rumba434
rumba434
Joined: May 13, 2020
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May 13th, 2020 at 3:52:40 PM permalink
Playtech's new game Sette e Mezzo runs like this:

Each round is a new 40 card deck: 1-7. Jack, Cavalier, King in 4 suits. Numerical cards are worth their value, picture cards are worth 0.5 except the king of coins who is wild. (turns 7.0 or x.5 into 7.5, turns x.0 into 7, where x is less than 7)

The player and the dealer both get 1 card. The player is obligated to take more cards until his total is at least equal to the dealer, then he can choose to hit or stand. After the player stands, if the dealers card is 4 or less the dealer plays and will keep taking cards until he has 5 or higher. The aim is to get 7.5. If you get 8 or higher you bust. If player busts he loses regardless of dealer hand. If player and dealer end up on 5.0 dealer wins, if player/dealer match any other score it's a push, otherwise highest non bust hand wins.

What's best strategy for this game?
Mission146
Mission146
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May 13th, 2020 at 5:37:02 PM permalink
Quote: rumba434

Playtech's new game Sette e Mezzo runs like this:

Each round is a new 40 card deck: 1-7. Jack, Cavalier, King in 4 suits. Numerical cards are worth their value, picture cards are worth 0.5 except the king of coins who is wild. (turns 7.0 or x.5 into 7.5, turns x.0 into 7, where x is less than 7)

The player and the dealer both get 1 card. The player is obligated to take more cards until his total is at least equal to the dealer, then he can choose to hit or stand. After the player stands, if the dealers card is 4 or less the dealer plays and will keep taking cards until he has 5 or higher. The aim is to get 7.5. If you get 8 or higher you bust. If player busts he loses regardless of dealer hand. If player and dealer end up on 5.0 dealer wins, if player/dealer match any other score it's a push, otherwise highest non bust hand wins.

What's best strategy for this game?



Sette e Mezzo is actually Italian for seven-and-a-half and is a centuries old game, so this modification is based on that player-v-player game. The player-v-player version would essentially have no house edge (but would be skill-based) with the deal getting passed around from one player to the next.

The way the home version of this game worked is that each player (as well as the dealer) would have one face down card. Each player would take as many cards as they like, but they must show the face down card if they either bust or achieve 7.5, otherwise the face down card remains hidden until the dealer completes his play.*

After all the players had went, the dealer would reveal his face down card and would then draw as many cards as he wishes, until either busting or hitting the 7.5 total. In the home version of the game, the dealer not only wins with a higher score, but also wins all ties. Paying winning 7.5 at 2-1 was sometimes done and sometimes not, and a natural (two card) 7.5 beating one consisting of three, or more, cards was sometimes done and sometimes not.**

The rules of the Italian version allow for some degree of bluffing. A player, for instance, could not take any extra cards (even with a low card face down) in the hopes that the dealer would think it was a high card and try to score a higher total.

Anyway, my gut instinct is usually you would stand on 5.5 (or higher) and hit anything else against a lower dealer card*******, but I'm betting there are some exceptions. Even the exceptions will have exceptions that might be based on remaining deck composition if there are a lot of low cards out. I'll look at some of these individually tomorrow morning.

*https://en.wikipedia.org/wiki/Sette_e_mezzo
**http://www.blackjackforumonline.com/content/history-of-blackjack.htm

*******WRONG!!! TRUST THE MATH, NOT YOUR GUT!!!!
Last edited by: Mission146 on May 14, 2020
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
Wizard
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Wizard
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Thanks for this post from:
ksdjdj
May 13th, 2020 at 8:09:40 PM permalink
I saw this game in Uruguay. Here is my page on Siete y Media (seven and a half).
It's not whether you win or lose; it's whether or not you had a good bet.
ksdjdj
ksdjdj
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May 13th, 2020 at 8:10:49 PM permalink
According to the site below the RTP is: 99.31% (when playing optimal strategy*** ).

***: I don't know what that is at the moment, but the "gut instinct" strategy in the post by Mission seems logical to me.

https://www.livedealer.org/blog/2020/05/ever-played-sette-e-mezzo/
Last edited by: ksdjdj on May 13, 2020
rumba434
rumba434
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May 14th, 2020 at 3:38:15 AM permalink
Quote: Wizard

I saw this game in Uruguay. Here is my page on Siete y Media (seven and a half).



