# Keno - FAQ

Mr. Wizard, your site is truly informative. There is a keno game here where we can bet on HEAD, TAIL or EVEN. HEAD meaning 11 numbers or more in the first forty numbers, TAIL meaning 11 numbers or more in the last forty numbers. EVEN meaning 10 numbers each in the first forty and last forty respectively. There are 20 numbers drawn each time. What are the odds of each bet winning? One more thing, since the house is negative according to you (for some online casinos), does that mean that a player can consistently win in the long run in the game of blackjack?

Tony from Malaysia

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%.

Is there an advantage in playing same numbers in caveman keno, or playing different numbers every time. or changing one number at a time?

Mike from Mesa, USA

It doesn’t make any difference.

I visit the casinos quite frequently and have noticed that people seem to do fairly well on the video keno quarter machines. Do you have any suggestions as to what numbers to play. I have noticed that there are some that come up more frequently than others.

Karen from Antioch, USA

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

Dear Sir, We are avid Keno players. It is our intuitive belief that if we play two or more keno machines using the same numbers that our chances for hitting those numbers are significantly increased. Can you enlighten us with some statistics to support our intuition? Thank you.

Gene & Rosie from Bayside, WI

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.

Which are the most and least volatile games?

Anonymous

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

Does the RNG in a keno machine pick the numbers and if they come up you win or just determines if you win or lose and the numbers are just for show?

Anonymous

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.

I have seen a keno game with the following side bets. What is the scoop on these bets?

**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**

Anonymous

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%.

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This shows the probability of 11 to 20 heads is 39.84%, for a house edge of 20.32%. The probability of exactly 10 is 20.32%, for a house edge of 18.70%.

Dear wonderful Wizard, firstly THANK YOU VERY MUCH for your wonderful site! I have spent hours upon hours exploring all that your superbly done site has to offer, and I am grateful for your truly invaluable advice, so THANK YOU! I have a question regarding a side bet of Keno here in Australia called "Heads and Tails". The board is divided in half, numbers 1 to 40 are heads and 41 to 80 are tails. If the majority of numbers drawn are low (1 to 40) then heads wins, if the majority are high (41-80) then tails wins. Both bets pay 1 to 1. There is also a bet called Evens, which pays 3 to 1 if 10 numbers are low and 10 are high. My question is, what is the house edge of each bet?

Anonymous

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

Heads | Combinations | Probability |
---|---|---|

0 |
137846528820 |
0.000000039 |

1 | 5251296336000 |
0.0000014854 |

2 |
88436604204000 |
0.0000250152 |

3 |
876675902544001 |
0.0002479767 |

4 |
5744053569793500 |
0.0016247638 |

5 |
26468598849608400 |
0.0074869114 |

6 |
89077015359259200 |
0.0251963366 |

7 |
224342112756653000 |
0.0634574402 |

8 |
429655207020554000 |
0.1215323297 |

9 |
632136396535987000 |
0.1788061862 |

10 |
718528370729238000 |
0.2032430317 |

11 |
632136396535987000 |
0.1788061862 |

12 |
429655207020554000 |
0.1215323297 |

13 |
224342112756653000 |
0.0634574402 |

14 |
89077015359259200 |
0.0251963366 |

15 |
26468598849608400 |
0.0074869114 |

16 |
5744053569793500 |
0.0016247638 |

17 |
876675902544001 |
0.0002479767 |

18 |
88436604204000 |
0.0000250152 |

19 |
5251296336000 |
0.0000014854 |

20 | 137846528820 |
0.000000039 |

Total |
3535316142212170000 |
1 |

This shows the probability of 11 to 20 heads is 39.84%, for a house edge of 20.32%. The probability of exactly 10 is 20.32%, for a house edge of 18.70%.

Sir, I recently read in a book about odds that the odds of hitting all 20 numbers in keno are a quintillion to one. The book described this by saying that if there were one drawing per week and everyone on earth always bought a ticket, it would take 5 million years to produce a winner. My question is, is there a prize for hitting all 20, and if so, has anyone ever hit it? I’ve heard that no one has ever hit keno in the history of Vegas, it this true?

