Lunar Poker

I read your artical about Lunar Poker. The thing that I would like to know is, from the part where you wrote: "It concludes the house edge, as measured by the ratio of the expected loss to the mandated two initial wagers, is 4.90%". So does that mean if I bet 10$ on ante and 10$ on super wager bet, my total bet being 20$, then is my expected loss 49c from 10$ ante bet or 98c fron both bets tougheter.

Thank you for your time, I am looking forward to hear your answer.

The House Edge is the percentage edge that the house has in terms of a ONE unit bet (the starting bet).
In this case it is 4.90%, which (with $10 bet) means it "costs" 49 cents to play, but this is a bit misleading because multiple units will be bet. With a $5 total starting bet size, the house edge will cost 24.5 cents to play, as 4.9% of ONE $5 initial bet is 24.5 cents.

So, the Element of Risk is used to show the edge when considering the total bets made.
This is the view of the house edge of ALL units wagered as the "main bet", so that 49 cents is really just 2.38% of the full action for a round of play, - since you'll play more than the starting bet amount.