more paytables often used in blackjack side bets.
I'll play around with the spreadsheet and see how it works exactly.
Do you think it is possible to consistently have an edge over the house (in terms of pair bets)?
Quote: Magical
Do you think it is possible to consistently have an edge over the house (in terms of pair bets)?
No.
Quote: MagicalWhy do you say that? As the cards are used during play, the probability of a pair appearing will change. Is it not possible that the chance of a pair appearing will be be higher than 1/11? Or is it a very rare occurrence; rendering it useless? (like card counting in baccarat)
There is no simple +- tags that you can use, as there is in blackjack. My guess is it is a very rare occurrence even with perfect play.
-BW