Tim from Greenville, SC

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.

My wife and father in law went to Las Vegas a couple of months ago and she asked where the Keno games are (sot the Keno slot machines) and was told that most hotels didn’t have keno any more. Is that true? And if so do you know why Mr. Wizard?

Bill from Malibu

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.

There is an interesting way that one can play keno, though not in the way the state intended. Bet that at least 11 of the 20 numbers will appear in 3 rows; horizontal, vertical or a combination. Stress that there are 18 rows. Many times the sucker will play. A variant of this bet is that one row will be blank. I hope that you can use this. You have an excellent, informative site. Note that one needs a bankroll though not a large one. 10 to 15x the largest bet that you are going to cover is enough.

Richard M. from Silver Spring, MD

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

Prop | Probability of a Win |
House Edge |
---|---|---|

No row will have 5 or more hits | 53.47% | 6.94% |

Greatest number of hits in a column will be exactly 4 | 55.2% | 10.4% |

Every row to have at least one mark | 66.61% | 33.23% |

Number of empty columns will not be 1 | 54.08% | 8.15% |

Top/bottom to have 9 to 11 marks | 56.09% | 12.17% |

3 lines (rows and/or columns) will contain 12 or more marks | 53.68% | 7.36% |

Does it matter what numbers you pick in video keno at all? I understand that it is a RNG chip like any slot machine and that the numbers are just to give us the illusion of control. I’ve tried writing IGT but they won’t respond. Thanks!

Jari from Minnetonka, MN

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

Suppose you are playing standard 80-spot Keno with 20 drops, but the drops are "with replacement." That is, after a ball is dropped, its number is recorded and it is put back in the hopper and may be drawn again. Suppose you mark a card with 4 spots. What are the probabilities of 0, 1, 2, 3 and 4 distinct hits?

Eliot from Santa Barbara

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 page, problem 205.

What ratio of genes would I have in common with a full brother or sister, other than identical twin?

HotBlonde

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 .

What is the variance in Cleopatra Keno??

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

Event | Pays | Combinations | Probability | Return | Return^2 |
---|---|---|---|---|---|

0 | 0 | 2,558,620,845 | 0.088266 | 0.000000 | 0.000000 |

1 | 0 | 7,724,138,400 | 0.266464 | 0.000000 | 0.000000 |

2 | 0 | 9,512,133,400 | 0.328146 | 0.000000 | 0.000000 |

3 | 0 | 6,226,123,680 | 0.214786 | 0.000000 | 0.000000 |

4 | 3 | 2,362,591,575 | 0.081504 | 0.244511 | 0.733533 |

5 | 10 | 530,546,880 | 0.018303 | 0.183026 | 1.830259 |

6 | 56 | 68,605,200 | 0.002367 | 0.132536 | 7.422014 |

7 | 180 | 4,651,200 | 0.000160 | 0.028882 | 5.198747 |

8 | 1000 | 125,970 | 0.000004 | 0.004346 | 4.345661 |

Total | 28,987,537,150 | 1.000000 | 0.593301 | 19.530214 |

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 - 0.593301^{2} = 19.178208.

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

var(ax) = a^{2}x, 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 2^{2} × 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 2^{2} × 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 + 19.178208^{2} = 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 - 0.308198^{2} = 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

Event | Pays | Combinations | Probability | Return |
---|---|---|---|---|

0 | 0 | - | 0.000000 | 0.000000 |

1 | 0 | - | 0.000000 | 0.000000 |

2 | 0 | - | 0.000000 | 0.000000 |

3 | 0 | - | 0.000000 | 0.000000 |

4 | 3 | 472,518,315 | 0.753119 | 2.259358 |

5 | 10 | 132,636,720 | 0.211402 | 2.114019 |

6 | 56 | 20,581,560 | 0.032804 | 1.837010 |

7 | 180 | 1,627,920 | 0.002595 | 0.467036 |

8 | 1000 | 50,388 | 0.000080 | 0.080310 |

Total | 627,414,903 | 1.000000 | 6.757734 |

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:

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) =

0.021644 × 6.757734 × 14.239212 = 2.082719.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 .