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October 22, 2018
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A bum is down to his last dollar and decides to spend it on a Cash Five ticket. In his state, Cash Five is a 5/36 game with the following prize structure
5/5: $30,000
4/5: $300
3/5: $15
2/5: free quick pick ticket
Luckily for this bum, his Aunt Betty works for the lottery. She tells him that today's drawing will be fixed, but she can't tell him explicitly what the numbers will be. Instead, she gives him three clues.
The highest number will be five times the smallest number
The second-highest number will be three times the second-smallest number
The middle number will be equal to the average of all five numbers.
The bum realizes that isn't enough information to figure out what the winning combo will be, but he knows how to optimize his expected winnings. What numbers should the bum play if he can only buy one ticket?
Chino, CA United States
Member #196,351
March 3, 2019
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First time posting, I couldn't resist a good math puzzle concerning our favorite topic. By brute force I came up with three sets of five numbers meeting the criteria, {2 3 6 9 10}, {4 6 12 18 20}, and {6 9 18 27 30}. The numbers 6, 9 and 18 are common are shared by at least two of the sets, but the last one has all three. So that is the one the bum should play, since he would get a 2/5 prize for sure if either of the other two came up, or the jackpot if that is the one.
Hopefully he wouldn't stay a bum after that, but we know how that can go :) Just as an anecdote, I once (in the 90s) called out the CA Lottery on a mis-printed odds for one of the lesser prizes of a new game, and they did send back a reply. "A number of eagle-eyed observers caught this mistake, thank you."
Just that, no reward, I sure felt like a bum then. Ha ha
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October 22, 2018
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Quote: Originally posted by mathbum on Mar 3, 2019
First time posting, I couldn't resist a good math puzzle concerning our favorite topic. By brute force I came up with three sets of five numbers meeting the criteria, {2 3 6 9 10}, {4 6 12 18 20}, and {6 9 18 27 30}. The numbers 6, 9 and 18 are common are shared by at least two of the sets, but the last one has all three. So that is the one the bum should play, since he would get a 2/5 prize for sure if either of the other two came up, or the jackpot if that is the one.
Hopefully he wouldn't stay a bum after that, but we know how that can go :) Just as an anecdote, I once (in the 90s) called out the CA Lottery on a mis-printed odds for one of the lesser prizes of a new game, and they did send back a reply. "A number of eagle-eyed observers caught this mistake, thank you."
Just that, no reward, I sure felt like a bum then. Ha ha
Hello Mathbum, Welcome to Lottery Post and thank you for solving my math puzzle! I don't get many takers as you can see...
You got the right answer, for almost the right reason. There are actually six sets of number that work:
they all have an equal chance of winning the top prize, but the last set {6, 9, 18, 27, 30} has more 2/5 overlaps with the other sets and gives a higher chance of winning the free ticket than any of the other sets.
The CA lottery should have at least given you a free ticket.
Chino, CA United States
Member #196,351
March 3, 2019
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Hi oate, thanks for your gracious reply. I rushed and shot from the hip, just got lucky seeing that set of numbers. I was told back in grade school to watch my carelessness, guess it still gets me sometimes ;) I have seen many puzzles posted that are thought-provoking, so even without a response there is value in them. Remember the Monty Hall problem that caused some stir? Now the question is, pick your own numbers, use quick pick, use the Random site, or use a calculator (I have programs on my HP-50), but still emotions or convenience cause me to jump between all of these. I think we are left with the only good advice, "You can't win if you don't play." Best of luck - regards