Link to home
Start Free TrialLog in
Avatar of Jerry_Pang

asked on

Odds winning the lottery (with certain conditions)


Here's a question i was planning to answer
but too tired to think
too busy to answer
or too hard for me to answer
too lazy to research

Here's the question.

Lets say 6/49 lottery.

a.) whats is the odds of winning the lottery if we remove combinations that contains at least 3 consecutive number combinations?  
 1,2,3, 22,23,25
 41,42,43, 12,34,20

b.) also remove atleast  2(consecutive numbers combinations)
1,2, 44,45, 33,36
2,3, 10,11, 33,32
22,23, 25,26, 5,7

Jackpot (all six winning numbers selected)
There are a total of 13,983,816 different groups of six numbers which could be drawn from the set {1, 2, ... , 49}. To see this we observe that there are 49 possibilities for the first number drawn, following which there are 48 possibilities for the second number, 47 for the third, 46 for the fourth, 45 for the fifth, and 44 for the sixth. If we multiply the numbers 49 x 48 x 47 x 46 x 45 x 44 we get 10,068,347,520. However, each possible group of six numbers (combination) can be drawn in different ways depending on which number in the group was drawn first, which was drawn second, and so on. There are 6 choices for the first, 5 for the second, 4 for the third, 3 for the fourth, 2 for the fifth, and 1 for the sixth. Multiply these numbers out to arrive at 6 x 5 x 4 x 3 x 2 x 1 = 720. We then need to divide 10,068,347,520 by 720 to arrive at the figure 13,983,816 as the number of different groups of six numbers (different picks). Since all numbers are assumed to be equally likely and since the probability of some number being drawn must be one, it follows that each pick of six numbers has a probability of 1/13,983,816 = 0.00000007151. This is roughly the same probability as obtaining 24 heads in succession when flipping a fair coin!

we have 1/13,983,816 probability of winning 6/49
What are the Odds now?

thank you.
Avatar of Member_2_4694817

Link to home
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
Avatar of Member_2_4694817

BTW, a quick for-loop can confirm the theoretical results. I just checked the no-three-in-a-row variant this way because with its many subcases I was afraid I might have lost overview.
Avatar of Jerry_Pang


shouldnt i get higher probability if i dont bet on "No-2-in-a-row"?

i was expecting a higher probability with B than A. since there should be fewer "No-3-in-a-Row" than "No-2-in-a-row"

im trying to verify b.
i mean - higher probability betting on "No-2-in-a-row" than "No-3-in-a-row".
Avatar of Paranormastic
I've always been curious to opt out the no 3 in a row patterns and see how that turned out.  e.g. 1,2,3 / 1,3,5 / 1,4,7 / etc.
Link to home
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
i thought i read your answer wrong thehagman.
i was looking at your first answer as A and 2nd answer as B.
it would be interesting to test out all the possibilities. i mean removing sequential consecutive numbers like 1,3,5 / 1, 4, 7

i just wanted to see how far is the probability of winning in 2 in row number vs 3 in a row number.

thanks for the answers.