Question 346137
Let's calculate the odds of actually winning first.
P(1 on die)={{{1/6}}}
P(red card)={{{26/52=1/2}}}
P(1 and red)={{{(1/6)(1/2)=1/12}}}
So then 
P(not(1 and red))+P(1 and red)={{{1}}}
P(not(1 and red))=1-P(1 and red)
P(not(1 and red))={{{1-1/12}}}
P(not(1 and red))={{{11/12}}}
So the chances of not winning are {{{11/12}}}
In terms of odds, there is 1 chance of winning, 11 chances of not winning.
Odds are 11:1 against winning.