aceside
aceside
Joined: May 14, 2021
  • Threads: 0
  • Posts: 139
March 28th, 2022 at 8:19:23 AM permalink
Quote: ChesterDog

Quote: aceside

Quote: aceside

Quote: ChesterDog

Quote: DJTeddyBear

Thanks ChesterDog.

But I've hurt my brain trying to wrap my head around your formula, specifically, to make it into an Excel calculation.

For example, the calucation the Wiz uses is = 2 * [whatever].

I substituted = ( 2 + 2/13 + (3/13)*(2/13) ) * [whatever]. This is:

2 for the original split, plus...
2/13 since there's 2 chances of matching the card, plus...
3/13 since there's 3 chances of a match of the three splits times 2/13 which is the odds that there was the original split.

Similarly, the paired tens line uses = 2 + 8/13 + (12/13)*(8/13) * [whatever].

Interestingly, this DOES change the strategy. Paired 3s vs 2 doesn't split, while paired 9s vs 7 and 10s ve 3 thru 7 all split.

It also changes the edge from 0.484% to a player advantage of 0.132%. That can't be right.

Can anybody help me figure out what I'm missing?

Thanks.
link to original post



Rather than using the complicated formula for splitting up to three times, a more instructional approach to analyzing re-splitting is first to analyze infinite re-splitting.

As an example, let's use a pair of 3s vs dealer's 2. Let E1 stand for the EV of a pair of 3s allowing one split. We know the formula for E1 is:

E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]
where, for example, E(6) is the EV of hitting 6 vs 2 and E(10) is the EV of doubling 10 vs 2.

The formula for the EV of a pair of 3s vs dealer's 2 allowing infinite re-splitting, Einf., is just a little different. It's:
Einf. = (2/13) [E(soft 14) + E(5) + Einf. + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:
Einf. - (2/13)Einf.= (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Then:
(11/13)Einf. = (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Multiplying both sides by 13/11 yields:
Einf. = (2/11) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

A trick to make the Excel calculations easier is to replace the 0 with E(6) - E(6), where again E(6) is the EV of hitting 6 vs 2:
Einf. = (2/11) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13) - E(6)]

We can simplify the above using the fact that E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]. The final result is:

Einf. = (2/11) [(13/2)E1 - E(6)] = (13/11) E1 - (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare Einf. with E(6).

The above formula can be used for pairs of 2s through 9s. (Of course, we would use E(4) for 2s, for example, instead of E(6).) The formula for 10s would be a little different because the probability of getting a 10 is 4/13. And since aces cannot usually be resplit, we would just use the one-split process.
link to original post


I have followed yours and Dog Hand's instructions to get the expected results. They all seem good to me. However, to explore this a little more, I have two new questions and greatly hope you can help:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p - p^2

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2 - 6p^3 + 2p^4

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= what?

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = what?
link to original post


I figured out the last part myself.
The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



Yes, I agree that K =1/(1-2p) for the infinite re-splitting formula.

By the way, I find it interesting that 1/(1-2p) = 1 + 2p + 4p2 + 8p3 + 16p4 + . . . and that the first three terms of this infinite series match those of the formula for the three-splits formula, K = 1 + 2p + 4p2 - 6p3 + 2p4
link to original post


The problem is my coefficients Ks do not match to yours. Any thoughts? Based on this math, the coefficients Ks should be:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= 1 +2p +4p^2 ++(2p)^n

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
ChesterDog
ChesterDog
Joined: Jul 26, 2010
  • Threads: 8
  • Posts: 1042
March 28th, 2022 at 5:21:09 PM permalink
Quote: aceside

Quote: ChesterDog

Quote: aceside

Quote: aceside

Quote: ChesterDog

Quote: DJTeddyBear

Thanks ChesterDog.

But I've hurt my brain trying to wrap my head around your formula, specifically, to make it into an Excel calculation.

For example, the calucation the Wiz uses is = 2 * [whatever].

I substituted = ( 2 + 2/13 + (3/13)*(2/13) ) * [whatever]. This is:

2 for the original split, plus...
2/13 since there's 2 chances of matching the card, plus...
3/13 since there's 3 chances of a match of the three splits times 2/13 which is the odds that there was the original split.

Similarly, the paired tens line uses = 2 + 8/13 + (12/13)*(8/13) * [whatever].

