4 ms·

That's precisely why you should always let the computer pick your numbers for you - your expected value is (slightly) higher because the odds of a shared jackpo

by eric_h 11y ago

That's precisely why you should always let the computer pick your numbers for you - your expected value is (slightly) higher because the odds of a shared jackpot go (slightly) down.

tzs 11y ago
Even better, instead of letting the lottery sales computer pick your numbers let your computer pick them, and filter out combinations that human pickers favor. Also, once your computer has generated a set of numbers for you only play those once. If you want to play again later roll a new set of numbers. The reason for this is that if you get into the habit of playing the same numbers every time you play, and you do not play every time, you face the possibility of seeing your numbers win a drawing you did not enter. It could be really annoying to know that you would have won if only you had bought a ticket that week.
- logfromblammo 11y ago
  No. Don't do that. Sharing a jackpot is still better than not winning. Your EV would drop more by abandoning that part of the probability space than it would drop by the possibility of splitting the pot. The only valid reason not to use the lottery computer's random picks would be mistrusting its random number generator.
  - 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.
  - digler999 11y ago
    > abandoning that part of the probability space this is hogwash. ALL numbers are equally likely to come up.
  - wongarsu 11y ago
    How does a bad random number generator or a smaller probability space change the EV? The numbers 1-2-3-4-5-6 are just as likely to be the winning numbers as the output of a perfect random number generator. The only remaining variable in the EV is the liklyhood of sharing the winning numbers, and that probability can be reduced by decreasing the probability space in smart ways.
    - deleted 11y ago
      [deleted]
      - tzs 11y ago
        If I use a ticket generator that always generates the ticket "1 2 3 4 5 6", I've excluded 292201337/292201338 of the probability space, so by your argument I'm lowering my chances from 1 in 292 million to approximately 1 in 85 quadrillion. You argument applies to any particular ticket, and thus if right would show that the odds of any particular ticket winning are 1 in 85 quadrillion.
digler999 11y ago
Then thats precisely why you should always pick your own numbers, and specifically choose combos containing numbers above 31. You would get (slightly) better EV than even the RNG-generated ones, because those can include both +/- 31
- squozzer 11y ago
  I prefer quick picks but for a different reason. I used to have a set of favorite numbers, but then I heard a story - urban legend most likely - about someone in Ireland who forgot to play their lucky numbers. Of course, those numbers won that time, so the now distraught Irishman committed suicide.

by eric_h 11y ago

That's precisely why you should always let the computer pick your numbers for you - your expected value is (slightly) higher because the odds of a shared jackpot go (slightly) down.

tzs 11y ago

Even better, instead of letting the lottery sales computer pick your numbers let your computer pick them, and filter out combinations that human pickers favor. Also, once your computer has generated a set of numbers for you only play those once. If you want to play again later roll a new set of numbers. The reason for this is that if you get into the habit of playing the same numbers every time you play, and you do not play every time, you face the possibility of seeing your numbers win a drawing you did not enter. It could be really annoying to know that you would have won if only you had bought a ticket that week.

logfromblammo 11y ago
No. Don't do that. Sharing a jackpot is still better than not winning. Your EV would drop more by abandoning that part of the probability space than it would drop by the possibility of splitting the pot. The only valid reason not to use the lottery computer's random picks would be mistrusting its random number generator.
- 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.
- digler999 11y ago
  > abandoning that part of the probability space this is hogwash. ALL numbers are equally likely to come up.
- wongarsu 11y ago
  How does a bad random number generator or a smaller probability space change the EV? The numbers 1-2-3-4-5-6 are just as likely to be the winning numbers as the output of a perfect random number generator. The only remaining variable in the EV is the liklyhood of sharing the winning numbers, and that probability can be reduced by decreasing the probability space in smart ways.
  - deleted 11y ago
    [deleted]
    - tzs 11y ago
      If I use a ticket generator that always generates the ticket "1 2 3 4 5 6", I've excluded 292201337/292201338 of the probability space, so by your argument I'm lowering my chances from 1 in 292 million to approximately 1 in 85 quadrillion. You argument applies to any particular ticket, and thus if right would show that the odds of any particular ticket winning are 1 in 85 quadrillion.

digler999 11y ago

Then thats precisely why you should always pick your own numbers, and specifically choose combos containing numbers above 31. You would get (slightly) better EV than even the RNG-generated ones, because those can include both +/- 31

squozzer 11y ago
I prefer quick picks but for a different reason. I used to have a set of favorite numbers, but then I heard a story - urban legend most likely - about someone in Ireland who forgot to play their lucky numbers. Of course, those numbers won that time, so the now distraught Irishman committed suicide.