Ask Mission

Honestly, you HOPE that you end up betting more than 2x your initial bet, because that is a very good sign for you. Part of what makes the EV in favor of splitting (rather than hitting) is the fact that it results in sometimes winning more on the hands that you do win.

The reason that this happens is because the splitting now sometimes leads to situations that are +EV. You should want to get into a situation where you have to double. You split 8's against a nine or ten and one of your eights draws a three, so now you can double against a dealer nine, which is a great situation for the player.

However, the calculator even says it's a split with no DAS (double after split), eight decks and only being allowed to split once. It doesn't really get any worse than that for 8-8 and is still better than surrendering and much better than just taking a card. The only way hitting becomes better is those same rules against a ten and the dealer does not check for Blackjack, so you can't eliminate the dealer from having an ace, and in fact, the other rules don't matter and you would still only take one card if the dealer doesn't check (unless you could surrender).

So, how can it be better than hitting with no DAS? Let's break this one down with minimal math:

Hit 16: You have eight ranks that bust you and the hand ends immediately. You have five ranks (A-5) where you make a standing hand.

Split 8's:

1. For each eight, you have six ranks where you make a standing hand (9, 10, J, Q, K, A), so your prospects for at least making a hand that can win or push a playing dealer hand are immediately better.
2. Each eight can draw a two or three, which even without DAS, gives the opportunity for a hit on those hands that literally can not bust and can lead to making hands that can win or push the dealer.
3. With exception to another eight, all the other ranks (4-7) lead to you having the opportunity to hit to a different hand total (12-15) that is less likely to bust than your sixteen was.

One other thing you can use the calculator to do is look at a player Hard Eight v. Dealer Ten for a close enough comparison. If you double the negative EV on the Hard Eight hand's best decision (hitting, obviously) it still comes out to a lower Expected Loss than does hitting sixteen against a dealer ten. Actually, doubling the -EV on hitting the Hard Eight has a lower Expected Loss (slightly) than does surrendering a Hard 16, but again, that's because you cannot immediately bust hitting a Hard Eight and also that Hard Eight can turn into a good hand to draw to, such as eleven.

Why this comparison? Well, I think the ability to DAS makes the decision obvious, but even if you couldn't...two hands that are each Hard Eight have a lower combined expected loss than one hand that would see you either hit or stand on 8-8. If you can't DAS, then you're basically splitting 8's to make two separate hands that are both Hard Eight.

So, I hope this explanation helps without getting too far into the weeds. I also don't like Blackjack, so Blackjack discussion isn't really my strong point.

Eight Decks: 416 Cards

Player 8-8: (Two Non-Ten Cards)

We will say the dealer's upcard hasn't come out yet. There are (16 * 8 = 128) ten value cards remaining of 414 total remaining cards. 414-128 = 286

nCr(286,3)*nCr(128,0)/nCr(414,3) = 32.8611% (Rounded)---Zero Dealer Tens

nCr(286,2)*nCr(128,1)/nCr(414,3) = 44.4319% (Rounded)---One Dealer Ten

nCr(286,1)*nCr(128,2)/nCr(414,3) = 19.7995% (Rounded)---Two Dealer Tens

nCr(286,0)*nCr(128,3)/nCr(414,3) = 2.90761% (Rounded)---Three Dealer Tens, even though that can't happen.

However, by the time this information actually matters (when it comes time for the player to make a decision) the dealer's upcard either is already a ten or is not a ten. For that reason, you either remove the ten (or not a ten), as well as any other known player cards, and do the math accordingly.

For one example, let's say the dealer's card is a nine.

Eight Decks: 416 Cards

Player 8-8: (Two Non-Ten Cards)

Dealer Upcard: 9

nCr(285,2)*nCr(128,0)/nCr(413,2) = 0.4756811396600766 or 47.5681% that the next two cards are not tens. It will be slightly more likely neither are tens if a ten is the dealer's upcard.

Anyway, none of this is really pertinent to anything. If the dealer's upcard is a nine, then 8-A all end dealer actions and we do not even see the dealer draw any cards.

