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dgreening last won the day on October 30 2015

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About dgreening

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    Maryland [DC area]

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  1. This is very interesting, but I think there is a more concrete answer.
  2. My first thought was like Capt Ed. BUT thinking about it some more.
  3. No dispute here.
  4. Aha! I just didn't follow the same process through to get the supper precision devices! Thanks
  5. Curious about your result. If you average over many trials are your expected winnings positive? As I mentioned, I only did a small number of runs [about 100]. BTW - I like your title "Retired Expert"! I think that if you do a large number of runs, you eventually get to about 50%. I was quoting bonanova solution/answer, not my proposed strategy. Your strategy is not applicable, you can't bet more money than you have, if your current money is 50$, your original money is 100$, so 2*100-50 = 150$, which you don't have. Now I explain, my proposed strategy. Hidden Content yes, if you [and the house] are willing to bet micro, nano or pico dollars, then the game can go on forever. But your money is gone, so it is theoretically correct, but not very practical.
  6. An answer for part 1
  7. Perhaps some more description of the task would help. What is the context for this question??
  8. Now I am confused. I think I followed the strategy you described yesterday, when you postulated that it would deliver 100% chance of winning. But, now your simulation it gives you a 50% chance of winning. Did I miss something??
  9. I was working on the strategy that you describe: Bet half my money [rounded to an integer]; orbet the "ceiling" [twice my original money - Current money].Testing this in a spread sheet [with only about 100 trials] I am winning about 62% of the time.
  10. The hint about primary numbers did not help, but I noticed a pattern
  11. I would be interested in how you calculated your answer. I came up with something slightly different
  12. While driving home last night I realized that I had made the problem much more complicated than it needed to be
  13. If we assume that all 1000 bottles might be used ... one approach
  14. Maybe I am looking at this too simplistically