CSP or UTH for JACKPOT?

Hi,

Two interesting questions: At which of the two games do I have a better chance of winning a jackpot? And at what amounts is it profitable to play for the three progressive jackpots?

Here is the situation: in a casino there´s one CSP (Caribbean Stud Poker) and one UTH (Ultimate Texas Hold´Em Poker). Both tables are linked to the same jackpot. That is, on both tables you have a chance to win any of the progressive jackpot-prizes or fixed bonus-prizes. How is the jackpot built up? The bet is $2,5 and the prizes are as follows:

Jackpots (progressive)
Jackpot A: $25.000 (or more) for royal or straight flush of diamonds
Jackpot B: $2.500 (or more) for royal or straight flush of spades/hearts/clubs
Jackpot C: $500 (or more) for four of a kind AA/KK/QQ/JJ

Bonuses (fixed)
Bonus D: $125 for four of a kind TT-22
Bonus E: $62,50 for full house
Bonus F: $50 for flush
Bonus G: $37,50 for straight
Bonus H: $12,50 for three of a kind

Please note that at UTH the jackpots and bonuses are only paid on the combination of the 2 cards you are dealt and the flop. At CSP it is paid on the 5 cards you are dealt. Now because both games are working with 5 cards to qualify for the jackpot, it seems to me that the chances of winning one of the jackpot-prizes are the same:

Diamond royal or straight flush at CSP OR flopping that at UTH: 1:259.896.
Royal or straight flush spades/hearts/clubs at CSP OR flopping that at UTH: 1:86.632.
Four of a kind AA/KK/QQ/JJ at CSP OR flopping that at UTH: 1:13.536,25.

Please correct if I am wrong on my math.

Now the first question is: At which of the two games do I have a better chance of winning a jackpot? Let´s assume there are seven players at both tables and one hand at CSP takes exactly the same amount of time as one hand at UTH. My answer would be: it does not matter.

But wait, maybe it DOES matter! What about the fact that all seven players at UTH are dealt the same flop? If the flop is favorable for a jackpot combination, isn´t the chance of someone having it higher? And what about CSP, where all players get “their own flop” so to speak? All seven players can be dealt a jackpot combination, while if a non-favorable flop happens at UTH a jackpot combination is out of the question for all players! What do you think?

Once we have determined what game is best for playing the jackpot, the second question is: At what amounts is it profitable to play for the three progressive jackpots? I am very curious what you think of it. Here is my answer: the chance of any of the three jackpot combinations is 1:11.202. The cost to play is $2,50. Therefore the jackpots added up must be at least 11.202 x $2,50 = $28.006. Makes sense?

Many thanks in advance for reading, thinking about it, making calculations and replying!

Greetings,
P3t3r P3n

My apologies for not introducing myself. It was certainly not my intention to practice bad forum etiquette. My name is Peter Pen and I am a croupier/dealer in The Netherlands. In the casino I work, they have decided to merge the jackpots for CSP and UTH. Many players now ask themselves which game is better to hit a jackpot. And how high must it be to see it as a +EV-situation.

Ideas anyone? A chance to outsmart the Wizard himself maybe? His initial reaction for me was "Ultimate X is a complicated game"...

Greetings,
P3t3r P3n

I am not a mathematician, but I’d guess that UTH should have a better chance of hitting a jackpot, because it uses more cards to complete one round. This is similar to the cut card effect as discussed in another thread about the blackjack EV loss from using a cut card.

Technically two people could win the Jackpot under both games.

When using the flop it would have to be QJT with one playing having AK and another 98, or similar down to 543 with players having 76 and A2. Unless the three cards are suited and running there can't be two winners. Similarly if the flop is (say) AAA then only the player with the last Ace can win the Jackpot.

When using 5-cards from each player, it's quite possible for multiple players to have a jackpot.

Quote: aceside

I am not a mathematician, but I’d guess that UTH should have a better chance of hitting a jackpot, because it uses more cards to complete one round. This is similar to the cut card effect as discussed in another thread about the blackjack EV loss from using a cut card.
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The cut card effect only happens in BJ because the number of cards dealt per round depends strongly on whether the cards are high or low cards. I have never played CSP, but I don't see how the player can change the number of cards per round in UTH based on which cards come out.

I think you are right. I probably said something wrong about these two games. Actually CSP uses a lot more cards per round to play. For one player playing heads-up with a dealer, CSP uses 10 cards to complete one round, but UTH uses 9 cards. The probability of drawing a royal flush hand is 4/C(52,5) =4/2598960 for both games. This formula is based on a five-card draw from a 52-card deck, so the math should be valid for both CSP and UTH because both games use player’s first five cards to determine a royal flush.

However, in actuality, each round draws more than five cards and I am not sure if this affects the odds or not. I hope charliepatrick can run a simulation to clear my doubts.

