charliepatrick
charliepatrick
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November 29th, 2021 at 12:05:36 AM permalink
^ As you say there will be so many combinations that have to be looked at (one advantage of DO loops!); looking just at the second hand of split 6s the results I got earlier matches their strategy for Set 2 (2 3 4 5 6 8). I can imagine it could get different for different sets/multipliers. This is probably a game where a computer might know the strategy but no-one in practice would.
Quote: charliepatrick

...Multiplier for this hand is x5

D: 6 Spl: A2346789X H: 13 (13 13 12 12 11 11) h: 7 (7 7 8 7 7 7[u/]) D: D16 D15 D14 D13 D12 D11 D10 D9 D8 D7 D6 D5 D4 D3 D2 D1 d: d10 d9 d8 d7 d6 d5 d4 d3
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charliepatrick
charliepatrick
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November 29th, 2021 at 1:20:30 PM permalink
After working out the splits (there's actually more to them than in the Evolution strategy) I've had to make a few assumptions, and there's no guarantee that I'm correct, but I played around with the value of passing hands on.
For fun I wondered what would happen if there was no value of passing any hand to the next phase, for a regular game it gave the regular strategy and a house edge of 0.792% (this assumed you didn't pay the "Ante"). Using similar logic you get a range of EVs for various multipliers (0.499 for 2x, up to 16.639 for 25x).

Then I plugged those values into the carry forward and, after a quick play around, found if Set 1 thru Set 6 were picked with 50% 20% 12.5% 7.5% 5% 5% I got a House Edge of 0.36% (i.e. payback of 99.63%).

I've yet to work through the strategy to see if it matches.

About splits - I looked at all the splits as on occasions it mattered which cards you had split (for instance splitting 9s might make it easier to get to 19 than splitting 6s). So the H1: and h1: lines give the hit/stand point for pairs of 2 3 4 5 6 7 8 9 X (Note: some of these values are impossible, e.g. you can't really get a soft total where you might hit from splitting 10s or 9s., also you never split 5s.) I'm not sure why I've got more hitting than they have for the first hand, I only checked one case so might revisit this.

