[Route66]

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I agree that the hand is drawing dead, the only way that it could win would be to catch the remaining 5's/7's.. but how would you count the outs?

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Assuming 2 pair is good for the win, you have to catch 2 of the six remaining 5s and 7s... on the turn there are 47 unseen cards, 6 of which will give you a redraw to 2 pair or trips (41 cards will leave you drawing dead if someone has a pair higher than 7s):

(47 - 41) / 47 = 12.8% or about 6.8 to 1.

Then you have to hit one of the remaining five 7s or 5s on the river out of 46 unseen cards:

(46 - 41) / 46 = 10.9% or about 8.2 to 1.

Now you have to multiply the odds of BOTH of those events happening:

.128 * .109 = 1.4% or about 70 to 1, which makes it worth less than a 1 outer.

You also have a backdoor two gap straight draw that's also worth less than 1 out.