Is this 21 Variant Beatable?

[Romerio]

Amongst friends, we play a game that we call pontoon, although it doesnt have much in common with any casino based pontoon game I've ever seen,

Played with a standard deck of cards,
Dealer and player/s both get 1 card face down
Player looks at his card and has the option to bet 1 unit, or 2 units
Dealer looks at his first card and has the option to force the player to double his bet
Each player then gets a second card.
If the players total is under 12, then he must 'buy' a card for at least 1 unit, and at most his original amount.
The player must do this until his total is over 12
Once the players total is over 12 he may continue to 'buy', he may 'flip' meaning the card gets turned face up.
Once the player is close enough to 21 and happy with his hand, he sits.
The dealer then opens his cards, and may take cards, one at a time until he thinks his hand beats the player/s, or until his total is more than 21 (busts)

Anytime a player gets 5 cards, and a total under 21 (5 and under), he is payed double.
Anytime a player gets a 2 card 21 (blackjack) he is payed double.
If the dealer gets Blackjack or 5 and under, the players still in the hand (not busted out) must pay double whatever there current stake is.

any players whose first 2 cards total 14, can double their stake, and 'burn' recieving 2 new cards, if these cards are 14, they may double the stake and 'burn'

How would I even begin to work out if this is favourable to the player, and what optimal strategy would be?

Thanks
- Adam

[Pokerfarian]

Eeeww, so many options, so hard to work out an optimal strategy. Actually it depends on how many players there are since the dealer will alter his strategy depending on the other players in the hand & not just you. So there isn't even going to be a "complete mathematical optimal strategy" My gut instinct says its not beatable in general.