These are for set 1.
Parms: ndx:8 m17:0.499345 m18:0.499345 m19:1.089821 m20:1.724817 m21:2.374416 mBJ:3.031365 mIN:0 vow:1 Time:20:46:56:782
D: 1 Spl: A2346789 H: 17 (17 17 17 17 16 16) h: 8 (8 8 8 8 8 8) D: d: H1: 17 17 17 17 17 17 17 17 17 h1: 8 8 8 8 8 8 8 8 8
D: 2 Spl: A2346789 H: 15 (15 16 16 15 13 12) h: 8 (8 8 8 8 7 7) D: D11 D10 d: d8 H1: 16 16 16 16 16 16 16 16 16 h1: 8 8 8 8 8 8 8 8 8
D: 3 Spl: A2346789 H: 15 (15 16 16 14 13 12) h: 8 (8 8 8 8 7 7) D: D11 D10 d: d8 H1: 16 16 16 16 16 16 16 16 16 h1: 8 8 8 8 8 8 8 8 8
D: 4 Spl: A2346789 H: 14 (14 15 15 14 12 11) h: 8 (8 8 8 8 7 7) D: D11 D10 d: d8 d7 H1: 15 15 15 15 16 16 16 16 15 h1: 8 8 8 8 8 8 8 8 8
D: 5 Spl: A2346789 H: 14 (14 15 15 13 12 11) h: 8 (8 8 8 8 7 7) D: D11 D10 D9 d: d8 d7 H1: 15 15 15 15 15 15 15 15 15 h1: 8 8 8 8 8 8 8 8 8
D: 6 Spl: A2346789 H: 14 (14 15 15 13 12 11) h: 8 (8 8 8 8 7 7) D: D11 D10 D9 d: d8 d7 d6 H1: 15 15 15 15 15 15 16 15 15 h1: 8 8 8 8 8 8 8 8 8
D: 7 Spl: A2346789 H: 16 (16 16 16 16 16 16) h: 8 (8 8 8 7 7 7) D: D11 D10 d: H1: 16 16 16 16 16 16 16 16 16 h1: 8 8 8 8 8 8 8 8 7
D: 8 Spl: A2346789 H: 17 (17 17 17 17 16 16) h: 8 (8 8 8 8 8 7) D: d: H1: 17 17 17 17 17 17 17 17 17 h1: 8 8 8 8 8 8 8 8 8
D: 9 Spl: A2346789 H: 17 (17 17 17 17 16 16) h: 8 (8 8 8 8 8 8) D: d: H1: 17 17 17 17 17 17 17 17 17 h1: 8 8 8 8 8 8 8 8 8
D: X Spl: A234789 H: 17 (17 17 17 16 16 16) h: 8 (8 8 8 9 8 8) D: d: H1: 17 17 17 17 17 17 17 17 17 h1: 8 8 8 8 8 8 8 9 9
Overall Result: (Multiplier=0)
Total EV:0.6319449252624262
Parms: ndx:8 m17:0.499345 m18:0.499345 m19:1.089821 m20:1.724817 m21:2.374416 mBJ:3.031365 mIN:2 vow:1 Time:20:46:56:800
D: 1 Spl: A2346789 H: 17 (17 17 17 17 16 16) h: 8 (8 8 8 8 8 8) D: D11 D10 d: H1: 17 17 17 17 17 17 17 17 17 h1: 8 8 8 8 8 8 8 8 8
D: 2 Spl: A2346789 H: 14 (14 15 15 13 12 12) h: 8 (8 8 8 8 7 7) D: D13 D12 D11 D10 D9 D8 d: d9 d8 d7 d6 d5 d4 d3 H1: 15 15 15 15 15 15 15 15 15 h1: 8 8 8 8 8 8 8 8 8
D: 3 Spl: A2346789 H: 14 (14 14 14 13 12 12) h: 8 (8 8 8 8 7 7) D: D13 D12 D11 D10 D9 D8 d: d9 d8 d7 d6 d5 d4 d3 H1: 14 14 14 15 15 15 15 15 14 h1: 8 8 8 8 8 8 8 8 8
D: 4 Spl: A2346789 H: 13 (13 14 14 13 12 11) h: 8 (8 8 8 8 7 7) D: D13 D12 D11 D10 D9 D8 d: d9 d8 d7 d6 d5 d4 d3 H1: 14 14 14 14 14 14 14 14 14 h1: 8 8 8 8 8 8 8 8 8
D: 5 Spl: A2346789 H: 13 (13 13 13 12 11 11) h: 8 (8 8 8 8 7 7) D: D13 D12 D11 D10 D9 D8 D7 d: d9 d8 d7 d6 d5 d4 d3 H1: 13 13 14 14 14 14 14 13 13 h1: 8 8 8 8 8 8 8 8 8
D: 6 Spl: A2346789 H: 13 (13 13 13 12 11 11) h: 8 (8 8 8 8 7 7) D: D13 D12 D11 D10 D9 D8 D7 d: d9 d8 d7 d6 d5 d4 d3 H1: 14 14 14 14 14 14 14 14 13 h1: 8 8 8 8 8 8 8 8 8
D: 7 Spl: A2346789 H: 16 (16 16 16 16 16 16) h: 7 (7 8 8 7 7 7) D: D11 D10 D9 d: d8 d7 d6 H1: 16 16 16 16 16 16 16 16 16 h1: 7 7 7 7 7 7 7 7 7
D: 8 Spl: A2346789 H: 17 (17 17 17 16 16 16) h: 8 (8 8 8 8 8 7) D: D11 D10 D9 d: d8 d7 H1: 17 17 17 17 17 17 17 17 17 h1: 8 8 8 8 8 8 8 8 8
D: 9 Spl: A2346789 H: 17 (17 17 17 16 16 16) h: 8 (8 8 8 8 8 8) D: D11 D10 d: d8 H1: 17 17 17 17 17 17 17 17 17 h1: 8 8 8 8 8 8 8 8 8
D: X Spl: A2346789 H: 16 (16 17 17 16 16 15) h: 8 (8 8 8 8 8 8) D: D11 D10 d: H1: 17 17 17 17 17 17 17 17 17 h1: 8 8 8 8 8 8 8 8 8
Overall Result: (Multiplier=2)
Total EV:1.122538316229665
gordonm888
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MrCasinoGames
November 29th, 2021 at 3:05:50 PM permalink
Yes, I agree, Evolution's strategy for the first split hand does not account for whether you are splitting Tens or Sixes - and one might wish to be less aggressive hitting stiff hands when splitting Tens than when splitting Sixes because the second hand is more likely to readily get you a high multiplier when the second split hand is a Ten (and less likely when the second split hand is a Six.)

This is what I referred to in a previous post when I said that I think that Evolution missed something when they developed their strategy for Hit/Stand decisions on split hands.

