Division one, three and four exclude the bonus number. Thus only the first six balls rolled out of the barrel matter. How many combinations are there? Using the standard combinatorial formula C(n,r) = n!/(r!(n-r)!) I have worked the out the total number of possibilities. Six balls chosen from forty balls without replacement works out as:
C(40,6) = 3 838 380 possible combinations
For Division Two and Five the bonus number is included and the order of the draw is important because it is always the seventh ball out of the barrel.
When the bonus number is present there are many more possible outcomes. There are two ways of working this out.
Method 1.
There are C(40,6) ways to select the six balls from 40 which leaves 34 left. Thus there are 34 ways of choosing the bonus number. The total number of possibilities are therefore 34 x C(40,6)
Method 2.
Choose the bonus number first. There are 40 possibilities. Then choose the next 6 from the remaining 39. Total possibilities are therefore 40 x C(39,6).
The result of the calculation for both methods are 130 504 920.
There is only one winning combination for division one and that is if exactly the same six numbers which I choose come in.
Pr (Division One Win) = 1/3 838 380 = 2.6053 x 10^-7
This is exactly six numbers chosen correctly on my game board, one of which is the bonus number. There is only 1 way of choosing the correct bonus number and 6 ways of choosing the remaining five. Total winning combinations are therefore 6*1 = 6.
Pr (Division Two Win) = 6 / 3 838 380 = 1.5632 x 10^-6
One might say:
>This is exactly six numbers chosen correctly on my game board, one of >which is the bonus number. There is only 1 way of choosing the >correct bonus number and 6 ways of choosing the remaining five. >Total winning combinations are therefore 6*1 = 6.
That is, 6 possible combinations *on your score card* which will win. This is *not* the number of *outcomes of the draw* which result in your winning Division Two. You should work with numbers either of combinations on the score card, or of outcomes of the draw; do not confuse these with each other.
> Pr (Division Two Win) = 6 / 130 504 920 = 4.5975 x 10^-8
Incorrect. Probability of winning division 2 is 1/639730, as shown below:
Method 1:
Considering combinations on the score card. Number of combinations on your score card which win division 2 = 6 Total number of possible combinations on your score card = 3838380 Probability of winning division 2 = 6 / 3838380.
Method 2:
Considering outcomes of the draw. Number of outcomes of the draw which make your score card win division 2 = 6 x 34 (six possible bonus numbers, 34 possible unmatched numbers) Total number of possible outcomes of the draw = 130504920 Probability of winning division 2 = 6 x 34 / 130504920
This is only five winning numbers, excluding the bonus number. One number selected is wrong. There are 34 or (40 - 6) wrong numbers. There are 6 ways to select 5 numbers from the six winning numbers. You leave one of the 6 numbers out in turn. Therefore there are 6 * 34 = 204 winning combinations of 5 numbers.
Pr (Division Three Win) = 204 / 3 838 380 = 5.31474 x 10^-5
This is four winning numbers excluding the bonus number. There are C(6,4) = 15 ways to choose 4 of the winning numbers from the six. There are 34 incorrect numbers. We want to choose 2 of these. There are C(34,2) = 561 ways to select these. Therefore there are 15 x 561 = 8415 winning combinations.
Pr (Division Four Win) = 8415 / 3 838 380 = 2.19233 x 10^-3
There are three winning numbers plus the bonus number. Four numbers are needed to match in all.
Again looking only a winning combinations there is only one way to choose the bonus number. There are three balls to choose from the winning six and then 2 to choose from the remaining 33 losing ones.
Winning Combinations = 1 x C(6,3) x C(33,2) = 10560
20 x 528
Pr (Division Five Win) = 10560 / 3 838 380 = 2.7512 x 10^-3
Pr (Division One Win) = 1/3 838 380 = 2.6053 x 10^-7 Pr (Division Two Win) = 6 /3 838 380 = 1.5632 x 10^-6 Pr (Division Three Win) = 204 /3 838 380 = 5.3147 x 10^-5 Pr (Division Four Win) = 8415 /3 838 380 = 2.1923 x 10^-3 Pr (Division Five Win) = 10560 /3 838 380 = 2.7512 x 10^-3
Probability of a Division One Win = 1/3 838 380 = 2.6053 x 10^-7 = 0.000026 % Probability of a Division Two Win = 6 /3 838 380 = 1.5632 x 10^-6 = 0.000156 % Probability of a Division Three Win = 204 /3 838 380 = 5.3147 x 10^-5 = 0.00531 % Probability of a Division Four Win = 8415 /3 838 380 = 2.1923 x 10^-3 = 0.219 % Probability of a Division Five Win = 10560 /3 838 380 = 2.7512 x 10^-3 = 0.275 %
40 numbers can be chosen (correspond to 40 balls) Each number can only occur once. (Sampling without replacement).
Six numbers will be selected to be winning.
Total possible ways to select 6 distinct numbers from the numbers 1 to 40 = C(40,6) = 3 838 380
This will also be called the TOTAL POSSIBLE OUTCOMES
Single selection means I fill out only one selection of 6 numbers. Each number may be between 1 and 40.
For the sake of this exercise assume I choose the numbers 1 2 3 4 5 6
There is only ONE winning combination of 6 numbers that win. This is: 1 2 3 4 5 6
One number is wrong. There are 34 (40 - 6) wrong numbers.
There are 6 ways to select 5 numbers from the six winning numbers. You leave one of the 6 numbers out in turn.
Therefore there are 6 * 34 = 204 winning combinations of 5 numbers.
204 ways of getting exactly 5 correct
+ 1 way of getting 6 correct
------
205 ways of getting at least 5 correct
There are C(6,4) = 15 ways to choose 4 of the winning numbers from the six.
There are 34 incorrect numbers. We want to choose 2 of these. There are C(34,2) = 561 ways to select these.
Therefore there are 15 * 561 = 8415 winning combinations.
There are C(6,3) = 20 ways to choose 3 of the winning numbers from the six.
There are 34 incorrect numbers. We want to choose 3 of these. There are C(34,3) = 5984 ways to select these.
Therefore there are 20 * 5984 = 119680 winning combinations.
There are C(6,r) ways to choose r of the winning numbers from the six.
There are 34 incorrect numbers. We want to choose r of these. There are C(34,r) ways to select these.
Therefore there are C(6,r) * C(34,6-r) winning combinations.
Now looking at only the winning combinations there is only one way to select the correct bonus number. There are 6 ways of choosing which of the five numbers to choose. There are 34 or (40-6) ways of choosing the remaining number.
Winning Combinations = 1 * C(6,5)* 33 = 198
Again looking only a winning combinations there is only one way to choose the bonus number. There are three balls to choose from the winning six and then 3 to choose from the remaining 33 losing ones.
Winning Combinations = 1 * C(6,3) * C(33,3) = 109 120
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