Here’s a ‘bridge’
to winning tickets

Feb 28, 2005 1:13 AM

Let’s consider the same ticket we did a couple of weeks ago, the simple way ticket of 10 spots, grouped 3-3-2-2. If we split the ticket into twos and threes, we’ll find that the threes will make up a one-way-six and two-way-three, while the deuces will make up a one-way-four and two deuces. Putting this in our table and adding the columns and rows like last week gives us this result:

6

3

3

4

10

7

7

2

8

5

5

2

8

5

5

We know from Pascal’s Triangle that since there are four groups on the ticket, we will find 15 total ways, and indeed there are 15 entries on our table. We have a one-way 10, a two-way 8, a two-way 7, a one-way 6, a four-way 5, a one-way 4, a two-way 3 and a two-way 2.

An interesting feature of the bridge system is that it doesn’t matter how we split the ticket, as long as it is split in two. For instance, suppose we split the same ticket into 3-2 and 3-2. Now a three and a two produce a one-way-five, a one-way-three and a one-way-two. In this case our table will look like this:

  5 3 2
5 10 8 7
3 8 6 5
2 7 5 4

This produces the same totals as above. Or we could split the ticket into 3-3-2 and a 2. Now the 3-3-2 produces a one-way-8, a one-way-6, a two-way-5, a two-way-3 and a one-way-two, while the 2 produces only a one-way-two. Construction of our table produces this:

8

6

5

5

3

3

2

2

10

8

7

7

5

5

4

Note that the same totals are produced as in the first two instances.

Since any split produces the same results it is purely arbitrary which split we choose to use, and this choice is normally made on which split is the most convenient.

Now what I have described above is the standard bridge system. It soon became apparent to some checkers that these numbers could be expressed in a fractional format (although these are not true fractions in the normal mathematical sense) for brevity, and this produced the abbreviated bridge. Using the same split as our first example, 3-3 and 2-2, our split ticket may be represented as 1/6 and 2/3, and 1/4 and 2/2. Our abbreviated table now looks like this:

 

1/6

2/3

1/4

 

 

2/2

 

 

Now by multiplying the numerators and adding the denominators of each column and row entry, we produce this final table:

1/6

2/3

1/4

1/10

2/7

2/2 2/8 4/5

This produces the same results as the standard bridge system. The abbreviated bridge system proved to be one of the most powerful tools for the keno checker.

Well, that’s it for now. Good luck! I’ll see you in line!