Recent content by ericfarmer

Thank you, Ken, for maintaining this extremely useful site. I have learned something new pretty much every time I came here.

Very interesting write-up. This is an important question-- rephrasing slightly, "How do we know that a 2 billion round simulation is good enough to calculate the expected value of a blackjack round?" I think the initial response must be another question: "How good do you want it to be?"...

Hmmm, I may have misunderstood your initial question, then. This sounds like you would like to *sim* a bunch of shoe compositions, but *compute* EV (via CA) for each of those shoes, and look at those trends. If so, then I think we are already on the same page; the blue points in the plot in my...

I think the last statement above is the key. That is, I think 10^5 samples is simply not enough to get a good handle on estimated expected value. If the standard deviation for a round is approximately 1 (and probably higher; WizardOfOdds tells me that it's 1.1418 for a 6-deck round), then even...

I tried something like this many years ago as something of a science project. I yanked the chunk of Americium out of a First Alert brand smoke detector, and placed it next to a radiation monitor ((Dead link: http://www.aw-el.com/)) that was linked to the serial port on my PC. Each detection...

I suppose I can (I'll take a look tonight)... but I think it would take a lot more than 1000 shoes to get a reasonable comparison. Your 0 pen example illustrates this most clearly; for the expected return it is just a single point, but for the actual return it will be a spread of values...

The data were generated using my blackjack CA (see here: https://sites.google.com/site/erfarmer/), with a wrapper to automate the Monte Carlo trips through the 1000 shoes. I made the plot using Mathematica.

Usually no more than 5-10%, but possibly as much as 25%. The attached plot shows the distribution of expected return (in % of initial wager) vs. penetration (in % of the 104 cards in the shoe), assuming CDZ- strategy with ridiculously optimistic rules: S17, DOA, SPL4, RSA, with surrender...

Your algorithm seems to be working fine. As you mentioned later in this thread, my latest version will compute both CDP and CDP1 for comparison. I think this highlights the challenge we encounter when trying to *optimize* a strategy, as opposed to the relative ease with which we can...

Although I do not know for sure that iCountNTrack's values are correct, remember that he is computing optimal strategy (i.e., perfect for the round), which may entail making a decision not to re-split at some point. We, on the other hand, assume that once we initially split a pair, we also...

Yes, that was very clear, thanks! I did a quick spot-check with a full single deck, where CDZ-, CDP1, and CDP actually differ (vs. dealer 2 is a good place to look for differences). We agree across the board, where your optimal strategy corresponds to my CDP1 option. Cool!

Ok, I think I see your point. I think we have a partial answer, though: it is important to remember that, in my and MGP's case at least (and I think for k_c's as well?), and for CDZ-/CDP/CDP1/CDPN in general, SPLn means that strategy is to always split at every opportunity up to n times, even...

Ah, I think I understand better now. And I think I can reproduce your results for those cases; this idea of taking a strategy optimized for one shoe and applying it to a different shoe is one of the things that I wanted to make easy in my CA. For example, following is what I get for CDBS-...

I get -0.29490884073 for SPL2; this option is not available in the pre-built executable, but it is available in the code; I can provide an example if anyone wants to know how to do this. (Actually, you can specify a different maximum number of hands for each split rank. I have no idea why this...

I think there are still some outstanding CDZ- problems as well. In addition to the (2,0,0,0,0,2,0,0,0,48) shoe mentioned above, I also see values from MGP that I don't think I agree with for the earlier shoe (1,1,1,1,1,1,1,4,4,8). The largest departure that I found was 10-10 vs. 7, yielding...