CSP/OSP with "exchange one card" option - optimal strategy ?

Hi I hope somebody can help me with this question. I might be able to calculate this myself but if somebody allready have these numbers I don't want to waste my time doing it.

The game is CSP/OSP
You have the option to exchange one card. This cost you an amout equal to the ante.
The payouts on your bet are(if you beat the dealers quallifying hand of AK or higher):
RSF 100:1
SF 50:1
4 of a kind 20:1
Full House 7:1
Flush 5:1
Straight 4:1
3 of a kind 3:1
2 pairs 2:1
1 pair 1:1
AK 1:1

When the dealer doesn't quallify you push on your bet and get paid 1:1 on the ante.
If you chose to exchange a card you don't quallify for any bonus payouts.

What I'm looking for is the correct "exchange strategy" for the player and info on how much this "exchange one card" option reduces the house edge on the game if used correctly.
The wizard of odds page tells me that the house edge for this game without the exchangeoption is 5,22%

An example of my calculations:

You have 4,9,10,J,Q (no flush possible).

Lets play this hand 46 times. (there is 46 unseen cards left since you have 5 and know one of the dealers)

If you do not exchange a card it is correct to fold � you lose 46xANTE

3 cards gives you a pair of nines
3 cards gives you a pair of tens
3 cards gives you a pair of jacks
3 cards gives you a pair of queens

4 cards gives you a Q-high straight
4 cards gives you a K-high straight

26 cards gives you nothing and you fold.

If the dealer does not qualify:
26 times you loose 2xANTE since you paid for a card and still folded = -52xANTE
20 times you push overall(you win on the ante but paid the same amout to exchange a card)

So if the dealer don�t qualify you lose 52xANTE instead of 46xANTE

If the dealer qualifies it get a litte more complicated.
3 cards gave you a pair of 9�s. If you loose to the dealers hand you lose 3x4xANTE=-12xANTE
3 cards gave you a pair of 9�s. If you beat the dealers hand you win 3x3xANTE=9xANTE
50% of the dealers qualifying hands beats a pair of 9�s, so (0,50*-12)+(0,50*9)= -1,5xANTE
(the above percentage and the others that follow are a quick estmate and might be a little off)

3 cards gave you a pair of 10�s if you loose to the dealers hand you lose 3x4xANTE=-12xANTE
3 cards gave you a pair of 10�s if you beat the dealers hand you win 3x3xANTE=9xANTE
44% of the dealers qualifying hands beats a pair of 10�s, so (0,44*-12)+(0,56*9)= -0,25xANTE

3 cards gave you a pair of J�s if you loose to the dealers hand you lose 3x4xANTE=-12xANTE
3 cards gave you a pair of J�s if you beat the dealers hand you win 3x3xANTE=9xANTE
38% of the dealers qualifying hands beats a pair of J�s, so (0,38*-12)+(0,62*9)= +1xANTE

3 cards gave you a pair of Q�s if you loose to the dealers hand you lose 3x4xANTE=-12xANTE
3 cards gave you a pair of Q�s if you beat the dealers hand you win 3x3xANTE=9xANTE
32% of the dealers qualifying hands beats a pair of Q�s, so (0,32*-12)+(0,68*9)= +2,25xANTE

4 cards gave you a Q-high straight if you loose to the dealers hand you lose 4x4xANTE=-16xANTE
4 cards gave you a Q-high straight if you beat the dealers hand you win 4x9xANTE=36xANTE
1% of the dealers qualifying hands beats a Q-high straight, so (0,01*-16)+(0,99*36)= +35,5xANTE

4 cards gave you a K-high straight if you loose to the dealers hand you lose 4x4xANTE=-16xANTE
4 cards gave you a K-high straight if you beat the dealers hand you win 4x9xANTE=36xANTE
1% of the dealers qualifying hands beats a K-high straight, so (0,01*-16)+(0,99*36)= +35,5xANTE

26 cards gave you nothing and you lose 26x2xANTE=-52xANTE on these hands.

Total when the dealer qualifies: +20xANTE

Dealer qualifies 54% of the time and you win 20x ANTE
Dealer don�t qualify 46% of the time and you lose 52x ANTE

Return when exchanging a card instrad of folding: (0,54*20)-(0,46*52)= -13xANTE
Return when not drawing a card and just folding -46xANTE

I don't think the "exchange option" is worth anything because you have to buy the card.According to my own calculations, there is strictly no situation where this is justified.

Francis Salmon

My calculations show me that I shoul exchange a card when I have four cards to a Royal, straight flush, flush, and openended-straight draw(lowest card 5).
I don't believe an you can get an overall edge in the game but correct use of the exchange one card option will reduse the overall house edge.
In my post I showed an exsample of my calculations on the open ended straight draw that showed that buying a card would cause you to lose 33 antes less in 46 hands. Can you pointout any mistakes in my calculations ?

Sorry for the bad spelling. English is not my first language.

RW

I agree that buying a card with a 4-flush or an open straight is correct as long as you're not destroying a pair.
But still - as you mentioned - even with optimal strategy you cannot offset the casino edge.So why would you want to play a losing game?
By the way the casinos in my region don't allow folding after buying.

Francis Salmon

I don't think I will ever play the game. But as a former dealer I hate not knowing the "right way" to play the game. I allways liked the fact that when people asked I was able to tell the correct BS-move for blackjack, or the house edge on the different games/bets. The only thing I didn't know was the correct exchange strategy for this game and of course how it changes the overall house edge.

RW