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Subject: Losses when more units present than loss table shows? rss

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Mike Owens
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9.11.6 says to consult the loss table for combat losses that include fractions. The greatest number of losses in the loss table is 2 units.

So does a force of 8 units that takes 1/3 losses only lose 2 units?
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bertrand d
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In your case, 8=6+2

6/3=2 losses. Nobody challenges that.

The fraction only manages the remaining 2, for which there can be never ending discussions about rounding.
And 2 for a 1/3 results is 1 loss, according to the table.
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Mike Owens
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Lugbar wrote:
In your case, 8=6+2

6/3=2 losses. Nobody challenges that.

The fraction only manages the remaining 2, for which there can be never ending discussions about rounding.
And 2 for a 1/3 results is 1 loss, according to the table.
That is not what the table says. The table refers to number of units in a force. There are 8 units in the force. Unless you are saying that a force of 8 units is actually a force of 6 and a force of 2, which is not how a force is defined in 1.2.

I can determine from the pattern in the loss table that the intent is that fractional losses are rounded up (FRU). So why don't the rules just say that?!?

My beef with this rule is that the loss table is, at best, too clever by half; and at worst, insulting my intelligence as a wargamer that I can't be trusted to calculate 1/3 FRU.
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bertrand d
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Well don't forget this game was made in the 80s.
Maybe the homo wargamicus was not that evolved at that time,
It was certainly not an insult in those old days.

With 1 to 6, you get all the possible roundings for 1/3, 1/2 and 2/3 in the same table.
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