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Your EV would increase if you removed numbers that are picked more often. Your chance of winning is still the same but the expected payout is higher because th

by nimos 11y ago

Your EV would increase if you removed numbers that are picked more often. Your chance of winning is still the same but the expected payout is higher because the expected number of identical tickets is lower.

logfromblammo 11y ago
Your chance of winning is not the same if you never pick a number that could be drawn with the winning combination. If you never pick 7, the winning combination could still have a seven in it. By never picking seven, you are voluntarily abandoning 1/69 of the white balls and 1/26 of the red balls. There are (68 C 4) x 25 + (69 C 5) such combinations out of the (69 C 5) x 26 possible combinations, or 31598138 out of 292201338. You lower your chances of winning in the first place to 89% of normal by avoiding just one number.
- tzs 11y ago
  What you've actually shown is that someone who bought all possible tickets that do not contain 7 has less chances of winning than someone who bought one each of all possible combinations that do not contain 7 has less chance of winning that someone who bought one each of all possible combinations. For someone buying a single ticket excluding 7 has no effect at all on their chances that their ticket will win.
  - logfromblammo 11y ago
    Checking the math: P(7) = 31598138 / 292201338 ~= 0.108 P(~7) = 260603200 / 292201338 ~= 0.892 P(win|7) = 0 P(win|~7) = 1 / 260603200 ~= 3.84e-9 P(win) = P(win|7) * P(7) + P(win|~7) * P(~7) = 0 + 260603200 / 292201338 * 1 / 260603200 = 1 / 292201338 You are correct, sir. I withdraw my incorrect statement. Nevertheless, I presume that most of the surge players, who only buy tickets for very high jackpots, will not have favorite numbers, and thus you would only be avoiding splitting the pot with the regulars playing their usual numbers. If someone plays his birthday every week, but bought an extra 10 tickets for the record jackpot, those extras are likely to be completely random draws. Judging by the increase in the pot from the last draw, there are a lot of people buying more than the usual amount of tickets. So in the EV calculation, if the amount attributable to not splitting the pot is not greater than the actual value of your time in filling out the form to play specific numbers, you are better off taking the random draw.
    - nkurz 11y ago
      Wow! Your comments on the site are usually great, and higher in this thread is the first time I've read something you've written and said "no, that's completely wrong!". Coming to the bottom and finding that you've already done the math to prove yourself wrong is wonderfully enheartening. I hope that I can have the grace and ability to do the same the next time I make a similar mistake.
      - logfromblammo 11y ago
        Your complimentary comment made me slightly uncomfortable, for reasons that I can't quite discern. Rather than allowing myself to pretend it never existed.... Thank you for recognizing in me a quality you would wish to nurture in yourself. It takes some of the sting out of being wrong.

by nimos 11y ago

logfromblammo 11y ago

Your chance of winning is not the same if you never pick a number that could be drawn with the winning combination. If you never pick 7, the winning combination could still have a seven in it. By never picking seven, you are voluntarily abandoning 1/69 of the white balls and 1/26 of the red balls. There are (68 C 4) x 25 + (69 C 5) such combinations out of the (69 C 5) x 26 possible combinations, or 31598138 out of 292201338. You lower your chances of winning in the first place to 89% of normal by avoiding just one number.

tzs 11y ago
What you've actually shown is that someone who bought all possible tickets that do not contain 7 has less chances of winning than someone who bought one each of all possible combinations that do not contain 7 has less chance of winning that someone who bought one each of all possible combinations. For someone buying a single ticket excluding 7 has no effect at all on their chances that their ticket will win.
- logfromblammo 11y ago
  Checking the math: P(7) = 31598138 / 292201338 ~= 0.108 P(~7) = 260603200 / 292201338 ~= 0.892 P(win|7) = 0 P(win|~7) = 1 / 260603200 ~= 3.84e-9 P(win) = P(win|7) * P(7) + P(win|~7) * P(~7) = 0 + 260603200 / 292201338 * 1 / 260603200 = 1 / 292201338 You are correct, sir. I withdraw my incorrect statement. Nevertheless, I presume that most of the surge players, who only buy tickets for very high jackpots, will not have favorite numbers, and thus you would only be avoiding splitting the pot with the regulars playing their usual numbers. If someone plays his birthday every week, but bought an extra 10 tickets for the record jackpot, those extras are likely to be completely random draws. Judging by the increase in the pot from the last draw, there are a lot of people buying more than the usual amount of tickets. So in the EV calculation, if the amount attributable to not splitting the pot is not greater than the actual value of your time in filling out the form to play specific numbers, you are better off taking the random draw.
  - nkurz 11y ago
    Wow! Your comments on the site are usually great, and higher in this thread is the first time I've read something you've written and said "no, that's completely wrong!". Coming to the bottom and finding that you've already done the math to prove yourself wrong is wonderfully enheartening. I hope that I can have the grace and ability to do the same the next time I make a similar mistake.
    - logfromblammo 11y ago
      Your complimentary comment made me slightly uncomfortable, for reasons that I can't quite discern. Rather than allowing myself to pretend it never existed.... Thank you for recognizing in me a quality you would wish to nurture in yourself. It takes some of the sting out of being wrong.