Thanks for the link. Do you think the fact its a 1-deck game, the presence of the king of coins, and the dealer win on ties of 5 would change the strategy any?
Mission146
Mission146
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May 14th, 2020 at 3:49:03 AM permalink
Quote: Wizard

I saw this game in Uruguay. Here is my page on Siete y Media (seven and a half).



It looks like one key difference is the player loses ties with a total of five in the Playtech version.

Another very important difference is that the player automatically must hit in the Playtech version if the player has less in total than the dealer. Because of that, some of the stands listed on that page the player literally could not do even if he wanted to. The player only has an actual decision to make if he has more than the dealer.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
Mission146
Mission146
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May 14th, 2020 at 4:42:22 AM permalink
Quote: rumba434

Playtech's new game Sette e Mezzo runs like this:

Each round is a new 40 card deck: 1-7. Jack, Cavalier, King in 4 suits. Numerical cards are worth their value, picture cards are worth 0.5 except the king of coins who is wild. (turns 7.0 or x.5 into 7.5, turns x.0 into 7, where x is less than 7)

The player and the dealer both get 1 card. The player is obligated to take more cards until his total is at least equal to the dealer, then he can choose to hit or stand. After the player stands, if the dealers card is 4 or less the dealer plays and will keep taking cards until he has 5 or higher. The aim is to get 7.5. If you get 8 or higher you bust. If player busts he loses regardless of dealer hand. If player and dealer end up on 5.0 dealer wins, if player/dealer match any other score it's a push, otherwise highest non bust hand wins.

What's best strategy for this game?



Okay, so I feel like Wizard's page is going to give us a good starting point, so let's look at the situations where a player would stand under those rules:

2 v. 2

3 v. 1-3

3.5 v. 2-3

4 v. .5-4

4.5 v. 2-4

5 v. .5-4

Not included are situations where players would stand against a higher dealer total, because that's not allowed in this version.

The first one that is interesting to me given the push on five ties rule is the situation of 5 v. .5-4. We know that we would definitely choose to hit on 5 v. 5, because staying on that is definitely losing. But, if the dealer reaches a total of five, then we lose if we have stood on five.

Player (Natural) 5 v Dealer 4

---Player Stands

Okay, so the player has decided to stand pat on five, so let's look at all the different possible results for the dealer. Fortunately, this is relatively simple, because the dealer can only take a maximum of two more cards (otherwise would have more than five).

Remaining Deck:

1-4
2-4
3-4
4-3
5-3
6-4
7-4
J-4
C-4
K3 + WILD

Okay, so I assume the Wild King is automatically winning for the dealer. That leaves us with three possibilities for the first card: The dealer can win immediately with WILD or 1-3, the dealer busts with four or greater and the dealer draws again with 0.5.

Win Immediately: (13/38) = 0.34210526315

Lose Immediately: (14/38) = 0.36842105263

Draw Again: (11/38)---SEE BELOW

Okay, so if the dealer has to draw again, then what we are left with in the deck are 37 cards, but now 23 of those cards cause the dealer to win (.5 is now winning for the dealer with the five tie) whilst 14 of the cards remaining would cause the dealer to lose.

Draw Again-WIN: (11/38) * (23/37) = 0.17994310099

Draw Again-LOSE: (11/38) * (14/37) = 0.10953058321

Probability Dealer WIN: 0.34210526315 + 0.17994310099 = 0.52204836414

Probability Dealer LOSE: 0.36842105263 + 0.10953058321 = 0.47795163584

On the face of it, the dealer has a greater than 50% chance of winning, so now we have to see if drawing more cards would be worse for the player. For these purposes, I'm going to assume that the player is going to stand on a total of 5.5, or more.

Remaining Deck:

1-4
2-4
3-4
4-3
5-3
6-4
7-4
J-4
C-4
K3 + WILD

PLAYER HITS:

The first thing that we are going to point out is that the player will automatically lose with 3-7, so that's going to be 18/38 remaining cards right off the bat.

(18/38) = 0.47368421052

As we can see, that's a pretty substantial likelihood of the player hitting and losing immediately. With that, let's look at the second most likely scenario, the player hits for a .5 point card, stands and the dealer draws:

We start with the probability of hitting for a half point card at (11/38), so now we have the following remaining deck composition for the dealer to hit his four, I'm going to pretend the player got a Jack just because it really doesn't make a difference which one it is.