Let's go with the dealer showing nine and look at more precise numbers:

1.) The undercard is already a ten:

128/413 = 30.9927%

2.) The undercard is 8, 9, or A and the hand ends anyway (Two eights and a nine are gone already):

93/413 = 0.22518159806 or 22.5182%

3.) The hole card is 2, 3, 4, 5, 6 or 7 and the dealer takes (at least) one other card, which is a ten:

192/413 = 46.4891% (Overall)

Ten: (192/413 * 128/412) = 0.14443216812 or 14.4432%

Not Ten: (192/413 * 284/412) = 0.32045887303 or 32.0459%

Therefore, in an eight-deck game and with the dealer showing a nine, the probability of the dealer either having a ten undercard OR having an undercard that would require the dealer to draw AND drawing a ten on the next card is: .309927 + .144432 = .454359 or 45.4359% (Based on rounded numbers)

A dealer could also end up with a ten in a sequence such as 9 + 3 (Undercard) + 2 (Drawn Card) + 10...or any other number of ways that I am not going to go through.

I hope that helps. Either way, it's less than 50/50 mainly because the dealer could have (not a ten) and still not have to draw any cards.

This is actually a simpler problem than you would think. The first thing that we want to do is look at this.

I'm just leaving the rules that are already in there the same. 6L5 has a House Edge of 1.78895% and 3:2 is 0.43096%, so:

1.78895% - 0.43096% = 1.35799% is the difference.

With that, we're going to take a trip back in time and concern ourselves with the Least Common Denominator, that way, we will index 6/5 and 3/2 to being the same thing. Let's go ahead and do that:

We start with the fact that our Least Common Denominator is 10, so that makes the adjusted fractions thus:

15/10 (3:2) and 12/10 (6/5)

The next thing we do is ask ourselves, "What would this adjusted fraction be for 9:5 Blackjack?" That's easy: It is 18/10.

We see that 1.35799% is the difference between 15/10 (3:2) and 12/10 (6:5) Blackjack, and relative to the Least Common Denominator of 10, the difference in our numerator (top number) is three.

The next thing we posit is that the probability of a winning player natural does not change, so any changes in the pays are unaffected by strategy.

Our 18:10 Blackjack goes three up in the numerator compared to the base game instead of three down, as a result, 1.35799% is subtracted from the House Edge.

0.43096 - 1.35799 = -0.92703 or 0.92703% Player Advantage

As we can see, it's the same thing as making it 6:5 Blackjack, just the other way.

1. This method would NOT work for most strategy-dependent casino games. The only reason it works here is because the probability of a winning player natural cannot be changed by strategy. This would work for adjusting pays on side bets, for instance, provided there are no strategy changes that would make a particular result more or less likely. For most side bets, there are no such changes as many of them are just based on whatever it is you are dealt.
2. But, I do hope that this was a useful example of how to use what you already know to get to what you want to figure out.
3. You also know now that, having indexed it to ten, every 1/10th unit pay up or down changes it 1.35799/3 = 0.45266333333...at least for the rules linked above. (Only number of decks will change this)

Someone later would confirm this with some kind of software, which I don’t think is as fun.

MY RESPONSE:

I think we can handle this with just a little bit of combinatorics:

The first thing I am going to assume is that two decks were cut off, which would be pretty high, but not altogether unusual. Therefore, that leaves us with the following:

Cards: 312

Starting Aces: 24

Finishing Cards: 104

Finishing Aces: 13

Okay, so here is the combinatorial formula we can use to determine that probability:

First, we determine the total number of ways to select 208 of 312 cards:

nCr(312, 208) = 8.46496974×10^84

The next thing we have to determine is the probability of selecting eleven aces as well as 197 other cards.

nCr(24, 11)*nCr(288, 197) = 1.32766836×10^83

Now, to get the probability we have to take the total number of ways the eleven aces can happen and divide it by the total number of ways that anything can happen:

(1.32766836*10^83)/(8.46496974*10^84) = 0.01568426587

Which I did using this scientific calculator, though you have to put it in a little differently.

