A Hexadecimal Keno Perspecitve

Reply #8

Datapile — Thu, 25 May 2017 04:02:24 GMT

Below, we see that there is a direct correlation between how many bits are in the hex, and how often it occurs. The frequency was determined by counting occurrences over total draws for a single slot. The calculated average dist between hits is 1 divided by the frequency. (I have confirmed that all of the slots have identical frequency and distance behaviors for each hex.)

Hex Frequency Calculated Avg Dist Between Hits Bit Count

0 30.8% 3.2 0

1 10.8% 9.3 1

2 10.8% 9.3 1

4... [ More ]

Reply #7

Datapile — Sun, 21 May 2017 22:21:50 GMT

(Note: The following was counted using the first 1,454,405 million Michigan Club Keno drawings.)

As previously shown, the most common hex ratio is 6-9-4-1-0, occurring 14.9% of the time, or 216,717 times out of the draw history.

Now, lets look at the ratios of the hex values involved.

CHB0 consists only of Hex 0, so 6 of these will always be the same.

CHB1 can be any combination of 1s,2s,4s, or 8s. and their ratios will need to add up to 9. There are 220 possible ratios to do this, b... [ More ]

Reply #6

Datapile — Sun, 07 May 2017 16:13:30 GMT

Now we have constrained ranges for each hex bit-sum category. We can establish minimums and maximums for each, and combine them- working toward building a working model.

First, we establish the boundaries:

CHB0 CHB1 CHB2 CHB3 CHB4

Min 3 3 1 0 0

Max 9 14 8 3 1

Now we begin laying out the table with the range of possible hexadecimal values, without consideration for order. The first 7 hex values will be used to fill in the minimum values (from min\max table above). i.e. T... [ More ]

Reply #5

Datapile — Mon, 24 Apr 2017 22:51:51 GMT

Thanks dr san- I've worked with hex so long, I have forgotten not knowing it.

Reply #4

dr san — Mon, 24 Apr 2017 11:54:28 GMT

0hex = 0dec = 0oct 0 0 0 0

1hex = 1dec = 1oct 0 0 0 1

2hex = 2dec = 2oct 0 0 1 0

3hex = 3dec = 3oct 0 0 1 1

4hex = 4dec = 4oct 0 1 0 0

5hex = 5dec = 5oct 0 1 0 1

6hex = 6dec = 6oct 0 1 1 0

7hex = 7dec = 7oct 0 1 1 1

8hex = 8dec = 10oct 1 0 0 0

9hex = 9dec = 11oct 1 0 0 1

Ahex = 10dec = 12oct 1 0 1 0

Bhex = 11dec = 13oct 1 0 1 1

Chex = 12dec = 14oct 1 1 0 0

Dhex = 13dec = 15oct 1 1 0 1

Ehex = 14dec = 16oct 1 1 1 0

Fhex = 1... [ More ]

Reply #3

Datapile — Sun, 23 Apr 2017 21:58:08 GMT

Now that we have reduced the ratio count from 184 down to 25, lets work on eliminating the impossible...

Reference each hex sequence with the one that came before it, for repeats in the exact same slot:

Repeat Chance

0 2.76%

1 12.29%

2 23.69%

3 26.63%

4 19.50%

5 9.99%

6 3.77%

7 1.09%

8 0.23%

9 0.04%

10 0.01%

11 0.00%

This looks pretty good, but perhaps we can narrow it down further by separating the times when a 0 repeat... [ More ]

Reply #2

Datapile — Sun, 23 Apr 2017 17:55:54 GMT

5 groups and 5 different percentages...

Reply #1

Datapile — Sun, 23 Apr 2017 15:56:26 GMT

If we count the hexadecimal digits by bit group (0|1,2,4,8|3,5,6,9,A,C|7,B,D,E|F) labeling them Count Hex Bit (CHB#), we can track their ratio to each other. There are about 184 possible ratios, but the top 25 will be good for just over 97% of the drawings.

CHB0 CHB1 CHB2 CHB3 CHB4 Chance

6 9 4 1 0 14.9%

7 7 5 1 0 11.5%

5 10 5 0 0 10.7%

6 8 6 0 0 10.1%

5 11 3 1 0 8.7%

7 8 3 2 0 6.4%

4 12 4 0 0 5.4%

7 6 7 0 0 4.3%

8 6 4 2 0 4.2%

6 10 2 2 0 4.0... [ More ]

A Hexadecimal Keno Perspecitve

Datapile — Sun, 23 Apr 2017 15:43:28 GMT

For the purpose of this post, I'm talking about 20 numbers drawn from a pool of 80.

If we look at it as 0 for numbers not drawn, and 1 for numbers that are drawn, we can generate a drawing as an 80 digit binary number. Take a step back, and we can express every 4 digits as a hexadecimal. If we compare the hex digits against each other, each digit has the same chance for each hex digit occurring.

Hex Chance

0 0.308

1 0.108

2 0.108

3 0.035

4 0.108

5 0.035