What is the expected payoff using Ron's strategy?

The answer is supposed to be 1.20. Please show work/explain why.

Back in the Gryffindor common room, Ron was showing Harry the latest magic from the Weasley shop. ... and these are the dice we should use when gambling." "Hold on, Ron," replied Harry. "They look normal to me." "Not when you roll them. The way they work, if both dice come up as the same number, then one of the dice will magically change to a different number, with an equal chance of the other numbers. Remember that dice game we played the other day -- betting on the sum of the two dice? As you recall, in a fair game, here's the price to play, probabilities, and payoffs:" Sum to Two B-sided Dice 2 3 4 5 6 7 8 9 10 11 12 Price to Play $1.00 $1.00 $1.00 $1.00 $1.00 $1.00 $1.00 $1.00 $1.00 $1.00 $1.00 Probability 1/36 2/36 336 4/36 5/366/36 5/364/36 3/3 Winning Payoff $36.00 $18.00 $12.00 99.00 7.20 $6.00 $72059.00 $12.00 $18.00 $36.00 "We can use these dice to our advantage." "Let me see if I understand this," said Harry, scribbling on the Daily Prophet. "If my calculations are right, for example, the sum can never be 12." "You got it. Since a 12 is a '6' and '6', one of the '6s' will magically change to a '1', '2', '3', '4', or a '5'. Same is true for a sum of 2. Won't happen since both dice would have to be '1. You can still get sums like 8. '5' and '3'. '2' and '6'. But not with a '4' and '4'. One of the '4s' will change to a '1', '2', '3', '5', or a '6'. '4' and '4' will become '4' and something else, with a sum different from 8. "What's the plan, Ron?" "Well, just to avoid being too obvious, we'll go short on 12' and long '11