Loss Rebate sim Q for Fat Freddie

Hi FF,

I decided to start a new thread since the old one is running off the page.

I worked out a formula for the loss rebate problem. You mentioned your "approximate simulation" to MONKEY's problem. Could you tell me what was simulated, exactly, so I can compare my results to yours? Did you:

A) Sim actual blackjack hands with the stipulated rules
B) Sim a coin toss with an even payoff, but biased to lose with p=0.497
C) Sim a fair coin toss, but with unequal payoffs resulting in EV = -0.006, or
D) Something else?

Thanks in advance,
ETF

The other key limitation was that I used a single-purpose spreadsheet simulation, which basically choked my underpowered computer. So I could only simulate about 3,000 hands per run, and I averaged 10 runs for a total of 60,000 hands per result. This low number of hands surely resulted in some inaccuracy.

Later I thought of a way this could be done in closed form (without a sim) but my statistics are a little rusty, and I don't think the equation would be exact, (though it should be close). So I am very interested in seeing your results.

... (B) would have been my guess, based on your restuls for +1/-1, but I wasn't sure. I do have a closed form solution for (B). I want to double check my results, and get back to you tomorrow, ok?

ETF

For a biased coin that pays one unit when you win, and one unit when you lose, let p = probability of a win, q = 1-p = probability of a loss, B = the loss rebate per units lost, and G = a +/- Goal (player will quit once reaching +G units or -G units, but not before), then the expectation for the strategy is:

E(+G/-G) = G x ((p/q)^G - 1 + B)/((p/q)^G + 1)

It's easy to show algebraically that this is correct for G = 0, 1, and 2 (the last requires a little power series trick). I used induction to show it holds for all positive integers.

However, as usual, I've reinvented the wheel. I did a little research and can now produce formulas for E(+G/-H) with G<>H, and EVEN (hold your breath) for uneven payoffs, such as single number roulette bets. I'm not posting these results yet, because I'm not sure they've ever been connected to the loss rebate problem, so I'm taking my time, making lots of chicken scratches on 8x11s, following Gauss' dictum "few but ripe."

Here are some results, to six digits, for your parameters, p=0.497, q=0.503, and B=0.1:

E(+1/-1) = 0.044300
E(+2/-2) = 0.077201
E(+3/-3) = 0.098705
E(+4/-4) = 0.108816
E(+5/-5) = 0.107541
E(+6/-6) = 0.094886
E(+7/-7) = 0.070861

I also checked some other strategies, like E(+5/-4), but E(+4/-4) is in the lead so far.

Since my theory and your sims don't match perfectly, it's incumbent on me to do a sim. I'll keep you posted on that within a week or so.

Griffin has a formula, in "Extra Stuff" for E(n), so a fairly complete solution to the rebate problem is there, even with a wager requirement. Play only if E(n) is positive, and once you've waged the required amount, continue play only if you are inside the optimum +/- boundaries.

Complete solution EXCEPT for the dirty little detail that blackjack has multiple payoffs. The above formula can be adjusted to work when ties are allowed (p + q <> 1), but not for doubles, splits, etc. For that, I think a sim is needed.

ETF

I obtained very similar values coming from a totally different direction, a fact that strongly validates your equation and results. Your approach and formula are clearly more rigorous. My formula is based on the empirical observation from previous sims that the average number of hands played to reach the goal (G) is approximately G^2 (for sim results that led to this observation, see the post entitled "Table of Variations and Outcomes" from the previous thread on the Loss-Rebate problem. The top post in the thread is referenced below).

I've used your nomenclature where possible.

G = Goal (quit at achieving either +G or -G)
B = Rebate as a fraction of the loss
p = probability of a win
EV = Expected Value of Base Game = 2p-1
EVS = Expected Value of Strategy (i.e. playing to +G or -G)

Here is the approach I used:

Average Hands/week = G^2
Average Gain/week (pre-rebate) = (Hands/week)EV =(G^2)(EV) = (G^2)(2p-1)
EVS pre rebate = [Average Gain/week (pre-rebate)]/G = G(2p-1)
Fraction of Winning Weeks = 0.5+((EVS pre rebate)/2) = 0.5+G(p-0.5)
Fraction of Losing Weeks = 0.5-((EVS pre rebate)/2) = 0.5-G(p-0.5)
EVS post rebate = 1(Fraction of Winning Weeks) - ((1-B)(Fraction of Losing Weeks)
. . = (0.5+G(p-0.5)) - (1-B)(0.5-G(p-0.5))
Win per Week = (Amount risked per week)(EVS post rebate)
. . = G [(0.5+G(p-0.5)) - (1-B)(0.5-G(p-0.5))]

For the values we have been using (p=0.497, B=0.10), the above formula gives the following results:
E(+1/-1) = 0.044300
E(+2/-2) = 0.077200
E(+3/-3) = 0.098700
E(+4/-4) = 0.108800
E(+5/-5) = 0.107500
E(+6/-6) = 0.094800
E(+7/-7) = 0.070700

These results match your results very closely. The result for E(+1/-1) match exactly, then the differences increase to a maximum of 0.000161 for E(+7/-7). I suspect the difference arises because my formula makes the simplifying assumption that you play the same average number of hands during winning weeks and loosing weeks. This is not strictly correct, and the difference increases as G increases. This might explain why the difference between our results increases as G gets larger.

Your formula is more rigorous, and your approach is more general than what I did. I look forward to seeing how it can be applied to other situations. Congratulations on an excellent piece of work.

that, as foolish as I may be, that since you are unable to get past the coin-toss stage two weeks after the initial problem was posed, that maybe my approach had rather more wisdom in it than you suggested. The promotion may well have expired by now. The expected return from an expired promotion is zero.

I'd can't imagine a better illustration of the impotence of endless theoretical hair-splitting.

I'm SURE no one will ever offer such a promo again. But of course, YOUR approach was practical... stop at +1, and +1 only.

Like I said, every time you open your mouth. [ROTF!] Good cover, though. Making it appear to the casinos you reside on some other planet.

Thanks for the chuckle. :-) Have a nice day.

ETF

That's a pretty interesting alternate approach you came up with. I'll have to check to see if G^2 is the exact average hands/week. It is for p=0.5, I think. Maybe there's a self-correcting aspect to your two simplifying assumptions!?

I haven't even looked in Feller to see if he mentions the problem. I just wanted the fun of hacking it out. If it's not in Feller, he probably thought it was too obvious to bother. One of those space-wasting "quick consequences," you know.

L8R,
ETF

I'm also relatively sure that if p=0.5, then average number of hands (N) is exactly G^2. The sims suggest that when p~0.5, N is very close to G^2. However this probably breaks down as p approachs 1 (or 0), since when p=1 (or p=0) then N=G.

BTW, what is Feller? A book I should add to my collection?

"An Introduction to Probability Theory and its Applications", William Feller, 1950.

Wow, maybe the ghost of Feller reads this board!? (Post above)

Yes, Feller's is probably the most often referenced book on probability. There were several editions after the 1950 one.

ETF