1-4
2-4
3-4
4-3
5-3
6-4
7-4
J-3
C-4
K3 + WILD

Okay, so now the player has a 5.5 against the dealer four. In addition to 4-7, the dealer also loses immediately if he draws a 1 because that would give the dealer a total of five. If the dealer draws a 2, 3 or wild, then the dealer wins immediately. Play, once again, only continues beyond this if the dealer draws a .5 card.

Dealer wins immediately:

(11/38) * (9/37) = 0.07041251778

So, we look at this: 0.07041251778 + 0.47368421052 = 0.5440967283

Okay, so our probability of the dealer winning if we stand was only 0.52204836414 and the probability of the player busting if he hits + the player hitting a .5 card and then the dealer winning on the very next card combines to be more than the probability of the dealer winning if we stand.

Similarly, we will demonstrate that the player should not hit on the 5.5 in this situation because of the probability of busting on the following card, which would be a 3 or higher causing a bust:

Player 5.5-Dealer 4, player hits again:

(11/38) * (18/37) = 0.14082503556

Once again, just given the possibility of the player busting and ignoring everything else that could happen, the player ends up with a greater probability of losing (combined with the first hit) than if the player had just stood on 5 v. 4 in the first place. We thought that maybe ties on five losing could make the difference compared to the Wizard's write up on the other version of this game, but it does not.

Multi-Card 5's v. Dealer's 4

There is no need to do anything with this because a multi-card five will generally make it more likely that the dealer busts by hitting his four. The reason why is because more cards that do not bust the dealer will have been eliminated from the deck. Remember, 4-7 busts the dealer immediately and we were already starting with a five in the original example. The only multi-card five that takes away a dealer bust card (and then still only one) is the combination 4-.5-.5 or 4-1. All other multi-card fives would actually increase the number of cards the dealer can bust on, thereby making the situation better for the player.
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Mission146
Mission146
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May 14th, 2020 at 5:33:17 AM permalink
Okay, the next situation we are going to look at will involve a Player 4 v. a Dealer 4. This is going to be fairly simple to analyze for the player standing because we simply have the dealer taking only one card, or taking a half point card and then another card and there are no other possibilities:

Player 4 v. Dealer 4--Player Stands.

Remaining Deck Composition:

1-4
2-4
3-4
4-2
5-4
6-4
7-4
J-4
C-4
K3 + WILD

The probabilities for the dealer are actually exactly the same, because all we have done is exchanged a five for a four, which from a probabilistic standpoint, doesn't change anything. We can copy from above:

Okay, so I assume the Wild King is automatically winning for the dealer. That leaves us with three possibilities for the first card: The dealer can win immediately with WILD or 1-3, the dealer busts with four or greater and the dealer draws again with 0.5.

Win Immediately: (13/38) = 0.34210526315

Lose Immediately: (14/38) = 0.36842105263

Draw Again: (11/38)---SEE BELOW

Okay, so if the dealer has to draw again, then what we are left with in the deck are 37 cards, but now 23 of those cards cause the dealer to win (.5 is now winning for the dealer) whilst 14 of the cards remaining would cause the dealer to lose.

Draw Again-WIN: (11/38) * (23/37) = 0.17994310099

Draw Again-LOSE: (11/38) * (14/37) = 0.10953058321

Probability Dealer WIN: 0.34210526315 + 0.17994310099 = 0.52204836414

Probability Dealer LOSE: 0.36842105263 + 0.10953058321 = 0.47795163584

On the face of it, the dealer has a greater than 50% chance of winning, so now we have to see if drawing more cards would be worse for the player. For these purposes, I'm going to assume that the player is going to stand on a total of 5.5, or more.

Player 4 v. Dealer 4-Player hits

Okay, so in this scenario, the player is going to take a card. The first thing that we want to consider is the probability that the player does this and simply busts immediately. This is the same as for the dealer (4-7), consists of 14 cards and can be taken from above:

Lose Immediately: (14/38) = 0.36842105263

There are fewer busts in this scenario, but what we do know from the previous thing is that the player should stand on any total of five, or greater, because Player 5 v. Dealer 4---STAND would not be true otherwise, and we know it is true.

So, with that, let's look at the most likely hit for the player, which is hitting for a half point card, and analyze whether the player should stand or hit again:

***Player 4 + .5 v. Dealer 4***

***Player Hits***

1-4
2-4
3-4
4-2
5-4
6-4
7-4
J-3
C-4
K3 + WILD

Okay, so the first thing is that the player had a probability of 11/38 to even get a half point card in the first place. At this point, 4-7 will still bust the player, so let's take a look at that:

(11/38) * (14/37) = 0.10953058321

Okay, so now we add that to the 0.36842105263 probability that the player just loses right away and end up with 0.47795163584.

