Quote:DJTeddyBearThanks 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

Let me check if your coefficients are correct or not.

E_2 -E_0= (1+2/13 -(1/13)(1/13)) (E_1 -E_0).

Do we pick a maximum among E_0 and E_1 first? My logic is not very clear here.

The EV of the 2-max-split pair is the maximum of E_0 and E_2.

I don’t know how. Just a suggestion. Let ChesterDog come back.

Perhaps I can help you to apply ChesterDog's equation correctly.

Let's consider the case of 9,9 vs. 7. From the Wiz's spreadsheet, if you don't split, then the non-splitting highest EV option is to Stand (see cell G23 on the "split" worksheet), with an EV of 0.39955: this is what ChesterDog calls the E

_{0}value.

The EV for a single split, what ChesterDog calls the E

_{1}value is given in cell G9 as 0.37. Since E

_{1}< E

_{0}, the Wiz's spreadsheet tells us not to split.

Now let's apply ChesterDog's equation!

Since we are contemplating a 9 split, p = 1/13: that is, the probability of drawing a 9 is 1 out of 13 for the infinite-deck case. That means the value of K is this:

K = 1 + 2p + 4p

^{2}- 6p

^{3}+ 2p

^{4}= 1 + 2/13 + 4/13

^{2}- 6/13

^{3}+ 2/13

^{4}= 1.17485...

(Naturally, the K value is the same for splitting everything except 10's, since p = 1/13 for any rank other than 10.)*

Now, if we split 9's three times vs. the dealer's 7, then the EV is

E

_{3}= E

_{0}+ K*(E

_{1}- E

_{0}) = 0.39955 + 1.17485*(0.37 - 0.39955) = 0.36483

So, since E

_{3}< E

_{0}, B.S. says NOT to split 9's vs. a 7.

By the way, if the Wiz's spreadsheet says NOT to split for his Sp1 case, then for Sp3 you also will not split.

Hope this helps!

Dog Hand

*Footnote: for a 10-split, p = 4/13 and so

K = 1 + 2p + 4p

^{2}- 6p

^{3}+ 2p

^{4}= 1 + 2*(4/13) + 4*(4/13)

^{2}- 6*(4/13)

^{3}+ 2*(4/13)

^{4}= 1.8372255...

Quote:DJTeddyBearThanks 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 E

_{1}stand for the EV of a pair of 3s allowing one split. We know the formula for E

_{1}is:

E

_{1}= (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, E

_{inf.}, is just a little different. It's:

E

_{inf.}= (2/13) [E(soft 14) + E(5) + E

_{inf.}+ E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:

E

_{inf.}- (2/13)E

_{inf.}= (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)E

_{inf.}= (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:

E

_{inf.}= (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:

E

_{inf.}= (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:

E

_{inf.}= (2/11) [(13/2)E

_{1}- E(6)] = (13/11) E

_{1}- (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare E

_{inf.}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.

E_3= E_0.

This cannot be done in iPhone?

_{sub}[/sub] may help?

(Typed on my phone. It's a hassle without a keyboard.)

And dieter beat me to the s

_{u}b question. 🤪

Yeah, it’s a PITA on a phone, but doable.

Check the format codes link below the post button to learn how to do it.

Quote:DJTeddyBear

And dieter beat me to the s_{u}b question. 🤪

link to original post

I left some [sup]

^{sup}[/sup] for you if you're hungry?

Quote:ChesterDogQuote:DJTeddyBearThanks 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 E_{1}stand for the EV of a pair of 3s allowing one split. We know the formula for E_{1}is:

E_{1}= (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, E_{inf.}, is just a little different. It's:

E_{inf.}= (2/13) [E(soft 14) + E(5) + E_{inf.}+ E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:

E_{inf.}- (2/13)E_{inf.}= (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)E_{inf.}= (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:

E_{inf.}= (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:

E_{inf.}= (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:

E_{inf.}= (2/11) [(13/2)E_{1}- E(6)] = (13/11) E_{1}- (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare E_{inf.}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?

Quote:acesideQuote:ChesterDogQuote:DJTeddyBear

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 E_{1}stand for the EV of a pair of 3s allowing one split. We know the formula for E_{1}is:

E_{1}= (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, E_{inf.}, is just a little different. It's:

E_{inf.}= (2/13) [E(soft 14) + E(5) + E_{inf.}+ E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:

E_{inf.}- (2/13)E_{inf.}= (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)E_{inf.}= (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:

E_{inf.}= (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:

E_{inf.}= (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:

E_{inf.}= (2/11) [(13/2)E_{1}- E(6)] = (13/11) E_{1}- (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare E_{inf.}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)

Quote:acesideQuote:acesideQuote:ChesterDogQuote:DJTeddyBear

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 E_{1}stand for the EV of a pair of 3s allowing one split. We know the formula for E_{1}is:

E_{1}= (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, E_{inf.}, is just a little different. It's:

E_{inf.}= (2/13) [E(soft 14) + E(5) + E_{inf.}+ E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + 4E(13)]

Rearranging yields:

E_{inf.}- (2/13)E_{inf.}= (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)E_{inf.}= (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:

E_{inf.}= (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:

E_{inf.}= (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:

E_{inf.}= (2/11) [(13/2)E_{1}- E(6)] = (13/11) E_{1}- (2/11)E(6)

And to determine if we should split 3s vs 2, we just compare E_{inf.}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 + 4p

^{2}+ 8p

^{3}+ 16p

^{4}+ . . . and that the first three terms of this infinite series match those of the formula for the three-splits formula, K = 1 + 2p + 4p

^{2}- 6p

^{3}+ 2p

^{4}