# Bingo probability question

• Jun 2nd 2010, 02:32 AM
NTAHS
Bingo probability question
Hello all,

I have visited the wizzard of odds to get some answers on bingo probabilities but I have a new version that isn't included on that site. I was wandering if anyone could help me with my problem.

First of all, how bingo is played (90 ball). Each player has a card with 15 numbers on it. The bingo caller starts calling numbers and when one player has all his 15 numbers on his card called he wins. So the earliest someone can win is on the 15th ball call.

This new variant I am trying to figure out has 91 balls, numbers 1-90 and one extra ball (let's call it the X ball). To win a special prize, you have to match 15 numbers on your card and have the X ball being called the second before last winning call. Therefore, the earliest someone can win is on the 16th ball call (15 numbers + X ball).

What I am trying to figure out is: The probability in a game of getting Bingo AND the X ball being called the second before last.

I thought that even if you have 1 player with 1 card, he will eventually get bingo, even if it gets to the 91st ball call. Therefore the probability of getting Bingo in a bingo game is 100%. The probability of getting the X ball in a specific calling position (before the last ball call) = 1/91. Therefore the overall probability is 1/91 (1 * 1/91). What do you think?

For the probability of getting this winning combination by a specific number of ball calls, I used the folllowing formula:

Combin(16,16)*(combin(75,ballcall-16)/combin(91,ballcall))

So for example, for the 30th ball call the formula would be:
Combin(16,16)*(combin(75,30-16)/combin(91,30)

Am I on the right track? Should be a piece of cake for anyone with basic knowledge on statisics.

• Jun 2nd 2010, 04:17 PM
awkward
Quote:

Originally Posted by NTAHS
Hello all,

I have visited the wizzard of odds to get some answers on bingo probabilities but I have a new version that isn't included on that site. I was wandering if anyone could help me with my problem.

First of all, how bingo is played (90 ball). Each player has a card with 15 numbers on it. The bingo caller starts calling numbers and when one player has all his 15 numbers on his card called he wins. So the earliest someone can win is on the 15th ball call.

This new variant I am trying to figure out has 91 balls, numbers 1-90 and one extra ball (let's call it the X ball). To win a special prize, you have to match 15 numbers on your card and have the X ball being called the second before last winning call. Therefore, the earliest someone can win is on the 16th ball call (15 numbers + X ball).

What I am trying to figure out is: The probability in a game of getting Bingo AND the X ball being called the second before last.

I thought that even if you have 1 player with 1 card, he will eventually get bingo, even if it gets to the 91st ball call. Therefore the probability of getting Bingo in a bingo game is 100%. The probability of getting the X ball in a specific calling position (before the last ball call) = 1/91. Therefore the overall probability is 1/91 (1 * 1/91). What do you think?

For the probability of getting this winning combination by a specific number of ball calls, I used the folllowing formula:

Combin(16,16)*(combin(75,ballcall-16)/combin(91,ballcall))

So for example, for the 30th ball call the formula would be:
Combin(16,16)*(combin(75,30-16)/combin(91,30)

Am I on the right track? Should be a piece of cake for anyone with basic knowledge on statisics.

As for the probability of winning, if there are n players (and n bingo cards), then any one person's probability of winning is 1/n.

As for the probability that the next to last ball drawn is X, the probability is 1/90. We know the last ball drawn is not X (if I understand the rules correctly), so that leaves 90 balls that could be next to last, and they are all equally likely. One of the balls is X, so its probability is 1/90.
• Jun 3rd 2010, 12:20 AM
NTAHS
Hello,

I think you are right about the chance of a player winning but I am interested in the fact of ANYONE winning, same as saying occuring bingo in a bingo game, which has to be 100%. Right?

Now about the X ball. The X ball is a normal ball that can be called anytime. To win the specific bingo variant you have to get the X ball before the last called ball. It doesn't mean that the X ball cannot be called last. Therefore it should be 1/91, right?

So if we have to say what is the probability in a bingo game to have the X ball before the last ball call, wouldn't that be : (someone getting bingo - 100%) * (getting the X ball before the last ball call - 1/91) = 1/91. What do you think?

Regarding the excel formula to calculate the probability per ball call, do you agree?
• Jun 3rd 2010, 01:04 AM
downthesun01
Why isn't it:

$\displaystyle \frac{{15\choose 14}{76\choose 0}}{{91\choose 14}}$$\displaystyle *\frac{1}{77}$$\displaystyle *\frac{1}{76}$

+

$\displaystyle \frac{{15\choose 14}{76\choose 1}}{{91\choose 15}}$$\displaystyle *\frac{1}{76}$$\displaystyle *\frac{1}{75}$

+

$\displaystyle \frac{{15\choose 14}{76\choose 2}}{{91\choose 16}}$$\displaystyle *\frac{1}{75}$$\displaystyle *\frac{1}{74}$

+

...

+

$\displaystyle \frac{{15\choose 14}{76\choose 74}}{{91\choose 88}}$$\displaystyle *\frac{1}{3}$$\displaystyle *\frac{1}{2}$

+

$\displaystyle \frac{{15\choose 14}{76\choose 75}}{{91\choose 89}}$$\displaystyle *\frac{1}{2}$$\displaystyle *\frac{1}{1}$

It's the summation of the probability of getting 14 correct numbers out of the first X numbers called multiplied by the probability of getting the "special ball" out of the balls remaining multiplied by the probability of getting the your last needed number out of the remaining number of balls.
• Jun 3rd 2010, 01:13 AM
NTAHS
Makes sense to me. What results do you get? Here are my results:

Ball call Probability (inverse)
16 23,702,123,506,231,200,000
17 1,394,242,559,190,070,000
18 154.915,839,910,008,000
19 24,460,395,775,264,400
20 4,892,079,155,052,890
..
89 134
90 110
91 91
• Jun 3rd 2010, 01:48 AM
downthesun01
There's something very wrong with what I typed. I need to look at it more. There, I think I fixed my previous post.

