Blue Efficacy said:
Now, what would really make things interesting, is if they allowed the player to receive the 3:2 payout still if you doubled down and received a 10.
That payout would happen 4 times out of 13. 5 times (a,2,3,4,5) you need the dealer to bust to get paid or you've given the house twice as much money (30% of the time or so). I'd estimate the last (6,7,8,9)4 cards you can get would give the house a 65/35 advantage, since they're weighted towards getting 20 and you're evenly distributed among 17-20.
so 4/13 3:2 on 2 bets
. 5/13 .3*2 bets, .7*loss
. 4/13 .35*2 bets, .65*loss
4/13*(1.5*2)+5/13*(.3*2)+4/13*(.35*2)
=1.4 bets return on your 2 bets.
If my math's right, i can't think of a scenario in which doubling down on a non-DoubleExposure game is a good idea.