Interestingly, this DOES change the strategy. Paired 3s vs 2 doesn't split, while paired 9s vs 7 and 10s ve 3 thru 7 all split.

It also changes the edge from 0.484% to a player advantage of 0.132%. That can't be right.

Can anybody help me figure out what I'm missing?

Thanks.
link to original post



Rather than using the complicated formula for splitting up to three times, a more instructional approach to analyzing re-splitting is first to analyze infinite re-splitting.

As an example, let's use a pair of 3s vs dealer's 2. Let E1 stand for the EV of a pair of 3s allowing one split. We know the formula for E1 is:

E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]
where, for example, E(6) is the EV of hitting 6 vs 2 and E(10) is the EV of doubling 10 vs 2.

The formula for the EV of a pair of 3s vs dealer's 2 allowing infinite re-splitting, Einf., is just a little different. It's:
Einf. = (2/13) [E(soft 14) + E(5) + Einf. + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:
Einf. - (2/13)Einf.= (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Then:
(11/13)Einf. = (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Multiplying both sides by 13/11 yields:
Einf. = (2/11) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

A trick to make the Excel calculations easier is to replace the 0 with E(6) - E(6), where again E(6) is the EV of hitting 6 vs 2:
Einf. = (2/11) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13) - E(6)]

We can simplify the above using the fact that E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]. The final result is:

Einf. = (2/11) [(13/2)E1 - E(6)] = (13/11) E1 - (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare Einf. with E(6).

The above formula can be used for pairs of 2s through 9s. (Of course, we would use E(4) for 2s, for example, instead of E(6).) The formula for 10s would be a little different because the probability of getting a 10 is 4/13. And since aces cannot usually be resplit, we would just use the one-split process.
link to original post


I have followed yours and Dog Hand's instructions to get the expected results. They all seem good to me. However, to explore this a little more, I have two new questions and greatly hope you can help:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p - p^2

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2 - 6p^3 + 2p^4

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= what?

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = what?
link to original post


I figured out the last part myself.
The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



Yes, I agree that K =1/(1-2p) for the infinite re-splitting formula.

By the way, I find it interesting that 1/(1-2p) = 1 + 2p + 4p2 + 8p3 + 16p4 + . . . and that the first three terms of this infinite series match those of the formula for the three-splits formula, K = 1 + 2p + 4p2 - 6p3 + 2p4
link to original post


The problem is my coefficients Ks do not match to yours. Any thoughts? Based on this math, the coefficients Ks should be:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= 1 +2p +4p^2 ++(2p)^n

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



I was thinking about our differences in the values of K on and off all today. Finally, I thought of a simple thought experiment that would help settle the problem.

Suppose we have a new game. We have an infinite deck of cards consists of 75% aces and 25% tens. Any hand of one ace and one ten is worth 1, and all other hands are worth 0.

Let's find the values of the hand TT if we stand (E0), split to a maximum of once (E1), or split to a maximum of twice (E2). Let p = the probability of pulling a ten = 0.25

Then we can compare the directly calculated value of E2 with E0 + (1 + 2p)(E1 - E0) and E0 + (1 + 2p - p2)(E1 - E0) to see if either matches.
aceside
aceside
Joined: May 14, 2021
  • Threads: 0
  • Posts: 139
March 29th, 2022 at 11:42:55 PM permalink
Quote: ChesterDog

Quote: aceside

Quote: ChesterDog

Quote: aceside

Quote: aceside

Quote: ChesterDog

Quote: DJTeddyBear

Thanks ChesterDog.

But I've hurt my brain trying to wrap my head around your formula, specifically, to make it into an Excel calculation.

For example, the calucation the Wiz uses is = 2 * [whatever].

I substituted = ( 2 + 2/13 + (3/13)*(2/13) ) * [whatever]. This is:

2 for the original split, plus...
2/13 since there's 2 chances of matching the card, plus...
3/13 since there's 3 chances of a match of the three splits times 2/13 which is the odds that there was the original split.

Similarly, the paired tens line uses = 2 + 8/13 + (12/13)*(8/13) * [whatever].

Interestingly, this DOES change the strategy. Paired 3s vs 2 doesn't split, while paired 9s vs 7 and 10s ve 3 thru 7 all split.

It also changes the edge from 0.484% to a player advantage of 0.132%. That can't be right.

Can anybody help me figure out what I'm missing?

