Expected winnings of a slot machine Of the three wheels on a slot machine where each wheel has ten items:


*

*First wheel has 2 flowers, 4 dogs, 4 houses.

*Second wheel has 6 flowers, 3 dogs, 1 house. 

*Third wheel has 2 flowers, 3 dogs, 5 houses.


You win $20$ dollars per dollar if $3$ flowers show, $10$ dollars if $3$ dogs show, and $5$ dollars if $3$ houses show (and lose otherwise). We're asked to find the expected winning per dollar spent.
I tried solving it this way:  
$p(\text{flowers})=(2/10)(6/10)(2/10) = 3/125$
$p(\text{dogs})=(4/10)(3/10)(3/10) = 9/250$
$p(\text{houses})=(4/10)(1/10)(5/10) = 1/50$
$p(\text{rest})=1-[p(\text{flower})+p(\text{dogs})+p(\text{houses})]=23/25$
So expected winning is;  
$20(3/125)+10(9/250)+5(1/50)-1(23/25)=0.92$  
I thought I did it correctly, but the answer key says it is $0.94. Did I do something wrong?

It concerns purely the winning on a dollar spent.
$p_{f}=\frac{2}{10}\frac{6}{10}\frac{2}{10}=\frac{24}{1000}$
$p_{d}=\frac{4}{10}\frac{3}{10}\frac{3}{10}=\frac{36}{1000}$
$p_{h}=\frac{4}{10}\frac{1}{10}\frac{5}{10}=\frac{20}{1000}$
$\text{expected winning}=20\times\frac{24}{1000}+10\times\frac{36}{1000}+5\times\frac{20}{1000}=\frac{940}{1000}=0.94$

If it had concerned 'rendement' in the sense of winning minus loosing then the correct answer would have been $0.94-1.00$. In all cases you must pay a dollar. Not only when you do not win.
