The standard (mechanical) way that a Pick 3 digit is selected is to draw one ball from a machine that contains a pool of ten balls. This is done 3 separate times (there are 3 machines that each contain ten balls) to give us our winning Pick 3 combination. Since there are ten balls in each machine we can calculate the total possible combinations by multiplying each of the machines ten balls by the other machines ten balls and so on. When this is done we get 10x10x10, which equals 1000 possible combinations. Additionally, we can easily see that the odds of getting a single particular digit out of a particular machine is 1 in 10. For example, the odds of a 5 being drawn out of the first machine is 1 in 10 or the odds off getting a 5 out of the second machine is 1 in 10 and so on.
How about the odds of correctly guessing a digit from the first machine AND correctly guessing a digit from the second machine? Say we were betting on 4 and 7 as the first two Pick 3 numbers (front pair). The odds for the 4 to be drawn from the first machine would be 1 in 10 and the odds for the 7 being drawn from the second machine would also be 1 in 10. The overall odds of correctly guessing those first two digits would be 1 in 100 because 1/10 times 1/10 equals 1/100. Most everyone who plays Pick 3 knows the above situation and knows that any digit has a 1 in 10 chance of appearing in a selected position.
Now suppose that televised Pick 3 drawings were held out of the viewers watching eyes. By this I mean that each machine is covered with a black cloth and viewers cannot see the action inside of each machine. Now the viewer can hear the machines running and knows that the person performing the drawing is actually operating the machines and that balls are in fact being drawn, but no one can actually see what numbers were selected until the host reveals it to the audience by removing the black cloths one machine at a time.
For the first drawing the host reveals the first digit and it’s a 5, he then reveals the second digit and it is also a 5. What are the odds that the third machine will reveal a 5 once the cloth is removed? Since we know that there are ten balls in that last machine and only one of those ten balls is a 5 we assume the odds to be 1 in 10.
In another drawing a similar situation is played out, but only this time the host reveals the third machine first and it’s a 2. He next reveals the first machine and it also drew a 2. Now its time to reveal the second or middle machine and mathematically you know that there is supposed to be a 10% chance that it will also show a 2.
In yet another drawing the host reveals the third machine first. It produced a 7. He then reveals the second machine and it too produced a 7. Lastly, the host grabs the cloth off of the first machine. What are the odds that the first (position one) machine will also have produced a digit 7? The odds should be 1 in 10 or 10% - or should they????
Imagine that every single time that your state draws a combination that contains AT LEAST 2 repeated digits that they are revealed in the above manner. The machines and ball pools are separate. If any two particular machines produced the same digit (i.e. 00, 11, 22, 33, etc…) the last machine revealed should in fact still have a 1 in ten or 10% chance of also producing the necessary digit for a Triple.
In thinking the above way you would statistically expect to have one Triple Digit combination (all three balls the same digit) drawn for every ten times a Double Digit Pair is drawn. Mathematically this would seem to make sense, however it just can’t be possible. In 1000 combinations there are 270 doubles and only 10 triples. In a statistically perfect run of 10,000 games you should have something like 2800 combinations that contain at least 2 repeated digits and 280 of those would be triples if each Double Digit Pair had at a 10% chance of becoming a Triple Digit combination.
In the real world, a statistically perfect run of 10,000 Pick 3 games would in fact have 2800 3 digit combinations that contain AT LEAST 2 repeated digits, but 2700 will be Double Digit combos and only 100 will be Triples.
Stop for one second at look at the following combinations. The X represents an unknown digit or separate machine that contains 10 balls with the digits 0-9 in it.
11X, 1X1, 22X, X22, 7X7, 99X, 4X4, 55X, X00
Since each “X” has a 10% chance of being the digit needed to make all 3 numbers the same, you would think that we would see Triples more often.
The question I have for anyone who knows the answer is this: What is the overall chance that Double will become a Triple and how is it calculated?
I know that each Double Digit pair can appear within 28 different straight combinations and that only one of those 28 is a Triple (leaving 27 doubles). Also, the 27 straight Double Digit combos also represent 9 Boxed Double Digit combos (1 Triple is 10% when adding the 9 boxed for a total of 10).
Are the odds of a Double becoming a Triple 1 in 28 or a 10% chance of a 27% chance? (there are 270 doubles in the Pick 3 = 27%.) Some of my data seems to support the later – i.e. .027 or 2.7%.
GOOD SHOW !!