Thanks.
link to original post



Rather than using the complicated formula for splitting up to three times, a more instructional approach to analyzing re-splitting is first to analyze infinite re-splitting.

As an example, let's use a pair of 3s vs dealer's 2. Let E1 stand for the EV of a pair of 3s allowing one split. We know the formula for E1 is:

E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]
where, for example, E(6) is the EV of hitting 6 vs 2 and E(10) is the EV of doubling 10 vs 2.

The formula for the EV of a pair of 3s vs dealer's 2 allowing infinite re-splitting, Einf., is just a little different. It's:
Einf. = (2/13) [E(soft 14) + E(5) + Einf. + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:
Einf. - (2/13)Einf.= (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Then:
(11/13)Einf. = (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Multiplying both sides by 13/11 yields:
Einf. = (2/11) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

A trick to make the Excel calculations easier is to replace the 0 with E(6) - E(6), where again E(6) is the EV of hitting 6 vs 2:
Einf. = (2/11) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13) - E(6)]

We can simplify the above using the fact that E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]. The final result is:

Einf. = (2/11) [(13/2)E1 - E(6)] = (13/11) E1 - (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare Einf. with E(6).

The above formula can be used for pairs of 2s through 9s. (Of course, we would use E(4) for 2s, for example, instead of E(6).) The formula for 10s would be a little different because the probability of getting a 10 is 4/13. And since aces cannot usually be resplit, we would just use the one-split process.
link to original post


I have followed yours and Dog Hand's instructions to get the expected results. They all seem good to me. However, to explore this a little more, I have two new questions and greatly hope you can help:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p - p^2

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2 - 6p^3 + 2p^4

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= what?

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = what?
link to original post


I figured out the last part myself.
The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



Yes, I agree that K =1/(1-2p) for the infinite re-splitting formula.

By the way, I find it interesting that 1/(1-2p) = 1 + 2p + 4p2 + 8p3 + 16p4 + . . . and that the first three terms of this infinite series match those of the formula for the three-splits formula, K = 1 + 2p + 4p2 - 6p3 + 2p4
link to original post


The problem is my coefficients Ks do not match to yours. Any thoughts? Based on this math, the coefficients Ks should be:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= 1 +2p +4p^2 ++(2p)^n

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



I was thinking about our differences in the values of K on and off all today. Finally, I thought of a simple thought experiment that would help settle the problem.

Suppose we have a new game. We have an infinite deck of cards consists of 75% aces and 25% tens. Any hand of one ace and one ten is worth 1, and all other hands are worth 0.

Let's find the values of the hand TT if we stand (E0), split to a maximum of once (E1), or split to a maximum of twice (E2). Let p = the probability of pulling a ten = 0.25

Then we can compare the directly calculated value of E2 with E0 + (1 + 2p)(E1 - E0) and E0 + (1 + 2p - p2)(E1 - E0) to see if either matches.
link to original post


This problem is tricky. To verify the coefficient K, I have tried your simplified mathematical model using the infinite deck of aces (75%) and (25%) tens. In the EV formula of a non-aces pair split to a maximum of 2 times,
E_2 = E_0 + K*(E_1 - E_0),
I got the coefficient K = 1 + 2p-2p^2. It is neither 1+2p, nor 1 + 2p-p^2.

This also makes me think that the coefficient K in the infinite-time split case may have some problem. You used a pair of 3s vs dealer's 2 to demonstrate the derivation (E_1 stand for the EV of a pair of 3s allowing one split, and E_0 stand for the EV of a pair of 3s without splitting) and showed
E1 = (2/13) [E(soft 14) + E(5) + E_0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)],
but the derivation from E_1 to E_2 is still not clear to me. This part is my headache.

How did you get the coefficient K = 1 + 2p + 4p^2 - 6p^3 + 2p^4 , when a 3-time-maximum split is allowed?
ChesterDog
ChesterDog
Joined: Jul 26, 2010
  • Threads: 8
  • Posts: 1042
March 30th, 2022 at 8:50:37 PM permalink
Quote: aceside

Quote: ChesterDog

Quote: aceside

Quote: ChesterDog

Quote: aceside

Quote: aceside

Quote: ChesterDog

Quote: DJTeddyBear

Thanks ChesterDog.

But I've hurt my brain trying to wrap my head around your formula, specifically, to make it into an Excel calculation.

For example, the calucation the Wiz uses is = 2 * [whatever].

