what happens if more than 1 players in the hall call bingo?
September 20th, 2014 at 10:54:08 PM
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Hi all,
Today I got a bingo cage kit from a store. I don't have the bingo cards so I generated the card in computer and playing the game with my friends. We use 75 balls, 5x5 (24 numbers and 1 free spot) card. We have 8 people playing the game. It will call out all 75 numbers until bingo. After some game, it happens that 3 of use call out the bingo at the same time. I wonder in casino or other online system, will all 3 claimed as the legal winner or the very first one call out bingo will be the winner? If it is the very first one will be counted as the winner, how does the banker tell who is the first one if they call out at about the same time?
Ok. Besides bring my above question here. I am thinking something on the math. I am pretty new to bingo game, I didn't play that before but the way the game play is pretty attracting. I decide to calculate the probability of winning a bingo. We set the game rule to be
1) using 75 balls
2) 5x5 cards, 24 numbers and 1 wild in the center
3) 12 winning pattern: 5 columns, 5 rows, two diagonals
4) each game draw at max 55 numbers
To start with, I did a simulation and it turns out that the winning probability with 55 numbers drawn is 0.844. So is this the correct number for a single player to win the game?
Anyway, after playing the game with 8 people today. I am pretty sure that I won't win the game at that high probability. I think the reason is for each game, within 75 draw, other players have the same probability to win the game. Whoever other than myself win the game before me will cause my losing the game. So the real probability of winning a bingo game against 8 people is not that high. I have been thinking on this math for whole night, I think this seems to be a time-dependent issue because it depends who is the first one win the game. So I wonder if the only way to figure the probability is by simulation?
Today I got a bingo cage kit from a store. I don't have the bingo cards so I generated the card in computer and playing the game with my friends. We use 75 balls, 5x5 (24 numbers and 1 free spot) card. We have 8 people playing the game. It will call out all 75 numbers until bingo. After some game, it happens that 3 of use call out the bingo at the same time. I wonder in casino or other online system, will all 3 claimed as the legal winner or the very first one call out bingo will be the winner? If it is the very first one will be counted as the winner, how does the banker tell who is the first one if they call out at about the same time?
Ok. Besides bring my above question here. I am thinking something on the math. I am pretty new to bingo game, I didn't play that before but the way the game play is pretty attracting. I decide to calculate the probability of winning a bingo. We set the game rule to be
1) using 75 balls
2) 5x5 cards, 24 numbers and 1 wild in the center
3) 12 winning pattern: 5 columns, 5 rows, two diagonals
4) each game draw at max 55 numbers
To start with, I did a simulation and it turns out that the winning probability with 55 numbers drawn is 0.844. So is this the correct number for a single player to win the game?
Anyway, after playing the game with 8 people today. I am pretty sure that I won't win the game at that high probability. I think the reason is for each game, within 75 draw, other players have the same probability to win the game. Whoever other than myself win the game before me will cause my losing the game. So the real probability of winning a bingo game against 8 people is not that high. I have been thinking on this math for whole night, I think this seems to be a time-dependent issue because it depends who is the first one win the game. So I wonder if the only way to figure the probability is by simulation?
September 21st, 2014 at 1:16:21 AM
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I am still working on the second question but didn't have any clue. Before that I would like to study the probability of winning the BINGO within N draws (N<75). To start with, I read the article found on http://www.durangobill.com/BingoHowTo.html
Up to now, everything looks good to me. I wrote the code to test it and I obtain the same number as above. The author also said that if we want to find the probability of BINGO after 10 numbers out of 75 drawn, it is given that
where the numbers before the equal sign are from the above table. So the first formula interprets as
1) the probability of that we have 4 hits on a card and those 4 hits happens to make a bingo pattern is 0.000376
2) there is total COMBIN(75, 10) ways to draw 10 numbers out of 75
3) in the 10 numbers drawn from 75, the number of ways that 4 of them filled the 4-number pattern and other 6 are not on the card (i.e. 6 are drawn from 51 numbers which are the numbers not on the 24 numbers in a card) should be COMBIN(24, 4) * COMBIN(51, 6)
Based on that explanation, we add all above 7 numbers will get the probability to win the BINGO on or before 10th draw. I understand all that math now. But Let's extend above calculation to different case, if I ask what's the probability to win a BINGO after 60 numbers have been called, we will calculate as follows
As you will see, COMBIN(51, 60-4) will give you invalid result because 60-4>51, so does it mean the method given in that page only works for maximum number of draws less than or equal to 55? If so, how does the author obtain the table in here (http://www.durangobill.com/BingoStats.html)?
