I thought the same when I saw it.

Weirdly enough, though, the average was at about $4.40 when I saw it earlier today, and it is $4.27 now.

Wonder if this technique might be driving some people to pay more (I will do it as soon as I get home), but not enough to offset all the people that just want lots of games for no money at all.

There is also the possibility that it is cannibalizing their higher-paying customer base. For instance, perhaps if I hadn't been presented with that option I would have paid $10 instead. In that case it would drive the average down rather than up.

Presumably that's why they're A/B testing it. :-)

If the average is currently $X after N purchases, that means they have taken in $(N * X). When the next person pays $(X + 1), the total is now $((N * X) + (X + 1)) == $(((N + 1) * X) + 1). The new average is $(X + (1 / (N + 1))), which is guaranteed to be higher than the previous average, even if $(X + 1) < $10.

Yes, the average after purchase (AAP) would be greater than the average before purchase (ABP). However, the AAP of $10 is greater than the AAP of $X+1.

The question is to what extent the $X+1 purchases depress greater purchases, as well as to what extent the $X+1 purchases increase smaller purchases.

The answer can only be determined by actually testing it.

If the average is currently $X after N purchases, that means they have taken in $(N * X). When the next person pays $(X + 1), the total is now $((N * X) + (X + 1)) == $(((N + 1) * X) + 1). The new average is $(X + (1 / (N + 1))), which is guaranteed to be higher than the old average.