I substituted = ( 2 + 2/13 + (3/13)*(2/13) ) * [whatever]. This is:

2 for the original split, plus...
2/13 since there's 2 chances of matching the card, plus...
3/13 since there's 3 chances of a match of the three splits times 2/13 which is the odds that there was the original split.

Similarly, the paired tens line uses = 2 + 8/13 + (12/13)*(8/13) * [whatever].

Interestingly, this DOES change the strategy. Paired 3s vs 2 doesn't split, while paired 9s vs 7 and 10s ve 3 thru 7 all split.

It also changes the edge from 0.484% to a player advantage of 0.132%. That can't be right.

Can anybody help me figure out what I'm missing?

Thanks.
link to original post



Rather than using the complicated formula for splitting up to three times, a more instructional approach to analyzing re-splitting is first to analyze infinite re-splitting.

As an example, let's use a pair of 3s vs dealer's 2. Let E1 stand for the EV of a pair of 3s allowing one split. We know the formula for E1 is:

E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]
where, for example, E(6) is the EV of hitting 6 vs 2 and E(10) is the EV of doubling 10 vs 2.

The formula for the EV of a pair of 3s vs dealer's 2 allowing infinite re-splitting, Einf., is just a little different. It's:
Einf. = (2/13) [E(soft 14) + E(5) + Einf. + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:
Einf. - (2/13)Einf.= (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Then:
(11/13)Einf. = (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Multiplying both sides by 13/11 yields:
Einf. = (2/11) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

A trick to make the Excel calculations easier is to replace the 0 with E(6) - E(6), where again E(6) is the EV of hitting 6 vs 2:
Einf. = (2/11) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13) - E(6)]

We can simplify the above using the fact that E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]. The final result is:

Einf. = (2/11) [(13/2)E1 - E(6)] = (13/11) E1 - (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare Einf. with E(6).

The above formula can be used for pairs of 2s through 9s. (Of course, we would use E(4) for 2s, for example, instead of E(6).) The formula for 10s would be a little different because the probability of getting a 10 is 4/13. And since aces cannot usually be resplit, we would just use the one-split process.
link to original post


I have followed yours and Dog Hand's instructions to get the expected results. They all seem good to me. However, to explore this a little more, I have two new questions and greatly hope you can help:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p - p^2

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2 - 6p^3 + 2p^4

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= what?

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = what?
link to original post


I figured out the last part myself.
The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



Yes, I agree that K =1/(1-2p) for the infinite re-splitting formula.

By the way, I find it interesting that 1/(1-2p) = 1 + 2p + 4p2 + 8p3 + 16p4 + . . . and that the first three terms of this infinite series match those of the formula for the three-splits formula, K = 1 + 2p + 4p2 - 6p3 + 2p4
link to original post


The problem is my coefficients Ks do not match to yours. Any thoughts? Based on this math, the coefficients Ks should be:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= 1 +2p +4p^2 ++(2p)^n

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



I was thinking about our differences in the values of K on and off all today. Finally, I thought of a simple thought experiment that would help settle the problem.

Suppose we have a new game. We have an infinite deck of cards consists of 75% aces and 25% tens. Any hand of one ace and one ten is worth 1, and all other hands are worth 0.

Let's find the values of the hand TT if we stand (E0), split to a maximum of once (E1), or split to a maximum of twice (E2). Let p = the probability of pulling a ten = 0.25

Then we can compare the directly calculated value of E2 with E0 + (1 + 2p)(E1 - E0) and E0 + (1 + 2p - p2)(E1 - E0) to see if either matches.
link to original post


This problem is tricky. To verify the coefficient K, I have tried your simplified mathematical model using the infinite deck of aces (75%) and (25%) tens. In the EV formula of a non-aces pair split to a maximum of 2 times,
E_2 = E_0 + K*(E_1 - E_0),
I got the coefficient K = 1 + 2p-2p^2. It is neither 1+2p, nor 1 + 2p-p^2.

This also makes me think that the coefficient K in the infinite-time split case may have some problem. You used a pair of 3s vs dealer's 2 to demonstrate the derivation (E_1 stand for the EV of a pair of 3s allowing one split, and E_0 stand for the EV of a pair of 3s without splitting) and showed
E1 = (2/13) [E(soft 14) + E(5) + E_0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)],
but the derivation from E_1 to E_2 is still not clear to me. This part is my headache.

