Ayecarumba
Ayecarumba
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January 24th, 2018 at 9:41:42 PM permalink
The dealer gets three hole cards, and I also get three, but have to discard one before five board cards are revealed.

After 100 rounds, how many times will my best five of seven be better than the dealer’s five of eight?
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Mission146
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January 24th, 2018 at 10:08:14 PM permalink
Gosh darn it.

It sucks because I have a way of doing this were the hands of the player and the dealer not directly correlated. Given your suggestion, I don't even know where to start. I think it would probably take a simulation.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
JB
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January 24th, 2018 at 11:11:36 PM permalink
There are 1,674,391,363,444,800 outcomes to analyze to know the exact answer. Shortcuts would cut that down, but it's still a lot of work.

Even a simulation would require a strategy for the discard. With some 3-card combinations the best discard might seem obvious, but you would need precise rules to cover every possible starting hand unambiguously, which would be a bit of an undertaking. For example, suited A76: do you discard the 6 or the Ace? Choosing a discard at random wouldn't be a realistic strategy either, and would produce unreliable results.

Is this a game you have seen at a casino?
SM777
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January 25th, 2018 at 7:45:22 AM permalink
This is a game Shuffle Master has shown at G2E the last few years. I forget the name.
Romes
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January 25th, 2018 at 7:45:56 AM permalink
Quote: JB

There are 1,674,391,363,444,800 outcomes to analyze to know the exact answer. Shortcuts would cut that down, but it's still a lot of work.

Even a simulation would require a strategy for the discard. With some 3-card combinations the best discard might seem obvious, but you would need precise rules to cover every possible starting hand unambiguously, which would be a bit of an undertaking. For example, suited A76: do you discard the 6 or the Ace? Choosing a discard at random wouldn't be a realistic strategy either, and would produce unreliable results.

Is this a game you have seen at a casino?

Agree with everything except the example =P... You would always keep A-7 over 7-6 =).
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CrystalMath
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January 25th, 2018 at 8:58:14 AM permalink
Quote: SM777

This is a game Shuffle Master has shown at G2E the last few years. I forget the name.



HIC (High Island Creations) Holdem
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JB
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January 25th, 2018 at 3:36:51 PM permalink
Quote: Romes

Agree with everything except the example =P... You would always keep A-7 over 7-6 =).


Yeah, that was a bad example. I was trying to quickly think of a hand where the best choice isn't obvious or intuitive. I can't seem to think of one now.
Ayecarumba
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January 25th, 2018 at 5:04:46 PM permalink
Quote: JB

Yeah, that was a bad example. I was trying to quickly think of a hand where the best choice isn't obvious or intuitive. I can't seem to think of one now.

Thanks for the input. What if the hand is 89 suited with an Ace off suit?

The game is in trials at the California hotel downtown.
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Mission146
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January 25th, 2018 at 5:16:57 PM permalink
Quote: Ayecarumba

Thanks for the input. What if the hand is 89 suited with an Ace off suit?

The game is in trials at the California hotel downtown.



I could be wrong, but I'd say A-9, no-brainer as potential to win on the high pair. I assume that this is played with one deck, and as we know the five cards are correlated with both hands, so you have an Ace that the dealer can't have. That gives you the advantage if you and the dealer both finish with a pair, two pair or trips. Technically quads and boats, too, but I should imagine that would be unlikely and would not come into play as often.

Again, could be wrong, but I think Aces would be very valuable in a game like this.

A much closer decision might be something like KQ, QJ or J10 with an off-suit ace, because you don't get punished as badly on high pairs or trips when it comes to the other stuff. A pair of nines (or two pair with nines high) is pretty low.
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CrystalMath
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January 25th, 2018 at 7:40:52 PM permalink
Quote: Ayecarumba

Thanks for the input. What if the hand is 89 suited with an Ace off suit?

The game is in trials at the California hotel downtown.



I'm pretty sure you go with the 89 suited. Do you recall if there was a flop and a turn for the community cards or were all 5 cards revealed in one step?
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Mission146
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January 25th, 2018 at 7:54:59 PM permalink
Quote: CrystalMath

I'm pretty sure you go with the 89 suited. Do you recall if there was a flop and a turn for the community cards or were all 5 cards revealed in one step?



Like I said, I'm pretty sure you go with the 8-9 suited. ;)

If CrystalMath says 8-9 suited, that's enough to convince me.
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Mission146
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January 25th, 2018 at 7:58:21 PM permalink
My reasoning was basically that, in Texas Hold 'Em, you're better off to have an Ace as opposed to another player having one (or both) cards that have a higher value than both of your cards with the same 8-9 suited. The difference is so great that I assumed it wouldn't change THAT much given the rules of the new game. The dealer is expected to have a ten or better, given three cards.

ADDED:

Okay, so we take 8-9-A out of a single deck, ignoring suits.

Remaining are 49 cards of which the dealer gets three. Of those 49 cards, 30 have a rank of nine or lower. Obviously, a pair of nines for the dealer (or eights) is EXTREMELY bad for the player and the player would definitely have been better off with A-9, but we'll ignore that for now.

The probability of the dealer getting three cards that are nine, or less, is:

(30/49 * 29/48 * 28/47) = 0.22036474164

Even ignoring the possibility of the dealer pairing up in what the player is holding, the dealer has a probability of greater than 77%, very close to 78%, of getting AT LEAST one card that is a ten or better.

