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if whet randbetween

asked 2017-02-23 18:07:24 +0100

sad_dog gravatar image

updated 2017-02-23 18:23:15 +0100

karolus gravatar image

why dose

  IF((B4/2)*$N$52>=RANDBETWEEN(1,100),"!!! JA !!!",
  IF((B8)*$N$51>=RANDBETWEEN(1,100),"!!! JA !!!","NEJ")))

not work it returns #VALUE!
i used same spreadsheet in google spreadsheet and it worked there

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Dear sad_dog, are you sure that to say JA (rarely) and NEJ (frequently) when filling two cells you need so complicated formula?

JohnSUN gravatar imageJohnSUN ( 2017-02-23 18:48:06 +0100 )edit

it's part of a much bigger spreadsheet whet conditions when to even display the certain values

sad_dog gravatar imagesad_dog ( 2017-02-24 16:03:06 +0100 )edit

N51 and N52 can only be 1 or 0, so thy act as a switch not a value

sad_dog gravatar imagesad_dog ( 2017-02-24 16:59:44 +0100 )edit

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answered 2017-02-23 20:23:08 +0100

CupcakeMedia gravatar image

Ok, so I played around with the formula. The reason why the formula might return a #VALUE error is if one of your cells (B4, B8, N52 or N51) contains a non-number value. For instance - if one of those four cells contain letter (A-Z) or if one of the cells contains exclamation marks, quotation marks etc - then the formula will return an error. Make sure that B4, B8, N52 or N51 only contain numbers.

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ok found it. Seem's like it can't handle "" as in empty. google spreadsheet just ser "" as no value rater then a text

sad_dog gravatar imagesad_dog ( 2017-02-24 16:04:24 +0100 )edit

this fixed it =IF(AND(B4="",B8="")=1,"",IF((IF(ISTEXT(B4),0,B4)/2)$N$52>=RANDBETWEEN(1,100),"!!! YES !!!",IF(IF(ISTEXT(B8),0,B8)$N$51>=RANDBETWEEN(1,100),"!!! YES !!!","No")))

sad_dog gravatar imagesad_dog ( 2017-02-24 16:41:00 +0100 )edit
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Asked: 2017-02-23 18:07:24 +0100

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Last updated: Feb 23 '17