Bingo Mathematics / Formula Help

Hi Everybody,

I came here from the Ask The Wizard companion site - and i've been reading through the forums and searching, lots of interesting info here - but I have a question that has my basic skills stumped (this isn't homework!!)

There's a bingo game here in the UK, using a standard card with 15 numbers in 3 rows of 9 columns - out of a total of 90 balls (1 - 90). The last game sold just over 23,000 cards.

All ok so far, but they guarantee a single winner every time (no ties). Apparently, this is all transparent and the numbers are picked at random until a winner is found - under the presence of independent observers. They are so confident that only 1 winner will be picked for each prize, that they have up to a £1.5million insurance claim against another player claiming the prize at the same time.

So my question is - how can this be? If multiple cards share even 1 number, all other numbers could be called before that number - then when that number is called, multiple winners could occur, so therefore only 6 totally unique cards are possible? Is this assumption correct?

So how does this work? Are the card permutations special in some way that reduces the chance of ties happening to nearly zero, or does the bingo number drawing system use a sneaky method of drawing the numbers in such a way that only 1 card ever wins?

Thanks in advance!

Hmmmm.... from wikipedia:

Quote: WikiPedia

A typical bingo ticket is shown to the right. It contains twenty-seven spaces, arranged in nine columns by three rows. Each row contains five numbers and four blank spaces. Each column contains either one, two, or very rarely three, numbers:

The first column contains numbers from 1 to 9,
The second column numbers from 10 to 19,
The third 20 to 29 and so on up until the last column, which contains numbers from 80 to 90 (the 90 being placed in this column as well).

If the game is of the blackout variety where all 15 numbers must be chosen, it would be very easy to make this claim. After all, there are a total of 45,795,673,964,460,800 combinations of cards, and with only 23,000 cards sold, the odds of a duplicate are about 1,991,116,259,324,380:1 so it would be very reasonable to make this claim.

Thanks for the quick reply - I'm probably just being thick, but from this page

Quote:

American bingo, which is based on a 24-number card (plus a free square) and 75 balls (different system I know, but principles are the same)

...

Ties are common in all bingo games, including coveralls. The greater the number of cards, and the easier the pattern is to cover, the more ties you will see. The following table shows the averge number of people that will call bingo accoring to the pattern and number of cards.

And the expected number of players to call Bingo (Coverall / Blackout) with 10,000 cards in play is 1.38

So even though entire duplicate tickets in 23,000 are probably impossible, say you have 2 cards that share a single number, all 14 other numbers on those cards are different, but all are called before that shared number (28 numbers called) - then the shared number is called, causing both those 2 cards to win at the same time? and with 20,000 tickets in play there's going to be a lot of shared numbers?

Thanks for any more help!

I also visited the wiki article. It says that the UK national bingo lottery only uses the coverall rule.

To parse the question:
1) tickets are sold with 15 numbers printed on them, chosen out of 90, under the following restriction: there must be between 1 and 3 numbers in each group of 10.
2) numbers are then called until there is a coverall winner (all 15 numbers called).
3) what is the probability that two tickets will win at the same time, given ~23000 tickets are sold?

Even calculating the possible number of tickets is not as easy as US bingo. An exact answer will take a fair amount of work, simulation is probably easier. My intuition is failing me, I can't even come up with a good estimate for the answer.

That's exactly the problem - I imagine they have a formula to generate tickets to minimize the likely hood of 2 winners, instead of purely random tickets, but what exactly could this be? I've run simulations, but with semi random tickets you get something like (this is off the top of my head)

20,000 tickets in play every game (same tickets)

out of 20,000 games
~2,000 games 2 tickets won
~800 games 3 tickets won
~300 games 4 tickets won
~100 games 5 tickets won

So for instance, if I purchase the first 6 tickets - the system could generate me 6 unique tickets. This is how normal uk bingo tickets kind of work - check this image out, 6 per view tickets use all numbers 1-90 across 6 tickets ().

