endy
endy
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March 22nd, 2020 at 4:41:14 PM permalink
I'm wondering if somebody could help me with a math question for the video keno game: Back 2 Back Keno. In this keno game, you get paid for keno wins in the usual way according to how many spots you mark, how many you hit and how much you bet (the pay tables vary a little bit for max betting vs. bet 1 and 1 < bet < max). In addition, if you win a game, you have a chance to win a "back 2 back" pay in the next game if you hit enough of the numbers you hit in the game before.

Example: in the first game, you mark 6 spots, and hit 3 and it pays 2 times your bet. On the next game, you play the same numbers, and if you hit the same 3 numbers (and no other winning numbers), then you'd get two pays: 2 times your bet (for hitting 3 this game), and 27 times your bet (for hitting the same 3 numbers this game that you hit in the game before).

So my question is: given you marked 6 spots and hit 3 this game, what are the odds of hitting those same 3 numbers the next game? More generally, how do I compute those odds for all the various mark X, catch Y, and catch Y back 2 back scenarios.

Thank you for your assistance!
ksdjdj
ksdjdj
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March 22nd, 2020 at 8:19:03 PM permalink
Hi and welcome to the forum,

Below is my attempt to answer your first question: "...you marked 6 spots and hit 3 this game, what are the odds of hitting those same 3 numbers the next game?..."

For the example you provided, I think it is about 1/555 chance of "3 hits from 6 spots and then hitting those same 3 numbers in the next draw from the 6 original numbers"

"Proof":
Chance of getting 3 hits from 6 spots = 12.98...%
Chance of getting 3 hits from 3 spots =1.387...%

12.98...% x 1.387...% = 0.1801...% (about 1/555)


Note: I didn't know how to work the figures above, so I just used the "keno calculator", see link below

https://wizardofodds.com/games/keno/calculator/

Lastly, even though I think I am correct in what I said above, you may want to wait until someone either confirms this, or answers your question(s) more completely.

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Edit (1 am, Pac time)

Sorry, after having another look, I now have it as about a 1/593 overall chance of happening and a 1.298...% chance if you have already hit 3 out of 6 in the first game (again, would recommend you wait for another person to confirm if this is correct, or not).

"proof (new)":

Chance of getting 3 hits from 6 spots = 12.98...%
Chance of getting the same 3 hits from 6 spots =1.298...% ***

12.98...% x 1.298...% = 0.1685...% (about 1/593)

***: The reason why i think this is correct, is there are 10 combinations/ways of getting 3 hits from 6 numbers, and you are looking for just 1 of those ways (the order shouldn't matter, and I hope I used the right terms).
Last edited by: ksdjdj on Mar 23, 2020
endy
endy
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ksdjdj
March 23rd, 2020 at 1:01:07 PM permalink
Thanks for your reply! I will noodle over these calculations.
ksdjdj
ksdjdj
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March 23rd, 2020 at 2:06:49 PM permalink
Quote: ksdjdj

(snip)
Edit (1 am, Pac time)

Sorry, after having another look, I now have it as about a 1/593 overall chance of happening and a 1.298...% chance if you have already hit 3 out of 6 in the first game (again, would recommend you wait for another person to confirm if this is correct, or not). (snip)


Sorry for giving you the wrong info (I needed to divide the figures above by 2).

So, it has about a 1/1,187 overall chance, and
about 0.649% chance if you have already hit 3 out of 6 in the first game (since there are 20 combinations in the first game, and you are just looking for 1 of those to match in the second game)

Here are some helpful links for working out the combinations

Note: I used numbers 1,2,3,4,5 and 6 as my "6 keno picks", but it works for any "picks" (when n is "the number of picks" and r is "how many numbers are used/hit").

https://www.mathsisfun.com/pascals-triangle.html
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html ***


***:
Combinations without repetition (n=6, r=3)
Using Items: 1,2,3,4,5,6

List has 20 entries.
{1,2,3} {1,2,4} {1,2,5} {1,2,6} {1,3,4} {1,3,5} {1,3,6} {1,4,5} {1,4,6} {1,5,6} {2,3,4} {2,3,5} {2,3,6} {2,4,5} {2,4,6} {2,5,6} {3,4,5} {3,4,6} {3,5,6} {4,5,6}

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Since I made so many mistakes in my first reply, I think you should wait for someone to confirm this (for "peace of mind").
endy
endy
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March 23rd, 2020 at 3:47:55 PM permalink
Thanks again for your updated reply! I will continue noodling.
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