Bets, bets and .. bets.

I had a couple of situations where I was not 100% sure if I could improve my bet or not after all, so I think would be a good idea to post them here as it was.

All cases are me BR2 last to act on the last hand and Bet Min/Max 100/1000 and Surrender allowed and re-split all except aces, plus double after split allowed.

1) BR1 2150 chips and bet 852. - BR2 2001 chips, did a bet of 550.

2) BR1 2200 chips and bet 802. - BR2 2001 chips, did a bet of 400.

3) BR1 2200 chips and bet 750. - BR2 2001 chips, did a bet of 350.

4) BR1 2200 chips and bet 499. - BR2 2001 chips, did a bet of 100.

5) BR1 2200 chips and bet 498. - BR2 2001 chips, did a bet of 947.

6) BR1 2150 chips and bet 500. - BR2 2001 chips, did a bet of 200. (think the correct bet range for BR2 is 163-201 for taking surrender low if BR1 surrender too)

- How bad would have been a bet of 650 or 325 for BR2 instead ?

7) BR1 2150 chips and bet 399. - BR2 2001 chips, did a bet of 100.

8) BR1 2150 chips and bet 398. - BR2 2001 chips, did a bet of 946. - A bet of 747 would have been a better choice for BR2 ? (and how bad was BR1 bet ?)

9) BR1 2150 chips and bet 397. - BR2 2001 chips, did a bet of 944. - But 745 instead for BR2 ? And how bad was this semi-trap for BR1 instead of 148 or 296 ?

 

The first few times I read this, my eyes glazed over due to the number of cases. I finally read your post in detail and was relieved to see that, in all cases, BR1 has given up the low to you. This makes the cases easier to analyse, since it is then given (I think) that you should take the low in each case. It just remains to compute the proper bet for each case.

In general, when surrender is available, and my opponent has given up the low, I like to bet the max such that my surrender beats his surrender, if possible. If it's not possible not then I simply leave more unbet chips than my opponent. Let's see if your examples reveal anything more complex.

Here, according to the principle above, I would bet 452. Your bet of 550 is enough to beat BR1 by doubling, but 502 would have been sufficient for that. How did you come up with 550?
edit: my arithmetic is bad. I would bet 552.

Similarly here, I would bet 600. I see no reason to bet any less.
edit: more bad arithmetic. I would bet 402

Here we agree on a bet of 350.

Here again, we agree on a bet of 100.

Here is the first case where you can not bet small enough for your surrender to beat BR1's surrender. However, in my opinion, your next priority is still to take the low. BR1 has left 1702 unbet, so I think that you should leave 1703 unbet and bet 298. A bet of 648 is enough to take BR1 high and take his double with your own double. If I were going high in the heat of battle, I would make the easy max bet. Can you explain how you came up with 947?

Once again we agree on a bet of 200. Any bet in the range 100-200 would be equivalent, but there is no reason to bet less than 200.

325 preserves the low at the expense of surrendering vs BR1's surrender, but doesn't gain you anything that I can see. If you're going to go high then, you should bet at least 650. Once again, in the heat of battle, on the final hand I would make the easy max bet.

Agreed.

Here you can only bet to surrender for a tie vs BR1's surrender. That still has value over just taking the low by betting 248 and I still believe that taking the low is still superior to any form of taking the high.

BR1's mistake was in giving up the low in the first place. He should have bet 248 or less. Since 248 doesn't get him anything on the high side, I think he should have bet 148 to keep the low outright.

Here you can not bet to beat BR1's surrender with your own surrender. Is this what you mean by a semi-trap? As above, I still think BR2 should take the low with a bet of 247. As above, BR1 should have bet 148.

 

1) Well, here I really do not understand why you would bet less than 502 ? Any bet in the range of 502-550 surrendered beats his 852 surrender, and also if doubled can take the high. A bet of 452 do the same for surrender, but can not take the high if doubled ? So I still not see why anything else than 502-550 ?

I`ve come up with 550 because I was thinking that if he gets bad cards he may be thinking at surrender and if he does not does his exact chips calculation and he actually surrender his 852 bet, he gives me a guaranteed win by 1 chip. If I would have bet 502, the difference would have been much more visible I think.