That means that the probability of EXACTLY eleven aces coming out given 208 cards of 312 drawn is about 1.568426587%

The probability of EXACTLY eleven aces coming out given three consecutive trials is:

.01568426587^3 = 0.00000385826

That's about 1 in 259,184, but by itself, it doesn't signify anything. Let's instead take a look at all of the other possibilities that could have happened involving eleven, or FEWER, aces and add them together:

nCr(24, 10)*nCr(288, 198) = (4.79435795×10^82)/(8.46496974×10^84) = 0.00566376265

nCr(24, 9)*nCr(288, 199) = (1.44553506×10^82)/(8.46496974×10^84) = 0.00170766713

nCr(24, 8)*nCr(288, 200) = (3.61835495×10^81)/(8.46496974×10^84) = 0.00042745042

nCr(24, 7)*nCr(288, 201) = (7.45484894×10^80)/(8.46496974×10^84) = 0.00008806704

nCr(24, 6)*nCr(288, 202) = (1.24862569×10^80)/(8.46496974×10^84) = 0.0000147505

nCr(24, 5)*nCr(288, 203) = (1.67044557×10^79)/(8.46496974×10^84) = 0.00000197336

nCr(24, 4)*nCr(288, 204) = (1.74004747×10^78)/(8.46496974×10^84) = 0.000000205558617

nCr(24, 3)*nCr(288, 205) = (1.35808583×10^77)/(8.46496974×10^84) = 0.0000000160435993

nCr(24, 2)*nCr(288, 206) = (7.46168074×10^75)/(8.46496974×10^84) = 0.00000000088147754

nCr(24, 1)*nCr(288, 207) = (2.5702912×10^74)/(8.46496974×10^84) = 0.000000000030363856

nCr(24, 0)*nCr(288, 208) = (4.17054462×10^72)/(8.46496974×10^84) = 0.00000000000049268276

You can actually pop the second part into Google and then, when it became necessary, I used this converter.

Okay, so now we have to add all of these together to get the probability of eleven, OR FEWER, aces:

(0.01568426587)+(0.00566376265)+(0.00170766713)+(0.00042745042)+(0.00008806704)+(0.0000147505)+(0.00000197336)+(0.000000205558617)+(0.0000000160435993)+(0.00000000088147754)+(0.000000000030363856)+(0.00000000000049268276) = 0.02358815948

Okay, so the combined probability of seeing eleven, or fewer, aces under the circumstances described is 2.358815948%, so three times in a row:

0.02358815948^3 = 0.00001312448 or 1/0.00001312448 = 1 in 76,193.4949402

Well, it's MUCH more likely than a dealt royal, and one of those just happened at Downtown Grand recently.

Other Observations

1. This all assumes that two decks were cut off, when dealing with numbers like these, the number of decks cut off can substantially change the probabilities. It's also possible the casino cuts off more than exactly two decks, which would make this event more likely.
2. It assumes that your friend was counting the aces accurately, which may or may not be the case.
3. It assumes that the story is true.

Answer: The first card can be anything.

Six-Decks: (1 * 5/311 * 4/310) = 0.00020744736 or 1/0.00020744736 = 1 in 4,820.50

Eight Decks: (1 * 7/415 * 6/414) = 0.00024445608 or 1/0.00024445608 = 1 in 4,090.7143729

The reason it is more likely with eight decks is because the effect-of-removal isn't as great. Let's say the card is an Ace of Spades, the percentage of Aces of Spades to start the show is always 0.01923076923, or 1.923076923%, but with six decks:

5/311 = 0.01607717041, so 1.607717041% of the remaining cards are Aces of Spades after one has been taken away.

4/310 = 0.0129032258, so 1.29032258% are after two have been taken away.

With eight decks:

7/415 = 0.01686746987, so 1.686746987% are after one is taken away.

6/414 = 0.01449275362 so 1.449275362% are after two have been taken away.

Even small differences become pretty meaningful when the Ace of Spades becomes the specific card you want.

