ExhaustedTech
ExhaustedTech
Joined: Feb 16, 2022
  • Threads: 1
  • Posts: 7
February 16th, 2022 at 7:44:44 PM permalink
I'm trying to make heads or tails of the Wizard of Odds site but can't figure it out. Essentially, I'm trying to calculate the odds of winning dual dab in 35,36,37,38 numbers with 100 cards, as well as 8,000 cards for each.
Where I play, there are roughly 200 people per session and around 40 cards for each player and I buy 100 cards. What are my chances personally and for at least one person in the session to win?

Does anyone know how to calculate this exactly?
ddloml
ddloml
Joined: Dec 5, 2009
  • Threads: 1
  • Posts: 78
February 16th, 2022 at 9:39:51 PM permalink
Do you really buy 100 Dual Dab cards at a session? That game usually costs $1 or $2 per card at most casinos, costing you at least $100 to play just that one game.

Yes, most players will buy 40 or more cards for the regular games. The blue cards, which pay about $50 per game, cost about $0.15 per card. And each card is usually used for two or three games (eg. single hardway into double hardway into full card), making the cost per game even lower.

However, Dual Dab is a different animal. That card is only used for one game at that much higher price. I would think that most players buy a handful of cards, at most, for Dual Dab. Personally, I buy two of them, and Iíve won the $250 consolation prize once or twice.
ExhaustedTech
ExhaustedTech
Joined: Feb 16, 2022
  • Threads: 1
  • Posts: 7
February 16th, 2022 at 10:41:20 PM permalink
If the jackpot is worth it, yes. I generally spend $400-500 per session. Roughly 300 for regular games and another 200 for jackpot games.
Sometimes I only buy 50, if it's less than $20k, then spend the rest on higher jp games.
Since nearly all casinos have DD jackpots, that's usually the one I buy the most of. I've hit the JP once and it was fairly low, around 15k, in only 32 numbers. I know that is a one-off and very rare, that's why I'm asking about actual odds for 35-38 calls with 100 cards.
Vegasrider
Vegasrider
Joined: Dec 23, 2017
  • Threads: 78
  • Posts: 844
February 16th, 2022 at 11:47:47 PM permalink
Is this at an Indian Casino. Most places have a max on the progressive card buy ins. However I have been known to hire shills to sit in front of the machines that I loaded.
ExhaustedTech
ExhaustedTech
Joined: Feb 16, 2022
  • Threads: 1
  • Posts: 7
February 17th, 2022 at 7:39:29 AM permalink
For the particular one I'm about to go to, yes, it's an Indian casino. Though that is generally how I'd play in Vegas as well. Some places have a 50 card Max and others are 90
Vegasrider
Vegasrider
Joined: Dec 23, 2017
  • Threads: 78
  • Posts: 844
February 17th, 2022 at 9:04:18 AM permalink
Generally for DD, I will not play them until it gets to 36 numbers and 56 for coveralls. I always max out the cards too but itís important and try to select a session that would draw the least amount of people. Either early morning or very late weekday sessions. This is because itís important to hit the consolation prize from time to time so you donít take a bath burying yourself chasing the progressive.

In Vegas, where there are so many Bingo rooms, I try to stay away from the large cash ball or large progressive sessions and focus playing at other venues when those games are in progressive. The rooms are emptier.

The Pandemic was the best thing for Bingo. I took advantage of the occupancy level cap and they never adjusted the max on the cards. There were times when I played there were less than 20 people playing and I hit about 40-50% of the Bingos.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 110
  • Posts: 5412
February 17th, 2022 at 11:01:40 AM permalink
Quote: ExhaustedTech

Does anyone know how to calculate this exactly?
link to original post


The problem with "exactly" is, you have to make some assumptions about what is on everybody else's cards.

Here's something that should be "close enough."
First, calculate the probability of getting a coverall in 35 (or fewer) numbers.
To simplify it, instead of using numbers, I will use colors. There are 25 colors; 24 of them have 2 balls each, representing the 2 numbers in the same square on the card, and the 25th color has 27, representing the 27 numbers not on the card. You need to pull out 35 balls such that you have at least one of each of the first 12 colors.
If you draw 1 of each square color, there will be 11 of the 25th color drawn, so the number of ways to do it is 2^24 x C(27,11).
If you draw 2 of one square's color and 1 of each of the other 23, there are C(24,1) ways to select the one 2-ball square, 2^23 ways to draw 1 ball of each of the other 23 square colors, and C(27,10) ways to select the 10 balls not in any square.
If you draw 2 of 2 squares' colors and 1 of each of the other 22, there are C(24,2) ways to select the two 2-ball squares, 2^22 ways to draw 1 ball of each of the other 22 square colors, and C(27,9) ways to select the 9 balls not in any square.
And so on, down to having 2 of 11 different squares' colors. You cannot have 12 squares where you draw both balls as you would need another 12 balls to complete the coverall, but that's a total of 36 balls drawn.
The probability of winning = the sum of these numbers divided by the number of ways to draw 35 balls out of 75, which is C(75,35).
The value is about 1 / 79,404.
Since you have 100 cards, the probability that at least one is a winner is about 1 / 794.
However, you then need to multiply this by the probability that none of the other 8000 cards do; this is (1 - 1 / 794)^8000, or about 0.90416. The final answer is about 1 / 878.
The probabilty that at least one of the 8100 cards wins is 1 minus the probability that none do; this is about 1 - (1 - 1 /79,404)^8100, or about 1 in 10.3.

(Did I say I get about 1/20 in simulation? When I simulate it a different way, it is close to 1 in 10.3.)
Last edited by: ThatDonGuy on Feb 17, 2022
ExhaustedTech
ExhaustedTech
Joined: Feb 16, 2022
  • Threads: 1
  • Posts: 7
February 17th, 2022 at 11:28:40 AM permalink
Well I go only a couple times a year and don't really have the luxury of choosing one with the least player base. I go when the trip is booked. If I lived near some Bingo halls, I would definitely do that! I'm very aware that I have less chances to win the more people there are, however, I'm also not going to waste a trip on only winning $50-$150 a game so there needs to be feasible jackpots to win for it to be worth it for me. I know the one I'm about to go to has 200 players per session on average and upto 350 during special events but it's the only one in a 4 hour radius.
ExhaustedTech
ExhaustedTech
Joined: Feb 16, 2022
  • Threads: 1
  • Posts: 7
February 17th, 2022 at 12:00:39 PM permalink
Thank you!
Vegasrider
Vegasrider
Joined: Dec 23, 2017
  • Threads: 78
  • Posts: 844
February 17th, 2022 at 12:58:21 PM permalink
Quote: ExhaustedTech

Well I go only a couple times a year and don't really have the luxury of choosing one with the least player base. I go when the trip is booked. If I lived near some Bingo halls, I would definitely do that! I'm very aware that I have less chances to win the more people there are, however, I'm also not going to waste a trip on only winning $50-$150 a game so there needs to be feasible jackpots to win for it to be worth it for me. I know the one I'm about to go to has 200 players per session on average and upto 350 during special events but it's the only one in a 4 hour radius.
link to original post



You should come to Vegas or even Reno when the casinos host the big Bingo events. High dollar buy-ins and the bingo pays over a $1000 with $20k or 50k coveralls. Itís usually a 2 day event. Iím sure you have gotten mailers about those events. Waay to many players for me to play, despite the huge payouts.

  • Jump to: