Math Experts please help!

Hi, I am new to this forum but I have taken the suggestion of a friend and decided to post a question that I have.

Here is the situation:

An online book is offering a regular blackjack game. There is shuffling after each hand, and I believe the rules are standard, so there should be around a .6% disadvantage against the player. Please correct me if this number is wrong.

The player keeps all wins, and at the end of each week, if a player has a net loss on the blackjack game, he is refunded 10% of his net loss for that week.

So if I win $500, and lose $600, I am down $100 for the week. The casino refunds me 10% of the $100, so I have really only lost $90.

The question is, with this policy of the 10% refunds on net losses, does this take enough away from the .60% disadvantage to make the game playable, or perhaps profitable? If so, is there any good strategy one can take here in order to maximize profits? If profitable, perhaps one should play less hands, at bigger stakes, in order to try to avoid a long term disadvantage.

Thank you for your replies.

Monkey

To solve a problem like that, you need to know your exact advantage.
How many decks is it? What "standard rules" are you offering?

Typically on a cash-rebate system like that, you want a fairly short play session with a high variance. If you can give me some more information, I'll try my best to give you the optimum amount to play, duration, and net EV.

The question is, with this policy of the 10% refunds on net losses, does this take enough away from the .60% disadvantage to make the game playable, or perhaps profitable? If so, is there any good strategy one can take here in order to maximize profits? If profitable, perhaps one should play less hands, at bigger stakes, in order to try to avoid a long term disadvantage.

You should play till you actually win a hand, then stop. You have almost a 10% advantage, so basically, given the low table maximums online, you should bet the max.

If they gave you a 10% rebate on every individual losing bet, this would be advantageous to the tune of 10% of the probablity of losing, which as you say is 50.3% (0.6% disadvantage off the top). So the refund would give you 5.03%. But this is not what they're doing. They're giving you 10% of every losing bet in excess of the winning bets, not 10% of your total action. Your losing bets are going to outnumber your winning bets by only about 0.6%/49.7% = about 1.21%. So ten percent of that is only 0.121%. That's all the rebate does for you.

Math experts, let me know if (or should I say "that") I am wrong.

They're giving you 10% of every losing bet in excess of the winning bets, not 10% of your total action. Your losing bets are going to outnumber your winning bets by only about 0.6%/49.7% = about 1.21%. So ten percent of that is only 0.121%. That's all the rebate does for you.

I'm not sure what you are saying here-are you assuming some minimum number of hands has to be played? I've reread Monkey's post and I can't seem to find any requirement.
It follows that the more hands you play, the more your losses will approach infinity. 10% rebate on an infinite loss is worth nothing. Working backwards it follows that the best strategy is to bet big until you win a hand.

Let's put it another way, if I offered you a toin-coss with a 10% rebate, I'd be offering you 5/4 odds on an even-money proposition. That is a lot better than 0.121% obviously.

The player keeps all wins, and at the end of each week, if a player has a net loss on the blackjack game, he is refunded 10% of his net loss for that week.

(note; above emphasis is mine, and I assumed the base game, played without the rebate, has an EV of -0.6%.)

(1) If your goal is to maximize your percent return on dollars bet (in a single account), then the optimum strategy is to make a single bet each week. The expected return for this strategy is approximately 4.4% of the total dollars bet. This win rate is estimated as follows (assuming a raw win rate (win-or-lose) as opposed to the complex expectations for doubling, splitting etc).

0.044 = (0.497*1.00) - (0.503*0.90)

(2) If your goal is to maximize your dollar return with multiple accounts, then you should follow strategy #1, using maximum sized bets, under as many different accounts as possible (i.e. bankroll your friends to bet for you). Again your expected return will be about 4.4% of the total dollars bet.

(3) If your goal is to maximize your dollar return in a single account, then you should make maximum bets until you have either won or lost a total of three initial bet units. Then you should immediately stop betting for the week. In other words, if you are betting $10/hand, you should immediately quit the first time you are either up $30 or down $30. (Note that in my simulation, using a limit of four bet units came out almost the same as using a limit of three bet units). Under this strategy, your long term weekly dollar win rate will be about 2.5x your weekly dollar win rate for a single bet each week (strategy #1).

Strategy #3 is based on a rough spreadsheet simulation. I'm sure someone with much more brainpower than me can give you a more exact answer, but this should get you in the ballpark.

Be aware of other key assumptions: The bonus/promotion never expires. Time value of money is zero. You play standard basic strategy. The online casino is fair. The online casino pays you upon request. The online casino does not have any mimimum play requirement.

Of course you should use Kelly bet sizing principles to maximize the return on your bankroll. My guess is that #2 is preferable over #3 from a Kelly standpoint (although #3 may be preferable from a practical standpoint).

... almost 10%! May is right about that. Don't know where he got his advice to play until you win one hand, though.

Think of a fair coin toss, instead of BJ, to simplify. At the zero point, your edge is .5x(+bet) + .5x(-bet)x.9 = .05xbet. A 5% advantage. Suppose you win (or lose!) that flip. You are then one bet from the lucrative zero point. You have a 50/50 chance to get back to the sweet spot. So your edge is .5x5% = 2.5%. If you are TWO bets from zero, edge is .5x.5x5% = 1.25% etc. Edge overall = (1 + 1/2 + 1/4 + 1/8 + 1/16 + ...) x 5% = 10%! Only problem being you can't, and don't WANT to (trust me) play an infinite length of time.

How long you decide to play is a function of how much you value your time. You have to subtract the 0.6% house edge (roughly) from above figures. There may be other good bonuses waiting for you on other sites, much better than the 0.65% x bet available when you're two bets from zero.

You definitely don't want to quit at zero, though. You want some definite result. If there's a minimum action requirement, you might consider playing small at the beginning (to keep variance low, so you will wind up at the end near the zero point) and slowly increasing your bet, like a gambler, as you near the end if you are within a couple of max bets from zero.

I wouldn't want you to play the minimum until action req is met, then suddenly increase to max bets. That would draw attention and could conceivably kick in the fabled "cheat switch," if there is one at this joint.

ETF

Think of a fair coin toss, instead of BJ, to simplify. At the zero point, your edge is .5x(+bet) + .5x(-bet)x.9 = .05xbet. A 5% advantage. Suppose you win (or lose!) that flip. You are then one bet from the lucrative zero point. You have a 50/50 chance to get back to the sweet spot. So your edge is .5x5% = 2.5%.

If you win one unit you would be ahead. If you then bet the same amount, which would take you back to zero, I don't see how the loss rebate is worth anything, since the value of the rebate for losing nothing is nothing. I thought this would leave you with a .6% disadvantage.

It is possible I am misunderstanding how exactly the rebate works, but I can't see why offhand.

If you win one unit you would be ahead. If you then bet the same amount, which would take you back to zero, I don't see how the loss rebate is worth anything

You plan to lose at +1 unit every time? Should you get to zero, you plan to just stop??

Actually, your first post didn't say anything about winning one unit, you said: "You should play till you actually win a hand ..." which is a very different thing. But this new interpretation doesn't make any sense either.

I'm sorry, you just don't realize how foolish you sound when you try to post about math. Your title shows you literally don't know what EV means.

MONKEY, please listen to Fat Freddie. He's saying basically the same thing I am.

ETF

What the above shows is that, for infinite play, the loss bonus is worth 10% of a single bet, assuming all bets are equal. It's not worth 10% of total action -- that would be TGTBT.

ETF

You plan to lose at +1 unit every time? Should you get to zero, you plan to just stop??

I have no idea what these sentences mean.
I am not intending to lose at +1. I am intending to stop playing.

Once again. When you are 1 unit ahead, a 1 unit loss leaves you with a net loss of zero. The rebate is worthless in that situation and you would be making a negative EV wager. This is not desireable.

If you are recommending some other course of action then please explain it, but I sincerely doubt any one can draw a practical strategy from the abstract theoretical comments you have made.

Thanks for the input, you guys know some good math.

The casino tells me that there is no minimum number of plays or action requirement for the 10% rebate. Based on the info I have read and assumed myself to be logical, it sounds like the way to go is to play few hands, and bet the highest amount that one's bankroll will permit.

May wrote that if you win your first hand, any consecutive plays for THAT WEEK will be negative EV. There is truth to this, as in your first hand you are risking .90 units to win 1 unit, which is +EV. Once you are +1 unit, then you are really risking 1 unit to win 1 unit, which is slightly negative EV.

What has to be taken into account is the fact that in BJ, most of the +EV comes from the ability to hit BJ which pays 3:2. This means that you will likely be losing a significant majority of your hands (40-42% maybe), and getting enough BJ hands to overcome your losers.

My strategy would be this:

1. Bet 1 hand, if it is a winner, quit for the week. Or, as someone suggested, try getting a friend/s to bankroll you on this.

2. Bet 1 hand for $100. If it wins, you are up $100 (or up $150 if you hit BJ). At this point, if you're up $200, you could risk $200 on your next hand. If you win, you are up $300, lose you are down $100 but in reality you are down $90 if you factor in the rebate.

