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Table Games "Double Draw Poker" table game? strategy?

Discussion in 'Table Games' started by wrxrob, Mar 21, 2013.

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  1. wrxrob

    wrxrob High-Roller

    Oct 13, 2009
    Trips to Las Vegas:
    Delaware just introduced a new table game that I cannot even find on wizardofodds. Its called Double Draw Poker. Now, my google searches are only yielding video poker games, which this is not.

    One site lists the basics of the game found below

    problem is, I'm looking for a strategy, if there is one. I watched some people play this carnival game, and burn through their $100-200 in short order (betting the $5 minimum).:confused2: Therefore, I think I'll just wait this one out, and avoid that game for now. :thumbsup:

    I think the problem with this game is that you must have 2 pair or better in order to get paid. And the $5 minimum turns into $15+5 bonus if you play all the way through.
  2. WrongWayWade

    WrongWayWade VIP Whale

    Sep 20, 2010
    Atlanta, GA
    Trips to Las Vegas:
    Send an e-mail to the Wizard and I bet he'll try to work out a strategy.

    Here's what I'd do:
    I'd fold any initial hand that didn't have a pair or 4 to a straight or 4 to a flush. (Note the large bonus for straights).

    I'd fold if after the first draw I didn't have a paying hand, except 4 to a straight flush/royal flush.

    But that's just a WAG.

    Also, if the minimum is $5, it really is $10, because it seems you must make both the Ante and Bonus bet to start the game. Play to the end and you have $20 on the table.

    I predict you end up folding off the bat or after the first draw A LOT. The two jokers do help, however.
  3. mike_m235

    mike_m235 Tourist

    Aug 22, 2012
    Colorado Springs
    Trips to Las Vegas:
    That's a typo on the straight, right? You don't really get 12-1. Is that supposed to be 2-1? Because if it's 12-1, you should pretty much be playing for straights I think, especially if you've got a joker.
  4. topcard

    topcard Older than the Stardust!

    Aug 8, 2012
    Fort Worth
    Trips to Las Vegas:
    If this game shows up in Vegas, I'll play it.

    Essentially, you can have up to 9 cards to make a two-pair or better?

    I'd probably play every hand that started with a pair (or better) and draw all the way through.

    Playing 3-card straights and flushes would be a little tougher, as your first draw would only be two cards.

    So - I agree that if one does not have either an initial pair ~or~ 4 cards of a straight or flush, folding would be in order. (I would make an exception to the "4-to-straight or flush" rule if it were 3 cards to straight-flush, or if I had a joker.)

    If it's there on my next Vegas trip, (June?) I'll play it and report back here.
    Seems like forever from now, but the flights are booked, so it counts!
  5. tringlomane

    tringlomane STP Addicted Beer Snob

    Jan 21, 2013
    Trips to Las Vegas:
    At least you don't play the dealer in this game, so the casino math is less difficult.

    As Wade said, you're probably going to see at least one draw with any pair. Otherwise this game would require players to fold way, way too much. Also I am assuming straights pay 2-1 here on the bonus bet paytable.

    So lets look at 1 pair hands first. Also I am assuming Ace + Joker can be a pair of Aces since 5 Aces is a payout in the game.

    Place Ante and Bonus bets:

    If we fold immediately, we lose two units.

    If we pay one unit and draw three, here are the probabilities to improve in the worst case scenario, discarding one ace/joker:

    Four of a kind: 1*47 = 47 combos
    Full house: 2*1 + 1*10 + 9*4 + 2*9*6 + 2*2*3 + 2*1*10 = 188 combos
    Three of a kind: 2*(C(47,2) - 9*6 - 2*3 - 1*10) = 2022 combos
    Two Pair: 9*6*43 + 1*10*42 + 2*3*44 = 3006 combos
    No improvement: C(9,3)*4^3 + C(9,2)*4^2*2*3 + C(9;2)*4^2*1*5 + 9*4*1*5*2*3 + 9*4*3*3 + 5*3*3 = 13,161

    Total combos: 18,424

    Sanity check: C(49, 3) = 49*48*47/6 = 18424

    Now lets determine the payouts. Remember, we will always see the hand to completion if we improve.

    Four of a Kind: 47/18,424*(20 + 3) = 0.0586734694
    Full House: 18/18,424*(5+3) = 0.0816326531

    Now with Three of a Kind and 2 pair, we will have to take redraws of 1 card each.

    Three of a Kind: 2*5*42 = 420 combos (Ax or Joker-x) will give us 4 outs to a full house and one out to quads; 2*C(9,2)*4*4 + 2*9*4*2*3 = 1584 of these combos give us 3 outs to a full house and one out to quads. 2*3*3 = 18 of these combos give us 2 outs for a full house and one out to quads.

    420/18,424*(1/46)*(20+3) + 420/18,424*(4/46)*(5+3) + 420/18,424*(41/46)*(1+3) + 1584/18,424*(1/46)*(20+3) + 1584/18,424*(3/46)*(5+3) + 18/18,424*(42/46)*(1+3) + 1584/18,424*(1/46)*(20+3) + 18/18,424*(2/46)*(5+3) + 18/18,424*(43/46)*(1+3) = 0.2469227284

    Two pair: 1*10*42 = 420 of the combos will have 5 outs for a boat, 9*6*43 = 2322 of the combos will have 4 outs for a boat and 2*3*44 = 264 of the combos will have 3 outs for a boat.

    420/18,424*(5/46)*(5+3) + 420/18,424*(41/46)*(0+3) + 2322/18,424*(4/46)*(5+3) + 2322/18,424*(42/46)*(0+3) + 264/18,424*(3/46)*(5+3) + 264/18,424*(43/46)*(0+3) = 0.5613283241

    No improvement:
    13,161/18,424*(-3) = -2.1430199739

    Summing all of this together:
    4 of a Kind: 0.0586734694
    Full House: 0.0816326531
    Three of a Kind: 0.2469227284
    Two Pair: 0.5613283241
    No improvement: -2.1430199739
    Total from drawing: -1.194462799
    Total from immediately folding: -2.000

    Although drawing with a pair is not a winning play, it is better than the other option of folding and losing 2 units immediately. So draw every pair for at least one draw.

    Now, we probably should double check that it's not optimal to draw a single pair again with no improvement.

    So looking at the best no improvement scenario, holding an ace/joker and another card that has three outs available:
    (2/46)*(1+3) + (7/46)*(0+3) - (37/46)*(4) = -2.5869565217
    This is STILL better than folding after the first draw, which will cause you to lose 3 units.

    The best no improvement scenario without ace/joker:
    (2/46)*(1+3) + (6/46)*(0+3) - (38/46)*(4) = -2.7391304348

    If we already discarded one of our two pair outs on a previous draw, then the EV becomes:
    (2/46)*(1+3) + (5/46)*(0+3) - (39/46)*(4) = -2.840909
    STILL better than folding and automatically losing 3 units.

    If we already discarded two of our two pair outs (highly, highly unlikely): You have to have something like 77852 and redraw the 8, 5, and 2 AGAIN!
    (2/46)*(1+3) + (4/46)*(0+3) - (40/46)*(4) = -3.0434782609.
    So in this rare scenario, it's a fold, but this will happen so rarely, we are better off ignoring it.

    So yep, draw down with any pair the ENTIRE way. This is going to be a rough game variance wise for sure.

    I'll look into more of this later. Correcting my math since I forgot the damn "Bugs" was quite enough for one evening.
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