[malvim]

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.

[joshuacc]

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. :-)

[sheriff]

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.

[joshuacc]

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.

[sheriff]

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.