Keno - Boxes and Verticals

Hi. There's a Keno game at the bars in Nebraska in which one may bet for boxes, or one may bet for verticals. You can either fill out a ticket, or just tell the Keno tender that you want "boxes" or "verticals." Whichever you decide, it's $1.90 per game.
The official tickets have the same format as the play slips - here's an example of a play slip:

An example of winning a verticals game would be hitting 1, 11, 21, 31 as well as 48, 58, 68, 78. It must be 2 vertical rows of 4.
As far as boxes go, the numbers must start on the first or third line in each section, and with any odd number going across. An example would be hitting 1, 2, 11, 12 and 49, 50, 59, 60.
Examples of boxes that would NOT count are 11, 12, 21, 22 or 2, 3, 12, 13.

So, if you're playing verticals and you hit 2 rows of verticals (at $1.90 per game) you would win $1,000. If more rows come in, you get paid more. Same with boxes.

What I can't seem to figure is what are the actual odds of winning this game on a single ticket? How would the odds look if someone were to get one ticket that was good for 50 games (yes - a $95 ticket)?

If you are playing two rows of four then you are essentially playing an 8-spot. The frequency of the solid 8 is 230,115.

I once holed up in a work mission in Lincoln, Nebraska. I remember they had keno in the bars. If I remember correctly the numbers were drawn from a central location and broadcast on TV's in the bars.

Yeah, each city in the area has its own keno hall. You can play/drink/eat there, or go to bars that participate and receive the broadcast, print tickets, etc.

The thing about boxes and verticals though, is it's not just an 8 spot - as in having the same odds if you were to pick 8 numbers. There are multiple ways to hit verticals or boxes - but considering that they have to be in a certain order, I can't tell if the odds would be better or worse.

I get that 2 rows of 4 is an 8 spot - but when there are 20 different rows and 20 balls drawn, what are the odds there?

Quote: heyimjason

Yeah, each city in the area has its own keno hall. You can play/drink/eat there, or go to bars that participate and receive the broadcast, print tickets, etc.

The thing about boxes and verticals though, is it's not just an 8 spot - as in having the same odds if you were to pick 8 numbers. There are multiple ways to hit verticals or boxes - but considering that they have to be in a certain order, I can't tell if the odds would be better or worse.

I get that 2 rows of 4 is an 8 spot - but when there are 20 different rows and 20 balls drawn, what are the odds there?

Hopefully, I get this right. If I don't someone will jump in and tell us. If you were to add a third row then you could play three ways of 8. So dividing 230,115 by three would put your chances at 76,705. Adding a 4th row you would be playing 6 ways of 8 and the frequency would be 38,352.

With 20 rows you would be playing 190 ways of 8 (20X19/2). The frequecy would be 230,115/190 = 1211

Heyimjason, if you tell us what the cost of buying 20 rows is, and what the payoffs are, then we can tell you the house edge is in the game.

I did in the first post - when you play boxes or verticals, you get the whole ticket. It's $1.90 (minimum) per game. You're not betting on a specific box or vertical... so ANY 2 rows of 4 verticals will work on the ticket. For verticals - If you just get 1 row then 3 out of 4 on another, it's $10 - which doesn't matter to me. If you get 2 rows of 4, however, it's $1,000. I want to know what the odds are of getting 2 rows of 4 on these tickets (which have 20 rows, 80 numbers, 20 balls drawn).

So there are 190 ways to win. Essentially as other point-out its a win based on 8 spots... theres 190 of them at ONE PENNY apiece.

At $0.01 paying $1000 for 8/8 is 100,000 for 1 (1000/0.01). Same thing applies to the 7/8... the $10 payout PER PENNY is 1000 for 1.
Doing the math all 8 is 100,000/230,114.6 and 7/8 is 1000/6232.3
From what you posted on payments these two payments are a 59.5% return.

Ok, I think I got most of that. I want to ask if the odds of getting any of the 8 spots is 1 in 230,114, but considering that there are 190 different 8 spots on one game, I think it's closer to 1,200 - as someone mentioned earlier. Correct?
And, thanks for all the replies.

