si Joined: Nov 5, 2021
• Posts: 10
November 5th, 2021 at 4:04:21 AM permalink
I have a couple of questions for the mathematics guys out there, if that's ok...

1. Does anyone know if a statistical table exists that shows the expected win ratio of every Blackjack hand permutation?

For example;
if the dealer shows a 2 and the players initial cards are AA then, if basic strategy is followed, the player is expected to win x % of the time
if the dealer shows a 2 and the players initial cards are A2 then, if basic strategy is followed, the player is expected to win x % of the time
if the dealer shows a 2 and the players initial cards are A3 then, if basic strategy is followed, the player is expected to win x % of the time

all the way through to;

if the dealer shows an A and the players initial cards are KK then, if basic strategy is followed, the player is expected to win x % of the time

2. Also, if such a table exists, would there be a difference in the figures between American and European Blackjack in terms of a hole-card being present or not?
OnceDear Joined: Jun 1, 2014
• Posts: 6143
November 5th, 2021 at 4:30:43 AM permalink
Try this...
https://youtu.be/jCF-Btu5ZCk

Yes. Some slight differences. If you can follow that workbook, you can modify it.
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
Dieter Joined: Jul 23, 2014
• Posts: 2561
November 5th, 2021 at 4:45:58 AM permalink
Quote: si

2. Also, if such a table exists, would there be a difference in the figures between American and European Blackjack in terms of a hole-card being present or not?

Yes, as game rules change, basic strategy & probable outcomes get adjusted.

If memory serves, the adjustments are covered under peek vs no-peek and lose all vs OBBO (original & busted bets only).
May the cards fall in your favor.
si Joined: Nov 5, 2021
• Posts: 10
November 5th, 2021 at 5:53:21 AM permalink
Thanks for pointing me in that direction. It's a fascinating video and I managed to follow the process (not so much the more complex of the spreadsheet formulas).

What I'm finding tough to grasp is how I'm to extrapolate the data in that spreadsheet into the figures that I need.

For example, with a Hard 4 v the Dealer's 2, the EV of;
Standing = -0.292783727
Hitting = -0.114913328
Doubling = -0.585567454

So it's clear that the correct option is to hit.

Is it therefore possible for me to extrapolate that to produce an expected win % ? So that, in the event of a Hard 4 and the dealer has a 2, IF I follow the Basic Strategy, I can expect the hand to win x% of the time.

This isn't necessarily about EV and how much I can expect to win, but the % chance of the hand successfully beating the dealer ?

I'm missing something but I can't see the woods for the trees...
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5165
November 5th, 2021 at 6:56:31 AM permalink
Quote: si

Thanks for pointing me in that direction. It's a fascinating video and I managed to follow the process (not so much the more complex of the spreadsheet formulas).

What I'm finding tough to grasp is how I'm to extrapolate the data in that spreadsheet into the figures that I need.

For example, with a Hard 4 v the Dealer's 2, the EV of;
Standing = -0.292783727
Hitting = -0.114913328
Doubling = -0.585567454

So it's clear that the correct option is to hit.

Is it therefore possible for me to extrapolate that to produce an expected win % ? So that, in the event of a Hard 4 and the dealer has a 2, IF I follow the Basic Strategy, I can expect the hand to win x% of the time.

This isn't necessarily about EV and how much I can expect to win, but the % chance of the hand successfully beating the dealer ?

I'm missing something but I can't see the woods for the trees...

I don't think you can, as (a) you don't know how often you would push, and (b) assuming this is allowed, you have to allow for the possibility that, if you hit, your next card is a 7, so you would double on your 11.
You also didn't specify how pushes should be handled in your percentage (e.g. treat them as losses, since you "didn't beat the dealer," or just ignore them, and calculate wins divided by (wins + losses)).
If there were no pushes, or pushes counted as 1/2 win and 1/2 loss, and you assume that you can't double with three or more cards, then you could do it; an EV of -0.1149 has a win chance of (50 - 0.1149 / 2) %, or 49.94255%
si Joined: Nov 5, 2021
• Posts: 10
November 5th, 2021 at 7:16:50 AM permalink
If I understand your post correctly, you're saying that if pushes weren't a factor, then we can extrapolate the -0.1149 into an 'expected win' of 49.94255%, from that Hard 4 v 2 scenario?

As pushes are a factor in the normal course of BJ, any idea how this could be taken into account?
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5165
Thanks for this post from: November 5th, 2021 at 8:40:44 AM permalink
Quote: si

If I understand your post correctly, you're saying that if pushes weren't a factor, then we can extrapolate the -0.1149 into an 'expected win' of 49.94255%, from that Hard 4 v 2 scenario?

As pushes are a factor in the normal course of BJ, any idea how this could be taken into account?

Only if you know the probability of a push in that situation - in which case, you should also already know the probability of a win and of a loss.

Assume you know the probability of a push and the ER, but not the probability of a win or loss (and also assume that you can't double on three or more cards).
Let P be the probability of a push: the probability of a win or a loss is 1 - P.
Let W be the probability of a win, and L the probability of a loss: EV = W - L, and W + L = 1 - P.
Add them up: EV + 1 - P = (W - L) + (W + L) = 2W, so W = (EV + 1 - P) / 2. Note that, in this case, EV is negative.

Question for those more familiar with casino blackjack: how common is being allowed to double on three (or more) cards?
OnceDear Joined: Jun 1, 2014
• Posts: 6143
November 5th, 2021 at 9:17:24 AM permalink
Quote: ThatDonGuy

Question for those more familiar with casino blackjack: how common is being allowed to double on three (or more) cards?

Never seen online, nor in UK B&M
We usually have Double after split except on aces. Unlimited splits in UK B&M

Nice analysis, thanks.
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
si Joined: Nov 5, 2021
• Posts: 10
November 5th, 2021 at 9:38:25 AM permalink
Quote: ThatDonGuy

Quote: si

If I understand your post correctly, you're saying that if pushes weren't a factor, then we can extrapolate the -0.1149 into an 'expected win' of 49.94255%, from that Hard 4 v 2 scenario?

As pushes are a factor in the normal course of BJ, any idea how this could be taken into account?

Only if you know the probability of a push in that situation - in which case, you should also already know the probability of a win and of a loss.

Assume you know the probability of a push and the ER, but not the probability of a win or loss (and also assume that you can't double on three or more cards).
Let P be the probability of a push: the probability of a win or a loss is 1 - P.
Let W be the probability of a win, and L the probability of a loss: EV = W - L, and W + L = 1 - P.
Add them up: EV + 1 - P = (W - L) + (W + L) = 2W, so W = (EV + 1 - P) / 2. Note that, in this case, EV is negative.

It sounds like we've come full circle. The probability of a win from a specific starting hand is precisely what I'm trying to ascertain.

I'm just struggling to find out whether such information is available. Or even possible to calculate.
Dieter Joined: Jul 23, 2014
• Posts: 2561
November 5th, 2021 at 10:11:45 AM permalink
Quote: ThatDonGuy

Question for those more familiar with casino blackjack: how common is being allowed to double on three (or more) cards?