aceside
aceside
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March 25th, 2022 at 10:02:36 PM permalink
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


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稚 know how. Just a suggestion. Let ChesterDog come back.
Last edited by: aceside on Mar 25, 2022
DogHand
DogHand
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Thanks for this post from:
aceside
March 26th, 2022 at 6:22:14 AM permalink
DJTeddyBear,

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 E0 value.

The EV for a single split, what ChesterDog calls the E1 value is given in cell G9 as 0.37. Since E1 < E0, 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 + 4p2 - 6p3 + 2p4 = 1 + 2/13 + 4/132 - 6/133 + 2/134 = 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

E3 = E0 + K*(E1 - E0) = 0.39955 + 1.17485*(0.37 - 0.39955) = 0.36483

So, since E3 < E0, 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 + 4p2 - 6p3 + 2p4 = 1 + 2*(4/13) + 4*(4/13)2 - 6*(4/13)3 + 2*(4/13)4 = 1.8372255...
ChesterDog
ChesterDog
Joined: Jul 26, 2010
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Thanks for this post from:
aceside
March 26th, 2022 at 6:38:44 AM permalink
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.
aceside
aceside
Joined: May 14, 2021
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March 26th, 2022 at 8:34:41 AM permalink
Wonderful work! Thank you guys! I am still learning to type subscripts and superscripts. Let me try now.
E_3= E_0.
This cannot be done in iPhone?
Dieter
Administrator
Dieter
Joined: Jul 23, 2014
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March 26th, 2022 at 9:37:59 AM permalink
Forum formatting tags like [sub]sub[/sub] may help?

(Typed on my phone. It's a hassle without a keyboard.)
May the cards fall in your favor.
DJTeddyBear
DJTeddyBear
Joined: Nov 2, 2009
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March 26th, 2022 at 10:08:47 AM permalink
I知 still trying to wrap my head around this, but thanks guys.


And dieter beat me to the sub question. 🤪

Yeah, it痴 a PITA on a phone, but doable.

Check the format codes link below the post button to learn how to do it.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Dieter
Administrator
Dieter
Joined: Jul 23, 2014
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March 26th, 2022 at 10:51:25 AM permalink
Quote: DJTeddyBear


And dieter beat me to the sub question. 🤪

link to original post



I left some [sup]sup[/sup] for you if you're hungry?
May the cards fall in your favor.
aceside
aceside
Joined: May 14, 2021
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March 27th, 2022 at 2:05:06 AM permalink
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?
Last edited by: aceside on Mar 27, 2022
aceside
aceside
Joined: May 14, 2021
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March 27th, 2022 at 11:00:49 PM permalink
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)
ChesterDog
ChesterDog
Joined: Jul 26, 2010
  • Threads: 8
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March 28th, 2022 at 6:57:26 AM permalink
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

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