Now - here's another 8 spot related question. There are 80 spots, so I could pick every single number on a ticket (basically, just order 10 different 8 spots). This would be 10 cents per game minimum (1 cent for each set of 8), and catching 8/8 would pay $1,000. What are the odds that I would get an 8 spot during one game while playing 10 of them?

Here's partly why I'm asking (besides just being a generally curious person - who happens to be terrible with math): When it comes to slow games like keno, I'm a "set it and forget it" kinda guy. It's why I love my crock pot. I like to order a ticket, then go socialize, shoot pool, drink, all that. I believe the maximum for a ticket is $100, and I know the maximum length is 100 games. I could order a ticket for boxes or verticals good for 50 games ($95), then just check it at the end of the night or the next day. Or, I could order a ticket for ten 8 spots good for 100 games (only $10) and check that the next day.

So.. yeah, back to the 2nd question - how do the odds and returns look like on playing ten 8 spots? Considering the much lower price (fewer ways to win), I'm inclined to believe that the odds are the same.

1 in 1211 for 230114/190. That will cost you $2300+ on the average.

Quote: mickeycrimm

Hopefully, I get this right. If I don't someone will jump in and tell us. If you were to add a third row then you could play three ways of 8. So dividing 230,115 by three would put your chances at 76,705. Adding a 4th row you would be playing 6 ways of 8 and the frequency would be 38,352.

With 20 rows you would be playing 190 ways of 8 (20X19/2). The frequecy would be 230,115/190 = 1211

My results agree with those of MickeyCrimm, if you break the eights down into semi-columns, my method is:

nCr(4,4)*nCr(76,16)/nCr(80,20)=0.0030633923038986335

nCr(4,4)*nCr(72,12)/nCr(76,16)=0.0014185779146125218

0.0030633923038986335*0.0014185779146125218= 0.00000434566

1/0.00000434566 = 1 in 230114.643115

Like Mickey said, there are 190 possible combinations of semi-columns, with 1-11-21-31 working as the first semi-column and then going across before moving to the bottom half:

1,2-1,3-1,4-1,5-1,6-1,7,1-8,1-9,1-10,1-11,1-12,1-13,1-14,1-15,1,16-1,17-1,18-1,19,1,20 = (19 Pairs)
2,3-2,4-2,5-2,6-2,7-2,8-2,9-2,10-2,11-2,12-2,13-2,14-2,15-2,16-2,17-2,18-2,19-2,20 = (18 Pairs)

As you see, with each new starting column, there will be one fewer column pairs than the starting column before because it (as with 2,1 above) has already been included:

19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 190

Mickey's way is an easy way to account for the fact that you would not repeat pairs of semi-columns, but I wanted to show you the long way so you see why that works.

Okay, so if the probability of hitting two semi-columns for the solid eight is 0.00000434566, then the probability of it not hitting is (1-0.00000434566) and you would be covering all columns and that is the same as it not hitting 190 consecutive times:

(1-0.00000434566)^190 = 0.99917466358

Which means the probability of that hitting is: 1/(1-0.99917466358) = 1 in 1211.62713261

CONTINUED

Okay, so now we have to look at the possibility of hitting more than two semi-columns, because you get paid more as a result.

The probability of hitting three semi-columns is the same as hitting 12/12, which probably isn't going to happen:

nCr(4,4)*nCr(76,16)/nCr(80,20)=0.0030633923038986335

nCr(4,4)*nCr(72,12)/nCr(76,16)=0.0014185779146125218

nCr(4,4)*nCr(68,8)/nCr(72,12)=0.00048114775610182836

0.00048114775610182836*0.0014185779146125218*0.0030633923038986335= 2.09090488e-9 or 0.00000000209090488

1/0.00000000209090488 = 1 in 478261832.7429605

Okay, now that covers the probability of hitting 12,12, so the question is how many of those are covered?

1,2,3-1,2,4-1,2,5-1,2,6-1,2,7-1,2,8-1,2,9-1,2,10-1,2,11-1,2,12-1,2,13-1,2,14-1,2,15-1,2,16-1,2,17-1,2,18-1,2,19-1,2,20 (18 combinations)

(18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1) = 171 total combinations---which is (19*18)/2)---

The probability of it not happening is (1-0.00000000209090488) and it would need to not happen 171 consecutive times, so:

(1-0.00000000209090488)^171 = 0.9999996424553267

Okay, if this happened on Columns 1, 2 and 3, you'd get paid on 1+2, 1+3 and 2+3, so you would get $3000 in this case.

