2-1 win & 5-1 blackjack win PAYOFF

#1
I live in the UK and play the UK blackjack, 6 decks, S17, DAS, Respilt all, DA2, no hole card.
One local casino offers a promotion called the joker. For 1 game only, you get 2-1 if you win and it applies to doubles, but only goes on the first hand after splits. You get 5-1 if for a blackjack win. The joker remains when there is a push. Corrections

I have only starting gambling and playing blackjack for 2 months. I am a slow counter, but I am getting a lot of money from it. The maximum stake on it is £25 and use the full amount at all times. Maximum 12 jokers a week per person, sometimes more if I use my friends'. I guess the potential maximum would be about 20.

I know I would always win in this game. If I can ace track and card counting at the same time, I would make a fortune. In the last 6 weeks, I made £1008 just from that and I wasnt using all the jokers I could have.

1st Q - My simple maths, everytime I play the joker I would gain £12.5 approximately, times 20 a week equals £250. However I dont know how to work out the Epected return after doubling, splitting and blackjack. Can someone help me?

One downside is that most of the time only the CSM table is open at the promotion times but it is still good enough. I then follow the basic strategy offered by wizard of odds.

http://wizardofodds.com/blackjack/strategy/european.html

One day I was hit with a 5,6 vs T, The chart tells me not to (I think it is because dealer BJ takes all not only the initial bet) but I did consider to double given the high payoffs but I did not. I want to know the maths.
2nd Q - I was thinking I should generate a new strategy. I have casino verite but I cant do much with it. Can anyone show me a chart or tell me how to work them out from 1st principles or with a software.

3rd Q - how do I calculate Ev, RoR, SD, etc.. for myself. What is SCORE?
 
#2
If any win pays you 2:1, you'd be better off doubling less, much less. Because doubling decreases your chances of winning a hand; you won't get to take another card if you need one.
 

Nynefingers

Well-Known Member
#3
Automatic Monkey said:
If any win pays you 2:1, you'd be better off doubling less, much less. Because doubling decreases your chances of winning a hand; you won't get to take another card if you need one.
Is this correct? I would have thought we would double more since we are risking 1 more bet to win 2 more. Of course, I understand that we still have to compare the EV of doubling to the EV of hitting, and the EV of hitting is also higher than usual because of the 2:1 payout, but I would expect the EV of doubling to improve dramatically as well.
 
#4
Nynefingers said:
Is this correct? I would have thought we would double more since we are risking 1 more bet to win 2 more. Of course, I understand that we still have to compare the EV of doubling to the EV of hitting, and the EV of hitting is also higher than usual because of the 2:1 payout, but I would expect the EV of doubling to improve dramatically as well.
OK, the OP is not clear if the 2:1 bonus applies to the doubled portion of the bet or just the original bet. It only applies to one hand on a split and has a GBP 25 max stake so my assumption would be that the maximum amount that receives the bonus is GBP 25. If this is the case then you'd be getting paid 3 bets for your 2 instead of 4.
 

Nynefingers

Well-Known Member
#5
Automatic Monkey said:
OK, the OP is not clear if the 2:1 bonus applies to the doubled portion of the bet or just the original bet. It only applies to one hand on a split and has a GBP 25 max stake so my assumption would be that the maximum amount that receives the bonus is GBP 25. If this is the case then you'd be getting paid 3 bets for your 2 instead of 4.
You and I must be reading this statement differently:

For 1 game only, you get 2-1 if you win, it applies to doubles and the promotion goes on the first hand after splits.
It looks to me like it explicitly states that the promo does pay 2:1 on the double amount also. Maybe the OP can clarify?
 
#9
pomkon said:
What about 10 v 10?? How do I work out the percentages?
Just one more time, to get it clear in my monkey mind: you bet £25 on a hand and you double it, for a £50 total bet. If you win, the dealer will pay you £100 out of his rack, not £75. And if you push, you keep your money but lose the joker. Check?

If so, you'd want to get as much money on the table as possible in most circumstances. You'd probably want to double hands like 6 vs. 6, and possibly even hands like 12 vs. 6.

Here's how you do the math: you make a spreadsheet for every possible 2-card vs. upcard hand, and using a simulator like CVData generate the win rate for each hand without doubling any hand as well as the probability of winning a hand without doubling. Double the cash value of the wins alone and add it to the win rate. Repeat the spreadsheet, this time with doubling every hand. Compare the two modified win rates for each hand to get your strategy. Being CVData doesn't allow for doubling hard hands it might be a little tricky; you might have to fake it by setting up a complicated bonus structure that pretends soft doubles are hard doubles. This part might be easier with other software tools. But the promotion is stable enough the work is surely worth it for you.
 