Even in this scenario, the player has a probability of 10/37 of just hitting for another half point card, which results in a total of five that we know the player stands. Obviously, that's also going to be pretty bad news for the player. Let's look at that:

(11/38) * (10/37) = 0.07823613086

Okay, so that's the probability that the player is going to hit and get two half point cards in a row. Our remaining deck composition will be as such for the dealer to draw:

1-4
2-4
3-4
4-2
5-4
6-4
7-4
J-2
C-4
K3 + WILD

The dealer draws to a four, so the possibilities are as follows:

Lose Immediately: (14/36) = 0.38888888888
Win Immediately: (13/36) = 0.36111111111
Draw Again: (9/36) = 0.25

As we predicted, with more small cards removed from the deck, the probability of the dealer hitting and losing immediately is increased. In this specific case, however, we have only removed 'Draw again' half point cards, so the probability of the dealer winning immediately also increased a little bit.

Now, we will imagine that the dealer draws another jack, just for fun we'll make them all jacks, and then has to draw again. At this point, half point cards result in the dealer winning with a total of five. 4-7 still cause a bust:

(9/36 * 14/35) = .1

(9/36 * 21/35) = .15

Okay, now we have to work back in the probability of any of this happening in the first place after we do our total probabilities in this specific situation.

Player 4 v. Dealer 4, Player Hits Gets .5, Player Hits Again Gets .5:

Dealer Wins: .15 + 0.36111111111 = .511111111

Dealer Loses: .1 + 0.38888888888 = .48888888888

Probability of Player 4 + .5 + .5 = 0.07823613086 (From Above)

0.07823613086 * .511111111 = 0.03998735576 (Dealer Wins)

0.07823613086 * .488888888 = 0.03824877501 (Dealer Loses)

0.47795163584 + 0.03998735576 = 0.5179389916 (In this scenario of the player hitting 4 and busting, hitting 4 getting .5 and busting or hitting 4, getting .5, hitting again and getting .5 and then losing by standing, we are so close to the probability of the dealer winning if we just stand and let the dealer go that the dealer beating other totals such as 5.5 or 6.5 is pretty clearly going to exceed our probability of losing if we just let the dealer try to bust, instead). Hitting the 4.5 is a bad decision and we need go no further with that.

Player 4 v. Dealer 4, Player Hits and Gets .5---Stands

Okay, so we're now going to have the player stand on 4.5 instead, so let's look at the remaining deck composition:

***Player Stands***

1-4
2-4
3-4
4-2
5-4
6-4
7-4
J-3
C-4
K3 + WILD

Once again, the player had a 11/38 probability of getting a half point card in the first place, so that's going to be our starting point. We also know that the player had a:

Lose Immediately: (14/38) = 0.36842105263

So, we're going to look specifically at the player hitting, getting a .5 card and standing with 4.5 against the dealer 4. We're making the .5 card a Jack again. It's always a Jack. The Jack has a 100% probability of coming out. (LOL)

Okay, so the dealer is in the same situation he's always in. The dealer will now have the following probabilities:

Win Immediately: (13/37) = 0.35135135135
Lose Immediately: (14/37) = 0.37837837837
Hit Again (0.5): (10/37) = 0.27027027027

Okay, so now we just have to look at the probability of the dealer drawing the .5 card and then winning or losing. As you'll recall, another half point card is winning for the dealer:

Dealer Wins (10/37 * 22/36) = 0.16516516516
Dealer Loses (10/37 * 14/36) = 0.1051051051

Total Dealer Wins: 0.35135135135+0.16516516516 = 0.51651651651
Total Dealer Loses: 0.37837837837+0.1051051051 = 0.48348348347

Okay, now we just need to work back in our probability of the player getting a half point card in the first place:

(11/38) * 0.51651651651 = 0.14951793899 (Dealer Wins)
(11/38) * .48348348347 = 0.13995574521 (Dealer Loses)

So, we add the dealer win probability to our probability that the player takes that first hit and busts to begin with and get:

0.36842105263+0.14951793899 = 0.51793899162

How about that!? The player is equally likely to lose this way as the player is to hit the 4.5, get another .5 card, and then stand on 5. That's pretty wild, isn't it?

And, as we have already established, the player always stands on a total of five against a dealer four.