Ok. I just ran it through the OpenOffice spreadsheet program and got an answer of about 16.26%.

Maybe someone else can weigh in on how right or wrong my way of approaching the problem and solution is? Thanks

nevermimd, that's way wrong too. I'll give it another shot after I get some sleep.
• Jun 3rd 2010, 04:20 PM
awkward
OK, I think I understand your question a little better now, so let me rephrase it: You would like to know the probability of completing your card of 15 numbers when the nth ball is drawn, pretending that balls continue to be drawn until you complete the card. This is NOT your probability of winning, because you win only if yours is the first card to be filled.

To that end, it seems easier to compute the probability that you will fill your card on or before the nth draw. So suppose n balls have been drawn. There are $\displaystyle \binom{91}{n}$ possible combinations, all of which are equally likely. Let's count the number of combinations where your card is filled. Each arrangement must include all 15 of your chosen numbers, which can be done in only one way; then the remaining n-15 balls must be drawn from the remaining 76 numbers (including X), which can be done in $\displaystyle \binom{76}{n-15}$ ways. So the probability that your card is filled on or before the nth draw is
$\displaystyle F(n) = \frac{\binom{76}{n-15}}{\binom{91}{n}}$ for $\displaystyle n \geq 15$.

If you want the probability that your card is filled on the nth draw (exactly), that is given by

$\displaystyle F(n) - F(n-1)$,

the probability that the card is filled on or before the nth draw minus the probability that it is filled before the nth draw.
• Jun 4th 2010, 12:06 AM
NTAHS
Hello!

I think my excel formula is slightly different to yours.

Combin(16,16)*(combin(75,ballcall-16)/combin(91,ballcall)) * 1/91

Firstly I calculate all the posibile combinations of the remaining 75 balls abstracting n-16, not n-15. Should it be n-16 since there must be 16 balls called to get bingo? (15 numbers + X ball).

Secondly the numbers on the card can be drawn in any possible way, therefore combin(16,16), which is 15 numbers and the X ball. But because the X ball has to be drawn in an exact position (second to last), I multiply by 1/91.

What do you think?

And one other thing.. The 15 numbers on the card can be called in any combination. But we also need the X ball to be second last called ball. What is the right formula to calculate that? combin(15,15) * 1/91 or combin(16/16) * 1/91 ?

Thanks agaiN!
• Jun 4th 2010, 12:40 AM
downthesun01
After thinking about this problem for a while, I can't come up with a way of doing it short writing down the entire sample space and calculating it that way. That seems like it'll take a long time to work out.

And just a quick point, the probability of getting the "special ball" isn't $\displaystyle \frac{1}{91}$, it's dependent on how many balls are left, so $\displaystyle \frac{1}{x}$, where x is the number of balls remaining.
• Jun 4th 2010, 12:44 AM
NTAHS
The probability of getting the X ball will always be 1/91 imo.

For example, X ball being drawn on the 87th ball call:

90/91 (all balls are drawn except for the X)*
89/90(all balls are drawn except for the X)*
88/89(all balls are drawn except for the X)*
1/88 (the X ball)

= 1/91
• Jun 4th 2010, 02:02 AM
downthesun01
For example, if you've called 85 balls (none of them the x ball), then there's a $\displaystyle \frac{1}{6}$ that the "x ball" will be called next. It's not a static probability. It varies with how many balls have been called.
• Jun 5th 2010, 09:39 AM
awkward
Here is how to find the probability that you will draw the X ball on your n-1st draw and then complete your card on the nth draw.

Notation: $\displaystyle P(n,m) = \frac{n!}{(n-m)!}$ is the number of permutations of n objects taken m at a time.

If n balls have been drawn, there are P(91,m) possible sequences, all of which are equally likely.

Let's see if we can find the number of these in which you draw the X ball on draw n-1 and then complete your card of 15 numbers on draw n, where n > 15. There are 15 possibilities for the nth ball. The n-1st ball is X. Then the preceding n-2 balls must consist of the other 14 balls in your card and n-16 "miscellaneous" balls. Without regard to order, there are $\displaystyle \binom{75}{n-16}$ possible sets of n-2 balls; taking order into account, there are $\displaystyle \binom{75}{n-16} (n-2)!$ possible sequences. So all together, there are
$\displaystyle 15 \binom{75}{n-16} (n-2)!$
possible sequences for the first n balls.

Therefore the probability that you will draw the X ball on the n-1st draw and then complete your card on the nth draw is

$\displaystyle \frac{15 \binom{75}{n-16} (n-2)!}{P(91, n)}$

Earlier I wrote
Quote:

Originally Posted by awkward
[snip]
As for the probability that the next to last ball drawn is X, the probability is 1/90. We know the last ball drawn is not X (if I understand the rules correctly), so that leaves 90 balls that could be next to last, and they are all equally likely. One of the balls is X, so its probability is 1/90.

Having worked out the details, I now see this was an error. The problem is that the n-1 balls before the winning ball must include 14 of the numbers from your card. That makes it less likely that the X ball will be in that set.

If you crank out the numbers, I think you will find your chances of drawing the X just before you complete your card are not very good.