How did you get the coefficient K = 1 + 2p + 4p^2 - 6p^3 + 2p^4 , when a 3-time-maximum split is allowed?
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Regarding my new game with an infinite deck of 25% tens and 75% aces (where AT has an EV of 1 and all other hands have EV of 0), I get these numbers for the hand TT:

EV of standing on TT: E0 = 0

EV of splitting once: E1 = 2 EV(T hit) = 2[0.75 EV(AT) + 0.25 EV(TT stand)] = 1.5

EV of splitting up to twice: E2 = 0.75 [EV(AT) + 0.75 EV(AT) + 0.25 EV(TT split once)] + 0.25 [EV(TT split once) + EV(T hit)] = 0.75 [1 + 0.75(1) + 0.25(1.5)] + 0.25 [1.5 + 0.75] = 2.15625

Let's see which of the following formulas give 2.15625 as the EV of splitting TT up to two times:
1. E2 = E0 + (1 + 2p - p2) (E1 - E0)
2. E2 = E0 + (1 + 2p) (E1 - E0)
3. E2 = E0 + (1 + 2p - 2p2) (E1 - E0)

1. E2 = 0 + (1 + 2(0.25) - 0.252) (1.5 - 0) = 2.15625
2. E2 = 0 + (1 + 2(0.25)) (1.5 - 0) = 2.25
3. E2 = 0 + (1 + 2(0.25) - 2(0.252)) (1.5 - 0) = 2.0625

So, neither 2 nor 3 can be the correct formula, and 1 might be correct.

(In my next post, I'll show my derivation of formula 1.)
ChesterDog
ChesterDog
Joined: Jul 26, 2010
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March 31st, 2022 at 9:21:43 PM permalink
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Thanks ChesterDog.

But I've hurt my brain trying to wrap my head around your formula, specifically, to make it into an Excel calculation.

For example, the calucation the Wiz uses is = 2 * [whatever].

I substituted = ( 2 + 2/13 + (3/13)*(2/13) ) * [whatever]. This is:

2 for the original split, plus...
2/13 since there's 2 chances of matching the card, plus...
3/13 since there's 3 chances of a match of the three splits times 2/13 which is the odds that there was the original split.

Similarly, the paired tens line uses = 2 + 8/13 + (12/13)*(8/13) * [whatever].

Interestingly, this DOES change the strategy. Paired 3s vs 2 doesn't split, while paired 9s vs 7 and 10s ve 3 thru 7 all split.

It also changes the edge from 0.484% to a player advantage of 0.132%. That can't be right.

Can anybody help me figure out what I'm missing?

Thanks.
link to original post



Rather than using the complicated formula for splitting up to three times, a more instructional approach to analyzing re-splitting is first to analyze infinite re-splitting.

As an example, let's use a pair of 3s vs dealer's 2. Let E1 stand for the EV of a pair of 3s allowing one split. We know the formula for E1 is:

E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]
where, for example, E(6) is the EV of hitting 6 vs 2 and E(10) is the EV of doubling 10 vs 2.

The formula for the EV of a pair of 3s vs dealer's 2 allowing infinite re-splitting, Einf., is just a little different. It's:
Einf. = (2/13) [E(soft 14) + E(5) + Einf. + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:
Einf. - (2/13)Einf.= (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Then:
(11/13)Einf. = (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Multiplying both sides by 13/11 yields:
Einf. = (2/11) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

A trick to make the Excel calculations easier is to replace the 0 with E(6) - E(6), where again E(6) is the EV of hitting 6 vs 2:
Einf. = (2/11) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13) - E(6)]

We can simplify the above using the fact that E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]. The final result is:

Einf. = (2/11) [(13/2)E1 - E(6)] = (13/11) E1 - (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare Einf. with E(6).

The above formula can be used for pairs of 2s through 9s. (Of course, we would use E(4) for 2s, for example, instead of E(6).) The formula for 10s would be a little different because the probability of getting a 10 is 4/13. And since aces cannot usually be resplit, we would just use the one-split process.
link to original post


I have followed yours and Dog Hand's instructions to get the expected results. They all seem good to me. However, to explore this a little more, I have two new questions and greatly hope you can help:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p - p^2

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2 - 6p^3 + 2p^4

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= what?

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = what?
link to original post


I figured out the last part myself.
The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



Yes, I agree that K =1/(1-2p) for the infinite re-splitting formula.

