Do it Yourself BINGO question

San Manuel casino here in San Bernardino California has Do it Yourself BINGO every night except Thursday. The jackpot starts at $15,000 and grows each week if there are no winners. You must pick 8 numbers from 1 to 75. the casino draws numbers at random. if you match all 8 of your numbers in 20 numbers or less you win the jackpot. if no one does this then the casino continues drawing numbers until someone matches all 8 for the first consolation prize. There are two consolation prizes of $1,000 each. You can play as many sets of numbers as you want. A single set of 8 cost $1 each and multiple sets of 7 cost $5 each. My question is how do I figure the odds of winning for this game? I would also like to know the odds of winning within 8 numbers called, 9, 10 etc... I understand how to calculate the possibilities, for example the number of possible sets of 8 from 75 (75C8) is 16,871,053,725 etc. I do not think the odds of winning playing one set cannot be 1 in 16,871,053,725 as you have to match 8 out of 20 and not 8 out of 8. I also believe the odds change the more sets you play and that the number of players also affect odds. Any help would be appreciated.

Select Match Odds To 1

8 0 13.8563731865619
8 1 4.15691195596856
8 2 3.06298775702947
8 3 4.25414966254092
8 4 10.2099591900982
8 5 41.477959209774
8 6 293.11091174907
8 7 3956.99730861244
8 8 133929.139676113

Keno 75/20 Spot=: 8 (Same as 75 Bingo numbers, 20 drawn, must match 8: the last number is what you're looking for)

from this calculator site, where you can plug in the numbers you're wondering about (8 numbers called, 9, 10, etc...from your post above).

And here's how the numbers are calculated:

There are (75)C(20) = 803,167,998,494,073,240 ways to draw 20 numbers out of 75.
Color the balls of your eight chosen numbers red, and the other 67 numbers white.

For 8 out of 8, the 20 balls drawn are 8 red and 12 white.
There is (8)C(8) = 1 way to draw 8 red balls out of 8, and (67)C(12) = 5,996,962,277,488 ways to draw 12 white balls out of 67.
The probability of 8 out of 8 = (1 x 5,996,962,277,488) / 803,167,998,494,073,240 = 1 / 133929.139676113.

For 7 out of 8, the 20 balls drawn are 7 red and 13 white.
There are (8)C(7) = 8 ways to draw 7 red balls out of 8, and (67)C(13) = 25,371,763,481,680 ways to draw 13 white balls out of 67.
The probability of 7 out of 8 = (8 x 25,371,763,481,680) / 803,167,998,494,073,240 = 1 / 3956.99730861244.

Quote: ThatDonGuy

And here's how the numbers are calculated:

There are (75)C(20) = 803,167,998,494,073,240 ways to draw 20 numbers out of 75.
Color the balls of your eight chosen numbers red, and the other 67 numbers white.

For 8 out of 8, the 20 balls drawn are 8 red and 12 white.
There is (8)C(8) = 1 way to draw 8 red balls out of 8, and (67)C(12) = 5,996,962,277,488 ways to draw 12 white balls out of 67.
The probability of 8 out of 8 = (1 x 5,996,962,277,488) / 803,167,998,494,073,240 = 1 / 133929.139676113.

For 7 out of 8, the 20 balls drawn are 7 red and 13 white.
There are (8)C(7) = 8 ways to draw 7 red balls out of 8, and (67)C(13) = 25,371,763,481,680 ways to draw 13 white balls out of 67.
The probability of 7 out of 8 = (8 x 25,371,763,481,680) / 803,167,998,494,073,240 = 1 / 3956.99730861244.

Thanks, TDG! Wasn't sure anyone else was going to answer, so I directed him to a calc. Much prefer that you proofed it out.

Thanks for the quick response. I never thought about KENO most likely because I never played. one thing to note is that with BINGO they will continue to call numbers past 20 until two people win and unlike KENO you always have to match all eight numbers. There is only one winner in 20 numbers or less. There was on game in which they called 40 numbers before anyone called BINGO and then another 5 numbers before the second BINGO. I regularly play BINGO and I often play 100 sets or more in the Do It Yourself BINGO hoping to get better odds. The information ThatDonGuy and yourself provided is great. I am currently looking into what the minimum number of sets I would need to play in order to guarantee a win at 20 numbers even though the number would be to great to be physically possible. I am just curious to know what that number actually is. I also think that if I could come up with said set and played from it I would have better odds but that is probably just wishful thinking.

Quote: jbeck2862

I am currently looking into what the minimum number of sets I would need to play in order to guarantee a win at 20 numbers even though the number would be to great to be physically possible. I am just curious to know what that number actually is. I also think that if I could come up with said set and played from it I would have better odds but that is probably just wishful thinking.

This sort of problem is well-known, and while an "exact" formula still escapes mathematicians, somebody came across a lower bound for the value; in this case, you need at least 275,737 tickets for at least one 8-number hit and 64,966 tickets for at least one 7-number hit.

The lower bound (i.e. the actual number >= this) for a draw of p numbers out of n where each ticket has k numbers and you want to match at least t on at least one ticket is:

combin(n,k) x (n - p + 1) / (combin(p-1, t-1) * combin(k, t) * (n - t + 1))

In this case, n = 75, p = 20, k = 8, and t = 7 or 8 as appropriate.

See theorem 1.16 here

You could always partner up with a casino worker and cheat the game like they did at my local.

http://www.jsonline.com/news/milwaukee/112642749.html