Probability Dealer Busts

At what true count will the dealer break more than 50% of the time when showing a 4, 5, or 6? Three separate questions. Let's say it's an 8 deck shoe and the dealer stands on soft 17.

Quote: oboylere

At what true count will the dealer break more than 50% of the time when showing a 4, 5, or 6? Three separate questions. Let's say it's an 8 deck shoe and the dealer stands on soft 17.

If you are trying to count a side bet based on dealer busting, there are much more efficient counts than hi-lo. The easiest way to see it is to think about what the effect of removal of an Ace would be.

I actually just want to know if a hard 12-16 will ever yield a positive expected return based on the true count (using hi-lo as you mentioned) and dealer up card. Of course this is an equivalent statement to the original post. Essentially I want to know when it is profitable to "keep" a hard 12-16 if given the chance to get rid of it (like when the dealer makes a mistake and the supervisor gives everyone at the table the option to take back their original wager).

Quote: oboylere

I actually just want to know if a hard 12-16 will ever yield a positive expected return based on the true count (using hi-lo as you mentioned) and dealer up card. Of course this is an equivalent statement to the original post. Essentially I want to know when it is profitable to "keep" a hard 12-16 if given the chance to get rid of it (like when the dealer makes a mistake and the supervisor gives everyone at the table the option to take back their original wager).

I just did a simulation of 15 million shoes with 90% penetration, although I only dealt the dealer hands (so this does not take into account what cards may have been removed by player play, which could affect how likely each TC is), and I couldn't get any bust percentage better than 43% for any of the three up cards.

I appreciate that DonGuy. I think someone will have to run a similar simulation with a "high card rich" shoe. I don't have the ability to do that. My guess is that at a true count of +6, the dealer will break more than 50% of the time when showing a 5 or 6 (that guess is based on no real evidence). I'm curious to see if someone might be able to find the answer.

Quote: oboylere

I appreciate that DonGuy. I think someone will have to run a similar simulation with a "high card rich" shoe. I don't have the ability to do that. My guess is that at a true count of +6, the dealer will break more than 50% of the time when showing a 5 or 6 (that guess is based on no real evidence). I'm curious to see if someone might be able to find the answer.

I have a feeling one of us isn't quite understanding what the other one is saying.

I just ran another simulation of 25 million 8-deck shoes (again, just the dealer hands), in which there were about 2.86 million hands with a TC of +6 and the dealer's up card was 5, and another 2.86 million hands with a TC of +6 and the dealer's up card was 6. The bust rate was about 40% for the 5 up card, and 41% for the 6. In fact, all of the TCs between zero and +20 had a bust rate below 42% for both 5 and 6 up.

Ok I appreciate that as well. I am not going to say that your results are wrong, however intuitively they don’t make sense to me. Let’s look at what we know. 1) The dealer busts between 41.5% and 42.5% of the time when showing a 5 or 6 in an 8 deck shoe with a TC of 0. 2) If only 10 value cards remained in the shoe, the dealer would bust 100% of the time when showing a 5 or 6. Therefore I think we can safely assume that there’s a relationship between the true count and how often the dealer busts when showing a 5 or 6. I don’t know what the graph of this relationship would look like, however I am not convinced that it is constant between 0 and 20 TC and then makes a huge jump at say +25 or higher. Now I do realize that even if we removed all of the 2-6 cards, there would be plenty of aces, 7s, 8s, and 9s. But think about how having only aces, 7s, 8s, 9s, and lots of 10s would interact with the starting card (5 or 6). There’s a lot of combinations that lead to a dealer bust.

The numbers don't lie. Doesn't make sense? He must have a bad computer, huh?

Quote: oboylere

At what true count will the dealer break more than 50% of the time when showing a 4, 5, or 6? Three separate questions. Let's say it's an 8 deck shoe and the dealer stands on soft 17.

oboylere,

Here is a link to results for a 6D, H17 game with 75% penetration.

https://www.blackjacktheforum.com/showthread.php?18218-Dealer-Upcard-Probability-Bust-Rate-and-Bust-Frequency-vs-HiLo-TC&p=198590#post198590

The middle graph shows that at a HiLo TC of +9, the dealer's upcard of 5 or 6 will produce a dealer bust just over 50% of the rounds. For a 4 upcard, a HiLo TC of +13 is needed for the bust rate to exceed the 50% threshold.

I realize you asked about an 8 deck game, but the results will be substantially the same as for a six-decker.

Hope this helps!