2) If BR2 bets more than 402 he will no longer be able to surrender in order to beat BR1 802 surrender. But yes, a bet between 501 and up to 600, doubled will beat BR1 single win ! But a bet of 600 doubled will not beat a BR1 blackjack, so what bet should be after all ? (I think that I am still pretty uncleared on this)

5) Yes I wanted to take the high, and 947 is the minimum that can beat BR1 blackjack win for his bet of 498. - But if a bet of 248 is better, I`ve got it, thanks !

6) I would say that 163 is slightly better than anything smaller, because with a 4 way bet (split + double etc) of minimum 163, BR2 can beat BR1 single win.

- I was thinking 650 because it gives me the straight high and also allow me to surrender for the low if BR1 stands. And at exact the opposite for a 325 bet.

 

Take a look at this old thread - https://www.blackjacktournaments.com/threads/6245.

I suppose if BR1 can surrender to greater than BR2's total (i.e. if BR1 has bet no more than 'twice the lead, minus a chip') then that makes a big difference, as BR1 can prevent BR2 from advancing via a push.

 

Sorry -- flaky arithmetic on my part. If I have it right this time, BR2 can bet up to 552 and still surrender to beat BR1's surrender.

sigh -- more bad arithmetic. My bet would be 402.

I actually said 298 but, given what Colin has posted, perhaps some bad advice on my part. This is a case where BR1 can surrender to beat BR2's push. Taking the high might be better. I trust S. Yama's work, but I will verify it by simulating. It's always nice to get the same result via two different methods.

OK, I can see this. It is technically better for that reason. In live play, however, unless you are very quick with your calculations, you can probably get away with only considering your opponent's 2-bet results. Anything more requires a successful split+double which is pretty rare.

I'm going to recheck my previous post to see if there are more errors.

 

No worries, I had a couple of mistakes too, in my posting. Actually you are right that 552 is the maximum BR2 can bet for case 1).

As for case 5) yes, you did wrote 298 ideed, but for some reason (hope is just tiredness!) I have read it 248 .. even if I did read it twice.

If you run a sim for case 5), please can you run a sim for case 6) all 3 BR2 bet responses (200, 325, 650) ? (I would like to see how big is the difference)

 

For case 5, taking the high is indeed the correct play.

BR2 response of 298: BR2 wins 42.68% of the time
BR2 response of 947 or more: BR2 wins 46.06% of the time

It also looks like BR1's ability to surrender vs BR2's push does make things quite a bit worse for BR2 than in S. Yama's example.

Sims for case 6 are under way.

 

My eyes glazed over too, and to be honest I haven't properly looked through all the examples, nor all your answers.

PlayHunter, I think it would be better if you just ask one question at a time, and wait until that has been dealt with as fully as possible before widening the scope with further variations. Otherwise it just gets too confusing (at least for me).

And I think it would also help if you provide a few words to describe the thought process that lies behind a question, rather than just the raw numbers.

I have never done the maths or run any sims, but my instinct has always been that if a bet can be made that offers two ways to win - surrender into the low or double into the high - then this ought to be superior to just leaving more unbet chips than the opponent. So I would bet enough so that my surrender beats BR1's unbet chips, rather than BR1's surrender total, if that is the only way to be able to double to a high enough total.

As far as I can see, the only drawback of routinely betting at least 2n-1 (where n is the bet you would make to take the low, either against BR1's current, unbet chip total, or BR1's possible total following a surrender) is that you immediately lose if the dealer has a blackjack. Otherwise if you surrendered 100% of the time you'd be in exactly the same position as if you just took the low in the first place. So you don't have to gain much from the extra leverage that betting nearly twice as much gives you in order for it to be worthwhile.

Sometimes it might be possible to cover BR1's possible surrender while still betting enough to be able to double into the high, but if not then the double seems more valuable to me.

I've also assumed, again without gathering the evidence, that this two-ways-to-win approach is better than taking the high (in situations where both are possible).

I'm a bit confused.

Surrender wasn't an issue in S. Yama's teaser. The surprising result that taking the high was slightly superior seemed to stem from the cases where BR2 will advance by pushing while BR1 loses. So the thinking behind my comment in my first post was that the impact of this factor will be blunted if BR1 has the ability to surrender to a total that prevents this, and hence taking the high might not then be superior for BR2.

But, of course, it is more complicated than that because the surrender option can open up new opportunities for both BR1 and BR2.

So at the moment I've got myself in a bit of a muddle, but I am thinking that, depending on the specific numbers involved, any of the possible BR2 reponses to a BR1 overbet might be best -

• Simply taking the high.
• Simply taking the low. (Against BR1's surrender total, if possible.)
• Taking a middle road, with a view to either surrendering or doubling in due course.

In the specific case of example 5, though, BR1 cannot in fact surrender to beat a BR2 push -

BR1 surrenders to 1951. So a BR2 push of 2001 wins. (BR1 would need to bet 397 to beat a push via surrender.)

And, to put my theory to the test, could you try a BR2 bet of 596 in your sim? That gives the option to either surrender or double.

 

Colin, yes I will try to come with only one (or maybe maximum two) at a time as I did until now .. but describing the thought process will be a bit harder for me ..

 

Case 6 Sims

BR2 response of 200: BR2 wins 51.03%
BR2 response of 325: BR2 wins 42.09%
BR2 response of 650: BR2 wins 45.33%

 

My answers are all based on the original principle of taking the surrender-low followed by the low, which has now been shown to be not always correct. So any of my answers could now turn out to be wrong.

Sorry. My bad again. After reading S. Yama's teaser I got it into my head that the overbet in case 5 was of the same nature. It turns out that surrender is not a factor at all, since BR1's loss beats even BR2's surrender with a bet of 947. I'm now guessing that the lower success rate of going high in case 5 (46.06%) vs the ~49% in S. Yama's example must be due to some loss of coverage of BR1's doubled/split hands.

Underway.

 

The rules in S. Yama's teaser did not allow surrender. I was able to come up with the approximate 49% figure just using Wong's table 4 - https://www.blackjacktournaments.com/posts/41978.
(though this didn't allow for the effect of optimal strategy variations by both BR1 and BR2.)

With no surrender, any overbet that presents a choice to BR2 of taking the low or the high is of the same nature. (But in practice, I guess an overbet will usually be to take the high away from BR2.)

Because the rules didn't allow it, I hadn't really noticed that surrender would make a difference, but it in fact would allow BR1 to beat BR2's push -

BR1 535, bets 40, surrenders to 515.
BR2 510, any bet, pushes to 510.

But there doesn't seem to be much, if any, benefit to BR2 deriving from the ability to surrender. If BR2 bets 20, rather than 10, all that means is that a double from BR2 could overtake a push from BR1 (550 to 535).

Thanks. If you are still in the mood, perhaps you could try the S. Yama teaser, with and without the surrender option, and with some of the BR2 bet sizes that were suggested.

 

BR2 response of 596: BR2 wins 41.48% of the time

 

I'm struggling to understand how that could be the case -

If BR2 just surrendered 100% of the time then the figure would not be much less than 41%. I'd have thought all the additional opportunities to double for the high would make a huge difference.

 

Just a theory ...

1. BR1's ability to surrender first really limits the power of BR2's surrender. i.e. BR2 cannot surrender at all if BR1 surrenders first. This is also true of both the 298 and and 947+ responses, so there really is no difference for BR2 with respect to surrender for all of these responses.
2. With 596, BR2's ability to double is not perhaps as strong as it appears, since BR1 can still beat it with a double of his own. BR2 also now must double to be BR1's single bet win.

Because of 1), the bet of 248 is then perhaps slightly better than 596, since BR2 can lose and still beat a loss by BR1

Because of 2), the bet of 947+ is then perhaps better than 596 since the high was given up in order to make possible a double which BR1 can still cover.

 

Some possibly flakey mathematics from me, but ....

I hadn't noticed that a BR1 double beats a BR2 double; that will certainly put a dent in the benefits of a 596 bet.

But I still find it hard to believe the 41.48% result -

We have a baseline figure for taking the low with a bet of 298: 42.68%. If BR2, perversely, decides not to take any of the benefit from the doubling opportunities, and always surrenders the 596 bet (effectively betting 298), then the only reduction in that baseline 42.68% figure comes from a dealer blackjack.

A ten-up blackjack is an automatic loss for the 596 bet, whereas it is an automatic win for the 298 bet. So that reduces the success rate to 42.68 - 2.37% = 40.31%.
(I've just realised the above calculation will be slightly out, since I haven't accounted for the times when BR1 and/or BR2 also get a blackjack. But it will be close enough, I think.)

