Answers.Body: <p>You can't really &quot;prove&quot; it. But, presumably, players want to play the best move, so which move is more often played may be an indication of which move the <em>players</em> think is best.</p>
  <p>I looked at two databases with master games, and the most popular moves after 1.e4 in each were c5, e5, e6, and c6, in that order. This supports the first half of your statement (since e6 and c6 are played less often than e5 and c5) but not the second half (since c5 is played more than e5.)</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-05-08T08:01:53.350Z
Answers.Text: I just wanted to comment, although I agree to this answer, that it is hard to compare c5 and e5 because although e5 might have less losses for black (and more draws), c5 definitely has more wins for black (as well as more losses). So depending on how aggressively you want to play for the win as black, different moves might be better or worse (statistically).
Answers.UserDisplayName: sleepy
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-05-08T02:54:48.173Z
Answers.LastActivityDate: 2021-05-08T02:54:48.173Z
Answers.UserDisplayName: D M
Body: <p>I heard that for the first move, 1. e4, 1. d4, 1. c4 and 1. Nf3 are equally good and none of them is better than others, even marginally. However, after 1. e4, both 1... e5 and 1... c5 are marginally better than 1... e6 and 1... c6; but none of 1... e5 and 1... c5 is better than the other. I am wondering if there is a way to theoretically prove this, either using an engine, or using statistics, or using opening theories.</p>
  <p>So my question is: Can the following statement be proven?</p>
  <blockquote>
  <p>After 1. e4, both 1... e5 and 1... c5 are marginally better than 1... e6 and 1... c6; but none of 1... e5 and 1... c5 is better than the other.</p>
  </blockquote>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-05-07T20:30:22.520Z
Comments.Text: Yes, *theoretically* such a statement (if true) can be proven - once you solve chess completely. Then again, after a full prove of that kind, the situations cannot be marginally better: They are either won or lost or remis
Comments.UserDisplayName: Hagen von Eitzen
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-05-07T21:12:11.530Z
Comments.Text: What does it mean for a move to be "marginally better"? Other than "this move leads to a win/draw/lose", I can't think of a way to determine if a move is objectively better than other. Note that statistics can be biased. For example, if a move is popular among beginners, it will probably underperform
Comments.UserDisplayName: David
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-05-08T15:02:52.597Z
Comments.Text: Even when statistics are unbiased they can be misleading. If c5 is the go-to move for Black players who want to avoid a draw, then you will see fewer draws in those lines. That *might* mean that e5 is intrinsically more drawish than c5, but it could also be a reflection of the fact that many Black players take greater risks in the c5 lines.
Comments.UserDisplayName: John Coleman
Tags: <opening>
Title: Can one theoretically prove that these moves are marginally better?
ThreadText: # Can one theoretically prove that these moves are marginally better?
  
  (Asked by: username_0 on 2021-05-07)
  
  I heard that for the first move, 1. e4, 1. d4, 1. c4 and 1. Nf3 are equally good and none of them is better than others, even marginally. However, after 1. e4, both 1... e5 and 1... c5 are marginally better than 1... e6 and 1... c6; but none of 1... e5 and 1... c5 is better than the other. I am wondering if there is a way to theoretically prove this, either using an engine, or using statistics, or using opening theories.
  
  So my question is: Can the following statement be proven?
  
  > After 1. e4, both 1... e5 and 1... c5 are marginally better than 1... e6 and 1... c6; but none of 1... e5 and 1... c5 is better than the other.
  
  **Tags:** opening
  
  > **Comments:**
  > **username_1**: Yes, *theoretically* such a statement (if true) can be proven - once you solve chess completely. Then again, after a full prove of that kind, the situations cannot be marginally better: They are either won or lost or remis  
  > **username_2**: What does it mean for a move to be "marginally better"? Other than "this move leads to a win/draw/lose", I can't think of a way to determine if a move is objectively better than other. Note that statistics can be biased. For example, if a move is popular among beginners, it will probably underperform  
  > **username_3**: Even when statistics are unbiased they can be misleading. If c5 is the go-to move for Black players who want to avoid a draw, then you will see fewer draws in those lines. That *might* mean that e5 is intrinsically more drawish than c5, but it could also be a reflection of the fact that many Black players take greater risks in the c5 lines.  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_4 on 2021-05-08. Score: 1
  
  You can't really "prove" it. But, presumably, players want to play the best move, so which move is more often played may be an indication of which move the *players* think is best.
  
  I looked at two databases with master games, and the most popular moves after 1.e4 in each were c5, e5, e6, and c6, in that order. This supports the first half of your statement (since e6 and c6 are played less often than e5 and c5) but not the second half (since c5 is played more than e5.)
  
  > **Comments:**
  > **username_5**: I just wanted to comment, although I agree to this answer, that it is hard to compare c5 and e5 because although e5 might have less losses for black (and more draws), c5 definitely has more wins for black (as well as more losses). So depending on how aggressively you want to play for the win as black, different moves might be better or worse (statistically).  
  
  ---
  
  