By the way, I find it interesting that 1/(1-2p) = 1 + 2p + 4p2 + 8p3 + 16p4 + . . . and that the first three terms of this infinite series match those of the formula for the three-splits formula, K = 1 + 2p + 4p2 - 6p3 + 2p4
link to original post


The problem is my coefficients Ks do not match to yours. Any thoughts? Based on this math, the coefficients Ks should be:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= 1 +2p +4p^2 ++(2p)^n

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



I was thinking about our differences in the values of K on and off all today. Finally, I thought of a simple thought experiment that would help settle the problem.

Suppose we have a new game. We have an infinite deck of cards consists of 75% aces and 25% tens. Any hand of one ace and one ten is worth 1, and all other hands are worth 0.

Let's find the values of the hand TT if we stand (E0), split to a maximum of once (E1), or split to a maximum of twice (E2). Let p = the probability of pulling a ten = 0.25

Then we can compare the directly calculated value of E2 with E0 + (1 + 2p)(E1 - E0) and E0 + (1 + 2p - p2)(E1 - E0) to see if either matches.
link to original post


This problem is tricky. To verify the coefficient K, I have tried your simplified mathematical model using the infinite deck of aces (75%) and (25%) tens. In the EV formula of a non-aces pair split to a maximum of 2 times,
E_2 = E_0 + K*(E_1 - E_0),
I got the coefficient K = 1 + 2p-2p^2. It is neither 1+2p, nor 1 + 2p-p^2.

This also makes me think that the coefficient K in the infinite-time split case may have some problem. You used a pair of 3s vs dealer's 2 to demonstrate the derivation (E_1 stand for the EV of a pair of 3s allowing one split, and E_0 stand for the EV of a pair of 3s without splitting) and showed
E1 = (2/13) [E(soft 14) + E(5) + E_0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)],
but the derivation from E_1 to E_2 is still not clear to me. This part is my headache.

How did you get the coefficient K = 1 + 2p + 4p^2 - 6p^3 + 2p^4 , when a 3-time-maximum split is allowed?
link to original post



Below is derivation for the formula for E2, which is the EV of a pair in an infinite-deck blackjack game which allows splitting up to two times.

When a pair of 3s is split three things could happen:
1) Neither the first nor second 3 gets a 3 on its first hit.
2) The first 3 gets another 3 so the first hand can be re-split.
3) The first 3 does not get another 3, but the second 3 gets one so the second hand can be re-split.

Let p = the probability of drawing a 3. Then each of the events above have the corresponding probabilities:
1) (1 - p)2
2) p
3) (1 - p)p

The sum of probabilities is (1 - p)2 + p + (1 - p)p = 1- 2p +p2 +p + p - p2 = 1, as it should be.

The EVs of the resulting hands depend on whether a hit card after any splits can be a 3. If the hit card cannot be a 3, let's call the EV of the hand Eno. Then Eno = p[E(soft 14) + E(5) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)] / (1 - p)

Let Eyes stand for the EV of a hand that can have 3 as a hit card. Then Eyes = p[E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Here's the relationship between the two EVs, by the way:
(1 - p) Eno = Eyes - pE(6) = Eyes - pE0 because E(6) is the same as E0, the EV of a pair that can not be split

The total EV of the pair allowing up to two splits is:
E2 = (1 - p)2(Eno + Eno) + p(Eyes + Eyes + Eyes) + (1-p)p(Eno + Eyes + Eyes)

Rearranging:
E2 = Eno[2(1-p)2 + (1-p)p] + Eyes[3p + 2(1-p)p]
E2 = (1-p)Eno[2(1-p) + p] + Eyes[3p + 2p - 2p2]
E2 = (1-p)Eno[2 - p] + Eyes[5p - 2p2]
E2 = (Eyes - pE0)[2 - p] + Eyes[5p - 2p2]
E2 = Eyes[2 - p + 5p - 2p2] - E0[2p - p2]
E2 = Eyes[2 + 4p - 2p2] - E0[2p - p2]

Subtracting E0 from both sides:
E2 - E0 = Eyes[2 + 4p - 2p2] - E0[1 +2p - p2]

E2 - E0 = 2Eyes[1 + 2p - p2] - E0[1 +2p - p2]

E2 - E0 = [1 + 2p - p2] (2Eyes - E0)

Since 2Eyes = E1, the EV of splitting a pair once,
E2 - E0 = [1 + 2p - p2] (E1 - E0)

The formula for the EV of a pair in an infinite-deck game allowing three splits can derived using the same method.
aceside
aceside
Joined: May 14, 2021
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Thanks for this post from:
ChesterDog
April 2nd, 2022 at 7:20:21 AM permalink
Quote: ChesterDog

Quote: aceside

Quote: ChesterDog

Quote: aceside

Quote: ChesterDog

Quote: aceside

Quote: aceside

Quote: ChesterDog

Quote: DJTeddyBear

Thanks ChesterDog.

But I've hurt my brain trying to wrap my head around your formula, specifically, to make it into an Excel calculation.

For example, the calucation the Wiz uses is = 2 * [whatever].

I substituted = ( 2 + 2/13 + (3/13)*(2/13) ) * [whatever]. This is:

2 for the original split, plus...
2/13 since there's 2 chances of matching the card, plus...
3/13 since there's 3 chances of a match of the three splits times 2/13 which is the odds that there was the original split.

Similarly, the paired tens line uses = 2 + 8/13 + (12/13)*(8/13) * [whatever].

Interestingly, this DOES change the strategy. Paired 3s vs 2 doesn't split, while paired 9s vs 7 and 10s ve 3 thru 7 all split.

It also changes the edge from 0.484% to a player advantage of 0.132%. That can't be right.

Can anybody help me figure out what I'm missing?

Thanks.
link to original post



Rather than using the complicated formula for splitting up to three times, a more instructional approach to analyzing re-splitting is first to analyze infinite re-splitting.

As an example, let's use a pair of 3s vs dealer's 2. Let E1 stand for the EV of a pair of 3s allowing one split. We know the formula for E1 is:

E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]
where, for example, E(6) is the EV of hitting 6 vs 2 and E(10) is the EV of doubling 10 vs 2.

The formula for the EV of a pair of 3s vs dealer's 2 allowing infinite re-splitting, Einf., is just a little different. It's:
Einf. = (2/13) [E(soft 14) + E(5) + Einf. + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:
Einf. - (2/13)Einf.= (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Then:
(11/13)Einf. = (2/13) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Multiplying both sides by 13/11 yields:
Einf. = (2/11) [E(soft 14) + E(5) + 0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

A trick to make the Excel calculations easier is to replace the 0 with E(6) - E(6), where again E(6) is the EV of hitting 6 vs 2:
Einf. = (2/11) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13) - E(6)]

We can simplify the above using the fact that E1 = (2/13) [E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]. The final result is:

Einf. = (2/11) [(13/2)E1 - E(6)] = (13/11) E1 - (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare Einf. with E(6).

The above formula can be used for pairs of 2s through 9s. (Of course, we would use E(4) for 2s, for example, instead of E(6).) The formula for 10s would be a little different because the probability of getting a 10 is 4/13. And since aces cannot usually be resplit, we would just use the one-split process.
link to original post


I have followed yours and Dog Hand's instructions to get the expected results. They all seem good to me. However, to explore this a little more, I have two new questions and greatly hope you can help:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p - p^2

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2 - 6p^3 + 2p^4

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= what?

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = what?
link to original post


I figured out the last part myself.
The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



Yes, I agree that K =1/(1-2p) for the infinite re-splitting formula.

By the way, I find it interesting that 1/(1-2p) = 1 + 2p + 4p2 + 8p3 + 16p4 + . . . and that the first three terms of this infinite series match those of the formula for the three-splits formula, K = 1 + 2p + 4p2 - 6p3 + 2p4
link to original post


The problem is my coefficients Ks do not match to yours. Any thoughts? Based on this math, the coefficients Ks should be:

The formula for the EV of a non-aces pair split to a maximum of 2 times:
E_2 = E_0 + K*(E_1 - E_0), where K = 1 + 2p

The formula for the EV of a non-aces pair split to a maximum of 3 times:
E_3 = E_0 + K*(E_1 - E_0), where K = 1 + 2p + 4p^2

The formula for the EV of a non-aces pair split to a maximum of n times:
E_n = E_0 + K*(E_1 - E_0), where K= 1 +2p +4p^2 ++(2p)^n

The formula for the EV of a non-aces pair split to a maximum of infinite times:
E_inf = E_0 + K*(E_1 - E_0), where K = 1/(1-2p)
link to original post



I was thinking about our differences in the values of K on and off all today. Finally, I thought of a simple thought experiment that would help settle the problem.

