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.

P(2nd position) = 8/48 = 0.167

P(3rd position) = 7/47 = 0.149

P(4th position) = 6/46 = 0.130

P(5th position) = 5/45 = 0.111

P(6th position) = 4/44 = 0.091

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.

Que sera sera.

Paul McCoy - the Lottery Mine

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