pot odds question

[7stud]

I'm wondering about the correct approach to figuring out pot odds in a particular situation.

Assumptions:
1) $10-20 Holdem
2) Heads up on 4th street
3) Opponent bets $20 into a $60 pot (total = $80).
4) Opponent will bet 5th street 100% of the time.

Situation 1:
I have a draw to the nuts. I have to call a $20 bet now, and if I make my hand I will win $80 currently in the pot plus my opponent's $20 bet on 5th street. Therefore, the pot odds($80) + implied odds($20) are $100:$20 = 5:1. The $20 I put in the pot on 5th street to call is not at risk because I will only call if I make my hand, so it is excluded from the calculation.

Situation 2:
Same as above, but there is a pair on board. Now, I am suspicious that my opponent might have trips. Because of tells and what not, I decide there is only an 80% chance of winning if I make my hand. How do I calculate the pot odds now? Like this:

a) I have to call $20 to win the $80 in the pot. But I also have some implied odds on 5th street. On 5th street, I have an 80% chance of winning $20 or .8($20) = $16. On the other hand, I have a 20% chance of losing $20 more or .2($20) = $4. So, my expectation is that on 5th street, I will win an additional $16 - $4 = $12. As a result, I have to call $20 now to win $80(pot odds) + $12(implied odds), which are odds of $92:$20 = 4.6:1.

or

b) I risk $20 now and $20 on 5th street to win $80 in the pot now plus my opponents $20 bet on 5th street, giving me odds of $100:$40 = 2.5:1.

Which analysis is faulty and why?

TOP(p.52) says:
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your implied odds are the ratio of your total expected win when your card hits to the present cost of calling a bet.

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