Question about baccarat odds
February 9th, 2012 at 6:06:20 AM
permalink
Do somebody know what winning chances are in baccarat when the first cards are dealt, so 1 for the player and 1 for the banker.
example1: player 9 and banker 2 what is the chance that player will win?
example2 :player 5 and banker 8 what is the chance that banker will win?
example3: player 0 and banker 6 what is the chance that banker will win?
hope somebody can tell me.
best regards.
example1: player 9 and banker 2 what is the chance that player will win?
example2 :player 5 and banker 8 what is the chance that banker will win?
example3: player 0 and banker 6 what is the chance that banker will win?
hope somebody can tell me.
best regards.
February 9th, 2012 at 9:44:33 AM
permalink
Well, in the first two cases, the player wins 100% and the banker wins 100%, respectively, as the game ends if either the player or the bank is dealt an 8 or a 9.
In the third case, it depends in part on how many decks are in the game, but using an "infinite deck" method shouldn't be that far from the actual values.
The player has to draw a card, and the bank's decision is based on that card - since the bank has 6, it draws if the player draws a 6 or 7, and stands otherwise.
If the player draws 0-5 (there is a 9/13 probability of this happening), the player's total is less than 6, and the bank stands, so the bank wins.
If the player draws 8-9 (2/13 probability), the player's total is greater than 6, and the bank stands, so the player wins.
If the player draws 6 (1/13 probability), the player has 6, and the bank draws; the bank wins with a 1-3, loses with a 4-9, and draws with a 0 (so the probability of the bank winning is 1/13 x 3/13 = 3/169).
If the player draws 7 (1/13 probability), the player has 7, and the bank draws; the bank wins with a 2-3, loses with a 0 or 4-9, and draws with a 1 (so the probability of the bank winning is 1/13 x 2/13 = 2/169).
The total probability of the bank winning with a 6 and a player 0 = 9/13 + 3/169 + 2/169 = 122/169, or about 72.19%.
Now, all somebody has to do is to do this for every other combination of player and banker opening numbers...
In the third case, it depends in part on how many decks are in the game, but using an "infinite deck" method shouldn't be that far from the actual values.
The player has to draw a card, and the bank's decision is based on that card - since the bank has 6, it draws if the player draws a 6 or 7, and stands otherwise.
If the player draws 0-5 (there is a 9/13 probability of this happening), the player's total is less than 6, and the bank stands, so the bank wins.
If the player draws 8-9 (2/13 probability), the player's total is greater than 6, and the bank stands, so the player wins.
If the player draws 6 (1/13 probability), the player has 6, and the bank draws; the bank wins with a 1-3, loses with a 4-9, and draws with a 0 (so the probability of the bank winning is 1/13 x 3/13 = 3/169).
If the player draws 7 (1/13 probability), the player has 7, and the bank draws; the bank wins with a 2-3, loses with a 0 or 4-9, and draws with a 1 (so the probability of the bank winning is 1/13 x 2/13 = 2/169).
The total probability of the bank winning with a 6 and a player 0 = 9/13 + 3/169 + 2/169 = 122/169, or about 72.19%.
Now, all somebody has to do is to do this for every other combination of player and banker opening numbers...
