Keno - FAQ

The likelihood of drawing n numbers from the initial or latter 40, or any set of 40 numbers, is calculated as combin(40,n)*combin(40,20-n)/combin(80,20). Accordingly, the odds of achieving exactly 10 from the first 40 (and likewise from the last 40) result in combin(40,10)*combin(40,10)/combin(80,20) = 0.203243. The chance of one half winning over the other is 1 - 0.203243 = 0.796757. For a specific half to prevail, it would be half of that figure, or 0.398378. Should this bet settle at an even payout, the house edge is 20.32%. If the even bet provides a payout of 3 to 1, the house edge adjusts to 18.70%. A 4 to 1 payout would grant a slight advantage of 1.62% to the player. In positive expectation blackjack online, the more playtime a player has, the better the chances of achieving a net gain. Currently, the finest option appears to be Unified Gaming’s single deck, which offers a player edge of 0.16%. If a player were to flat bet a million hands, they would still face an 8.6% chance of being down. In Boss Media's single player game, the player edge is 0.07%, leading to a 27.5% likelihood of losing after a million hands.

It doesn’t make any difference.

I personally doubt that some numbers are inherently more favorable than others. My recommendation is to choose any numbers, as it does not affect your odds.

The expected payout remains consistent no matter how many games you engage in. While it is statistically more likely to hit a number when playing multiple machines, if all machines fail, your losses accumulate.

Pai Gow Poker is known to be the most stable, while Keno typically experiences the most volatility.

In Nevada and some other key gambling destinations in the U.S., the choice of balls is indeed random, leading to outcomes driven by those balls. However, class II slot machines, often found in Native American casinos, may operate differently.

The likelihood of winning the tie bet equates to combin(40,10)*combin(40,10)/combin(80,20) = 0.203243. With a 3 to 1 payout, the house edge becomes 18.703%. Conversely, the odds for winning HEADS (or TAILS) is (1 - 0.20343)/2 = 0.398378. At even money odds, the house edge sits at 20.324%.

Flattery can take you far. For n heads, the number of combinations is (40,n)*combin(40,20-n). This illustrates the various ways to select n numbers from the upper 40 and the remaining 20-n from the lower 40. The table below displays the probability of achieving 0 to 20 heads. combin The data indicates that the probability of obtaining 11 to 20 heads is 39.84%, resulting in a house edge of 20.32%. The odds for exactly 10 heads is 20.32%, leading to a house edge of 18.70%.

Recently, I read in a book about odds that the chances of hitting all 20 numbers is a quintillion to one. The author explained that if there were a single drawing each week and everyone on earth always purchased a ticket, it would take 5 million years to see a winner. I'm curious, is there a reward for hitting all 20? And, has anyone ever achieved this? I’ve come across claims that no one has ever won 20 in keno since the inception of the game in Las Vegas; is that accurate?

I beg to differ. I can't think of a significant casino on the Strip that lacks a keno lounge. Typically, the only establishments that don’t include keno are the local casinos located in the outskirts of Las Vegas, primarily because many locals recognize keno as a less favorable game.

There’s a unique strategy to playing keno that strays from official guidelines. One can bet that at least 11 of the 20 drawn numbers will appear in any of the 3 lines—horizontally, vertically, or a mix of both. Remember, there are 18 rows available. Many players fall into this trap often. A variant on this bet suggests that one of the rows will be blank. I trust this information may be useful. You have a wonderful, informative site. Just an aside, maintain a bankroll sufficient for your play; 10 to 15 times the maximum bet you intend to cover should suffice.

I hope this brings you joy; I've dedicated my entire day to this analysis. After conducting simulations, I calculated that the chance of any 3 lines having 11 or more marks is 86.96%! That number alone leaves little wiggle room for the competition. You can even increase your marks to 12 and still retain a winning probability of 53.68%, thus gaining an edge of 7.36%. However, I believe you've misunderstood the empty row bet's angle. The chance of having at least one empty row is merely 33.39%, so it’s wiser to back the opposite side with no empty rows. While examining this, I also derived many other probabilities, which I compiled into a new page.

The precise number of hits in a column will total exactly 4. keno props 3 lines (rows and/or columns) will contain 12 or more marks.

That presents a rather complex challenge. While determining the likelihood of how many times any of your four chosen balls may be drawn (including duplicates) is straightforward, the difficulty lies in ascertaining the probability for x distinct numbers appearing when y numbers are drawn. I provide the solution and explanation on my

If we liken the relationship to keno, everyone would possess 40 genes, represented by distinct keno balls. When two unrelated individuals mate, it's akin to combining 80 balls from both parents into a hopper and randomly selecting 40 genes for their offspring. MathProblems.info page, problem 205.

1/2.

This question sparked discussion in the forum of my other site.

As a note to our readers, Cleopatra Keno functions like traditional keno. However, if the last ball drawn matches a player's selected number and results in a win, they are awarded 12 free games with a 2x multiplier. These free games, however, do not yield additional free games.