Thanks.
Number of
Nbr. of hits Comb.with Total Probability there is at
on the card a Bingo Combinations least 1 Bingo(Col.2 / Col.3)
----------------------------------------------------------------------------------------------
0 0 COMBIN(24, 0) = 1 0.000000
1 0 COMBIN(24, 1) = 24 0.000000
2 0 COMBIN(24, 2) = 276 0.000000
3 0 COMBIN(24, 3) = 2,024 0.000000
4 4 COMBIN(24, 4) = 10,626 0.000376
5 88 COMBIN(24, 5) = 42,504 0.002070
6 912 COMBIN(24, 6) = 134,596 0.006776
7 5,928 COMBIN(24, 7) = 346,104 0.017128
8 27,102 COMBIN(24, 8) = 735,471 0.036850
9 92,520 COMBIN(24, 9) = 1,307,504 0.070761
10 244,092 COMBIN(24,10) = 1,961,256 0.124457
11 507,696 COMBIN(24,11) = 2,496,144 0.203392
12 841,100 COMBIN(24,12) = 2,704,156 0.311040
13 1,113,360 COMBIN(24,13) = 2,496,144 0.446032
14 1,174,620 COMBIN(24,14) = 1,961,256 0.598912
15 981,424 COMBIN(24,15) = 1,307,504 0.750609
16 644,445 COMBIN(24,16) = 735,471 0.876234
17 331,056 COMBIN(24,17) = 346,104 0.956522
18 133,428 COMBIN(24,18) = 134,596 0.991322
19 42,480 COMBIN(24,19) = 42,504 0.999435
20 10,626 COMBIN(24,20) = 10,626 1.000000
21 2,024 COMBIN(24,21) = 2,024 1.000000
22 276 COMBIN(24,22) = 276 1.000000
23 24 COMBIN(24,23) = 24 1.000000
24 1 COMBIN(24,24) = 1 1.000000
Up to now, everything looks good to me. I wrote the code to test it and I obtain the same number as above. The author also said that if we want to find the probability of BINGO after 10 numbers out of 75 drawn, it is given that
COMBIN(24, 4) * COMBIN(51, 6) / COMBIN(75,10) * 0.000376 = 0.000087
COMBIN(24, 5) * COMBIN(51, 5) / COMBIN(75,10) * 0.002070 = 0.000249
COMBIN(24, 6) * COMBIN(51, 4) / COMBIN(75,10) * 0.006776 = 0.000275
COMBIN(24, 7) * COMBIN(51, 3) / COMBIN(75,10) * 0.017128 = 0.000149
COMBIN(24, 8) * COMBIN(51, 2) / COMBIN(75,10) * 0.036850 = 0.000042
COMBIN(24, 9) * COMBIN(51, 1) / COMBIN(75,10) * 0.070761 = 0.000006
COMBIN(24,10)* COMBIN(51, 0) / COMBIN(75,10) * 0.124457 = 0.000000
Total = 0.000808
where the numbers before the equal sign are from the above table. So the first formula interprets as
1) the probability of that we have 4 hits on a card and those 4 hits happens to make a bingo pattern is 0.000376
2) there is total COMBIN(75, 10) ways to draw 10 numbers out of 75
3) in the 10 numbers drawn from 75, the number of ways that 4 of them filled the 4-number pattern and other 6 are not on the card (i.e. 6 are drawn from 51 numbers which are the numbers not on the 24 numbers in a card) should be COMBIN(24, 4) * COMBIN(51, 6)
Based on that explanation, we add all above 7 numbers will get the probability to win the BINGO on or before 10th draw. I understand all that math now. But Let's extend above calculation to different case, if I ask what's the probability to win a BINGO after 60 numbers have been called, we will calculate as follows
COMBIN(24, 4) * COMBIN(51, 60-4) / COMBIN(75,60) * 0.000376 = 0.000087
COMBIN(24, 5) * COMBIN(51, 60-5) / COMBIN(75,60) * 0.002070 = 0.000249
COMBIN(24, 6) * COMBIN(51, 60-6) / COMBIN(75,60) * 0.006776 = 0.000275
COMBIN(24, 7) * COMBIN(51, 60-7) / COMBIN(75,60) * 0.017128 = 0.000149
COMBIN(24, 8) * COMBIN(51, 60-8) / COMBIN(75,60) * 0.036850 = 0.000042
COMBIN(24, 9) * COMBIN(51, 60-9) / COMBIN(75,60) * 0.070761 = 0.000006
...