Dog Hand

Let's take a look at just one of the situations (Dealer shows a 6), and let's assume that this is the last low card (2-6) in the deck. For simplicity, we will also assume that there are an infinite, but proportional, number of aces, 7s, 8s, 9s, and 10s (by proportional I mean that for every 8 cards, on average there's one ace, one 7, one 8, one 9, and four 10s). There's a 50% chance that the dealer has a 10 value card underneath his 6 and a 12.5% chance for each of the remaining possible cards (A, 7, 8, 9). We'll go through these 5 situations:

10 underneath: only an ace saves the dealer = 12.5% chance to make a hand
9 underneath: dealer needs two consecutive aces = (0.125)^2 = 1.5625% to make hand
8 underneath: needs a 7 or 3 consecutive aces = 0.125 + (0.125)^3 = 12.6953%
7 underneath: 7, 8, A7, or AAAA are the possible ways for dealer to make hand = 26.5891%
A underneath: hand already made = 100%

Now we can multiply all of these probabilities of making a hand by the corresponding probability of each hole card:

10: (50%)*(12.5%) = 0.5*0.125 = 0.0625 = 6.25%
9: (12.5%)*(1.5625%) = 0.125*0.015625 = 0.001953 = 0.1953%
8: (12.5%)*(12.6953%) = ....... = ........ = 1.5869%
7: (12.5%)*(26.5891%) = ....... = ........ = 3.3234%
A: (12.5%)*(100%) = 12.5%

Now add (6.25% + 0.1953% + 1.5869% + 3.3234% + 12.5%) which equals 23.8556%. So there's a 23.8556% chance the dealer makes a hand in this hypothetical scenario. This means that there's a 76.1444% chance the dealer busts. This infinite scenario isn't exact but it's extremely close to exact if you remove all of the 2-6 cards in an 8 deck shoe. Here's the point, at a zero count the dealer will bust roughly 42% of the time when showing a 6. We now have shown that if you remove all of the 2-6 cards the dealer will bust more than 75% of the time when showing a 6. If you removed half of the 2-6 cards, the shoe would represent a TC of somewhere between 0 and 20 (definitely closer to 20). Intuitively it doesn't make sense that if we remove the first half of the low cards, nothing happens to the chances of the dealer busting. But if we remove the next half, all hell breaks loose and the chances of busting goes from 42% to 75%. You're right, the numbers don't lie.

Thank you DogHand, this is what I was looking for.

Quote: oboylere

Let's take a look at just one of the situations (Dealer shows a 6), and let's assume that this is the last low card (2-6) in the deck. For simplicity, we will also assume that there are an infinite, but proportional, number of aces, 7s, 8s, 9s, and 10s (by proportional I mean that for every 8 cards, on average there's one ace, one 7, one 8, one 9, and four 10s). There's a 50% chance that the dealer has a 10 value card underneath his 6 and a 12.5% chance for each of the remaining possible cards (A, 7, 8, 9).

If the only cards in the deck are 7s through Aces and there are an equal number of each, then, for every 8 cards remaining in the deck, the running count is +5. This means the TC is 5 x (52 / 8) = +32.5.

However, like I said earlier, I am not entirely convinced that my numbers aren't skewed by the fact that I'm not taking actual players into account. If they all play the same strategy that the dealer does (hit on all hands 16 and lower, and stand on all hands 17 and higher; never split; never double), then it shouldn't affect the numbers, but of course, that's not what happens in reality.

Yes that’s possible. And yes if you take away half of the 2-6 cards as I referenced, the TC is around 15 and if you take away all of them, I believe your number of 32.5. Thank you for your help.

Quote: DogHand

The middle graph shows that at a HiLo TC of +9, the dealer's upcard of 5 or 6 will produce a dealer bust just over 50% of the rounds. For a 4 upcard, a HiLo TC of +13 is needed for the bust rate to exceed the 50% threshold.

I changed my method a little - this time, I always dealt from a full 8-deck shoe, but removed a number of low cards to get the desired TC (for example, for TC 6, I removed 48 2-6 cards from an eight-deck shoe) and just dealt one hand - and got numbers close to these. The 50% points I get with 8 decks and S17 are +10 for a 4, +7 for a 5, and +8 for a 6.

The thing is, I don't know what effect having various numbers of 7/8/9 cards, which don't affect the count, will have.

I only did a quick simulation using UK rules - so the dealer won't draw any other cards if the Player has a BlackJack (or has bust all hands - which can't happen vs 5 or 6). These hands are counted as not having bust. I used 10m shoes of six decks, 83% penetration, stand soft 17 and got figures using HiLo count of
(six) 0: 38% - 15: 45%
(five) 0: 34% - 15: 41%
I'll see whether a longer simulation gets anything different, especially as the numbers fluctuate so much at the higher counts due to small sample sizes.

I also suspect if the dealer hits soft 17 then the bust number would go up.

I am not sure that I am reading your results correctly. The way I read them is that when the dealer shows a 6, at a zero count dealer busts 38% of the time and at a +15 TC dealer busts 45% of the time. And then for a 5, 34% at TC 0, and 41% at TC +15. If this is correct, then unfortunately I will have to disregard any further findings derived from using this same program (at least on this issue). The reason is quite simple, the dealer busts over 40% of the time when showing a 5 or a 6 at true 0. Although your sample may not be large enough to correctly comment on higher counts (as you mentioned), it is certainly large enough to correctly comment on what happens in zero counts. Because the simulation says that the dealer busts only 34% of the time when showing a 5 in a zero count after 10 million shoes, I have to assume that the simulation isn't doing what we want it to do for this specific question.

^ I had the exact same worry as the figures should be about 41% and 42% and did explain it was "quick". I've found the error was when the dealer had an Ace but busted! Also remember there are about 4.83% of hands that were Player BJs, so these might have busted as well! So taking this into account (i.e. ignoring the BJ hands) you get...e.g. at Count of 0 .399 hands bust but .048 hands are player BJs, so .399/.952=.419.

Ok. I am still trying to understand where this error came from. Obviously it wasn't just the possibility of the player having BJ that increased probability of busting in true zero from 34% to 41.9% when showing a 5. Are you saying that when an ace was underneath the 5 and the dealer proceeded to bust, the sim didn't count it as a bust? I'm not sure what you mean by the ace comment. Also, are the results you just posted today new results?

Yes - my code added 10 to the dealer’s total if they was an Ace, but also had set it to zero (so player always wins).if the dealer had bust. Usually, unless you double 8 and get a 2, this didn’t matter; but here it did!

I reran the tests, a small run, this morning.

Ok, I'm still not quite following but I trust that you fixed it. These new results make sense.

With a six there is a larger chance of getting 17 because there would be more Aces around. Here is the smoothed figures, ignoring any player BJs, from a slightly longer run than this morning's run - there is still noise at the higher counts - but it shows the trend. You will notice the values for the count=0 match up with what one might expect.

Count	Bust	17	18	19	20	21
00.0	42.305%	16.600%	10.602%	10.629%	10.140%	9.724%
00.5	42.664%	16.643%	10.470%	10.519%	10.052%	9.653%
01.0	42.999%	16.683%	10.342%	10.418%	9.969%	9.589%
01.5	43.326%	16.725%	10.225%	10.316%	9.881%	9.526%
02.0	43.649%	16.783%	10.108%	10.206%	9.799%	9.455%
02.5	43.991%	16.833%	9.982%	10.106%	9.710%	9.377%
03.0	44.328%	16.876%	9.870%	10.001%	9.618%	9.306%
03.5	44.656%	16.922%	9.761%	9.894%	9.532%	9.234%
04.0	44.971%	16.978%	9.635%	9.795%	9.445%	9.176%
04.5	45.277%	17.072%	9.503%	9.686%	9.364%	9.097%
05.0	45.573%	17.150%	9.392%	9.583%	9.264%	9.038%
05.5	45.957%	17.200%	9.275%	9.457%	9.162%	8.949%
06.0	46.348%	17.218%	9.134%	9.324%	9.081%	8.896%
06.5	46.661%	17.281%	9.022%	9.226%	8.980%	8.829%
07.0	46.972%	17.375%	8.923%	9.107%	8.897%	8.727%
07.5	47.318%	17.480%	8.792%	8.990%	8.806%	8.615%
08.0	47.567%	17.562%	8.675%	8.929%	8.714%	8.553%
08.5	47.882%	17.558%	8.574%	8.850%	8.678%	8.458%
09.0	48.329%	17.644%	8.435%	8.699%	8.509%	8.383%
09.5	48.577%	17.867%	8.338%	8.583%	8.424%	8.211%
10.0	48.699%	18.006%	8.202%	8.460%	8.430%	8.203%
10.5	49.062%	18.074%	8.024%	8.340%	8.322%	8.178%
11.0	49.654%	18.167%	7.827%	8.235%	8.104%	8.013%
11.5	50.026%	18.198%	7.664%	8.135%	8.019%	7.959%
12.0	50.223%	18.259%	7.594%	8.012%	7.988%	7.924%

Yes these results seem pretty good. Actual expected return of a hard 17 vs 6 at true 0 is very close to +0.9%; your results indicate +1.2%. Actual expected value of a hard 18 vs 6 at true 0 is +28.2%; your results show +28.4%. Actual EV of 19 vs 6 is +49.5%; your results show 49.6%. This makes me feel pretty confident in this simulation. Even though the results may not be precise, I know they are a very good indication of what to expect as the TC increases. Thank you.