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I believe the maximum for a ticket is $100, and I know the maximum length is 100 games. I like to order a ticket, then go socialize, shoot pool, drink, all that. 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. What are the odds that I would get an 8 spot during one game while playing 10 of them? This would be 10 cents per game minimum (1 cent for each set of 8), and catching 8/8 would pay $1,000. There are 80 spots, so I could pick every single number on a ticket (basically, just order 10 different 8 spots). Now - here's another 8 spot related question. 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. 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)? So, if you're playing verticals and you hit 2 rows of verticals (at $1.90 per game) you would win $1,000.
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It must be 2 vertical rows of 4.Īs far as boxes go, the numbers must start on the first or third line in each section, and with any odd number going across. The official tickets have the same format as the play slips - here's an example of a play slip:Īn example of winning a verticals game would be hitting 1, 11, 21, 31 as well as 48, 58, 68, 78. 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. There's a Keno game at the bars in Nebraska in which one may bet for boxes, or one may bet for verticals.