Is the following argument correct: A double six in a single turn in game B is 1/6 as likely as rolling a six in one turn in game A. But there are 6 times as many turns in game B as game A. Thus the two games are equally good bets.
Where game A is Pierre throws one die four times. He wins if at least once he rolls a six: .
and game B is he has 24 turns, and each time he throws two dice simultaneously. This times he wins if he rolls at least one “double six”
my working: P(not getting double 6)=$(36^1-35^1)/(26^1) = 1-(35/26)^1 = 0.3462 percent$
p(not getting a single 6) = $(6^1-5^1)=1/6=0.16%$
where do i go from here?