Hello,
In MGP's BJCA, with "Early surrender" and "American rules (OBO)" surrender EV against X or Ace is NOT -50%
How is it possible ?
Early surrender and ENHC or Late surrender and American rules (OBO) are OK
Phoebe,
Let's consider a particular case, so I can illustrate my point with numbers.
I ran MGP's BJCA for a SD game with ES (good luck finding THIS game!). When I checked the Composition-Dependent EV's for the hand of 8,9 vs. A, I got an Early Surrender EV of -0.257575757... or -17/66, if you prefer rational fractions.
So why didn't we get -0.5?
The answer is that the EV's shown for American-style rules are traditionally the "No Blackjack EV's": in other words, they're the EV's if the dealer does NOT have BJ. The reason for showing the NoBJEV's is that, with American-style rules, these are (typically) the EV's once the player gets to act on his hand. However, with ES the player gets to act BEFORE the dealer checks for BJ (just like in a ENHC game).
So, how are the EV's before and after checking for BJ related? Here's the equation:
EV = -1*Pdbj + NoBJEV*(1-Pdbj)
where "Pdbj" is the probability that the dealer has a BJ, "NoBJEV" is the EV once we know the dealer does NOT have a BJ, and "EV" is the EV BEFORE the dealer checks for BJ.
If we rearrange to solve for NoBJEV, we get:
NoBJEV = (EV + Pdbj)/(1-Pdbj)
For our case, EV = -1/2 (obviously), and Pdbj = 16/49 (since three non-X's have been removed from a single deck). Plugging these into the equation gives:
NoBJEV = (-1/2 + 16/49)/(1 - 16/49) = -17/66 =-0.257575757...
which is the value reported by the software.
Hope this helps!
Dog Hand
Thank you.
It will help Don, too.
Although it's a bit difficult for me to understand this convention even for ES as dealer can have BJ when you surrender (or not)
And it led me to play 7 vs Ace in Fun21 incorrectly until I generated indexes in CVData
Last edited by Phoebe; 06-09-2019 at 05:02 PM.
I checked this when we had our private discussion. But, for the special deck you described, and the missing tens, the probabilities didn't confirm the standard calculations that Dog Hand described. So, I concluded that something else had to be going on.
For a discussion on the methodology, see also BJA3, pp. 392-393. I tried this, but it didn't work for your situation. Don't know why.
Don
Basically, my problem was an EV for surrender different from -50%
When I surrender I expect to lose half of my bet.
Whew ! (in reality I made "ouf !")For our case, EV = -1/2 (obviously)
"EV" displayed by MGP's BJCA is a pseudo (IMHO useless) EV
Last edited by Phoebe; 06-07-2019 at 01:46 PM.
The number you want is 75%, and it is more correctly stated as "If you lose 75% of the time that you don't push ..." (since the EV of a push is zero, you must remove them from consideration).
This rule of thumb applies only to hands that you don't double or split. For those hands, the ratio of wins and losses is not useful, since they can be for different amounts. Of course, the universal criteria is EV <= -50%.
As others have already explained, in this case the EV reported by MGP's CA isn't wrong, it just requires interpretation. And I have no reason to suspect the accuracy of any EVs reported for any integral number of decks. However, as a side effect of another analysis, it's worth noting that there do appear to be some errors in the EVs reported by this CA for some depleted shoes. Unfortunately, since there isn't a nice scriptable interface to his CA to allow automated "searching" for discrepancies, I only have some anecdotal examples that yield incorrect results, and those anecdotes are all what I would describe as "pathological" . For example, consider a shoe with two aces, two 6s, and a bunch of 10s (the exact number doesn't seem to matter). I suspect that the bug is in the pair split portion of the algorithm, but I don't have any more insight at this point. I'll let you know what I find.
Question: would these "errors" be the result of how MGP et. al. computed split hands? If so, I was under the impression that MGP and Cacarulo had found out the optimal splitting algorithm to get exact results for all in-order split hands.
Now that I think about it: Certain split hand are dependent on the deck subset. Take for instance [22] vs 6; *assume SP1:
We get two hands: {2x, 2x}, we cycle though all x's from Ace to Face (1-10).
We then take the weighted expectations for each new [2x] hand and multiply by two. That is Split_22_6 = 2 * (EV_max[2A] + EV_max[22] + ... + EV_max[2T])
What becomes an issue is that the right-most [2x] hand has an immediate effect on the outcome of the left-most [2x] hand. That is, the conditional expectation changes per card removed (and as an effect, changes the conditional probabilities for each dealer hand for up-card rank 6.) This would result in different weighted expectations and a different overall expectation for each split decision and overall strategy expectation.
Sorry for the tangent. Just popped into my mind and would be interested in what others thought.
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