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Thread: Spanish 21 Jackpot

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    Spanish 21 Jackpot

    I was playing a Spanish 21 game with a $1 progressive jackpot. I was ignoring the side bets. Then, I got a pair of Aces v. a dealer Ace. And, the dealer had an Ace down! The progressive jackpot was over $9,000. So, it got me thinking, at some point that $1 jackpot is +EV. There is a 100% payout for four-of-a-kind (your first two cards match the dealer's first two cards). There are smaller payouts as well. Just thinking about the 100% payout on four-of-a-kind, at what point is that +EV for a $1 bet?

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    Quote Originally Posted by Craigrow View Post
    I was playing a Spanish 21 game with a $1 progressive jackpot. I was ignoring the side bets. Then, I got a pair of Aces v. a dealer Ace. And, the dealer had an Ace down! The progressive jackpot was over $9,000. So, it got me thinking, at some point that $1 jackpot is +EV. There is a 100% payout for four-of-a-kind (your first two cards match the dealer's first two cards). There are smaller payouts as well. Just thinking about the 100% payout on four-of-a-kind, at what point is that +EV for a $1 bet?
    I was playing a BJ with the $5 progressive jackpot. You get the jackpot when your two cards and dealer's upcard are all seven and the same suit. The next best thing is three cards all seven and the same color. Normally I play BJ with Zen main count, ace side count and 8/9 side count. But when I play this game, I replace the ace count with seven count. When remaining seven cards are very rich, I play the $5 side bet. One day on my first side bet that day, I got all three sevens of the same color. The jackpot is $83,000. So my bet paid me $8,300 (actually a little over $6,000 after tax). Since I did not play this side bet for 2 hours and hit it on my first bet, the dealer said I am the luckiest person he ever seen. I really want to tell him it is all skills. Luck has nothing to do with it.

    Once the jackpot was $180,000, the count almost reached my threshold to play but a little shy, I hesitated and decided to play on the next hand. And I hit three seven of the same rank. If I am a truly lucky person, I would have won that $180,000 jackpot. (Jackpot grew that big because no one has hit it for more than one year. It never happened again.)

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    Quote Originally Posted by BJGenius007 View Post
    ace side count and 8/9 side count.
    Well, why stop there? If you can side count Aces, 8's, and 9's on top of the normal counting, you mind well just side count the other few values and play like a perfect computer.
    Last edited by UncleChoo; 01-13-2022 at 07:49 AM.

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    Quote Originally Posted by BJGenius007 View Post
    ...Once the jackpot was $180,000, the count almost reached my threshold to play but a little shy, I hesitated and decided to play on the next hand. And I hit three seven of the same rank. If I am a truly lucky person, I would have won that $180,000 jackpot....
    I happen to be familiar with the 7s side bet and believe the b/e point to be around $113k

    ...sure enough checking the WoO webisite, the version you describe (table #1, ver #2) has a b/e point of $112k

    ...now granted counting 7s must also be considered, of which you say was 'close' to threshold, but also considering that the jackpot was 50% greater than the b/e point.....I think you 'missed your shot' on that one, buddy

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    Quote Originally Posted by UncleChoo View Post
    Well, why stop there? If you can side count Aces, 8's, and 9's on top of the normal counting, you mind well just side count the other few values and play like a perfect computer.
    First, 8/9 is counted as a combo, not separately. My goal is to cover every rank of the cards so I can insurance perfectly. Of course, it changes how I play based on ace and midcard side count.

    Maintaining three counts in the head is easier than most people realize. I used toes to count midcards and A-X to count aces.

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    There are 48 midcards of 8s and 9s in a six deck shoe, but there are only ten toes. How do you mark such a large number? The zen count itself is already beyond the capability of most people, I guess. What is your trick? And Tarzan’s?

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    INS is a very UNimportant index in Sp21.

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    Quote Originally Posted by aceside View Post
    There are 48 midcards of 8s and 9s in a six deck shoe, but there are only ten toes. How do you mark such a large number? The zen count itself is already beyond the capability of most people, I guess. What is your trick? And Tarzan’s?
    I order a Coke or Spirit when I sat down. 8/9 side count depends on the straw position and which toe touches the ground with force (start from the rightest to the leftest) and I drink a little bit and move the straw to the next direction (or casually touch the straw) after all ten toes touches ground once. It will become natural when you see an eight or nine and immediately move to the next toe. I also count the aces by thinking an object: apple, banana, chicken, duck, egg, fish, goose, ham, ice, juice, kiwi, lemon, mango, nectarine, orange, papaya, queen, rum, saki, tequila, utopia, vodka, wine and XO. Queen and utopia are cocktails. In the end, only the main Zen count uses numbers.
    Last edited by BJGenius007; 01-13-2022 at 02:12 PM.

