Keno House Edge

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Games available in most casinos are commonly called casino games. In a casino game, the players gamble cash or casino chips on various possible random outcomes or combinations of outcomes. Casino games are also available in online casinos, where permitted by law. Casino games can also be played outside casinos for entertainment purposes like in parties or in school competitions, some on machines that simulate gambling.

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See full list on wizardofodds.com. The house edge for Keno is between 20 per cent and 35 per cent. Pick the games with the smallest house edge. This means when you play, you want to win big. If you are lucky enough to win big, walk away from the game. Strategies to improve keno odds Use a. House Edge: This term is used to represent the percentage of bets that the house, i.e. The casino, can hope to win over time. The standard house edge is 30%, though this number is not necessarily the same for all casinos. Keno Glossary – Summing Up. The above are some of the terms that you will encounter when you start playing Keno. As you can see, the house edge gets greater and greater the further up the pay table you go. And most casinos cap the amount you can win, so if you were to hit 13 numbers and the keno was capped at $100,000, the house edge becomes astronomical. The truth is, keno has some of the biggest house edges in the casino.

There are three general categories of casino games: gaming machines, table games, and random number games. Gaming machines, such as slot machines and pachinko, are usually played by one player at a time and do not require the involvement of casino employees to play. Tables games, such as blackjack or craps, involve one or more players who are competing against the house (the casino itself) rather than each other. Table games are usually conducted by casino employees known as croupiers or dealers. Random number games are based upon the selection of random numbers, either from a computerized random number generator or from other gaming equipment. Random number games may be played at a table or through the purchase of paper tickets or cards, such as keno or bingo.

Some casino games combine multiple of the above aspects; for example, roulette is a table game conducted by a dealer, which involves random numbers. Casinos may also offer other type of gaming, such as hosting poker games or tournaments, where players compete against each other.

Common casino games[edit]

Notable games that are commonly found at casinos include:

See Full List On Wizardofodds.com

House advantage[edit]

Casino games typically provide a predictable long-term advantage to the casino, or 'house', while offering the players the possibility of a short-term gain that in some cases can be large. Some casino games have a skill element, where the players' decisions have an impact on the results. Players possessing sufficient skills to eliminate the inherent long-term disadvantage (the house edge or vigorish) in a casino game are referred to as advantage players.

The players' disadvantage is a result of the casino not paying winning wagers according to the game's 'true odds', which are the payouts that would be expected considering the odds of a wager either winning or losing. For example, if a game is played by wagering on the number that would result from the roll of one die, true odds would be 5 times the amount wagered since there is a 1 in 6 chance of any single number appearing, assuming that the player gets the original amount wagered back. However, the casino may only pay 4 times the amount wagered for a winning wager.

The house edge or vigorish is defined as the casino profit expressed as the percentage of the player's original bet. (In games such as blackjack or Spanish 21, the final bet may be several times the original bet, if the player double and splits.)

In American roulette, there are two 'zeroes' (0, 00) and 36 non-zero numbers (18 red and 18 black). This leads to a higher house edge compared to European roulette. The chances of a player, who bets 1 unit on red, winning is 18/38 and his chances of losing 1 unit is 20/38. The player's expected value is EV = (18/38 × 1) + (20/38 × (−1)) = 18/38 − 20/38 = −2/38 = −5.26%. Therefore, the house edge is 5.26%. After 10 spins, betting 1 unit per spin, the average house profit will be 10 × 1 × 5.26% = 0.53 units. European roulette wheels have only one 'zero' and therefore the house advantage (ignoring the en prison rule) is equal to 1/37 = 2.7%.

The house edge of casino games varies greatly with the game, with some games having an edge as low as 0.3%. Keno can have house edges up to 25%, slot machines having up to 15%.

The calculation of the roulette house edge is a trivial exercise; for other games, this is not usually the case. Combinatorial analysis and/or computer simulation is necessary to complete the task.

In games which have a skill element, such as blackjack or Spanish 21, the house edge is defined as the house advantage from optimal play (without the use of advanced techniques such as card counting), on the first hand of the shoe (the container that holds the cards). The set of the optimal plays for all possible hands is known as 'basic strategy' and is highly dependent on the specific rules and even the number of decks used. Good blackjack and Spanish 21 games have house edges below 0.5%.

Traditionally, the majority of casinos have refused to reveal the house edge information for their slots games and due to the unknown number of symbols and weightings of the reels, in most cases it is much more difficult to calculate the house edge than that in other casino games. However, due to some online properties revealing this information and some independent research conducted by Michael Shackleford in the offline sector, this pattern is slowly changing.^[1]

In games where players are not competing against the house, such as poker, the casino usually earns money via a commission, known as a 'rake'.

