Baccarat - start on 1 and win on 1

Just wondering what is the odds of starting with a score of 1 and staying on 1 and winning.

Some simulations I've done before show:
P(banker wins with score of 1)= .00495
P(player wins with score of 1)= .00488

And quick calculation shows:
P(score of 1 after two cards)=.09467

This isn't the exact answer but 0.00495*0.09467=0.000468639 (1 in 2,134) and 0.00488*0.09467=0.000462012 (1 in 2,164) should be close.

So if I make a side wager to do just that and it pays 100-1. House edge be 21% ?
Thx for the quick response

Quote: Deal2422

Just wondering what is the odds of starting with a score of 1 and staying on 1 and winning.

How do you "stay on one"? Do you mean that the player's dealt hand is worth 1, the player's third card was a 10/face, and the bank ended with a zero?

Quote: Deal2422

So if I make a side wager to do just that and it pays 100-1. House edge be 21% ?
Thx for the quick response

Not sure where you get that number from. Assuming Theo's numbers are correct, HE is around 50%:


Banker P Payout EV
win 0,00495 101 0,49995
loss 0,99505 -1 -0,99505
-HE -0,4951


Player P Payout EV
win 0,00488 101 0,49288
loss 0,99512 -1 -0,99512
-HE -0,50224

Quote: Deal2422

Just wondering what is the odds of starting with a score of 1 and staying on 1 and winning.

Me don't savvy.

In Baccarat you have to put your bet down before they draw any cards... once its down... the game proceeds irrespective of initial or subsequent scores until the hand is over.

You've already done the Banker, Player, Tie bit..... what's to be on after that?

Quote: ThatDonGuy

How do you "stay on one"? Do you mean that the player's dealt hand is worth 1, the player's third card was a 10/face, and the bank ended with a zero?

Assuming this is true, and it is an "infinite deck" shoe, here's what I get:
There are 16 ways out of 169 for the player to be dealt 1 (A/KQJT, 2/9, 3/8, 4/7, 5/6, 6/5, 7/4, 8/3, 9/2, KQJT/A)
For each of these, there is a 4/13 chance that the next card is a zero value, so the player still has 1
Since the player's third card has to be a zero, the banker stands on 4-9 and draws only on 0-3
There are 25/169 ways to be dealt 0 (A/9, 2/8, 3/7, 4/6, 5/5, 6/4, 7/3, 8/2, 9/A, KQJT/KQJT); for each one, there is a 4/13 chance of drawing a 0 card for a total of 0
There are 16/169 ways to be dealt 1 (A/KQJT, 2/9, 3/8, 4/7, 5/6, 6/5, 7/4, 8/3, 9/2, KQJT/A); for each one, there is a 1/13 chance of drawing a 9
There are 16/169 ways to be dealt 2 (A/A, 2/KQJT, 3/9, 4/8, 5/7, 6/6, 7/5, 8/4, 9/3, KQJT/2); for each one, there is a 1/13 chance of drawing a 8
There are 16/169 ways to be dealt 3 (A/2, 2/A, 3/KQJT, 4/9, 5/8, 6/7, 7/6, 8/5, 9/4, KQJT/3); for each one, there is a 1/13 chance of drawing a 7
The total probability is 16/169 x 4/13 x 148/2197 = about 1 / 510.

A payout of 100-1 is a HE of about 80.2% (since you win 100 1/510 of the time and lose 1 the other 509/510 of the time).

I've won 1 - Baccar many times, not sure about starting on 1 though.

I mean that either player or banker start with 1 and remains on 1 whilst the other side either doesn't improve zero or reduced to zero

Quote: Deal2422

I mean that either player or banker start with 1 and remains on 1 whilst the other side either doesn't improve zero or reduced to zero

In that case, the probability is the sum of the probabilities of four different possibilities:


The probability is about 1 / 230.
If the bet pays 100-1, the HE is 129 / 230 = 56.1%
If the bet pays 200-1, the HE is 29 / 230 = 12.6%

Note that when I simulate it with an 8-deck shoe and full penetration, the probability is about 1 / 233.