BTW, analyzng the splitting of a 10-10 pair is where the action is - because a 10-10 pair is roughly 16X more likely to occur than any other pair.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
charliepatrick
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MrCasinoGamesgordonm888
November 29th, 2021 at 3:15:26 PM permalink
Quote: gordonm888

... 10-10 pair is roughly 16X more likely to occur than any other pair.
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In my local casino they have to be identical ranks - so personally I'm currently assuming there are only 4 ways to get a pair of pictures - however this would probably only affect the EV rather than the strategy.
LC1
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November 30th, 2021 at 11:40:35 AM permalink
Quote: charliepatrick

For fun I wondered what would happen if there was no value of passing any hand to the next phase, for a regular game it gave the regular strategy and a house edge of 0.792% (this assumed you didn't pay the "Ante"). Using similar logic you get a range of EVs for various multipliers (0.499 for 2x, up to 16.639 for 25x).

Then I plugged those values into the carry forward and, after a quick play around, found if Set 1 thru Set 6 were picked with 50% 20% 12.5% 7.5% 5% 5% I got a House Edge of 0.36% (i.e. payback of 99.63%).

I've yet to work through the strategy to see if it matches.


You're clearly a lot further down the road with this than I am, but I have been giving some thought to how I might go about modifying a CA program to tackle this. In truth, I'm trying to restrain myself from diving in, because -
a) there are other things I am supposed to be doing
b) it looks like a lot of (difficult) work, and
c) there are clearly already some very able people on the case!

What you said in the quoted snippet reminds me of a question I have been pondering - Would an iterative approach work? The idea would be to start by generating a strategy with no value associated with the multiplier being carried forward, use its EVs to generate a modified strategy accounting for the value of the carried forward multiplier, use the EVs of the modified strategy to create yet another modified strategy, and so on until there are no more modifications arising.

Conceptually, the idea is to ask "If I were to apply strategy S0 to the next hand, what is the strategy, S1, I should I apply to the current hand in order to maximise the overall EV?". Then, "If I were to apply S1 on the next hand, what is the strategy S2...." etc.

Does that make sense / sound reasonable?


Also, when it comes to plugging the EVs into the carry forward, I think that it is only the excess EV above what would be won by applying the given strategy under normal BJ rules which should be counted in the carry forward. Otherwise, I think you end up counting this portion of the EV twice, once for each hand. [I have to admit my thoughts on this are all still a little fuzzy, though. I've been going round in circles to some degree.]
charliepatrick
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LC1
November 30th, 2021 at 11:58:30 AM permalink
Quote: LC1

...What you said in the quoted snippet reminds me of a question I have been pondering - Would an iterative approach work? The idea would be to start by generating a strategy with no value associated with the multiplier being carried forward, use its EVs to generate a modified strategy accounting for the value of the carried forward multiplier, use the EVs of the modified strategy to create yet another modified strategy, and so on until there are no more modifications arising.

Conceptually, the idea is to ask "If I were to apply strategy S0 to the next hand, what is the strategy, S1, I should I apply to the current hand in order to maximise the overall EV?". Then, "If I were to apply S1 on the next hand, what is the strategy S2...." etc.

Does that make sense / sound reasonable?


Also, when it comes to plugging the EVs into the carry forward, I think that it is only the excess EV above what would be won by applying the given strategy under normal BJ rules which should be counted in the carry forward. Otherwise, I think you end up counting this portion of the EV twice, once for each hand...

Yes I have tried the S0, S1 approach (which is how I got the results I did) but also if I carried on I got to a figure to 4 decimal places with 4 iterations. However it was a player advantage under the worst conditions so I suspect I'm missing a (-1) somewhere for paying the "Ante" to play the next hand.

Using S0 S1 can only be an estimate. That method assumes there is no value of a carry forward for the second hand, so you don't pay the "Ante" and also don't benefit from it - however this possibly means the strategy is slightly different - but it will give a rough idea of the correct EV. (It seems to give a strategy for the first hand close to theirs, so I suspect the carry forward EVs are fairly close and if I added in logic for paying for the "Ante", I might get some convergence.)

btw one reason for using coding is each iteration can create the evs[0]=nnnnn evs[1]=mmmm etc to plug into the next!
LC1
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November 30th, 2021 at 12:58:20 PM permalink
Thanks for the response.