Follow-Up Question: There are 24 sevens in the shoe. The number of ways to arrange 3 sevens out of 24 is combin(24,3)=2024. The probability is the number of winning combinations divided by total combinations, or 2024/5013320=0.0004, or about 1 in 2477.

Answer: You didn’t specify sevens.

Anyway, why are you using combinatorics for this? A much blunter tool will get the job done.

There are 312 cards in the shoe, 24 are sevens, but once a seven comes out, only five other sevens match it.

(24/312 * 5/311 * 4/310) = .00001595748

1/.00001595748 = 1 in 62,666.5363203

You’ll notice that’s roughly the six-deck result in my previous post for any trips multiplied by thirteen, and that’s because there are thirteen ranks.

You had the right general idea with your combinatorics to figure out three sevens, but your equation did nothing to specify they be suited.

Lesson: Don’t Play Side Bets, They’re Sucker Bets!

3.) Video Poker

Vpfree2.com is a great resource to get an idea about VP paytables.

Obviously, you want to play as small a denomination as you can stand, preferably single-line, if that's not too boring for you. The flashy newer games tend to require bets of more than five credits per hand, put the Variance through the roof, and often have different strategies than the base game...many of them unknown.

My advice is to look at return tables for various games.

And print out a strategy sheet based on those paytables.

Unlike the other games, you completely dictate the pace at which you play Video Poker. There is also no pressure from anywhere to make a decision quickly, so you can take your time looking over your strategy guide to make the right play, if you like.

I would suggest maybe finding a general game that you like, whether it be Jacks or Better, Deuces, Jokers, Double Double Bonus...whatever...and practice with this.

The reason why is so that you learn some general concepts for the game you like, that way you don't have to look up every single play all the time. While smart if you don't know them, that quickly gets boring. Feels more like work than fun. Best for 95%+ of your holds to essentially be automatic.

The lesson here applies to both Land Casinos as well as Online Gambling. The long and short of it is that, if you want your money to last for a particular amount of time, then you can look at your expected loss per hand (or per hour) by determining the minimum bet multiplied by the House Edge to determine an expected loss.

Using an online gambling example: If I wanted to deposit $100 at an online casino and play 9/6 Jacks or Better, then I can determine the return is:

0.99543904

Therefore, you expect to lose (1-0.99543904) = 0.00456096 every hand.

The simplest way to determine it is just to look at the amount you intend to bet per hand, assume you will play a certain number of hands per hour, and then see how long that should end up lasting you. You can even do it in that order. If I’m going to play $5/hand ($1 denomination), then I will expect to lose:

(0.00456096 * 5) = 0.0228048 per hand. I can now determine how many hands I should be able to play with that bankroll:

(100 * 0.0228048) = 4385.0417456

Okay, so at $4,385 in total bets, that represents 877 hands, which is just over one hour assuming 800 hands per hour.

Of course, that doesn’t tell the whole story. For example, if I were able to bet all $100 on one hand, I’m either going to win or lose that hand. While my expected loss may only be expected to lose 45.6096 cents on that hand enabling me to expect to play 100/.456096 = 219.252 hands...that’s probably not going to happen. I’m either going to win, break even or lose the entire $100 immediately after that one hand.

This is where we have to look at Standard Deviation, which for 9/6 Jacks or Better is 4.42, and that’s a very important number.

Standard Deviations work by way of Confidence Intervals. In other words, you can use Standard Deviations to determine a range of likely results for a given play. With a Standard Deviation of 4.42, that means over a sizable amount of hands, you can determine what is likely to happen based upon your average bet.

The first thing that you have to do is understand what Standard Deviation means with respect to Video Poker. It means that your expected results are going to fall within a certain range a given percentage of the time. How you will calculate Standard Deviation is simply to multiply the expectation by the Standard Deviation, and that gives you one Standard Deviation.