The logic is to think about a deck (or shoe) of cards as numbered 1, 2, 3 etc. Then consider the deck as a row of boxes, each containing one card. In the two games mentioned, since players have no choice to draw additional cards, which player gets which cards is pre-determined. So you could determine ahead of time which numbered boxes will make up the five you are going to consider.

Now whether you chose boxes {1,2,3,4,5} (as in 5-card poker), {1,3,5,6,7} (as in UTH assuming it's one for you one for me etc), or any other set of five numbers, the chances of any hand (like RF in Diamonds) remains the same. For every deck that has it in {12345}, there's an equivalent deck with it in {13567}.

Quote: P3t3rP3n

Here is the situation: in a casino there´s one CSP (Caribbean Stud Poker) and one UTH (Ultimate Texas Hold´Em Poker). Both tables are linked to the same jackpot. That is, on both tables you have a chance to win any of the progressive jackpot-prizes or fixed bonus-prizes. How is the jackpot built up? The bet is $2,5 and the prizes are as follows:

Jackpots (progressive)
Jackpot A: $25.000 (or more) for royal or straight flush of diamonds
Jackpot B: $2.500 (or more) for royal or straight flush of spades/hearts/clubs
Jackpot C: $500 (or more) for four of a kind AA/KK/QQ/JJ
Diamond royal or straight flush at CSP OR flopping that at UTH: 1:259.896.
Royal or straight flush spades/hearts/clubs at CSP OR flopping that at UTH: 1:86.632.
Four of a kind AA/KK/QQ/JJ at CSP OR flopping that at UTH: 1:13.536,25.
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Your frequency for 'Four of a kind AA/KK/QQ/JJ' does not correct for the fact that you need to use both cards in you hand. You are using the raw frequency for quads. Two out of five times you flop quads, the kicker will be among your hole cards. If this does not qualify you for the jackpot, then the frequency you are looking for is one in 22,560.

At UTH you get a jackpot if you FLOP it. So holding A2 in your hand and the flop is AAA is called flopping a four of a kind. It looks nicer when you´re holding AA and the flop is AA2, but for the jackpots of this game it´s all the same!

Quote: P3t3rP3n

At UTH you get a jackpot if you FLOP it. So holding A2 in your hand and the flop is AAA is called flopping a four of a kind. It looks nicer when you´re holding AA and the flop is AA2, but for the jackpots of this game it´s all the same!
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Thanks for clarifying the rules. Almost every bad beat jackpot requires that you use both cards in your hand to qualify.

Quote: charliepatrick

The logic is to think about a deck (or shoe) of cards as numbered 1, 2, 3 etc. Then consider the deck as a row of boxes, each containing one card. In the two games mentioned, since players have no choice to draw additional cards, which player gets which cards is pre-determined. So you could determine ahead of time which numbered boxes will make up the five you are going to consider.

Now whether you chose boxes {1,2,3,4,5} (as in 5-card poker), {1,3,5,6,7} (as in UTH assuming it's one for you one for me etc), or any other set of five numbers, the chances of any hand (like RF in Diamonds) remains the same. For every deck that has it in {12345}, there's an equivalent deck with it in {13567}.
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I am clear on this part but am still bothered by the probability calculation of another blackjack side bet, Blazing 7’s.

Wizard has researched this and showed that there are two versions of this side bet, which have slightly different probabilities, as follows:

Version #1. Wins are based on the first three player cards.
Version #2. Wins are based on the first two player cards and the dealer up card.

Version #1 probability of losing is 0.851843, but Version #2 probability of losing is 0.851833. Could you please take a look into this website and explain why the probability numbers are different for these two versions of games?

Thank you in advance.

So if a Royal Flush comes up once every 649,740 hands, how high should the jackpot be if putting down $1 per hand to win it? I don't know how much the JP increases per hand. If there's 5.5 players at the table plus the dealer, it would take 100,000 rounds to win a RF JP. How long would that take? If there's 30 hands per hour, it would take 33.3K hours to score the JP, or once every 7 months at 16 hours per day. How many extra straight flushes do you need to win to make up for missing a RF JP in 650K hands? And again, if the community cards are a Royal Flush, the Player does not win, except on the trips bet for UTH, The $1 progressive JP being split among the whole table is something I don't know about if the community cards are a Royal Flush. I don't know about the other game.

Deleted.

Quote: aceside

...Version #1. Wins are based on the first three player cards.
Version #2. Wins are based on the first two player cards and the dealer up card.

Version #1 probability of losing is 0.851843, but Version #2 probability of losing is 0.851833. Could you please take a look into this website and explain why the probability numbers are different for these two versions of games?

Thank you in advance.