Quote: charliepatrick


Using S0 S1 can only be an estimate. That method assumes there is no value of a carry forward for the second hand, so you don't pay the "Ante" and also don't benefit from it - however this possibly means the strategy is slightly different - but it will give a rough idea of the correct EV. (It seems to give a strategy for the first hand close to theirs, so I suspect the carry forward EVs are fairly close and if I added in logic for paying for the "Ante", I might get some convergence.)


Yeah, I glossed over some elements. I was sort of aware that there would be a need for an extra term for the EV arising from the average multiplier won by the second hand (I think I saw you mention this in an earlier post.). And, also, I am wondering if what is actually needed is the EV associated with not just the second hand, but with the average-length of winning streak.

How to correctly account for subsequent main + ante bets in relation to the first hand is a bit confusing to me at the moment. It seems like there is scope to inadvertently count them twice (a bit like the issue I mentioned with the carry-forward EV). With a simulation, you could just divide the total amount won or lost by the total of the initial bets of all the hands played, to get the per-hand EV. With an analytical approach the notional hand is simultaneously playing multiple roles (the 1st, 2nd, and maybe 3rd+ hands).
gordonm888
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November 30th, 2021 at 1:43:50 PM permalink
Quote: LC1

-snip-

What you said in the quoted snippet reminds me of a question I have been pondering - Would an iterative approach work? The idea would be to start by generating a strategy with no value associated with the multiplier being carried forward, use its EVs to generate a modified strategy accounting for the value of the carried forward multiplier, use the EVs of the modified strategy to create yet another modified strategy, and so on until there are no more modifications arising.

Conceptually, the idea is to ask "If I were to apply strategy S0 to the next hand, what is the strategy, S1, I should I apply to the current hand in order to maximise the overall EV?". Then, "If I were to apply S1 on the next hand, what is the strategy S2...." etc.

Does that make sense / sound reasonable?
link to original post



The approach you are describing has been used for years to develop poker theory - usually with some AI software together with a looping routine that runs the system several million times. See, for example, Modern Poker Theory by Acevedo.

However, poker decisions involve 'opponent behavior' and thus are not strictly optimizable by probability theory. This game does have the complication that how you play your current hand has implied effects on your effective value in the next hand, but I still think that combinatorial probability methods are the best approach here.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
LC1
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November 30th, 2021 at 10:57:16 PM permalink
Quote: gordonm888


The approach you are describing has been used for years to develop poker theory - usually with some AI software together with a looping routine that runs the system several million times. See, for example, Modern Poker Theory by Acevedo.

However, poker decisions involve 'opponent behavior' and thus are not strictly optimizable by probability theory. This game does have the complication that how you play your current hand has implied effects on your effective value in the next hand, but I still think that combinatorial probability methods are the best approach here.
link to original post


I think you have misunderstood what I was trying to say. See Charlie's reply to me - the approach I was describing does chiefly consist of combinatorial probability methods, and is essentially the same approach that he is taking.

That being said, I've been thinking some more, and have some new concerns/ideas which I will put in another post...
LC1
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December 1st, 2021 at 12:45:06 AM permalink
The double counting issue I've mentioned a couple of times is leading me to think I'm on completely the wrong track, and, by extension so might others be (since I thought we were on the same track!). [It is of course possible, likely even, that I have misunderstood some crucial detail of what's been described. Apologies if that's the case.]

What I hadn't previously considered is this -

Suppose we play two hands, A then B. We play B with a multiplier which we won in A.
If we assign an EV to B's blackjack bet, based on the multiplier, having also incorporated that EV into hand A ,then surely we have accounted for it twice! (The EV we assign to each hand includes both the winning of the multiplier and the playing of the multiplier.)

So I think we may need to do the following -

Use the sum of the hand EV and the carry-forward EV to rank the strategy decisions, but do not actually update the hand EV to include the carry-forward EV. (That way, the value of a multiplier is only counted when it is actually applied, but we make the right strategy decisions to exploit the multipliers.)

--------------------------------------

And, in general, the fundamental idea would be that only the amounts available to be directly won/lost on the current hand factor into the calculated EV for the hand. The additional EV that stems from carrying forward a multiplier to the next hand (and maybe beyond) only directly impacts the strategy decisions.


Does that make sense?

[Edit: On reflection, I'm not sure it does entirely. Seems like what is missing from the above is the need to determine the frequency with which each multiplier gets won, and use these to weight the contribution of the corresponding strategies to the total EV. Not sure if that can be done.

If I'm right that there is an issue regarding the double counting, maybe there are other, better ways to solve it??]
Last edited by: LC1 on Dec 1, 2021

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