Three at least, because otherwise it would be called a threesome, not a gangbang.
But depends on how much and many
she can take. For a "full-servicable" Lady the
minimum number
ImO would be
five though (one for her pussy, one for her mouth/throught, one for her a$$, and one in each of her hands
); but
eight would be
perfect in this case. (Also see "factor 2" below.
)
i have developed a
formula that helps you calculate the perfect number of
planned participants for a gangbang: 1. The basic factor is how many men a Woman desires to play with
at the same time. 2. Another factor is the difference between female and male sexuality: Both sexes typically need some startup-time, but, unlike Women, men typically need more time for recreation in between, so not all men present will be able to perform at a 100% rate. 3. The third factor is that - unfortunate but true - in most cases not every man who is scheduled for an event actually shows up.
To glue those factors together in a simple formula, let me give an example...
Given a Woman who desires to play with 3 men at the same time on average we will need to apply factor 1.5 to counterbalance lower male performance rates. Also we will have to apply the assurance factor of 1.25 for possible cancelations. This results in the number of 5.625 men for this example case. As men only come as a whole and not in pieces we finally have to round the result to the next higher integer number, giving us the final result... 6... as the
perfect number of
planned male participants in this Woman`s gangbang (because we expect that one guy won`t show up *lol*).
General formula (in maths):
gangplanned( W(x) ) = ceil( x * 1.5 * 1.25 )