Question on expected gain (probability) There are two padlocked boxes, one containing \$10 and the other \$1. There are two bunches of keys: A: 1 key opens the \$10 box, 1 key opens the \$1 box, 4 keys open neither box. Total 6 keys.
B: 2 keys open the \$10 box, 2 keys open the \$1 box, 5 keys open neither box. Total 9 keys.
There are two choices you can make:
1. You pay \$1. Choose one key from A and one of the two boxes. If the key opens the box, you keep whatever money is in the box. Otherwise you get nothing.
2. You pay $1.35. Choose one key from B and one of the two boxes. If the key opens the box, you keep whatever money is in the box. Otherwise you get nothing.  
Which choice should you make and why?  
Method 1: 
$Choice$ 1
Opening the \$1 box gives us a net gain of \$0. \$10 box gives us a net gain of \$9. Opening neither box gives us a net gain of -\$1.
So expected gain from choice 1 = $(\frac1{12})(0) + (\frac1{12})(9) + \frac8{12}(-1) \approx 0.0833$  
$Choice$ 2
Opening the \$1 box gives us a net gain of -\$0.35. \$10 box gives us a net gain of \$8.65. Opening neither box gives us a net gain of -\$1.35.
Expected gain from choice 2 = $(\frac2{18})(-0.35) + (\frac2{18})(8.65) + (\frac{10}{18})(-1.35) \approx 0.172$  
Therefore choice 2 is preferable.  
Method 2:
$Choice$ 1
Expected winnings = $(\frac1{12})(1) + (\frac1{12})(10) \approx 0.917$
Net gain = $0.917 - 1 = -0.083$
$Choice$ 2
Expected winnings = $(\frac2{18})(1) + (\frac2{18})(10) \approx 1.22$
Net gain = $1.22 - 1.35 = -0.13$ 
Therefore choice 1 is preferable.  
Which method is correct and why?  
EDIT: I realised Method 1 is wrong as the probabilities do not add up to 1. If instead I used $(\frac1{12})(0) + (\frac1{12})(9) + \frac{10}{12}(-1)$  I would have gotten $-0.083$. And for choice 2 I would have gotten $-0.13$.