Standard deviation[edit]

The luck factor in a casino game is quantified using standard deviations (SD).^[2] The standard deviation of a simple game like roulette can be calculated using the binomial distribution. In the binomial distribution, SD = √npq, where n = number of rounds played, p = probability of winning, and q = probability of losing. The binomial distribution assumes a result of 1 unit for a win, and 0 units for a loss, rather than −1 units for a loss, which doubles the range of possible outcomes. Furthermore, if we flat bet at 10 units per round instead of 1 unit, the range of possible outcomes increases 10 fold.^[3]

SD (roulette, even-money bet) = 2b√npq, where b = flat bet per round, n = number of rounds, p = 18/38, and q = 20/38.

For example, after 10 rounds at 1 unit per round, the standard deviation will be 2 × 1 × √10 × 18/38 × 20/38 = 3.16 units. After 10 rounds, the expected loss will be 10 × 1 × 5.26% = 0.53. As you can see, standard deviation is many times the magnitude of the expected loss.^[4]

The standard deviation for pai gow poker is the lowest out of all common casino games. Many casino games, particularly slot machines, have extremely high standard deviations. The bigger size of the potential payouts, the more the standard deviation may increase.

As the number of rounds increases, eventually, the expected loss will exceed the standard deviation, many times over. From the formula, we can see the standard deviation is proportional to the square root of the number of rounds played, while the expected loss is proportional to the number of rounds played. As the number of rounds increases, the expected loss increases at a much faster rate. This is why it is impossible for a gambler to win in the long term. It is the high ratio of short-term standard deviation to expected loss that fools gamblers into thinking that they can win.

Keno - Wizard Of Odds

It is important for a casino to know both the house edge and variance for all of their games. The house edge tells them what kind of profit they will make as percentage of turnover, and the variance tells them how much they need in the way of cash reserves. The mathematicians and computer programmers that do this kind of work are called gaming mathematicians and gaming analysts. Casinos do not have in-house expertise in this field, so outsource their requirements to experts in the gaming analysis field.

See also[edit]

References[edit]

1. ^'Michael Shackleford is the wizard of odds'. Observer. Retrieved 13 October 2015.
2. ^Hagan, general editor, Julian Harris, Harris (2012). Gaming law : jurisdictional comparisons (1st ed.). London: European Lawyer Reference Series/Thomson Reuters. ISBN978-0414024861.
3. ^Gao, J.Z.; Fong, D.; Liu, X. (April 2011). 'Mathematical analyses of casino rebate systems for VIP gambling'. International Gambling Studies. 11 (1): 93–106. doi:10.1080/14459795.2011.552575. S2CID144540412.
4. ^Andrew, Siegel (2011). Practical Business Statistics. Academic Press. ISBN978-0123877178. Retrieved 13 October 2015.

The probability of n numbers drawn in the first 40, last 40, or any given 40 is combin(40,n)*combin(40,20-n)/combin(80,20). So the probability of exactly 10 in the first 40 (and 10 in the last 40) is combin(40,10)*combin(40,10)/combin(80,20) = 0.203243. The probability of one half having more than the other is 1-.203243= 0.796757. The probability of a specific half having more is half this number, or 0.398378. If this bet paid even money the house edge would be 20.32%. If the even bet paid 3 to 1 then the house edge on that bet would be 18.70%. If it paid 4 to 1 the player would have a 1.62% edge. About positive expectation blackjack online the more the player plays the greater the probability of a net profit. The best game is currently Unified Gaming’s single deck, with a player edge of 0.16%. If the player flat bet one millions hands the probability of being down would still be about 8.6%. At Boss Media’s single player game with a player edge of 0.07% the probability of a loss after a million hands is about 27.5%.

Baccarat

I doubt certain numbers are more likely than others. My advice is to pick anything, it doesn’t make any difference.

Your overall expected return is the same regardless of how many games you play. Of course it is more likely to hit a number the more machines you play, but if they all miss you lose more money.

Keno House Edge

'Anonymous' .

Pai gow poker is the least volatile and on average keno is the most.

In Nevada, and I think other major gambling markets in the United States, the balls truly are random and the outcome determined by the balls. However in class II slots, sometimes found in Indian casinos, anything goes.

