Here's a crack at explaining why the playing of the infamous sequence
1-2-3-4-5-6 in a 649 lottery is such a bad idea.
One of the many ways to break down the 49 numbers into discrete groups
is also one of the most obvious. Of the 49 numbers, 40 are double-digit
(10-49) and 9 are single-digit (1-9). Pretty basic isn't it? In this
case, we could even ignore the fact that the paint represents numbers at
all...the difference is simply that 40 of the symbols use 2 characters
and 9 of the symbols use only 1 character.
The sequence in question is made up entirely of single characters. Let's
for a moment ignore the fact that they are also in a contiguous
numerical sequence. Calculating the probability of drawing such a
sequence is easy using the basic formulas.
The 1st position drawn must be any one of the single digits. The
probability of this is therefore 9/49 = 0.184. We continue until all the
positions have been drawn.
The overall P(1-2-3-4-5-6) = 0.184*0.167*0.149*0.130*0.111*0.091 =
~0.000006. If we invert this figure (1/P) we get the *relative* odds of
this event which is 166,474:1. This in fact would be the probability of
drawing *any* 6 single-digit numbers regardless of whether they are in
sequence or not. There are only 84 different ways that this outcome can
occur since C(9,6) = 84. However, since we are specifying one and only
one of those possible sequences, we must perform one more calculation to
get the *true* overall odds. The P(specific sequence) = 1/84 * P(any
sequence) = 0.012*0.000006 = 0.0000000715112384
When once again inverted, this figure magically becomes 13,983,816:1! No
real surprise since this merely validates that the previously calculated
probabilities are correct. Bear in mind that I have been using figures
here that have been rounded to a considerable degree. The actual P
values have many more decimal places that aren't shown here.
Also bear in mind that these calculations disregard the order that the
numbers are drawn in. If the problem was to calculate the probability of
drawing 1-2-3-4-5-6 in numerical order we would be seeing vastly smaller
P-values and consequently vastly larger odds of such an event occurring.
Now let's go back to what I said above, namely that the figure 166,474:1
is the *relative* odds of drawing such a combination. "Relative to
what?", I hear people asking. The answer is that this represents just
one of the 64 different ways that single-digit and double-digit numbers
can be drawn to form a 6 number combination. The P-value for each of
those can be calculated using the same methods outlined above. It turns
out that the P-value of any drawn combination being composed of 1
single-digit and 5 double-digits is ~0.4235. If we calculate
0.4235/0.000006 we can quickly see that this type of combination is
70,500 times more probable than the other!
Conclusion? While each and every combination always has a 13,983,816:1
chance of being drawn, it can be shown that certain types of
combinations are significantly more likely to occur when compared
relatively to others.
While it is clearly not impossible for the stated combination to appear
someday, it is remote enough that I wouldn't choose to waste money on
it. In spite of all the logic, apparently there are surprisingly large
numbers of people that regularly play this combination simply because it
is the longshot. If the sucker ever does show up, I would estimate that
the jackpot shares will be significantly diluted.