Australian Powerball
7, 5, 13, 4, 2, 14, 11 (main barrel)
FYI: 14 was the "Powerball number" (but it is not needed for the questions below)
1. For the "main barrel", what is the probability that the highest number would be 14 (or less)?
2. For the "main barrel", what is the probability that the highest number would be exactly 14? (edit/update: 145 pm)
"...Powerball is drawn from 2 barrels. 7 numbers are drawn from the main barrel (numbered from 1 to 35) and 1 Powerball number is drawn from the Powerball barrel (numbered from 1 to 20)."
The article below got me interested in asking the above questions.
https://www.msn.com/en-au/news/australia/one-lucky-winner-claims-massive-dollar40million-powerball-prize-but-some-who-missed-out-are-offended-by-the-ridiculous-winning-numbers/ar-BBUcIHa?li=AAgfLCP&ocid=mailsignout
Also, I don't know if this was the best spot to put this thread or "QUESTIONS AND ANSWERS - (gambling or math)", so Admin please feel free to move it, if necessary.
thanks
Quote: ksdjdj...1. For the "main barrel", what is the probability that the highest number would be 14 (or less)?
2. For the "main barrel", what is the probability that the highest number would be exactly 14? ...
For 1, I get about 1 in 1959.
And for 2, I get about 1 in 3919.
The probabilities aren't as small as I would have thought. That must be because only 35 numbers are in the "main barrel."
The more intuitive way is to think about each selection separately. Your initial draw will have a 14/35 chance of being a ball marked 14 or under. The second draw would have a 13/34 chance, then 12/33, and so on.
The chance of drawing 14 or less for all 7 balls in the next drawing would be: 14/35 x 13/34 x ... x 8/29 = .00051037 or about 1 in 1959.
The second way to do this is with combinations. (14 C 7)/(35 C 7) = .00051037
Quote: AyecarumbaThe main barrel contains 35 numbers, each with an equal chance of being selected. Why are people "offended"?
I don't know why (supposedly the person was joking though).
If that really did "offend" some people, then I would hate to see how "offended" they are if the main numbers were 1, 2, 3, 4, 5, 6 , 7.
There have been about 46 draws since the "7/35 Main and 1/20 Powerball" format was introduced.
So, the chance of "1. ...highest number would be 14 (or less)" over 46 draws is:
about a 1/43 chance that it happens at least once, right?
Quote: ksdjdj1. For the "main barrel", what is the probability that the highest number would be 14 (or less)?
2. For the "main barrel", what is the probability that the highest number would be exactly 14? (edit/update: 145 pm)
"...Powerball is drawn from 2 barrels. 7 numbers are drawn from the main barrel (numbered from 1 to 35) and 1 Powerball number is drawn from the Powerball barrel (numbered from 1 to 20)."
Look at the problems this way:
1. You have 35 balls; 14 (the ones numbered 1-14) are red, and the other 21 are white. You draw seven red balls; what is the probability that all of them are red?
There are (35)C(7) = 6,724,520 sets of 7 balls that can be drawn, and (14)C(7) = 3432 sets of seven red balls; the probability is 3432 / 6,724,520 = about 1 in 1959.36.
2. This time, one ball (#14) is blue, 13 are red, and 21 are white. What is the probability that you draw the blue ball and 6 of the red ones?
Again, there are 6,724,520 sets of 7 balls; there are (1)c(1) x (14)C(6) = 3003 sets of six red and one blue balls, so the probability is 3003 / 6,724,520 = about 1 in 2239.27.
Quote: ksdjdjThanks everyone
There have been about 46 draws since the "7/35 Main and 1/20 Powerball" format was introduced.
So, the chance of "1. ...highest number would be 14 (or less)" over 46 draws is:
about a 1/43 chance that it happens at least once, right?
Yes, but you might be missing the reason why. The odds of it not happening at all over 46 draws is (1958/1959)^46 = 97.7%
So the odds of it happening at least once is 1 - 97.7% = about 2.3% or 1 in 43.
Quote: ThatDonGuy...2. This time, one ball (#14) is blue, 13 are red, and 21 are white. What is the probability that you draw the blue ball and 6 of the red ones?
Again, there are 6,724,520 sets of 7 balls; there are (1)c(1) x (14)C(6) = 3003 sets of six red and one blue balls...
Thanks for your clear explanation. I see a typo, though--"(14)C(6)" should be "(13)C(6)" because there are 13 reds.
Quote: FTBI ran across an article about Idaho ending Powerball there. Apparently, lawmakers there want nothing to do with Powerball after its planned expansion into Australia this year and Britain next year.
Can you paste said article, because that makes no sense.
https://www.postregister.com/news/local/idaho-lawmakers-end-powerball-fearing-foreign-participation/article_40672079-afc5-5455-8cc2-0eeed865465b.html
