MathCurious
• Posts: 8
Joined: Nov 27, 2020
December 7th, 2020 at 6:34:24 PM permalink
Hi wizards of Vegas! I'm still at lost in Baccarat 1-deck. In the Wizard of odds the probability of winning a pair bet is 0.058824. But pair bet is defined as to which either side draws a pair right? Now how about betting a pair in a specific side? Pair bet to the player or banker. Now the order of drawing in a Baccarat game if first ball to player, 2nd ball to banker, 3rd ball to player and 4th ball to banker. Now how will you calculate the probability of the Player winning? The Banker?

My thoughts are, there are two cases for each side. Now suppose for a specific card, an ace. The probability for the first draw to player which is

(4/52).

Now when the banker gets the 2nd ball he also have the chance to get and ace, so we stumble on two cases, either the banker got an ace or not, hence the probability for the player is

(4/52)(3/50)+(4/52)(2/50)

we multiply this by 13 to get the probability for the whole deck. Is this correct? or is it the same as the wizard of odds which is 78/1326?

As always thank you so much for whoever responds to this
Last edited by: MathCurious on Dec 7, 2020
DogHand
• Posts: 1525
Joined: Sep 24, 2011
December 20th, 2020 at 12:48:05 PM permalink
MathCurious,

The probability of a Pair for the Player in one-deck baccarat is 1/17. Naturally, the probability for the Dealer is also 1/17.

You seem to be over-thinking the problem. Pretend the four cards are dealt face-down. Now we look at the Player's first card and see that it is an Ace. So what is the probability that the Player's second card is also an Ace? Well, right now the only information we have is that his first card is an Ace, so the other 51 cards contain 3 Aces, so the probability is 3/51 = 1/17.

Naturally, if the Player's first card is at other card, say a 4, then just like above the pair probability is 1/17.

Note that the Wizard's value of 78/1326 is equal to 1/17.

Hope this helps!

Dog Hand