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With the introduction of the new tie breaker, if the amount of stars is the same it will compare the average of the destruction % of best attacks. But what would happen if, for example, both clans get full stars (which is 100% average destruction for both clans)?

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    Then its a tie still – Ben Craig Oct 1 '15 at 23:48
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Average destruction is rarely ever tied, as the number goes up to four numbers, two on the right and left of the decimal. But assuming that the destruction averages are the same, it will be a draw as sourced from the Clash of Clans Wiki:

Should both clans deal the same amount of stars and total destruction, the war will be considered a draw. This, however can be considered rare, as total destruction dealt is different in each attack and hardly sync with the enemy clan.

So draws are possible... but much,much less likely.

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Your logic is quite confusing to follow here, however, I will do my best to explain.

Since your first and last statements contradict, I will tackle the first, and follow with the second.

If both clans tie in stars, the clan who has the highest total destruction on their best attack. There is no avaraging. The best attack is in theory determined by destruction, stars, level of attacker, level of defender ((town hall included)), as well as time used, clan castle troops killed, and remaining heroes. The clan with the highest percentile wins.

Now for your second statement, there seems to be a lot of confusion going around. Both clans having the same stars before the tiebreaker would cause a draw. After this update, whichever clan has the highest percentile of destruction in their clan best attack wins the war.

In short, it's not based on average destruction, it's based on each clans' best attacks.

The best attack is near about never 100%, as pointed out here.

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Assuming a straight up tie--every attack from both sides being a perfect 3 stars (average destruction still being 100%)--then it is still a draw.

Essentially, when it comes to percentages, it's likely that no building is a round figure. For example, assuming each building held the same value, 101 buildings holding the same weight would essentially be 100/101 = 99.0099% each. It's the .0099% that would ultimately cause deviations between the two "average destructions", even if they appear similar to the hundredths decimal point--one would outweigh the other to the thousandths and tenthousandths.

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    I've removed your insults from your answer. That took out three of your five paragraphs. If there was something useful in them, feel free to add them back, without the insults, please. – Frank Oct 14 '15 at 2:58

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