Since you didn’t specify the number of picks or the payout table, let’s examine the example of the 3-10-56-180-1000 pick-8 payout table. First, we need to calculate the return. Wizard of Vegas .

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Mr. Wizard, I truly appreciate the informative nature of your site. For instance, there's a keno game featured here that allows us to wager on HEAD, TAIL, or EVEN outcomes. The HEAD bet signifies that 11 numbers or more will be drawn from the first forty, TAIL indicates the same for the last forty, and EVEN means that there will be an equal distribution of 10 numbers from both halves. Each round draws a total of 20 numbers. Can you provide insight into the winning probabilities for these bets? Additionally, you mentioned that some online casinos have a negative house edge. Does that imply that players can actually maintain a long-term winning strategy in blackjack?

To determine the chances of n numbers appearing in the first or last forty, or in any specific 40, you can use the formula combin(40,n)*combin(40,20-n)/combin(80,20). Specifically, the likelihood of exactly 10 numbers appearing in both the first and last forty is calculated as combin(40,10)*combin(40,10)/combin(80,20) = 0.203243. The probability that one half exceeds the other is 1 - 0.203243 = 0.796757. If you want to find the chance that a particular half has more, it is half of this value, giving you 0.398378. If this specific bet offers even money payouts, the house edge stands at 20.32%. However, if it pays 3 to 1, the house edge drops to 18.70%. On the other hand, if it pays 4 to 1, players would enjoy a slight edge of 1.62%. When considering positive expectation blackjack online, the longer a player engages, the less likely they are to incur losses. Presently, Unified Gaming's single-deck game offers the best player advantage at 0.16%. Yet, even with a million hands played, there's still an 8.6% risk of ending up with losses. For Boss Media’s game, which has a player edge of just 0.07%, that loss probability can rise dramatically to around 27.5%.

Is there any strategic benefit to consistently playing the same numbers in caveman keno, or does varying the numbers each time or changing just one at a time work better?

I frequently visit various casinos and have observed that players tend to have success on the video keno quarter machines. Do you have any recommendations for numbers to consider? I've noticed that certain numbers get drawn more often than others.

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

I am skeptical about the idea that some numbers might be more likely to be drawn than others. My suggestion is simply to choose any numbers, as it won’t really make a difference.

Dear Sir, as enthusiastic players of Keno, we firmly believe that playing the same set of numbers across multiple keno machines significantly improves our chances of hitting those numbers. Could you provide us with some statistical evidence to support our belief? Thank you.

Your expected returns remain constant, regardless of how many games you engage in. While it's true that playing across more machines increases your chances of hitting numbers, if they all fail to display your selections, you end up losing more.

What games are considered to be the most and least volatile?

Pai gow poker ranks as the least volatile game, whereas keno generally exhibits the highest volatility.

Does a keno machine's RNG simply select numbers randomly such that if they appear, you win, or does it just dictate win/loss outcomes while the numbers serve merely for appearance?

In Nevada, as well as other principal gambling regions across the U.S., the results truly come from random ball draws. However, in Class II slots, particularly found in some tribal casinos, the outcomes may not be entirely reliant on randomness.²= 19.178208.

I came across a keno game that includes various side bets. What can you tell me about these?

var(ax) = a²x, where a is a constant.

HEADS - wagering that between eleven to twenty numbers from the upper half will show up, with even money payouts.

TAILS - betting that no more than nine numbers will appear from the upper half, also at even money.²EVENS - placing a wager that precisely ten numbers from the upper half will be drawn, which pays 3 to 1.²× 12 × 19.178208 = 920.554000.

The likelihood of winning the tie bet is combin(40,10)*combin(40,10)/combin(80,20) = 0.203243. When offering 3 to 1, the house edge stands at 18.703%. The probability of winning the heads (or tails) bet is calculated as (1 - 0.20343) / 2 = 0.398378, with a house edge of 20.324% when paying even money.

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

Dear esteemed Wizard, I want to extend my heartfelt thanks for your exceptional site! I have devoted countless hours exploring the myriad resources you provide, and I’m incredibly grateful for your invaluable insights, so thank you! I have a query regarding a side bet in Keno here in Australia called 'Heads and Tails'. The game divides the board into two halves: numbers 1 to 40 are classified as heads, while numbers 41 to 80 are tails. If the majority of drawn numbers fall within the lower range (1 to 40), the heads bet prevails, whereas the tails will win if the higher numbers dominate. Both bets yield a payout of 1 to 1. There is also an 'Evens' bet that pays 3 to 1 when there are ten low and ten high numbers. Could you tell me what the house edge for each of these bets is?²= 1123.309169.

Flattery will get you everywhere. The combinations for achieving n heads are calculated as (40,n)*combin(40,20-n). This provides the number of ways to select n numbers from the top 40 and 20-n from the bottom 40. A table showcasing not just heads but the probability of achieving counts from 0 to 20 heads will give more detailed insights.