This means we are getting +EV on each bet, since you are risking $90 to win $100 on hand #1, and then risking $190 to win $200 on hand #2. As I believe ET Fan said, your EV goes down with each hand you play each week. Bet #1 gives you the most edge, and if you keep doubling as you win, then your EV continues to be there, just not as much.

Ideally, we would like to play 2 hands, win them both, and then quit, as it might be risky to win $300 and risking it all plus more again, even though it is technically +EV to do so (unless I am incorrect about something).

Personally, I feel like the best way to go is to play 1 hand, and quit, regardless of what happens. If you win, you're up $100 (or possibly $150), lose, youre down $90. If there is another way to play which allows your bets to be +EV regardless, please explain it to me, or reiterate what you've already said, if you have already explained it. Remember, if you lose, you're down $100 UNTIL the rebate has cashed. If you continue to play after losing one hand, you're down $100, not $90. I don't know if there's a simple way around this, except to play just 1 hand.

Monkey

"I'm sorry, you just don't realize how foolish you sound when you try to post about math."

There must be something about John May that gets your hide. And in a bad way too!

Because you just don't realize how mean you sound when you try to comment about anything that May says.

Casino Operations Management by Kilby and Fox has a large section on just this topic. Only it is viewed as it's effect to a casino. In a nutshell, it states that the two deciding factors for the edge of a rebate program is length of play and volatility of betting. They use Baccarat as an example and say that with a house edge of 1.06% a player flat betting 50 hands with a 10% rebate only gives the house an edge of .48%. 100 hands is .63% and 1000 hands is .88%. Note that according to their chart only a 20% rebate and play of less than 50 hands will be positive expectation (.98%).

Hadn't heard of the book...is it one of those industry-orientated $200 type things?
With baccarat you can theoretically lock in a certain profit on this deal with a confederate betting on player while you bet on banker. Do they mention this?

and May seems to have the same problem whenever he talks about Grosjean.

BASIS FOR ANALYSIS

Analyzing this game m terms of EV (expected value) of individual bets can be difficult, since the EV of each bet is variable, and can be influenced by the outcome of the previous bet (and so forth back through time). Unless one is making single bets (or a limited and well defined number of bets), then it probably makes more sense to analyze the EV of the overall strategy (I'll call this EVS = Expected Value of Strategy - %) rather than the EV of individual bets. This is the approach used below

YOUR PROPOSED STRATEGY

You correctly pointed out that to maximize EVS, you should make a single bet and then quit for the week. In this case your EVS is

EVS = (Proportion of Gains x Magnitude of Gains) - (Proportion of Losses x Magnitude of Losses)

Assuming a basic strategy EV of -0.6% and a rebate of 10%, then
Proportion of Gains = 0.5 - (0.006/2) = 0.497
Proportion of Losses = 0.5 + (0.006/2) = 0.503
Magnitude of Gains = 1
Magnitude of Losses = 1 - Rebate = 1 - 0.1 = 0.9

So the EVS equation becomes
EVS = (0.497 x 1) - (.503 x 0.9) = 0.0443

This is how I obtained EV = 4.4% in my previous post. Please note that I conveniently left out the complicated payouts for blackjacks, doubling, pair splitting etc. Also, I assumed no re-betting of pushes, although it probably would be appropriate to do so. Overall I believe the above equation is a good approximation.

A high EVS is great, but it is just a percentage, and you can't put a percentage in your wallet and spend it. So we need to convert EVS into dollars. This is sometimes called Expected Win (EW), where:

EW = EVS x Dollars Risked

Let's assume you bet at the table limit is $100, and you follow the above strategy. Then your weekly expected win would be:
EW = 0.0443 x $100 = $4.43

AN IMPROVED STRATEGY

If your bankroll relatively small, (i.e. you cannot bet above the table maximum), then the strategy above is optimal, since it maximizes both EVS and by extension maximizes EW. However, if your bankroll is larger, then there are strategies that result in a higher EW, even though they have a lower EVS. This is achieved by increasing the weekly dollars risked to a value greater than the table limit.