If we play 1000 hands at $5 a pop, then our expected loss from above will be:

(1000 * 5 * .0228048) = 114.024

Getting the Standard Deviation is entirely different, though. What we have to do is multiply the Standard Deviation of the game (in general) by the square root of the number of hands, multiplied by the total credits bet (5) and the denomination ($1). Here we go, I like this calculator:

https://web2.0calc.com/

4.42*sqrt(800)*5*1 = 698.8633628972118324

Okay, now we can look at our general Standard Deviation Rules to determine what percentage of results will fall within certain ranges.

+/- 1 SD: 68% of the time.

+/- 2 SD: 95% of the time.

+/- 3 SD: 99.7% of the time.

That means that, when we factor the expected loss back in, your results will fall within:

+/- 698.8633628972118324 68% of the time.

Of course, most of this is going to be on the, “Lose $100,” side because, once you have lost the $100, you have no opportunity to recover. That means that you are going to bust more often than not. Beating Bonuses has a pretty good simulator for that.

My inputs were:

Deposit: 100

Bonus: 0

Wagering: 5000

Bet Size: 5

Simulations: 10,000

The first set of simulations yielded a bust rate of 83.53% and the second set 83.07%, so somewhere in the low-mid 80%’s.

The long story short is that, not only do you have an expected loss of over $100 to begin with, but you really don’t have the bankroll to expect to survive 1,000 hands. Let’s reduce it to 800 hands ($4,000 wagered and one hour of play at 800 HPH) and see what happens:

4.42*sqrt(800)*5*1 = 625.0823945689080116

So, you’re looking at roughly +/- $625 about 68% of the time, except, if you hit -$100, you’re done immediately. You don’t have anymore money!

For that reason, our likelihood of busting out doesn’t improve very much. Switching Wagering to 4,000, we’re still at 79.45% and 80.42% on two simulations of 10,000 players.

Should we be able to play quarters for 1,000 hands with any reliability?

4.42*sqrt(1000)*5*.25 = 174.7158407243029581

We already know that we fall +/- roughly $175 about 68% of the time. We also fall within roughly double that, or +/- $350 (approximately) 95% of the time, so I’m going to go out on a limb and say we probably bust out (-$100) a pretty substantial portion of the time.

After changing wagering to $1250 and bet size to $1.25, the two simulations had us busting out:

28.42% and 28.56% of the time.

If you want to go back to our 800 HPH analysis, here we go:

4.42*sqrt(800)*5*.25 = 156.2705986422270029

Again, we’ll be about +/- $312.50 about 95% of the time. I’m guessing that we bust out around 20-25% of the time still with a bet size of $1.25 and $1,000 total wagering:

I got 20.37% and 21.20%.

That means that, at 800 HPH and a $100 Deposit, we can expect to not even play for an hour about one in five times. Although, we do have a much better chance of lasting through 5,000 hands of total action with the reduced bet as compared to our $5 bets:

4.42*sqrt(4000)*5*.25 = 349.4316814486059162

The first thing we notice is that we expect to do +/- $698.86 (roughly) 95% of the time. The bad news is that we are still not expected to successfully wager $5,000 at $1.25/hand (4,000 hands) with a $100 bankroll. The good news is that our 2SD range with those numbers is less than $100 more than our 1SD range betting $5 per hand.

Changing Wagering to $5,000 and Bet Size to $1.25 on the Simulator:

The bust rate was 73.56% once and 73.86% once. The probability of having an overall profit also went up, because we lost everything less frequently.

Either way, the reduced Variance enabled us to survive $5,000 in total bets more often due to the reduced bet amounts.

Now, let’s take a look at how likely we are to be able to make 5,000 total hands ($1,250) betting $0.25/hand ($0.05 denomination):

4.42*sqrt(5000)*5*.05 = 78.1352993211135014

That means that we will finish +/- $156.28 (roughly) 95% of the time. I got 2.51% and 2.67% on the probability of busting imputing $1,250 in Wagering Requirements and a bet size of $0.25/hand to total 5,000 hands. The reason why is that, even when we do finish with an overall loss, we still haven’t busted out yet, most of the time.