(1-0.9999996424553267) * 3000 = 0.001072634019938512

Okay, so we see that your value on this particular result is just slightly greater than a tenth of a penny, so there is no real point in doing 4/4 semi-columns or 5/5 semi-columns.

CONCLUSION

Since the other column possibilities don't merit accounting for, we're going to say that you would cover 190 combinations of two semi-columns with a probability of winning of (1-0.99917466358) and a probability of losing of (0.99917466358) that means that your expected loss on $1.90 bet is:

((1-0.99917466358)*1000) - (1.9 *0.99917466358) = -1.0730954408019553

Since you said you don't care about the 6/7 pay, I'm not going to do it.

Okay, so we're going to take the probability of that solid eight not hitting on any combination of semi-columns and we want to see how often that happens fifty times in a row:

(0.99917466358)^50 = 0.9595567116313287

Thus, if you play fifty tickets, there is a 95.95567116313287% probability that all of them will fail to hit for the $1,000.

Do you want to know how many tickets you have to play to have a 50% chance of all of them losing?

(0.99917466358)^840 = 0.49978921172369594

BONUS

After you have played 840 games, you are slightly more likely to have won once (or more) than to have lost all of them.

Specifically, using the binomial equation:

Win only once: (0.500210788276304-0.153429416423046) = 0.34678137185

Win Twice: (0.153429416423046-0.03326460136617) = 0.12016481505

Win Thrice: (0.03326460136617-0.005538444042267) = 0.02772615732

Win Four Times: (0.005538444042267-0.000746135771811041) = 0.00479230827

If you want to know for any more than that number of times out of 840, or if you want to specify a different number of trials, go here:

http://stattrek.com/online-calculator/binomial.aspx

AND:

Put: 0.0008253364200000446 for probability of success on a single trial

Number of trials you want where it says, perhaps unsurprisingly, "Number of Trials."

And, then, the desired number of wins in that number of trials.

Have a pleasant day, welcome to WoV Forums.

Quote: heyimjason

Ok, I think I got most of that. I want to ask if the odds of getting any of the 8 spots is 1 in 230,114, but considering that there are 190 different 8 spots on one game, I think it's closer to 1,200 - as someone mentioned earlier. Correct?
And, thanks for all the replies.

Now - here's another 8 spot related question. There are 80 spots, so I could pick every single number on a ticket (basically, just order 10 different 8 spots). This would be 10 cents per game minimum (1 cent for each set of 8), and catching 8/8 would pay $1,000. What are the odds that I would get an 8 spot during one game while playing 10 of them?

Here's partly why I'm asking (besides just being a generally curious person - who happens to be terrible with math): When it comes to slow games like keno, I'm a "set it and forget it" kinda guy. It's why I love my crock pot. I like to order a ticket, then go socialize, shoot pool, drink, all that. I believe the maximum for a ticket is $100, and I know the maximum length is 100 games. I could order a ticket for boxes or verticals good for 50 games ($95), then just check it at the end of the night or the next day. Or, I could order a ticket for ten 8 spots good for 100 games (only $10) and check that the next day.

So.. yeah, back to the 2nd question - how do the odds and returns look like on playing ten 8 spots? Considering the much lower price (fewer ways to win), I'm inclined to believe that the odds are the same.

The odds are always exactly the sane no matter how many rows you are playing If you are playing 10 rows then it is 10X9 divided by or 450 Divide 239,115 by 450. You are not going to find a way to beat this game with the payoffs listed. All we can do is deliver you the truth.

One thing that has always pissed me off is fancy mathematicians with fancy equations. It can be explained in the simple so that any third grader can understand it. A fancy mathematian leaves out 95% of the public because they have no clue what the hell he is talking about I put equations in the raw so that even a third grader can understand it.

