Help with Odds calculation for craps...

[canada_dry]

If I were to make two place bets (6 and 8), then double it each time I lost, for a maximum of six doubles, what are the chances I would lose all six?

$12 on 6, $12 on 8 = $24
$24 on 6, $24 on 8 = $48
$48 ...
$96 ...
$192 ..
$384 ..

Therefore risking $1512 to win $14. Each roll i have 6/36 chance of losing and 10/36 chance of winning. Could someone help with the math.

I realize the inherent risks with a martindale system so i was trying to incorporate it with a betting pattern that has a greater chance of winning than losing (10 ways to hit a six or eight versus 6 ways to hit a seven).

Thanks..