Okay, now let’s imagine we want to play $5,000 in total action. Do we have a better chance with playing 20,000 hands a $0.25 a pop?

4.42*sqrt(20000)*5*.05 = 156.2705986422270029

The first SD is the same as the 2SD range was on the above calculation! It’s kind of funny how that one worked. Either way, our overall range on that $5,000 coin-in is going to be within $312.50, plus or minus, with 95% certainty. Let’s see what that does to our bust rate:

Changing the Wagering to $5,000, I get 55.23% and 55.07%.

The reason why the overall bust rate is still so high is because, as you get into larger numbers, the more likely a particular range of results closer to the expectation (5000 * .00456096) $22.8048 is going to be, but someone still has to be able to weather the swings for that to happen.

One thing that we notice on the simulator is that the, “Average Return,” is typically lower than the expected loss, and ironically, that’s because so many players bust out! See, when the entire $100 deposit is lost, those players do not have the opportunity to lose more than $100, which many of them would have. (See the Standard Deviations)

It is when nobody busts out that we come closer to the average return (unlimited bankroll) in simulations because the Standard Deviations fully work themselves out. Of course, even a simulation of 10,000 players is somewhat limited when we consider that about 9,970 of them will fall within 99.7% range of results.

Either way, this can give you a good idea of your probability of busting out if the goal is to play x number of hands. The simulator is an excellent tool to give you that probability, and when you want to weigh a deposit against how long you would like to play, you should use the simulator to make the determination. You could go for a 0.00% chance of bust, 25% chance, 50% chance, whatever you like!

You might ask, what about other games?

The good news for you is that you can make this same calculation for any game that you like, as long as you know the Standard Deviation. Many table games have a lesser Standard Deviation than does Video Poker. A good general rule is that, if there are no huge jackpots (like a Royal Flush) on the game, or anything even close, the Standard Deviation will likely be lower.

One example we can do is compare Pai-Gow Poker to Let it Ride Let’s assume that you want to play Live Dealer (or in a Brick-and-Mortar) and the minimum bet is $5. Based on that assumption and the Standard Deviation, we can see how likely we are to last at least 1,000 hands on a game, and what our range of results is:

Pai-Gow Poker: 0.75*sqrt(1000)*5 = 118.5854122563142249

Okay, so assuming NO SIDE BETS and after 1,000 $5 hands, we can assume we will be up or down within about $237 95% of the time. Let’s see how that works out with deposits of $100, and $500, respectively:

(The simulator assumes you can’t bank, so to get the House Edge right, I got cute with the winning probabilities. By doing so, I also made the Standard Deviation correct.)

WIN: 0.2923

Push: 0.4148 (Same)

LOSE: 0.2928

(Those aren’t correct, of course, but it accomplished what I needed it to. I needed the House Edge and Standard Deviation to be about the same)

The probability of busting was 61.41% and 60.97% with a $100 bankroll.

The probability of busting was 0.01% and 0.02% with a $500 bankroll.

The probability of busting was 10.62% and 11.32% with a $250 bankroll.

Now, for Let It Ride:

Let It Ride: 5.17*sqrt(1000)*5 = 817.4487751535260573

Okay, so we’re looking at +/- $1,634.90 (roughly) 95% of the time. I think you guys know what happens here, it’s going to have a larger bust rate than PGP in all cases and an extremely high bust rate on a $100 bankroll. Let’s make sure, though:

The probability of busting was 89.72% and 89.64% with a $100 bankroll.

The probability of busting was 33.68% and 34.56% with a $500 bankroll.

The probability of busting was 70.76% and 71.34% with a $250 bankroll.

Most practically, you may occasionally have a bonus or some sort of kickback that requires you put a certain percentage of total coin-in in order to qualify. When it comes to things of that nature, some Video Poker games (even with a worse RTP) might be better than some others if you have a limited bankroll. Simply put: If you ever find yourself in a situation in which you can guarantee a win, but might have to sacrifice a little value to do so, is it not sometimes smarter to guarantee a profit rather than lose your whole deposit and/or buy-in?