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Questions about Bingo |
In the Station casinos, there is the "Big 3" game in their bingo rooms. You must get 3 of the first 4 numbers to win a progressive jackpot. What are the odds of that? Thank you. – Joseph G. from Las Vegas
For the benefit of other readers, the Big 3 is a bingo side bet at all the Station Casinos and the Fiesta Rancho. The player is given a ticket, either paper or loaded into an electronic unit, with three random bingo numbers out of the 75 possible. If the first four bingo numbers called in that session contain all three of the player’s numbers, then the player will win a progressive jackpot. The jackpot starts at $1000 at grows by $200 a day until somebody wins. Every session, and property, has an independent jackpot.
The number of winning combinations is 72, because three of the balls must match, and the fourth can be any one of the other 72 balls. There are combin(75,4) = 1,215,450 possible combinations. Thus, the probability of winning is 72/1,215,450 = 0.000059. The player can buy 48 tickets for $10, thus the cost per ticket is 10/48 = 0.208333 dollars. The breakeven meter, where there is zero house edge, is (10/48)/(72/1,215,450) = $3,516.93.
Station Casinos indicate the Big 3 Jackpots on their Jumbo Bingo web site. There you will see the meter often will exceed $3517. When I answered this question on August 30, 2007, two of the eight properties had a player advantage, the Palace Station and Fiesta Rancho. This is one of the few bets in Las Vegas that often have a player advantage. Unfortunately, they limit the number of cards you can buy, making it not worth the bother to most people, including me, to make a special trip.
September 1, 2007
What is the probability that two bingo cards have no number in common? What is the probability they have every number in common? - Joe
The probability two bingo cards have no numbers in common is ( combin(10,5)/combin(15,5)) 4×(combin(11,4)/combin(15,4)) = 1 in 83,414.
The probability two bingo cards have all 24 numbers the same is (1/combin(15,5)) 4×(1/combin(15,4)) = 1 in 111,007,923,832,371,000.
December 26, 2006
Always, great site. I am hoping that you can settle an argument between myself and a friend of mine concerning online Bingo. The web-site allows you to buy bingo cards for 10 cents a piece. Assuming one has 5.00 to spend, my friend believes you have are better off buying 50 cards and playing one time instead of buying a single card each game at .50 and playing 10 times. I disagree and believe that since no matter how many cards you buy, they are .10 cents a piece, it makes no difference if you play them all at once or spread them out? – Ed from Indianapolis
Thanks for the compliment. The answer depends on how the jackpot is determined by the bingo site. If it is a percentage of total cards sold, which is usually the case, then it wouldn’t make any difference. However, if there is a fixed prize for the winner, then it would be better to play one game at a time, lest you compete against yourself.
October 4, 2006
At bingogala.com
they offer a $500 prize for a coverall within 54 calls. You
told me earlier that the probability of that at least 1 card
in 600 will get a coverall in 54 calls is 3.21%. So in 380
days (to date) at 8 sessions per day they should have 97.58
$500 winners, right? However I counted only 76 winners on
their home page. When I brought up my question about this in
chat my husband and I were both banned from the site which
really sent my antenna up? Sorry to be a pest but if they
are running an unscrupulous site I want to know how to
figure it out so that I can shout it far and wide with
facts. Thank you for any help you can give me in this
matter.
First, let me
explain that this is a rather old question that I put on
the back burner. According to their homepage, bingogala has now been in operation for
two years. The probability
of a coverall within 54 calls for a single card is
combin(74-24,54-24)/combin(54,54) = 0.000054. The
probability at least one card in 600 will get a coverall
in 54 call is 1-(1-.000054)600 = 0.032121. The
expected number of winners over 380 days at 8 sessions
per day is 97.65. The standard deviation is
(380*8*0.032121*(1-0.032121))1/2 = 9.72. So
this is (97.65-76)/9.72 = 2.23 standard deviations south
of expectations. The probability of 76 or fewer winners
in a fair game is 1.30%. So, this could either be
explained by bad luck on the part of the players or
fewer than 600 players on average. Perhaps they didn't
get as many in the early days. Ultimately, in my opinion, the evidence doesn't
warrant an accusation of foul play.
Sept.
24, 2002
What is the probability that out of 600 bingo cards at
least one will get a coverall within 54 calls?
The probability that any given card will have a
coverall with 54 calls is combin(51,30)/combin(75,54) =
114,456,658,306,760/2,103,535,234,151,140,000 =~ 1 in 18,738. The
probability of 600 cards not winning is (1-1/18738)600
=~ 96.79%. So the probability that at least one of the
600 players will hit is 3.21%.
Jan. 15, 2002
In your
bingo
section you
specify the probabilities of getting bingo in a particular
round, but what is the expected number of drawn numbers
before someone gets bingo?
Following is the
expected number of calls before somebody gets a bingo
according to the number of players.
1 player: 41.37
10 players: 25.51
50 players: 18.28
100 players: 15.88
200 players: 13.82
500 players: 11.56
1,000 players: 10.13 Jan.
2, 2002
Hi, I'm doing a
project on Bingo and I would like to know how to find the
probability of Bingo. The probability of getting a line,
horizontally, diagonally and vertically, a coverall, and the
four corners. I have already seen your probability table and
i wanted to know the formula you used. - Steph from Toronto,
Canada
The probability of
getting a bingo (5 in a row) is complicated to explain,
mainly due to the free square. I personally had to use a
computer to do it. Calculating the probability of a four corners win is much easier. The
probability of having 4 corners, given x marks on the card,
is combin(20,x-4)/combin(24,x). In other words, it is the number
of ways to put 4 of the marks in the corners and the rest
anywhere else divided by the number of ways to put all x
anywhere on the card. The probability of getting four
corners within y calls is the sum for i=4 to y of the
product of the probability that given y calls there will be
x marks on the card and the probability that these x marks
will form four corners (above). The probability of getting x
marks in y calls is
combin(24,x)*combin(51,y-x)/combin(75,y). Following this
logic you should be able to see the math for a coverall.
June
10, 2000
Consider a bingo game with 75 random cards. Draw 12
random numbers, according to standard bingo rules. Is the
probability of a bingo 75 * 0.00199521? (I got the
0.00199521 from your table of bingo probabilities for a
standard occurring within 12 numbers called) If not, what is
the probability that a bingo will occur? You have a great
page. - Charlie Burris of ?
You're right, according to my table of my probabilities
in bingo, the probability of any one person getting a
bingo within 12 numbers drawn is 0.00199521 . If the
probability of an event happening is p the probability that
it will happen at least once in n times is
1-(1-p)n. In this case the probability that at
least one person will get a bingo is 1-(1 -
0.00199521)75 = 1 - .9980048 75 = 1 -
.8608886 = .1391114 . March 4,
2000
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