An ace-up blackjack gets us into complicated insurance scenarios, for which I can't quite work out who has the edge for either size of bet, let alone how BR2's chances change from one to the other. (But I think BR2 may have a slight edge from acting after BR1??? )

If we assume that there is not much change in the success rate for the 298 bet versus the 596 bet when the dealer has an ace-up blackjack (or indeed just shows an ace, prompting BR1 and/or BR2 to consider insurance), then we have a success rate for BR2's always surrender strategy of somewhere around 40%.

If by playing optimally, doubling when that is best, the result for BR2 is only 41.48%, that means the ability to double is only worth around 1%. That doesn't seem right.

 

"effectively betting 298" isn't quite accurate because with a bet of 298, BR1 can still finish first by pushing while BR1 loses or winning while BR1 pushes. Always surrendering with a bet of 596 is equivalent to betting 298 and then always losing the hand. So the reduction from the baseline comes from more than just dealer blackjacks.

I can also tell you that my strategy generator does not yet consider insurance plays, so the 298 bet always wins and the 596 bet always loses for every dealer blackjack.
During the simulations, the dealer was dealt blackjack at the rate of 4.74%. Am I right in thinking that this represents a 9.48% swing in success rate?

I'm wondering now whether the extra reductions described above lower the expectation of the "always surrender strategy" enough to yield a more reasonable value of doubling?

 

Only the second of those two (winning while BR1 pushes) is a difference between 298 and a surrendered 596. But still, that's a good point, worth somewhere over 2%.

I'm not totally sure I've understood what you mean by a 9.48% swing, but I think the answer is no.

As I said, half of the 4.74% will be ten-up blackjacks, for which your sim will be performing correctly. When it comes to the ace-up, I've been struggling (and given up for the moment) to figure out the correct insurance strategies (whether to take it and then what to do next if the delaer does not have it) and the resulting success rates. I've blithely assumed that it won't be too far from 50:50 between BR1 and BR2, and that this won't change very much between the 298 and 596 bet sizes.

I've got myself thoroughly confused now, so that I am no longer sure what is reasonable, anyway!

I've gone back to basics and been consulting Wong. At first glance, he appears to contradict the S. Yama teaser and say that you should take the low if BR1 overbets (no surrender). [ P.69, 'BR2 on the Final Hand', and then Example 21. ] However, perhaps (as I think you suggested earlier) the unique point about the teaser is that BR2 can bet high enough to lock out a multi-bet win by BR1.

In Wong's Example 21 this is not possible -
BR1 540, bets 270.
BR2 500, bets 225 (leaving 275).

Interestingly, Wong says that if you play 'straight blackjack' in this example, you are a 53% favourite, whereas if you double when appropriate (the 'strong variation of Curt's Revenge) then you are still only a 55% favourite. So that does seem to indicate that the ability to double from the low to the high is only worth a couple of percent (at least when BR1 can pre-empt you by doubling first).

 

Just a brain cramp on my part. I thought that since one bet always wins and the other always loses, then a 4.74% rate of dealer blackjacks results in a differential of 2 x 4.74 = 9.48% between the two bets, but, as you have said, it's only 4.74% if insurance is not considered and probably something less if it is.

The rest of what you have written seems reasonable to me.

 

Playing on autopilot

Aside from the general academic interest, the reason I'm interested in this is that I've got into the habit of using the strategy I described, and it looks like I may have to either drop it altogether or see if there is a particular subset of scenarios to which it should apply.

For instance, I played the following final hand, despite what I have learned in this thread -

BR1 1800, bets 750, (leaving 1050).
BR2(me) 1451, bets 800, (leaving 651, surrender back to 1051).

Analysing it after the event, I can see that in this case -

• I couldn't have taken the high with a single max bet, so doubling is the only way to play for the high.
• I can't ever cover a double by BR1. (i.e. 2 * 1451 is less than the total BR1 can reach by doubling down.)

But presumably my best bet was in fact 400, rather than 800, and the success rates would be the same as for the previous example?

The only factor I can come up with in favour of my bet is a psychological one. The fact that I have (potentially) taken the low is disguised, meaning BR1 may be less likely to surrender. And BR1's big bet is suggestive of a concern with guarding against being beaten on the high side, which also hints that the possibility of surrender may not cross their mind (particularly if they can see that they currently have the low as well as the high).

As it happens, on this occasion I got away with it. BR1 did not surrender, I did, and the dealer obliged.