Nynefingers

Well-Known Member
#10
Partial solution

OK bear with me on this. I may be way off, but I think I found a way to get an idea of the correct double strategy for at least some hands. If anyone knows where I can find the push probability for each combination of player hand and dealer upcard, for hitting and for doubling, then I think I have a complete solution. At the very least, coming up with push percentages may be easier than what was proposed by Automatic Monkey (although maybe not by much). Anyway, hope you can follow along with my thought process here.

Initially, I thought AM's method was really the only way to go on this, but I thought I remembered seeing a chart with the necessary EV's on the Wizard of Odds site. Unfortunately, Appendix 1 only contained the necessary data for the usual case where bets are paid 1:1 on a win. I then wondered if pushes were consequential. Using the EVs given in the tables, we can calculate the probability of a win or loss, conditional on not pushing. With those probabilities, it seemed that it would be a simple matter to double the payout on the wins and calculate new EVs for the double payout case. Unfortunately, the p_push term would not go away. If we push 20% of the time, win 40%, and lose 40%, that's no different than pushing 10% and winning and losing 45%. Both are break even, but clearly they are different when you double the winning payouts. So. Back to the drawing board. My goal at this point was just to find a formula to relate the EV with the double payout case to the normal single payout EV. I was able to find the formulas, but the p_push term is still there. Here is my math up to this point:

Code:
probability of a push = p_push
probability of a win, conditional on not pushing = p
probability of a loss, conditional on not pushing = q
expected value for case where wins pay 1 unit = ev1
expected value for case where wins pay 2 units = ev2

p+q=1

so:

ev1 = (1-p_push)*(1*p - 1*q) + 0*p_push
ev1 = (1-p_push) * (p - q)
ev1 = (1-p_push) * (p - (1 - p))
ev1 = (1-p_push) * (2p - 1)
ev1 / (1-p_push) + 1 = 2p
ev1/(2*(1-p_push)) + 1/2 = p


ev2 = (1-p_push)*(2*p - 1*q) + 0*p_push
ev2 = (1-p_push)*(2p - q)
ev2 = (1-p_push)*(2p - (1 - p))
ev2 = (1-p_push)*(3p - 1)

substitute:

ev2 = (1-p_push) * [3 * (ev1/(2*(1-p_push)) + 1/2) - 1]
ev2 = (1-p_push) * [1.5*ev1/(1-p_push) + 1.5 - 1]
ev2 = (1-p_push) * [1.5*ev1/(1-p_push) + 0.5]
ev2 = 1.5*ev1 + 0.5 * (1-p_push)
While this was worked out for the hit case, it turns out that this formula is identical for the double down case.

Code:
ev2_h = 1.5*ev1_h + 0.5*(1-p_push_h)
ev2_d = 1.5*ev1_d + 0.5*(1-p_push_d)
We know that we will double if ev2_d>ev2_h, which can also be written as ev2_d-ev2_h>0. I start with taking the difference in the EVs:

Code:
ev2_d - ev2_h = 1.5*(ev1_d-ev1_h) + 0.5*((1-p_push_d)-(1-p_push_h))
ev2_d - ev2_h = 1.5*(ev1_d-ev1_h) + 0.5*(1-p_push_d-1+p_push_h)
ev2_d - ev2_h = 1.5*(ev1_d-ev1_h) + 0.5*(p_push_h-p_push_d)
Then I plug that into the inequality:

Code:
ev2_d-ev2_h > 0
1.5*(ev1_d-ev1_h) + 0.5*(p_push_h-p_push_d) > 0
0.5*(p_push_h-p_push_d) > -1.5*(ev1_d-ev1_h)
p_push_h - p_push_d > -3*(ev1_d-ev1_h)

So now we are at a point where we need to know some more about the push probabilities. I am assuming that we will never hit hard 17 or higher, although I haven't compared the EVs for that. So any single hit card that gives us a 17 or higher will give us the same push probability regardless of whether we hit or doubled down to get that card, since we will never want a second card. On the other hand, if we draw a small card after doubling, we will never push, while if we had hit, we might push after drawing a second card. That's an ineloquent way of showing logically that p_push_h>=p_push_d for all hard totals against any dealer up card.

For soft hands, I can't say that for sure (although I'm confident it is still true). If we get a soft 17, we will always want another card. That might result in eventually busting. The same is true with a soft 18 vs. 9, T, or A. I believe that p_push_h>=p_push_d is still probably true for soft totals, however, since we are more likely to end up with a stiff than with soft 17 or 18. That means that hitting a second time would increase the odds of a push.

I say all of this to say that p_push_h-p_push_d>=0 for at least all hard totals and probably most soft totals as well. Possible exceptions might be soft 17+.