Suppose we have a new game. We have an infinite deck of cards consists of 75% aces and 25% tens. Any hand of one ace and one ten is worth 1, and all other hands are worth 0.

Let's find the values of the hand TT if we stand (E0), split to a maximum of once (E1), or split to a maximum of twice (E2). Let p = the probability of pulling a ten = 0.25

Then we can compare the directly calculated value of E2 with E0 + (1 + 2p)(E1 - E0) and E0 + (1 + 2p - p2)(E1 - E0) to see if either matches.
link to original post


This problem is tricky. To verify the coefficient K, I have tried your simplified mathematical model using the infinite deck of aces (75%) and (25%) tens. In the EV formula of a non-aces pair split to a maximum of 2 times,
E_2 = E_0 + K*(E_1 - E_0),
I got the coefficient K = 1 + 2p-2p^2. It is neither 1+2p, nor 1 + 2p-p^2.

This also makes me think that the coefficient K in the infinite-time split case may have some problem. You used a pair of 3s vs dealer's 2 to demonstrate the derivation (E_1 stand for the EV of a pair of 3s allowing one split, and E_0 stand for the EV of a pair of 3s without splitting) and showed
E1 = (2/13) [E(soft 14) + E(5) + E_0 + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)],
but the derivation from E_1 to E_2 is still not clear to me. This part is my headache.

How did you get the coefficient K = 1 + 2p + 4p^2 - 6p^3 + 2p^4 , when a 3-time-maximum split is allowed?
link to original post



Below is derivation for the formula for E2, which is the EV of a pair in an infinite-deck blackjack game which allows splitting up to two times.

When a pair of 3s is split three things could happen:
1) Neither the first nor second 3 gets a 3 on its first hit.
2) The first 3 gets another 3 so the first hand can be re-split.
3) The first 3 does not get another 3, but the second 3 gets one so the second hand can be re-split.

Let p = the probability of drawing a 3. Then each of the events above have the corresponding probabilities:
1) (1 - p)2
2) p
3) (1 - p)p

The sum of probabilities is (1 - p)2 + p + (1 - p)p = 1- 2p +p2 +p + p - p2 = 1, as it should be.

The EVs of the resulting hands depend on whether a hit card after any splits can be a 3. If the hit card cannot be a 3, let's call the EV of the hand Eno. Then Eno = p[E(soft 14) + E(5) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)] / (1 - p)

Let Eyes stand for the EV of a hand that can have 3 as a hit card. Then Eyes = p[E(soft 14) + E(5) + E(6) + E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Here's the relationship between the two EVs, by the way:
(1 - p) Eno = Eyes - pE(6) = Eyes - pE0 because E(6) is the same as E0, the EV of a pair that can not be split

The total EV of the pair allowing up to two splits is:
E2 = (1 - p)2(Eno + Eno) + p(Eyes + Eyes + Eyes) + (1-p)p(Eno + Eyes + Eyes)

Rearranging:
E2 = Eno[2(1-p)2 + (1-p)p] + Eyes[3p + 2(1-p)p]
E2 = (1-p)Eno[2(1-p) + p] + Eyes[3p + 2p - 2p2]
E2 = (1-p)Eno[2 - p] + Eyes[5p - 2p2]
E2 = (Eyes - pE0)[2 - p] + Eyes[5p - 2p2]
E2 = Eyes[2 - p + 5p - 2p2] - E0[2p - p2]
E2 = Eyes[2 + 4p - 2p2] - E0[2p - p2]

Subtracting E0 from both sides:
E2 - E0 = Eyes[2 + 4p - 2p2] - E0[1 +2p - p2]

E2 - E0 = 2Eyes[1 + 2p - p2] - E0[1 +2p - p2]

E2 - E0 = [1 + 2p - p2] (2Eyes - E0)

Since 2Eyes = E1, the EV of splitting a pair once,
E2 - E0 = [1 + 2p - p2] (E1 - E0)

The formula for the EV of a pair in an infinite-deck game allowing three splits can derived using the same method.
link to original post


This derivation is neat and the numerical verification is creative. Everything is convincing to me. Thank you for your time. I am thinking about how to apply your E_yes and E_no technique to the infinite-time split case.

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