This game is no longer played! Instead, from game number 609 [18 October 2005] onwards Super7Ozlotto has taken over!
The following tables of frequencies for the Ozlotto game includes games from draw 1 held on 22/2/94 thru to draw held on 11/10/2005.
Table of frequencies for the Ozlotto game.
Table of frequencies for individual balls of the Ozlotto game.
Probability of getting all six numbers (no supplementary) with one game of six numbers is 1 in:
45*44*43*42*41*40*39! 45C6 = ----------------------- 6! * (45 - 6)!
Or about 1 in 8.145 million.
This is on the assumption that each ball has an equal likelihood of being drawn as first ball. However the table above indicates that this is not the case thus far, and that the likelihood of each ball being drawn varies from ball to ball, and from their order of being drawn.
So the probability of the first six numbers in the above table being drawn on 5/May/98 becomes:
42*40*39*38*37*37*6! 1 Pr(1'st 6) = ---------------------- = ----------- (1314)6 2,097,387.9
The general formula for any six numbers from the above table is:
Fr(n1)*Fr(n2)*...*Fr(n5)*Fr(n6)*6! Pr(Any six) = ----------------------------------- (1314)6 Where Fr(ni) is the Frequency of the i'th number chosen from the table.
When the above formula is applied to the middle six numbers (16, 2, 7, 22, 41, 3; 45 doesn't divide evenly so for the purposes of the exercise I've worked around 15, which is the 23'rd number in order of frequency) we get a probability of 1 in 12.03 million which is close to 1 in 8.145 million.
When the above formula is applied to the least six frequently occuring numbers (26, 32, 25, 23, 5, 42) the probability is 1 in 114.75 million. In other words, you may still win if you choose those numbers, but your odds of winning with those numbers as opposed to the six most frequently occuring numbers (according to the frequency table) is a lot worse. About 57 times worse.
Probabilities for sytems entries:
--------------------------------
In systems entries we play 7 or more numbers per a game and for the first division prize hope that the six numbers drawn will be ALL amongst those 7 or more numbers we played.
The formula then becomes: Pr(System N) =
6!*N!*(Weighted average of the frequencies of the N numbers chosen)6 ---------------------------------------------------------------------- (1314)6*6!*(N-6)!
For example a System 7 game using the most frequently occuring seven numbers (43, 9, 28, 30, 18, 29, 34) would have a probability of yielding the correct six winning numbers of:
6!*7!*((42+40+39+38+37+37+36)/7)6 1 ----------------------------------- = ----------- (1314)6*6!*(7-6)! 317,114.6
Note that
N! ---------- = NC6 6!(N-6)!
And for a System 7 game is simply equal to 7; for a System 8 game it is equal to 28, etc..
Take 4 and Take 5 games:
----------------------------
For a Take 4 game the probability of winning the first division prize with any 4 numbers from the table is;
4! * Fr(n1) * Fr(n2) * Fr(n3) * Fr(n4) Pr(Take 4) = --------------------------------------- (1314)4
For example, if the 4 most frequent numbers were played the probability of winning 1'st division would then have been:
4! * 42 * 40 * 39 * 38 1 Pr(Take 4) = ------------------------ = ---------- (1314)4 49,889.9
For Take 5 the formula would be;
5! * Fr(n1) * Fr(n2) * ... * Fr(n5) Pr(Take 5) = ------------------------------------ (1314)5
Discussion:
------------
One must bear in mind that 1 in 2 million is still very steep odds. In fact, if one were to try this strategy retrospectively, that is do a statistical analysis from draw 1 to 218, to determine the most likely six numbers in draw number 219, and so on backwards, one sees that analysis of draws 1 thru to 216 predict two numbers correctly for draw 217, then the following consequtive steps predict no balls for draws 218 & 219. Whether this is an improvement on just using random numbers (on the assumption that all balls are equally likely to be drawn as first ball) I have not yet looked at. But one thing is for sure, the above analysis greatly improves the Expectation value.
Expectation Value would be the above probability (of correctly predicting all six numbers, 1 in 2 million) multiplied by the expected value of the prize money (On the incorrect assumption that if you win, you won't share the prize). If sharing is taken into consideration the expectation value diminishes. For the coming draw (2/5/98), the 1'st division prize in OzLotto is $10 Million. Therefore the expectation value would be about $5 per a game. The cost of a single game is $1.20 in Victoria. So the expectation of winning $5 for $1.20 outlayed is positive. Therefore, with the above strategy a punt in this this weeks OzLotto is worthwhile. As a matter of fact it's still worthwhile if the odds were 1 in 8.145 million as the expectation value is nearly the cost of a single game.
Post Mortem (6May98):
--------------------------
The 5May98 draw has already been and gone. As predicted two of the drawn numbers were in the top six most frequent. See above discussion.
To come, further analysis to see if this strategy has any worthwhile advantages OVER random selection of numbers without any prior knowledge of their frequency.
Return to Agains't the Odds
Return to Home Page
Under Construction.