That code, while it found the main code error, will still sometimes show the error message.

$secretword=$words{$name};
if ($secretword eq ""){
  $secretword="stupid";
}
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If $name is not found in the hash %words, then $secretword will be undefined and will produce the error mentioned when used as if defined. If you however test for the definedness of $secretword you will be closer to the original aim.

if (defined $secretword) {
  $secretword = 'stupid';
}
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In perl style I'd probably write:

$secretword = $words{$name} || 'stupid';

But that is a matter of choice.