Reference - https://wizardofodds.com/games/blackjack/side-bets/blazing-7s/

Without going into the numbers if the payout is based on the first three cards the Player draws, then there are quite a few occasions when the Player either cannot (D=BJ) or doesn't want to draw a third card. Thus quite often there will only be two cards which might be 7s. As a simple example if the Player is dealt KQ then in Version #1 the Player will lose the sidebet, whereas in Version #2 the Player still has the small chance that the Dealer's up-card is a 7.

Personally I get the chances of a loss when the Dealer's up-card is included as being the chances that the first/next three cards from the deck are all non-7s. This is 288*287*286/312/311/310 = 78.59%.

Thank you for helping. First of all, I’ve never seen the Version #1 game anywhere, so let us forget about it. Second, there is another condition you might neglect in your above calculation, that is, wins one or two sevens are based on the player cards only. Taking this into consideration should increase the probability of a loss.

In other words, if the dealer up card is a 7, the player loses too if the hand does not include any 7. Therefore, the final probability of a loss is

288*287/(312*311)=85.1843%.

Perhaps I misread "Version 2: Wins are based on the first two player cards and the dealer up card. Wins for one or two sevens are based on the player cards only." as the second sentence contradicts the first. But now that you've explained it, the player loses if both their cards aren't a seven and the dealer's can be anything = 288*287*310/312/311/310 = 85.184269%.

I guess it's a coincidence that the value in the first version, which has the added value which allows the player to win based on their third card and the downside of not always drawing three cards, is so similar.

Personally I think it would be better to count the sevens in the first three cards as this is what Players would be used to if they'd played 21+3.

Quote: charliepatrick

The logic is to think about a deck (or shoe) of cards as numbered 1, 2, 3 etc. Then consider the deck as a row of boxes, each containing one card. In the two games mentioned, since players have no choice to draw additional cards, which player gets which cards is pre-determined. So you could determine ahead of time which numbered boxes will make up the five you are going to consider.

Now whether you chose boxes {1,2,3,4,5} (as in 5-card poker), {1,3,5,6,7} (as in UTH assuming it's one for you one for me etc), or any other set of five numbers, the chances of any hand (like RF in Diamonds) remains the same. For every deck that has it in {12345}, there's an equivalent deck with it in {13567}.
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So charliepatrick, based on your explanation of pre-determined cards, there is NO difference in the chance you get a jackpot hand in CSP or UTH (on the flop!)?

Quote: P3t3rP3n

Quote: charliepatrick

The logic is to think about a deck (or shoe) of cards as numbered 1, 2, 3 etc. Then consider the deck as a row of boxes, each containing one card. In the two games mentioned, since players have no choice to draw additional cards, which player gets which cards is pre-determined. So you could determine ahead of time which numbered boxes will make up the five you are going to consider.

Now whether you chose boxes {1,2,3,4,5} (as in 5-card poker), {1,3,5,6,7} (as in UTH assuming it's one for you one for me etc), or any other set of five numbers, the chances of any hand (like RF in Diamonds) remains the same. For every deck that has it in {12345}, there's an equivalent deck with it in {13567}.
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So charliepatrick, based on your explanation of pre-determined cards, there is NO difference in the chance you get a jackpot hand in CSP or UTH (on the flop!)?
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The only reason that there can be a cut-card effect in blackjack is that the dealer looks at the cards in the deck to determine how to proceed with the dealing of the subsequent cards. The dealer has to look at the cards to determine what he can or cannot do procedurally. For example, the dealer won't/can't give you another card after you have already busted.

The jackpot odds will be the same in any game where there is a jackpot based on five cards AND where the game is dealt without needing to look at the cards. The number of players and the order of the dealing cannot affect the odds.

The jackpot probabilities are dependent on the number of decks used. Since there are no five-of-a-kind jackpots here, I assume a single deck is used for these CSP and UTH games.

Quote: charliepatrick

Perhaps I misread "
I guess it's a coincidence that the value in the first version, which has the added value which allows the player to win based on their third card and the downside of not always drawing three cards, is so similar.
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I've considered this again and would like to clarify something here. Just like in Version #2, in Version #1 at least one 7 is required in the player’s hand for a win. If using basic strategy, the player will always draw a third card except upon a dealer blackjack. In other words, the player is capped at two cards if the dealer has a blackjack. In Version #2, we have

The probability of a non-7 hand (loss), 288*287/(312*311)=85.1843%;
The probability of a one-7 hand, 2*24*288/(312*311)=14.2468%;
The probability of a two-7 hand, 24*23*288/(312*311*310)= 0.5285%;
The probability of a three-7 hand, 24*23*22/(312*311*310)= 0.04037%.

The dealer blackjack in Version #1 will change the probability distribution, but only in the last two probability numbers. This makes Version #1 slightly different, but almost negligible.