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It should be noted that for the actual lottery, the advice given is wrong. Because in many real lotteries the first five digits can be in any order, 1 1 1 1 1 w

by maxjus 13y ago

It should be noted that for the actual lottery, the advice given is wrong. Because in many real lotteries the first five digits can be in any order, 1 1 1 1 1 would be less likely than, say, 13 23 42 33 2, because, though there is only one way to end up with 1 1 1 1 1, there are 5! == 120 ways to achieve the latter (I am ignoring the "bonus ball").

jstanley 13y ago
I think that's incorrect. If you secretly label the "1" balls with secondary numbers, do the odds change? There are still 5 separate "1" balls, so still 120 ways to draw 5 1's in a row. EDIT: Never mind, I get it. Removing the first "1" reduces the chances of getting the second "1", etc.
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- SEMW 13y ago
  > EDIT: Never mind, I get it. Removing the first "1" reduces the chances of getting the second "1", etc. That is a point, but it's not the point maxjus was making. Two different issues: First, whether order matters - whether a drawing of "17, 23, 31" is the same as a drawing of "23, 31, 17". Second, whether balls are replaced - whether you put back the "17" ball after it's been drawn, before you pick the next number. Assuming balls numbered 0 through 49: - If order matters and balls are replaced, then "1, 1, 1" is as likely as "1, 17, 23". - If order doesn't matter and balls are replaced, then it depends on how many of each number there are. If there's one of each number, "1, 1, 1" is six times less likely than "1, 17, 23", as there' six different ways to make the latter (since "1, 17, 23" and "23, 17, 1" are the same), but only one to make the former. If there's three balls of each number, they're equally likely. - If order matters and balls are not replaced, then it depends on how many balls of each number there are. If there's one of each number, "1, 1, 1" is impossible. With three of each number, "1, 1, 1" is still less likely than "1, 17, 23", as, e.g. for the second number, there's two "1"s but three "17"s. (More specifically, there's six ways to draw "1, 1, 1" (3x2x1), but 27 ways to draw "1, 17, 23" (3x3x3)). - If order doesn't matter and balls are not replaced, then it depends on how many balls of each number there are. If there's one of each number, "1, 1, 1" is impossible. If there's three of each number, "1, 1, 1" is way less likely than "1, 17, 23" - there's six ways to draw "1, 1, 1", but 162 ways to draw "1, 17, 23" (9x6x3). Edit: posted the above as an answer on stackexchange, since all answered posted so far have silently assumed that order matters and balls are replaced(!).
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- maxjus 13y ago
  Read the following, and see if you haven't changed your mind! http://www.datagenetics.com/blog/january42012/ http://www.datagenetics.com/blog/january42012/
daveungerer 13y ago
Deleted my comment after reading the rest of this thread. Thanks for giving me something to think about!

by maxjus 13y ago

jstanley 13y ago

I think that's incorrect. If you secretly label the "1" balls with secondary numbers, do the odds change? There are still 5 separate "1" balls, so still 120 ways to draw 5 1's in a row. EDIT: Never mind, I get it. Removing the first "1" reduces the chances of getting the second "1", etc.

deleted 13y ago
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SEMW 13y ago
> EDIT: Never mind, I get it. Removing the first "1" reduces the chances of getting the second "1", etc. That is a point, but it's not the point maxjus was making. Two different issues: First, whether order matters - whether a drawing of "17, 23, 31" is the same as a drawing of "23, 31, 17". Second, whether balls are replaced - whether you put back the "17" ball after it's been drawn, before you pick the next number. Assuming balls numbered 0 through 49: - If order matters and balls are replaced, then "1, 1, 1" is as likely as "1, 17, 23". - If order doesn't matter and balls are replaced, then it depends on how many of each number there are. If there's one of each number, "1, 1, 1" is six times less likely than "1, 17, 23", as there' six different ways to make the latter (since "1, 17, 23" and "23, 17, 1" are the same), but only one to make the former. If there's three balls of each number, they're equally likely. - If order matters and balls are not replaced, then it depends on how many balls of each number there are. If there's one of each number, "1, 1, 1" is impossible. With three of each number, "1, 1, 1" is still less likely than "1, 17, 23", as, e.g. for the second number, there's two "1"s but three "17"s. (More specifically, there's six ways to draw "1, 1, 1" (3x2x1), but 27 ways to draw "1, 17, 23" (3x3x3)). - If order doesn't matter and balls are not replaced, then it depends on how many balls of each number there are. If there's one of each number, "1, 1, 1" is impossible. If there's three of each number, "1, 1, 1" is way less likely than "1, 17, 23" - there's six ways to draw "1, 1, 1", but 162 ways to draw "1, 17, 23" (9x6x3). Edit: posted the above as an answer on stackexchange, since all answered posted so far have silently assumed that order matters and balls are replaced(!).
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maxjus 13y ago
Read the following, and see if you haven't changed your mind! http://www.datagenetics.com/blog/january42012/ http://www.datagenetics.com/blog/january42012/

daveungerer 13y ago

Deleted my comment after reading the rest of this thread. Thanks for giving me something to think about!