I'll walk through one example which is probably close to optimal, and which I described in my earlier post. Let's asssume you follow the following strategy: once a week you make repeated table limit bets until you are either up by 3x the table limit, or down by 3x the table limit. At that point you immediately quit for the week. If the table maximum is $100, then you are effectively betting $300 each week.... either you win $300 or you lose $300 (let's call this strategy "Stop at +3 or -3"). Based on my simulation (these are not exact numbers), the proportion of weekly wins and losses under this strategy would be approximately:

Proportion of wins = 0.493
Proportion of losses = 0.507

Therefore, going back to our original EVS equation, and using the above values:
EVS = (0.493 x 1) - (0.507 x 0.9) = 0.0367

So you only win about 3.7% under this strategy ("Stop at +3 or -3"), as compared to 4.4% under the single bet strategy ("Stop at +1 or -1). However you are now betting $300/week rather than $100/week. So your weekly expected win increases to:

EW = 0.0367 x $300 = $11.01

In other words, you are now winning $11.01/week with this improved strategy, rather than $4.43/week under the original strategy. This improved strategy ("Stop at +3 or -3") probably gives close to the optimal win rate. My simulations are showing that "Stop at +4 or -4" gives a slightly lower weekly expected win rate, while "Stop at +5 or -5" and "Stop at +2 or -2" both give you a significantly lower win rate. For reference, you will play an average of 9 hands per week under "Stop at +3 or -3", and an average of 16 hands per week using "Stop at +4 or -4"

It is possible you could get a higher win rate using an asymmetric strategy (i.e. Stop at +3 or -4), but I have not analyzed that. And of course all the assumptions described in my earlier post apply.

Thanks for the concise analysis. My last question would be, what is the maximum point in which you feel like the EV is not enough where you want to keep playing for that week? Obviously being +3 units is a good # to quit, but I am wondering if I play until I get down 5 units or up 5 units, how much worse will my EV be by doing that?

Also it seems like there is a lot of variance by using the rebate strategy, is this true? In this example I will assume a player plays 3 hands, then quits. For example, if you lose 3 in a row one week, and quit, you're down 2.7 units. If you win 3 in a row the next week, you're up 3 units. This would give you a profit of +.3 units, obviously, a nice return for playing 6 hands.

However, if you win 2 and lose 1, you're up 1 unit. If you lose 2 and win 1 the next time, you're down .9 units for a total profit of +.1 units, 1/3 of what you would make had your results been had you simply won or lost your hands in a different order.

So does this variance affect the way one should strategize his play, or is it something that must simply be lived with? I like your strategy of going for a result +3 or -3 units then quitting, since by doing that you're still getting very good EV and increasing your EW significantly by not just playing one hand.

Thanks very much for your help.

Monkey

If you win one $10 bet, you must then lose 2 $10 bets to get a $1 rebate. When you are even, you are essentially betting $9 to win $10, a huge edge. As soon as you win your first bet this advantage seems to disappear. And when you are losing, when you win a bet you give back 1/10 of a bet in rebate, also no edge. But the trick is, it doesn't entirely disappear in one bet.

Fat Freddie simulated above, but I get the same answer with some quick calculations. We lose about 48% of the hands in bj. When you are even at this promo, if the house edge is .6%, you have an edge of about (.1 * .48)-.006=.042 or 4.2%. But, you've won your first bet. Do you make another bet? Well, you have a .6% disadvantage but their is a 48% chance that you will get to make a bet with a 4.2% edge on the bet that follows. Ignoring pushes for simplicity at the moment, the value of the next bet after you win is: .48 * .042 -.006= 0.01416 or +1.416%, still positive expectation. Should you now make another bet after winning 2 in a row? Well, the expectation would be .48 * .48 * .042 - .006= 0.0036768 or +.00367%. Yes, it is still positive expectation. Now, you've won 3 bets. Should you make another bet? The value is .48^3 * .042 -.006= -0.001355136. No, you should quit after getting 3 bets ahead.

Now, suppose you're losing. In that case the formula is similiar to above except that you have only a .44 chance of winning and making the +4.2% bet again. So when even your edge is 4.2%. After one loss:

.44 * .042 -.006=0.01248 Still positive expectation, so you bet again. After you are two bets down, you get:

.44^2 * .042 - .006=0.0021312. This is positive, so you bet again when down two bets. Now you are three bets down:

.44^3 * .042 -.006= -0.002422272. This is negative expectation so you stop there. When 3 bets down you quit.

As long as there is sufficient probability that you will get back to the zero point you continue to bet. Once you are up or down 3 bets it will require too much ev to get back there so you will quit. This agrees with the answer Fat Freddie obtained via simulation.

Once you are already losing, you are no longer betting $9 to win $10. Because even though you get back 10% of each additional bet you lose, you also give back part of your rebate each time you win. Say you've lost 4 bets and have .4 bets in rebate coming so you are really down 3.6 bets. If you win a bet you are really only winning 9/10 of a bet. You are now down 3 bets with .3 in rebate coming, so you are essentially down 2.7 bets, compared to 3.6 bets prior to making this bet. Your win netted you .9 units. So you are essentially playing against the initial house edge at this point.