The data indicates that the probability of hitting 11 to 20 heads is 39.84%, indicating a house edge of 20.32%. The probability of getting exactly 10 positions, however, yields a house edge of 18.70%.²= 24.218253.

Sir, I recently delved into a book discussing odds, and I learned that the chances of matching all 20 numbers in keno are astronomically high, estimated at a quintillion to one. The book explained that if there was a single drawing every week with participation from everyone on Earth, it would take around 5 million years to see a winner. My inquiry is whether there is a reward for matching all 20 numbers, and if anyone has ever actually accomplished this feat? I've heard claims that no one has ever won a keno game in all of Las Vegas history; is this accurate?

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

The chances of matching all 20 numbers is 1 in combin(80,20) = 3,535,316,142,212,180,000. Thus, the odds are more closely aligned with 3.5 quintillion to one. Assuming a global population of 5 billion playing once weekly, there should be a winner roughly every 13.56 million years on average. Most casinos offer similar payouts for achieving near misses of 20; for example, the Las Vegas Hilton pays out $20,000 for hitting 17 or more correct guesses. I’ve yet to hear of anyone achieving 20 correct guesses, and I find it highly doubtful that it has ever happened.

A couple of months ago, my wife and father-in-law visited Las Vegas, and she inquired about the locations of keno games (not the slot machines) only to be informed that many hotels no longer offer keno. Is this claim accurate? If so, do you know the reasons behind this, Mr. Wizard?

I beg to differ. I can't pinpoint any prominent Strip casino that lacks a dedicated keno lounge. Typically, the absence of keno is noted only among local casinos situated in the suburban areas of Vegas, since most locals recognize keno as a less favorable game.

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

P.S. Subsequently, a reader wrote in to clarify that the New York New York casino in Las Vegas has indeed removed its keno lounge.

Interestingly, there is a unique approach to playing keno which may not align with the state's objectives. You can wager that at least 11 of the 20 numbers will appear across three rows, whether horizontally, vertically, or a mix thereof. Emphasize that there are 18 rows available. Many naïve players may fall for this. An alternative variant involves betting on whether one of the rows will remain blank. I hope you find this information useful; your site is truly excellent and informative. Just a note: having a bankroll is important, though it doesn’t need to be excessively large—between 10 to 15 times the largest bet you wish to place should suffice.

I trust you’re pleased with my effort; I dedicated an entire day to this. After conducting a simulation, I've discovered that the likelihood of any 3 lines containing 11 or more hits is an impressive 86.96%! This clearly provides an overwhelming advantage. You could even extend to 12 marks while maintaining a 53.68% probability of winning, yielding an advantage of 7.36%. However, I believe there’s a misunderstanding regarding the bet on having a blank row. The probability of having at least one empty row is only 33.39%, so it's wiser to take the contrary position of no empty rows. While conducting this analysis, I also explored various other probabilities and compiled them into a new page of resources.

0.021644 × 6.757734 × 14.239212 = 2.082719.

Here’s a brief overview of those and other profitable even-money bets.

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

The highest number of hits in any column tends to be exactly 4.

There’s a high likelihood that 3 lines (either rows or columns) will consist of 12 or more marks. Does it matter which numbers I select in video keno? I understand that the RNG operates like any other slot machine and that the numbers serve merely to create an illusion of choice. I’ve attempted to reach out to IGT, but they haven’t replied. Thank you!

Just like in conventional keno, the odds remain the same no matter how you choose, and they are independent of the numbers that the game randomly selects.

Imagine you’re engaging in a standard 80-spot Keno game with 20 draws, but with a twist: the draws are 'with replacement.' After dropping a ball, its number is documented and returned to the hopper for potential future draws. If you mark a card with 4 spots, what are the probabilities for achieving 0, 1, 2, 3, or even 4 distinct hits? Wizard of Vegas .

What percentage of genetic material would I share with a full sibling, other than an identical twin?

If we liken the situation to keno, each individual would possess 40 genes, each represented by a distinct keno ball. In the case of two unrelated individuals mating, it's akin to pooling 80 balls together from both parties into a hopper and randomly selecting 40 genes for their offspring.

So, during conception, you obtain half of the balls present in the hopper, while the remaining half are discarded. Your brother or sister, when conceived, would receive half from the balls drawn when you were born, alongside half that were not selected. Hence, you end up being 50% genetically identical—much like how if 40 numbers were drawn in keno, consecutive draws would typically share 20 common balls on average.

To remind our other readers, the game of Cleopatra Keno operates similarly to traditional keno, but if the last number drawn matches one of the player's selections and results in a winning combination, the player earns 12 free games with a 2x multiplier. Note that free games do not generate additional free games.

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

Imagine you’re engaging in a standard 80-spot Keno game with 20 draws, but with a twist: the draws are 'with replacement.' After dropping a ball, its number is documented and returned to the hopper for potential future draws. If you mark a card with 4 spots, what are the probabilities for achieving 0, 1, 2, 3, or even 4 distinct hits? Wizard of Vegas .