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    Quote Originally Posted by aceside View Post
    There are 48 midcards of 8s and 9s in a six deck shoe, but there are only ten toes. How do you mark such a large number? The zen count itself is already beyond the capability of most people, I guess. What is your trick? And Tarzan’s?
    I'm doing an ace side count with my normal. I'm not using Zen but an almost identical count system. Over time and with lots of practice, you learn to maintain both counts. I count my aces upwards rather than backward. Not sure why authors mostly suggest counting Aces backward. So if my normal count is say +7 and if I saw 3 Aces, mentally I'm thinking of two columns +7 3. It is easier to count aces forward rather than backward because all you have to do is look at the discard tray and you can easily see if poor or rich aces. Just my thoughts on the matter.

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    Quote Originally Posted by BJGenius007 View Post
    I order a Coke or Spirit when I sat down. 8/9 side count depends on the straw position and which toe touches the ground with force (start from the rightest to the leftest) and I drink a little bit and move the straw to the next direction (or casually touch the straw) after all ten toes touches ground once. It will become natural when you see an eight or nine and immediately move to the next toe. I also count the aces by thinking an object: apple, banana, chicken, duck, egg, fish, goose, ham, ice, juice, kiwi, lemon, mango, orange, papaya, queen, rum, saki, tequila, utopia, vodka, wine and XO. Queen and utopia are cocktails. In the end, only the main Zen count uses numbers.
    I must say this is quite creative BJGenius007

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    Ok, try to focus people! Stop feeding the Genious. He will use any provocation to brag about his system. Aceside will aceside, what can you do?

    Anyhow, I am also wanting to know the answer to OPs question. I don't understand progressive jackpots very well, but I haven't put much thought into it. From listening to GWAE, I know there are APs who go out and attack slot machines with progressive jackpots once they reach a certain point. It seems to me those people would understand this problem.

    Craigrow asked specifically how to calculate when it is +EV for a $1 bet to have 4 of a kind with the first two dealer and player cards. I'll assume the rank doesn't matter, therefore the first of the 4 cards is a freebee. Here's my quick & dirty approach:

    2nd card (assuming 6D) matches first card: 23 out o 311 chance (e.g., started with 24 of a given rank out of 312 cards in 6D).

    3rd card matches: 22 out of 310 chance.

    4th card matches: 21 out of 309 chance.

    Each of those probabilities is about 0.07. Multiply those three together, and I get 0.000357, or 1 out of 2804 chance. So I'm thinking that a $1 bet that wins 100% of the pot with this 4-of-a-kind would be +EV at a jackpot of $2804. Is this correct?

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    Quote Originally Posted by bejammin075 View Post
    Ok, try to focus people! Stop feeding the Genious. He will use any provocation to brag about his system. Aceside will aceside, what can you do?

    Anyhow, I am also wanting to know the answer to OPs question. I don't understand progressive jackpots very well, but I haven't put much thought into it. From listening to GWAE, I know there are APs who go out and attack slot machines with progressive jackpots once they reach a certain point. It seems to me those people would understand this problem.

    Craigrow asked specifically how to calculate when it is +EV for a $1 bet to have 4 of a kind with the first two dealer and player cards. I'll assume the rank doesn't matter, therefore the first of the 4 cards is a freebee. Here's my quick & dirty approach:

    2nd card (assuming 6D) matches first card: 23 out o 311 chance (e.g., started with 24 of a given rank out of 312 cards in 6D).

    3rd card matches: 22 out of 310 chance.

    4th card matches: 21 out of 309 chance.

    Each of those probabilities is about 0.07. Multiply those three together, and I get 0.000357, or 1 out of 2804 chance. So I'm thinking that a $1 bet that wins 100% of the pot with this 4-of-a-kind would be +EV at a jackpot of $2804. Is this correct?
    bejammin075,

    The OP mentioned that the game was Spanish 21, so the decks have no 10's. This means a 6D game has only 288 cards, rather than 312, so you need to change your denominators:

    P = (288/288)*(23/287)*(22/286)*(21/285) = 0.000454...

    or 1 in about 2,202.

    Note the OP mentioned Aces specifically. If the four of a kind must be in Aces, then the probability will be one-twelfth of the above P (because the first term will be 24/288), so the odds are 1 in about 26,418. This seems much more likely for the odds for a jackpot.

    Hope this helps!

    Dog Hand

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    Quote Originally Posted by Dog Hand View Post
    bejammin075,

    The OP mentioned that the game was Spanish 21, so the decks have no 10's. This means a 6D game has only 288 cards, rather than 312, so you need to change your denominators:

    P = (288/288)*(23/287)*(22/286)*(21/285) = 0.000454...

    or 1 in about 2,202.

    Note the OP mentioned Aces specifically. If the four of a kind must be in Aces, then the probability will be one-twelfth of the above P (because the first term will be 24/288), so the odds are 1 in about 26,418. This seems much more likely for the odds for a jackpot.

    Hope this helps!

    Dog Hand
    Good catch for the correction for Spanish decks.

    I interpreted Craigrow's post as that he started thinking about this situation when he happened to be dealt two Aces and the dealer had two Aces, but that the jackpot generally applied to any 4-of-a-kind. Perhaps Craigrow can clarify.

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