HEADS - bet that eleven to twenty numbers in the top half appear - even money
TAILS - bet that zero to nine numbers in the top half appear - even money
EVENS - bet that exactly ten numbers in the top half appear - pays 3 to 1

The probability of the tie bet winning is combin(40,10)*combin(40,10)/combin(80,20) = 0.203243. Paying 3 to 1 the house edge is 18.703%. The probability of the heads (or tails) bet winning is (1-0.20343)/2 = 0.398378. Paying even money the house edge is 20.324%.

Compliments will get you everywhere. The number of combinations for n heads is combin(40,n)*combin(40,20-n). This is the number of ways to choose n numbers out of the top 40 and 20-n out of the bottom 40. The following table shows the probability of 0 to 20 heads.

Probability of 0 to 20 Heads

More Keno House Edge Images

The probability of hitting all 20 is 1 in combin(80,20) = 3,535,316,142,212,180,000. So the odds are more like 3.5 quintillion to one. Assuming 5 billion people on earth, and they all played once a week, there would be one winner every 13.56 million years on average. Most casinos pay the same amount for hitting close to 20. For example the Las Vegas Hilton pays $20,000 for hitting 17 or more out of 20. I have never heard of anyone every hitting 20 out of 20, and doubt very much that it has ever happened.

I disagree. I can’t think of a single major Strip casino without a keno lounge. In general the only casinos without keno are the locals casinos in the suburbs of Vegas, because most of us locals know that keno is a sucker game.

P.S. A reader later wrote in to correct me, stating that the New York New York casino in Las Vegas removed their keno lounge.

I hope you’re happy, I spent all day on this. After writing and running a simulation I find that the probability that any 3 lines will contain 11 or more marks is 86.96%! That isn’t even giving the other side a fighting chance. You can go up to 12 marks and still have a probability of 53.68% of winning, or an advantage of 7.36%. However, I think you have the wrong side of the empty row bet. The probability of at least one empty row is only 33.39%, better to take the other side of no empty rows. While I was at this I did lots of other probabilities and put them in a new page of keno props. Here is a list from that page of these and other good even-money bets. The good side is listed.

Even Money Keno Props

Much like live keno the odds are the same regardless of what you pick, but they are independent of the balls the game draws.

That is actually a pretty hard problem. It is easy to get the probability of the number of times any of your four balls are drawn, including repeats. The tricky part is determining the probability that x distinct picks will be chosen, given that any pick was chosen y times. I indicate the answer and solution on my MathProblems.info page, problem 205.

1/2.

If we used keno as a comparison, everybody would have 40 genes, each represented by a keno ball. However, each ball would have unique number. When two people, who are not related, mate it is like combining 80 balls between the two of them into a hopper, and randomly choosing 40 genes for the offspring of the mating.

So when you were conceived, you got half the balls in the hopper, and the other half were wasted. When your brother or sister was conceived he/she got half from the balls drawn when you were born, and half that were not drawn. So you are 50% genetically identical. Much for the same reason that if 40 numbers were drawn in keno, two consecutive draws would average 20 balls in common.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Romes

As a reminder to our other readers, Cleopatra Keno plays like conventional keno, except if the last ball drawn matches one of the player's picks AND results in a win, then the player will also win 12 free games with a 2x multiplier. Free games do not earn more free games.

You didn't specify the number of picks or pay table, so let's use the 3-10-56-180-1000 pick-8 pay table, as an example. First, let calculate the return.

The number of ways to catch x balls out of y in keno is the number of ways to pick x balls out of 20 and y-x out of 60. This equals combin(20,x)*combin(60,y-x), to put it in Excel terms. As a further reminder, combin(x,y) = x!/(y!*(x-y)!). Finally x! = 1*2*3*...*x.

With that review out of the way, here is the return table for that pay table. The right column shows the expected square of the win, which we'll need later.

Pick 8 Keno

Next, let's calculate the average bonus. We can see from the table above that the average win, not counting the bonus, is 0.593301. In the bonus, the player gets 12 doubled free spins. Thus, the expected win from the bonus is 2×12×0.593301 = 14.239212.

Next, let's calculate the probability of winning the bonus. If the player catches four numbers, the probability the 20th ball is one of those 4 is 4/20. In general, if the player catches c, then the probability that the 20th ball contributed to the win is c/20.

The formula for winning the bonus is prob(catch 4)*(4/20) + prob(catch 5)*(5/20) + prob(catch 6)*(6/20) + prob(catch 7)*(7/20) + prob(catch 8)*(8/20). We know the probability of any given win from the return table above. So, the probability of winning the bonus is:

0.081504*(4/20) + 0.018303*(5/20) + 0.002367*(6/20) + 0.000160*(7/20) + 0.000004*(8/20) = 0.021644.