If p_push_h-p_push_d>=0, then p_push_h-p_push_d>-3*(ev1_d-ev1_h) is certainly true for any positive value of ev1_d-ev1_h. This means that any hand that is ordinarily a double, is still a double. The only time a hand would change from a hit to a double would be if hitting is ordinarily only a little better and the frequency of pushes is substantially less on a double down than on a hit. Some hands can clearly be ruled out, such as the previously suggested 6v6. In order to double that hand, we can calculate how high the push probability would have to be for the times we decide to hit.

Code:
p_push_h-p_push_d>-3*(ev1_d-ev1_h)
p_push_h-p_push_d>-3*(-.2819-(-.013))
p_push_h-0>-3*(-.2819+.013)
p_push_h>-3*(-.2689)
p_push_h>.8067
Since we know that the probability of eventually pushing when we hit our 6 against a dealer 6 is nowhere near 80%, we know we should not double this hand. Similarly, many doubles can be easily ruled out.

This won't give a final strategy, but it may at least reduce the amount of work necessary by giving a definite yes (double all normal doubles) or no (as shown above) for many decisions. I hope this helps, or at least gives you something to think about.

Now....do any stand/hit decisions change?
 
#12
Automatic Monkey said:
Just one more time, to get it clear in my monkey mind: you bet £25 on a hand and you double it, for a £50 total bet. If you win, the dealer will pay you £100 out of his rack, not £75. And if you push, you keep your money but lose the joker. Check?
Sorry that this is not clear. The joker remains after a push.

Nynefingers said:
Damn...after all that, I just realized this is a ENHC game... :flame:
Thanks for solving it partly and possibly doing it again. However do you input the code into some softwares to analyse?
 

miplet

Active Member
#13
Automatic Monkey said:
...
If so, you'd want to get as much money on the table as possible in most circumstances. You'd probably want to double hands like 6 vs. 6, and possibly even hands like 12 vs. 6.
...
I agree with AM. Doubling a hard 12 vs 6 has a greater ev ( about 0.44)than standing (about 0.27)
Double: win (39%) you get $100, lose (%56), you only lose $50.
Stand: win (%42) you get $50, lose (%58), you only lose $25.
Hope that makes sense.
 
#14
miplet said:
I agree with AM. Doubling a hard 12 vs 6 has a greater ev ( about 0.44)than standing (about 0.27)
Double: win (39%) you get $100, lose (%56), you only lose $50.
Stand: win (%42) you get $50, lose (%58), you only lose $25.
Hope that makes sense.
D= -17%
S= -16%

It seems like standing is better to me...
 

miplet

Active Member
#15
pomkon said:
D= -17%
S= -16%

It seems like standing is better to me...
That would be true if you were getting only 1 to 1, but you are getting paid 2 to 1 on winning hands.
For a $25 starting bet:
Code:
Double 12 vs 6:  win $100 39% = +$39
                lose $50  56% = -$28
                       total  = +$11

Stand 12 vs 6:   win  $50 42% = +$21
                lose  $25 58% = -$14.50
                       total  = +$6.50
 

Nynefingers

Well-Known Member
#16
miplet said:
I agree with AM. Doubling a hard 12 vs 6 has a greater ev ( about 0.44)than standing (about 0.27)
Double: win (39%) you get $100, lose (%56), you only lose $50.
Stand: win (%42) you get $50, lose (%58), you only lose $25.
Hope that makes sense.
It looks like you are correct. So far I have only looked at the hit vs. double decision, but not stand vs. double or stand vs. hit, etc. Where can I find win/loss/push percentages for each possible playing option? I know that CVData will give me the percentages, and I suppose I could create a playing strategy that always doubles every hand, another that always hits any hand, and another that always stand every hand, and another that always splits pairs, etc., and generate the data myself that way, but I'm sure it already exists somewhere. With that data, it's a relatively simple matter to find the optimal strategy for this game.
 

miplet

Active Member
#17
Nynefingers said:
... Where can I find win/loss/push percentages for each possible playing option? ...
I've searched hi and low and can't find any info online.

pomkon said:
...
1st Q - My simple maths, everytime I play the joker I would gain £12.5 approximately, times 20 a week equals £250. However I dont know how to work out the Epected return after doubling, splitting and blackjack. Can someone help me?
...
The joker is worth about £3.97 on the 5 to 1 BJ alone. You will have a winning BJ (and an extra £87.50) once in about 22 hands.
 
#18
miplet said:
I've searched hi and low and can't find any info online.


The joker is worth about £3.97 on the 5 to 1 BJ alone. You will have a winning BJ (and an extra £87.50) once in about 22 hands.
I dont know how to use the CV data, there are many options, especially I am not sure if they offer the ENHC rule. I want to know how to work out the EV and the strategy myself but I dont know what to do. Anyway the joker worths this little with a bet of £3.97?
 
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