Once again, we could do all of the work to factor in the player ending up with other totals (that are more than five) and still losing, but let's recall that the player standing on four results in a probability of:

0.52204836414 of the dealer winning, so if we do this:

0.52204836414-.51793899162 = 0.00410937252

So, we would only need combined probabilities of 0.410937252%, or more, of the player losing on a total of more than five to make the overall decision to hit on the four to begin with a worse decision. I could mathematically prove that this is the case, but it's so obvious already---and this is already taking forever---that I'm not going to waste the time doing that.

CONCLUSION

Player 4 v. Dealer 4

--The player should stand and hope for the dealer to bust.
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Mission146
Mission146
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May 14th, 2020 at 5:39:05 AM permalink
Let's take a quick breather and see where we are at so far:

Player Total x v. Dealer Total x:

The Player has no choice but to hit if the player's total is less than that of the dealer:

Player Total 5 v. Dealer Total 5:

Technically, the player does not have to hit, but should hit unless he wants to lose.

Player Total 5 v. Dealer Total 4:

The player should STAND and hope for the dealer to bust.

Player Total 4 v. Dealer Total 4:

The player should STAND and hope for the dealer to bust.

Player Total 4.5 v. Dealer Total 4:

The Player should STAND and hope for the dealer to bust. We didn't do the math on this, but because 4 v. 4 STAND is true and 5 v. 4 STAND is true, this must also be true. Also relevant, this removes an extra .5 card from the deck which is a card that does NOT bust the dealer, so that's good for the player.
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Mission146
Mission146
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May 14th, 2020 at 6:03:57 AM permalink
Okay, let's get a couple things out of the way really quick. We're going to demonstrate very quickly that the Player never hits a six unless he has no other choice:

Player 6 v. Dealer Less Than Six

Deck Composition (I'm giving the dealer a four, which is a best case scenario for the player other than a five, which would mean the player had already won)

1-4
2-4
3-4
4-3
5-4
6-3
7-4
J-4
C-4
K3 + WILD

Okay, so we start with the fact that 22 cards bust the player. (22/38) = 0.57894736842

That's more than half, so the player should stand. As we already know, the dealer has a greater probability of busting than the player's (1-0.57894736842) = 0.42105263158 probability of improving his hand, which doesn't even guarantee that the player wins. We will prove this because it is easy:

Dealer Busts or Totals 5 and loses immediately: (18/38) = 0.47368421052

There you have it, the dealer is more likely to bust than the player is to even improve his hand by taking a hit. Standing is clearly the correct decision.

Due to the high probability of the player busting by taking a card, the player standing on six will be the correct decision against any dealer total that is not greater than six.

Because the probability of the player busting is more than .5, the expected loss of that decision is more than zero. Because the expected loss of that decision is more than zero, the player should take the push on the total of Player 6 v. Dealer 6. Similarly, the player should take the Push on Player 7 v. Dealer 7.

With that, let's look at our strategy that we know so far:

Hitting Decisions:
Player .5 v. Dealer .5 (ANY)---The player can only improve his hand and cannot possibly bust.
Player x v. Dealer x (More than Player)---You must hit.
Player 5 v. Dealer 5---The player doesn't technically have to hit, but the alternative is automatically losing by standing.--So HIT. If the player hits and does not bust, then the player automatically wins, so the player would stand after this.

Standing Decisions:
Player 4 v. Dealer 4---The player should stand as previously demonstrated.
Player 4.5 v. Dealer 4---Because the player should stand on Player 4 v. Dealer 4, it must be true that the player should stand here.
Player 5 v. Dealer 4---The player should stand as previously demonstrated.
Player 5.5 v. Dealer 5---The player automatically wins and should stand.
Player 5.5 v. Dealer 4---Because standing on five against a Dealer 4 is true, and this is automatically better (Player wins against Dealer 5) this must be true.
Player 6 v. Dealer (6 or less)---The Player should stand. Just the probability of losing by hitting gives this play a negative expected value.
Player 7 v. Dealer (7 or less)---Obviously, the player should stand.

To Be Determined:

Player 5.5 v. Dealer .5-3

Player 5 v. Dealer .5, 1, 2 and 3

Player 4 v. Dealer .5, 1, 2 and 3

Player 3 v. Dealer .5, 1, 2 and 3

Player 2 v. Dealer .5, 1 or 2

Player 1 v. Dealer .5 or 1
Last edited by: Mission146 on May 14, 2020
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