COMBIN(24,24)* COMBIN(51, 60-24) / COMBIN(75,60) * 0.124457 = 0.000000
As you will see, COMBIN(51, 60-4) will give you invalid result because 60-4>51, so does it mean the method given in that page only works for maximum number of draws less than or equal to 55? If so, how does the author obtain the table in here (http://www.durangobill.com/BingoStats.html)?
Thanks.
September 21st, 2014 at 2:02:20 AM
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If bingo was won, they just split the prize. The catch rule normally used is it's only bingo if using the last number called. If you had bingo, but didn't call it on the last number out, you don't win. Thems the rules.
I am a robot.
September 21st, 2014 at 2:07:17 AM
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Less people playing probably does increase the ties. Think about playing Pigskin picks where the more playing, the more need for a perfect winning card to win first. Otherwise you'll have more identical picks tying because of mediocre players.
I am a robot.
September 21st, 2014 at 5:53:10 AM
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Come on, man.
September 21st, 2014 at 6:02:47 AM
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kong,
The first response nickel posted was correct for most real-money bingo games: ties split the prize, as long as the bingo includes the last number called. Some places, if you were a number behind and they got another number called before you said bingo, would still award the bingo to you but make you split it with any of those achieved on the new number. A few places would not accept a late bingo. Doesn't matter who said BINGO first as long as it's called timely.
The second response nickel posted was sarcasm and should be disregarded except for any entertainment value you might find in it.
The more players, the more likely there will be ties.
The first response nickel posted was correct for most real-money bingo games: ties split the prize, as long as the bingo includes the last number called. Some places, if you were a number behind and they got another number called before you said bingo, would still award the bingo to you but make you split it with any of those achieved on the new number. A few places would not accept a late bingo. Doesn't matter who said BINGO first as long as it's called timely.
The second response nickel posted was sarcasm and should be disregarded except for any entertainment value you might find in it.
The more players, the more likely there will be ties.
If the House lost every hand, they wouldn't deal the game.
September 21st, 2014 at 6:25:31 AM
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No not sarcasm whatsoever. I don't believe it's true the more players, the more likely bingos result in ties. Games should end sooner with more players, so the tie factor doesn't make sense to me though could be true.
I am a robot.
September 21st, 2014 at 6:28:33 AM
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Bingo on the last number called is a very common rule.
I've seen one church that has a unique twist to that rule: If you call a late bingo, but nobody else called bingo on that number, you win. If others have bingo, and have that last number, you're out of luck.
I've seen one church that has a unique twist to that rule: If you call a late bingo, but nobody else called bingo on that number, you win. If others have bingo, and have that last number, you're out of luck.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
September 21st, 2014 at 6:42:20 AM
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Quote: onenickelmiracleNo not sarcasm whatsoever. I don't believe it's true the more players, the more likely bingos result in ties. Games should end sooner with more players, so the tie factor doesn't make sense to me though could be true.
Ok, my apologies on thinking that was sarcasm. I think you're incorrect, however. The more cards, the more likely any given number will be duplicated on another card, and be the remaining number needed for more than one bingo, even if none of the other numbers coincide.
If the House lost every hand, they wouldn't deal the game.
September 21st, 2014 at 8:43:14 AM
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Quote: konglify
Ok. Besides bring my above question here. I am thinking something on the math. I am pretty new to bingo game, I didn't play that before but the way the game play is pretty attracting. I decide to calculate the probability of winning a bingo. We set the game rule to be
1) using 75 balls
2) 5x5 cards, 24 numbers and 1 wild in the center
3) 12 winning pattern: 5 columns, 5 rows, two diagonals
4) each game draw at max 55 numbers
At many Bingo halls 4 Corners also constitutes a BINGO.
A falling knife has no handle.
September 21st, 2014 at 10:24:12 AM
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I usually offer to wrestle the old ladies I win with for the whole thing.
"Man Babes" #AxelFabulous
September 21st, 2014 at 10:59:46 AM
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Quote: djatcI usually offer to wrestle the old ladies I win with for the whole thing.
Who usually wins that match?
September 21st, 2014 at 12:34:35 PM
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Ok, thanks all. I think it make sense that to split the pay for more than one winner. But what I concern more is about the math, seems no one give comment on the math to figure out the probability to win the game when 8 players playing it. Hopefully more comment will be there
September 21st, 2014 at 12:41:01 PM
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My wife won 4k at Foxwoods, or so she thought, until she had to split it with three others. The hall is so darn big it took several minutes before the "letdown".
Many people, especially ignorant people, want to punish you for speaking the truth. - Mahatma Ghandi