Quote: Mission146

My results agree with those of MickeyCrimm, if you break the eights down into semi-columns, my method is:

nCr(4,4)*nCr(76,16)/nCr(80,20)=0.0030633923038986335

nCr(4,4)*nCr(72,12)/nCr(76,16)=0.0014185779146125218

0.0030633923038986335*0.0014185779146125218= 0.00000434566

1/0.00000434566 = 1 in 230114.643115

Like Mickey said, there are 190 possible combinations of semi-columns, with 1-11-21-31 working as the first semi-column and then going across before moving to the bottom half:

1,2-1,3-1,4-1,5-1,6-1,7,1-8,1-9,1-10,1-11,1-12,1-13,1-14,1-15,1,16-1,17-1,18-1,19,1,20 = (19 Pairs)
2,3-2,4-2,5-2,6-2,7-2,8-2,9-2,10-2,11-2,12-2,13-2,14-2,15-2,16-2,17-2,18-2,19-2,20 = (18 Pairs)

As you see, with each new starting column, there will be one fewer column pairs than the starting column before because it (as with 2,1 above) has already been included:

19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 190

Mickey's way is an easy way to account for the fact that you would not repeat pairs of semi-columns, but I wanted to show you the long way so you see why that works.

Okay, so if the probability of hitting two semi-columns for the solid eight is 0.00000434566, then the probability of it not hitting is (1-0.00000434566) and you would be covering all columns and that is the same as it not hitting 190 consecutive times:

(1-0.00000434566)^190 = 0.99917466358

Which means the probability of that hitting is: 1/(1-0.99917466358) = 1 in 1211.62713261

CONTINUED

Okay, so now we have to look at the possibility of hitting more than two semi-columns, because you get paid more as a result.

The probability of hitting three semi-columns is the same as hitting 12/12, which probably isn't going to happen:

nCr(4,4)*nCr(76,16)/nCr(80,20)=0.0030633923038986335

nCr(4,4)*nCr(72,12)/nCr(76,16)=0.0014185779146125218

nCr(4,4)*nCr(68,8)/nCr(72,12)=0.00048114775610182836

0.00048114775610182836*0.0014185779146125218*0.0030633923038986335= 2.09090488e-9 or 0.00000000209090488

1/0.00000000209090488 = 1 in 478261832.7429605

Okay, now that covers the probability of hitting 12,12, so the question is how many of those are covered?

1,2,3-1,2,4-1,2,5-1,2,6-1,2,7-1,2,8-1,2,9-1,2,10-1,2,11-1,2,12-1,2,13-1,2,14-1,2,15-1,2,16-1,2,17-1,2,18-1,2,19-1,2,20 (18 combinations)

(18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1) = 171 total combinations---which is (19*18)/2)---

The probability of it not happening is (1-0.00000000209090488) and it would need to not happen 171 consecutive times, so:

(1-0.00000000209090488)^171 = 0.9999996424553267

Okay, if this happened on Columns 1, 2 and 3, you'd get paid on 1+2, 1+3 and 2+3, so you would get $3000 in this case.

(1-0.9999996424553267) * 3000 = 0.001072634019938512

Okay, so we see that your value on this particular result is just slightly greater than a tenth of a penny, so there is no real point in doing 4/4 semi-columns or 5/5 semi-columns.

CONCLUSION

Since the other column possibilities don't merit accounting for, we're going to say that you would cover 190 combinations of two semi-columns with a probability of winning of (1-0.99917466358) and a probability of losing of (0.99917466358) that means that your expected loss on $1.90 bet is:

((1-0.99917466358)*1000) - (1.9 *0.99917466358) = -1.0730954408019553

Since you said you don't care about the 6/7 pay, I'm not going to do it.

Okay, so we're going to take the probability of that solid eight not hitting on any combination of semi-columns and we want to see how often that happens fifty times in a row:

(0.99917466358)^50 = 0.9595567116313287

Thus, if you play fifty tickets, there is a 95.95567116313287% probability that all of them will fail to hit for the $1,000.

Do you want to know how many tickets you have to play to have a 50% chance of all of them losing?

(0.99917466358)^840 = 0.49978921172369594

BONUS

After you have played 840 games, you are slightly more likely to have won once (or more) than to have lost all of them.

Specifically, using the binomial equation:

Win only once: (0.500210788276304-0.153429416423046) = 0.34678137185

Win Twice: (0.153429416423046-0.03326460136617) = 0.12016481505

Win Thrice: (0.03326460136617-0.005538444042267) = 0.02772615732

Win Four Times: (0.005538444042267-0.000746135771811041) = 0.00479230827

If you want to know for any more than that number of times out of 840, or if you want to specify a different number of trials, go here:


AND:

Put: 0.0008253364200000446 for probability of success on a single trial

Number of trials you want where it says, perhaps unsurprisingly, "Number of Trials."

And, then, the desired number of wins in that number of trials.

Have a pleasant day, welcome to WoV Forums.

Woah. I didn't get half of that, but I appreciate the half I did get. Thanks for taking the time to do that.

Quote: mickeycrimm

The odds are always exactly the sane no matter how many rows you are playing If you are playing 10 rows then it is 10X9 divided by or 450 Divide 239,115 by 450. You are not going to find a way to beat this game with the payoffs listed. All we can do is deliver you the truth.

Oh, I don't expect there's any way to beat this game. I just want to know the odds of the games that pay well, because it's the only game offered at the bars.

I'm asking about the whole 8 spot thing because the first time I tried it, my buddy and I walked into the bowling alley, ordered our lane and shoes, grabbed a pitcher of beer, and went halvsies on a boxes ticket good for 5 games. The first or second game that came up after we got the ticket was boxes, so we ended up having pretty good night (when you're in your early 20's and fairly broke, $500 each is a nice chunk). Also, at one point that night, I was talking to a guy when I was getting more beer - he said he did the $10 tickets in which he'd simply play penny 8's with ten 8 spots - so it was good for 100 games. Every day or two, he'd come in, have a beer, and have them check/replay the same ticket. He said he'd been doing that for 3 years, and that he'd hit 3 times per year on average. In retrospect, if that was all true, he was probably just about breaking even. Hm.

I'm sure there are much better keno games... I dunno, maybe 10 or 25 cent 6 spots, or those strange keno tickets with groups of numbers that all play both independently AND with each other (to make larger groups). I talked to a guy that always played some weird ticket that was around 6 or 7 bucks per game, but it was something crazy like a 5 cent 8 spot, two 4 spots, four 2 spots, each with their own different wagers, and so on... but it was a lot more complex than that. I'm sure the odds were still against him, but I'd never seen the guy play more than 4 or 5 games, and it seems like damn near half the time he played, he walked out with something, and he seemed to have nice ($500+) wins fairly often. I really wish I remembered how his damn groups worked.

Quote: heyimjason

Quote: Mission146

My results agree with those of MickeyCrimm, if you break the eights down into semi-columns, my method is:

nCr(4,4)*nCr(76,16)/nCr(80,20)=0.0030633923038986335

nCr(4,4)*nCr(72,12)/nCr(76,16)=0.0014185779146125218

0.0030633923038986335*0.0014185779146125218= 0.00000434566

1/0.00000434566 = 1 in 230114.643115

Like Mickey said, there are 190 possible combinations of semi-columns, with 1-11-21-31 working as the first semi-column and then going across before moving to the bottom half:

1,2-1,3-1,4-1,5-1,6-1,7,1-8,1-9,1-10,1-11,1-12,1-13,1-14,1-15,1,16-1,17-1,18-1,19,1,20 = (19 Pairs)
2,3-2,4-2,5-2,6-2,7-2,8-2,9-2,10-2,11-2,12-2,13-2,14-2,15-2,16-2,17-2,18-2,19-2,20 = (18 Pairs)

As you see, with each new starting column, there will be one fewer column pairs than the starting column before because it (as with 2,1 above) has already been included:

19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 190

Mickey's way is an easy way to account for the fact that you would not repeat pairs of semi-columns, but I wanted to show you the long way so you see why that works.

Okay, so if the probability of hitting two semi-columns for the solid eight is 0.00000434566, then the probability of it not hitting is (1-0.00000434566) and you would be covering all columns and that is the same as it not hitting 190 consecutive times:

(1-0.00000434566)^190 = 0.99917466358

Which means the probability of that hitting is: 1/(1-0.99917466358) = 1 in 1211.62713261

CONTINUED

Okay, so now we have to look at the possibility of hitting more than two semi-columns, because you get paid more as a result.