But then imagine I purchase another ticket. This ticket must have shared numbers with previous tickets as there are no unique numbers left. If this ticket is randomly generated, it could well have 14 numbers shared with another ticket increasing the chance that this ticket and another ticket will win, so it can't simply be random tickets.

Being a bit of a gambler I'm curious and suspicious about the single winner promise - and I'm also interested in if there's a benefit to buying a lot of tickets early or late in the sequence (also there must be a sweet spot for number of tickets vs winning chance) - any thoughts?

Quote: dwheatley

Even calculating the possible number of tickets is not as easy as US bingo.

Here's how I figure it:
Since 10 of the 15 numbers must be from each of the ten "groups" (1-10, 11-20, ..., 81-90), it is a question of selecting the groups for the remaining five numbers subject to the condition that no group can be selected more than twice (since no group can have more than three numbers in it).
The distributions are:
2-2-1 - there are 9 choices for the 1, and (8 x 7) / 2 for the 2s = 280
2-1-1-1 - there are 9 choices for the 2, and (8 x 7 x 6) / 6 choices for the 1s = 560
1-1-1-1-1 - there are (9 x 8 x 7 x 6 x 5) / 120 choices = 252

Each 2-2-1 distribution is actually 3-3-2-1-1-1-1-1-1-1; each group with 3 has (10 x 9 x 8) / 3 = 240 different sets of numbers, each group with 2 has (10 x 9) / 2 = 45, and each group with 1 has 10, so the total = 280 x 240² x 45 x 10⁷ = 7,257.6 trillion (yes, trillion with a T).
Each 2-1-1-1 is actually 3-2-2-2-1-1-1-1-1-1; the total = 560 x 240 x 45³ x 10⁶ = 12,247.2 trillion
Each 1-1-1-1-1 is actually 2-2-2-2-2-1-1-1-1-1; the total = 252 x 45⁵ x 10⁵ = 4650.10875 trillion
The sum = 241,549.0875 trillion different possible tickets.

Of course, "different possible tickets" and "chance of two or more tickets filling in all of their numbers at the same time" are two vastly different things.

Quote: livesinabox

...they guarantee a single winner every time (no ties)...1 winner will be picked for each prize, that they have up to a £1.5million insurance claim against another player claiming the prize at the same time.

Sounds fishy to me - let's confine ourselves to the standard three rows of five numbers in nine columns (e.g. http://www.edwardthompson.com/home#8)!

When I played UK bingo there was club bingo which used a rotation of typically 12k (or 24k in larger clubs) of cards. So if yesterday they had sold books 8001-10234 they would then sell 10235 onwards. The "books" would come in boxes, and should be sold in order. Books were number 00001 to 96768 (or similar) and repeated permutations 1 to 12096. The guarantee from the manufacturers was against duplicates being issued, not that there would only be one winner. (NB Each set of 6 perms 1-6, 7-12 etc. would have all 90 numbers.)

I heard of one occasion when a club messed up and had somehow sold (say) 08001-10234 then 20001-21xxx. Thus two players had perm 08001. In this scenario the club has to pay both winners the full prize money each (and being legal here) was exceeding its APM [Added Prize Money] limit. If this had been the manufacturer's fault, they would have covered the loss, but it was the bingo club, so they had to pay.

Nowadays there are electronic devices, link games and all kinds of stuff; but the principle stays the same - no player has the same combination for the line, two lines or house (this doesn't apply to the National Game which is house only). I have seen lines won in six numbers where you see two of the possibles win.


Now to the sneaky bit - in theory the RNG/calling mechanism knows which cards are in play since (a) the opening and closing numbers are known (b) it knows the perms corresponding to those and (c) it used to alert the caller of a win (even though the player has to physically claim). I'm sure there is a method of calling the numbers so only a pre-determined book will win, but legally the RNG has to be random. Personally I would prefer it if the RNG could only send out numbers and was somehow separated from the process of checking cards.


In summary I am guessing you are referring to the guarantee against duplicate perms being in play rather than only having one winner.