With the probability of winning the bonus and the average bonus win, we can calculate the return from the bonus as 0.021644 × 14.239212 = 0.308198.

Not that we need to know, but the overall return for the game is the return from the base game plus the return from the bonus, which equals 0.593301 + 0.308198 = 0.901498.

Now, let's start getting into the actual variance. As a reminder, a general formula about variance is:

var(x + y) = var(x) + var(y) + 2*cov(x,y), where var stands for variance and cov stands for covariance. In this case of this game:

Total variance = var(base game) + var(bonus) + 2*cov(base game and bonus).

The fundamental formula for the variance is the E(x^2) - [E(x)]^2. In other words, the expected square of the win less the expected win squared.

That said, let's start with the variance of the base game. Remember when I said before when we would need that expected win squared from the first table. The lower right cell of that first table shows us the expected win squared is 19.530214. We already know the expected win is 0.593301. Thus, the variance of the base game is 19.530214 - 0.593301² = 19.178208.

Next, let's calculate the variance of the bonus (assuming it was already hit). For that, recall that:

var(ax) = a²x, where a is a constant.

Also recall that the variance of n random variables x is nx.

That said, if x is the base win in a bonus game, then the variance of the whole bonus is 2² × 12 × x. We know from above the variance of a single spin in the base game, not counting the bonus, equals 19.178208. So, the variance of the bonus, given a bonus was already hit, is 2² × 12 × 19.178208 = 920.554000.

However, what we need to know is the variance of the bonus before the first ball is drawn, including the possibility the bonus won't be won at all. No, we can't just multiply the variance of the bonus by the probability of winning it. Instead, recall that var(x) = E(x^2) - [E(x)]^2. Let's rearrange that to:

E(x^2) = var(x) + [E(x)]^2

We know the mean and variance of the bonus, so the expected win squared in the bonus is 920.554000 + 19.178208² = 1123.309169.

So, the expected square of the win from the bonus, before the first ball is drawn is the prob(bonus) × E(x^2) = 0.021644 × 1123.309169 = 24.313239.

We already calculated the expected win from the bonus, before the first ball, is 0.308198. So, the overall variance of the bonus, before the first ball, is 24.313239 - 0.308198² = 24.218253.

The next step is to calculate the covariance. 'Why is there a correlation between the base win and the bonus win?', you might ask. It's because the last ball drawn must contribute to a win to trigger the bonus. Given that the last ball contributed towards a win, the average win goes up. As a reminder, Bayes' formula of condition probability says:

P(A given B) = P(A and B)/P(B).

Let's then redo our return table for the base game, given that the last ball was a hit:

Pick 8 Keno given Last Ball Hit

The bottom right cell shows that assuming the last ball was a hit, the average win is 6.757734.

Next, recall from your college statistics class that:

Keno House Edge

cov(x,y) = exp(xy) - exp(x)*exp(y) .

In our case, let x = base game win and y=bonus win. Let's work on exp(xy) first.

Exp(xy) = prob(bonus won)*(average base game win given bonus won)*average(bonus win) + prob(bonus not won)*(average base game win given bonus not won)*average(bonus win given bonus not won). It's easy to say that average(bonus win given bonus not won) = 0, so we can rewrite as:

Exp(xy) = prob(bonus won)*(average base game win given bonus won)*average(bonus win) =

We already solved for E(x) and E(y), so the covariance is:

cov(x,y) = exp(xy) - exp(x)*exp(y) = 2.082719 - 0.593301 × 0.308198 = 1.899865.

Let's go back to the overall equation for the variance when covariance is involved:

Total variance = var(base game) + var(bonus) + 2*cov(base game and bonus) = 19.178208 + 24.218253 + 2×1.899865 = 47.196191. The standard deviation is the square root of that, which is 6.869948.

So, there you go. That one took me hours, so I hope you're happy.

This question is asked and discussed in my forum at Wizard of Vegas.

Upon doing some research, I found this isn't a side bet, but what the pick-20 ticket pays for catching zero. The following is my complete analysis of the Station Casinos pick 20 ticket.

Station Casinos Pick 20 Keno

The lower right cell shows the overall return of the ticket is 69.70%, which is typical for live keno.

To answer the question about catching 0, the probability column shows the probability of that is 0.001186 and at a win of 200 for 1, it returns 23.71% towards the return.