The probability of hitting three semi-columns is the same as hitting 12/12, which probably isn't going to happen:

nCr(4,4)*nCr(76,16)/nCr(80,20)=0.0030633923038986335

nCr(4,4)*nCr(72,12)/nCr(76,16)=0.0014185779146125218

nCr(4,4)*nCr(68,8)/nCr(72,12)=0.00048114775610182836

0.00048114775610182836*0.0014185779146125218*0.0030633923038986335= 2.09090488e-9 or 0.00000000209090488

1/0.00000000209090488 = 1 in 478261832.7429605

Okay, now that covers the probability of hitting 12,12, so the question is how many of those are covered?

1,2,3-1,2,4-1,2,5-1,2,6-1,2,7-1,2,8-1,2,9-1,2,10-1,2,11-1,2,12-1,2,13-1,2,14-1,2,15-1,2,16-1,2,17-1,2,18-1,2,19-1,2,20 (18 combinations)

(18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1) = 171 total combinations---which is (19*18)/2)---

The probability of it not happening is (1-0.00000000209090488) and it would need to not happen 171 consecutive times, so:

(1-0.00000000209090488)^171 = 0.9999996424553267

Okay, if this happened on Columns 1, 2 and 3, you'd get paid on 1+2, 1+3 and 2+3, so you would get $3000 in this case.

(1-0.9999996424553267) * 3000 = 0.001072634019938512

Okay, so we see that your value on this particular result is just slightly greater than a tenth of a penny, so there is no real point in doing 4/4 semi-columns or 5/5 semi-columns.

CONCLUSION

Since the other column possibilities don't merit accounting for, we're going to say that you would cover 190 combinations of two semi-columns with a probability of winning of (1-0.99917466358) and a probability of losing of (0.99917466358) that means that your expected loss on $1.90 bet is:

((1-0.99917466358)*1000) - (1.9 *0.99917466358) = -1.0730954408019553

Since you said you don't care about the 6/7 pay, I'm not going to do it.

Okay, so we're going to take the probability of that solid eight not hitting on any combination of semi-columns and we want to see how often that happens fifty times in a row:

(0.99917466358)^50 = 0.9595567116313287

Thus, if you play fifty tickets, there is a 95.95567116313287% probability that all of them will fail to hit for the $1,000.

Do you want to know how many tickets you have to play to have a 50% chance of all of them losing?

(0.99917466358)^840 = 0.49978921172369594

BONUS

After you have played 840 games, you are slightly more likely to have won once (or more) than to have lost all of them.

Specifically, using the binomial equation:

Win only once: (0.500210788276304-0.153429416423046) = 0.34678137185

Win Twice: (0.153429416423046-0.03326460136617) = 0.12016481505

Win Thrice: (0.03326460136617-0.005538444042267) = 0.02772615732

Win Four Times: (0.005538444042267-0.000746135771811041) = 0.00479230827

If you want to know for any more than that number of times out of 840, or if you want to specify a different number of trials, go here:

http://stattrek.com/online-calculator/binomial.aspx

AND:

Put: 0.0008253364200000446 for probability of success on a single trial

Number of trials you want where it says, perhaps unsurprisingly, "Number of Trials."

And, then, the desired number of wins in that number of trials.

Have a pleasant day, welcome to WoV Forums.

Woah. I didn't get half of that, but I appreciate the half I did get. Thanks for taking the time to do that.

It's a damn simple equation. It don't take a rocket scientist. For the life of me I can't figure out why they make it so complicated

It's a question of proofs, I agreed with your equations, of course, but I was intentionally doing it the long way to show why they work. Besides, he also wanted to know the probability of a win in fifty trials, so I did that.

Instead of playing this Keno game, he should send me $100 once a week and I'll send $70 back. Except, once a year, I'll send back $1000 and I'll end up taking a much lower percentage of his money than this Keno game.

The one potentially redeeming quality of Keno is that it can be about the ultimate in slow play. Potentially. The way they show them doing it in the movie 'Hard 8' for example, where IIRC they played one number for a buck, then waited for an outcome. Hard for the HE to kick in on betting about nothing.

I was going to do that in Harrington once, but asking a couple of employees where to make a bet, neither could tell me. Not making that up.

Quote: Mission146

It's a question of proofs, I agreed with your equations, of course, but I was intentionally doing it the long way to show why they work. Besides, he also wanted to know the probability of a win in fifty trials, so I did that.

Instead of playing this Keno game, he should send me $100 once a week and I'll send $70 back. Except, once a year, I'll send back $1000 and I'll end up taking a much lower percentage of his money than this Keno game.

Sorry, Mission. I would say that I was drunk. But that's no excuse. But I do have a pet peeve about fancy mathematicians. It goes back to my days of trying to figure gambling math out but couldn't get past the fancy code language.

Quote: odiousgambit

The one potentially redeeming quality of Keno is that it can be about the ultimate in slow play. Potentially. The way they show them doing it in the movie 'Hard 8' for example, where IIRC they played one number for a buck, then waited for an outcome. Hard for the HE to kick in on betting about nothing.

I was going to do that in Harrington once, but asking a couple of employees where to make a bet, neither could tell me. Not making that up.

I live in video keno heaven. I have several different video keno plays. There is a new play that just showed up. Its a regular nightmare analyzing the game. Its been a real pain in the ass but its a super strong play. You can play anywhere from two to ten spots. And each betting level has it's own payscale. When you make certain hits you get extra draws. So I'm having to do the math for 2,3,4,5,6,7,8.9. 10, 11, 12 13, 14, 15, 16, 17, 18 19, 20, extra draws. It's time consuming as hell.

Quote: mickeycrimm

I live in video keno heaven. I have several different video keno plays. There is a new play that just showed up. Its a regular nightmare analyzing the game. Its been a real pain in the ass but its a super strong play. You can play anywhere from two to ten spots. And each betting level has it's own payscale. When you make certain hits you get extra draws. So I'm having to do the math for 2,3,4,5,6,7,8.9. 10, 11, 12 13, 14, 15, 16, 17, 18 19, 20, extra draws. It's time consuming as hell.

I forgot to add that its a progressive play. But its not a money progressive. The progressives are free games, multipliers, and extra draws.

Quote: mickeycrimm

I forgot to add that its a progressive play. But its not a money progressive. The progressives are free games, multipliers, and extra draws.

Oh, that sounds juicy!

Quote: Mission146

It's a question of proofs, I agreed with your equations, of course, but I was intentionally doing it the long way to show why they work. Besides, he also wanted to know the probability of a win in fifty trials, so I did that.

Instead of playing this Keno game, he should send me $100 once a week and I'll send $70 back. Except, once a year, I'll send back $1000 and I'll end up taking a much lower percentage of his money than this Keno game.

It's one of those things... I'd rather be playing hold 'em, or possibly craps. But, it's at the bars, and that's the only game that's offered (or legal). When I play, I typically limit it to $20. I always plan on spending closer to $100, but the logical part of my brain just won't let me do that.
Then again, that logical part never shows up when I buy a weekly Powerball ticket. hah.

Big Red, was it? One of the straight games should have the lowest House Edge, but it still won't be very good. On my phone, now, but I'll check their site tomorrow for you.

Match all usually has better pay%'s. IIRC some of these joints pay $70 for all-3 on a $1.25 bet.

If I may go back to the OP of the 190 ways... given the payouts ( I also believe that 6/8 pays 50c) listed The player has about a 1 in 32 chance of winning $10 or more.

I'm at the bar right now, help me settle a bet.. What are the odds off hitting 3 out of 3 numbers in a game of keno?

Quote: heyimjason

I'm at the bar right now, help me settle a bet.. What are the odds off hitting 3 out of 3 numbers in a game of keno?

20 X 19 X 18/80 X 79 X 78 = 72.07

Quote: mickeycrimm

20 X 19 X 18/80 X 79 X 78 = 72.07

I took the shortcut. But I'm drinking righnow.

Total combinations

80 X 79 X 78/3 X 2 X 1= 82,160

Combinations that make a 3-spot

20 X 19 X 18/3 X 2 X 1 = 1,140


82160/1140 = 72.07