In the meantime:

(46/49 * 45/48 * 44/47) = 0.8239253148

The probability of the dealer getting three cards, none of which are aces, is 82.3925%...but I'm still going with CrystalMath unless someone smarter than me backs me up on this.
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SM777
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January 25th, 2018 at 8:23:32 PM permalink
Quote: CrystalMath

I'm pretty sure you go with the 89 suited. Do you recall if there was a flop and a turn for the community cards or were all 5 cards revealed in one step?



I think three cards first, and then the last two. But there's a pause for betting in there too.
Ayecarumba
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January 25th, 2018 at 8:52:17 PM permalink
- Players make an ante bet.
- The players and dealer each receive 3 cards.
- The players discard one of their cards.
- A randomizer lets players know if they can make a play bet of 2x, 3x, or 4x. Or, players can check.
- Three cards are flopped.
- Players that haven’t made a play bet can make a 1x bet or fold.
- The last two board cards are revealed.
- If the player’s best five card hand beats the dealer’s best five, the player wins even money on their ante and play wagers.
- If the player loses, they lose both.
- If the hands tie, both push.

There are also bonuses for certain strong hands, a side bet that pays for three of a kind or better, and an envy bonus.
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CrystalMath
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January 25th, 2018 at 9:42:34 PM permalink
Quote: Mission146


The probability of the dealer getting three cards, none of which are aces, is 82.3925%...but I'm still going with CrystalMath unless someone smarter than me backs me up on this.



I sure hope it's right, because I know who did the math ;)

It may have to do with the intermediate betting rounds, and not just the straight probability of winning when all 5 community cards are revealed. For instance, you might fold the 89 more often, but have stronger hands the times you don't fold. Of course, I'd have to spend a lot more time looking at the numbers to really figure it out, and time isn't something I have very much of these days.
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JB
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January 25th, 2018 at 9:56:35 PM permalink
Wow, that's an impressive analysis CM. I just put together an analyzer for the game described in the first post (where the only decision is which card to discard), and using as many optimizations as I could think of (including 3.5GB of lookup tables cached in memory) it still took 40 minutes to brute-force a single hand.

You must have had to do some serious shortcuts to analyze the actual game, which is a lot more complex than the one described in the first post.
Ibeatyouraces
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January 25th, 2018 at 9:58:15 PM permalink
7 cards vs 8... I'll pass!
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Mission146
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January 25th, 2018 at 10:05:04 PM permalink
Quote: CrystalMath

I sure hope it's right, because I know who did the math ;)

It may have to do with the intermediate betting rounds, and not just the straight probability of winning when all 5 community cards are revealed. For instance, you might fold the 89 more often, but have stronger hands the times you don't fold. Of course, I'd have to spend a lot more time looking at the numbers to really figure it out, and time isn't something I have very much of these days.



Yes, plus I wasn't aware of the bonuses for certain high-ranking hands until after I had made my post. I wouldn't even have ventured a guess, had I been.
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Mission146
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January 25th, 2018 at 10:06:30 PM permalink
Quote: JB

Wow, that's an impressive analysis CM. I just put together an analyzer for the game described in the first post (where the only decision is which card to discard), and using as many optimizations as I could think of (including 3.5GB of lookup tables cached in memory) it still took 40 minutes to brute-force a single hand.

You must have had to do some serious shortcuts to analyze the actual game, which is a lot more complex than the one described in the first post.



Was it for the A98? Does the suited 98 have a greater probability of winning, too?
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CrystalMath
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January 25th, 2018 at 10:21:36 PM permalink
Quote: JB

Wow, that's an impressive analysis CM. I just put together an analyzer for the game described in the first post (where the only decision is which card to discard), and using as many optimizations as I could think of (including 3.5GB of lookup tables cached in memory) it still took 40 minutes to brute-force a single hand.

You must have had to do some serious shortcuts to analyze the actual game, which is a lot more complex than the one described in the first post.



It wasn't a full combinatorial analysis. I don't think I have that many years to live. It was close enough that GLI agreed, although they used a slightly different method.
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JB
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January 25th, 2018 at 10:25:46 PM permalink
Yes, the results were:

Hold Win Lose Tie EV
A/9 9,871,905,681 14,719,377,449 663,488,566 -0.191943
A/8 9,738,090,759 14,813,188,636 703,492,301 -0.200956
98 8,943,758,176 15,537,036,380 773,977,140 -0.261071


But this was for the simple game where the only decision is which card to discard, not the actual (more complex) game.
Mission146
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January 25th, 2018 at 10:28:20 PM permalink
Quote: JB

Yes, the results were:

Hold Win Lose Tie EV
A/9 9,871,905,681 14,719,377,449 663,488,566 -0.191943
A/8 9,738,090,759 14,813,188,636 703,492,301 -0.200956
98 8,943,758,176 15,537,036,380 773,977,140 -0.261071


But this was for the simple game where the only decision is which card to discard, not the actual (more complex) game.



He shoots, he scores!!!

I didn't know the rules for the more complex game, I just wanted to say that A-9 would beat the dealer more frequently. I feel vindicated now, thanks!
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
CrystalMath
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January 25th, 2018 at 10:37:27 PM permalink
Quote: Mission146


He shoots, he scores!!!

I didn't know the rules for the more complex game, I just wanted to say that A-9 would beat the dealer more frequently. I feel vindicated now, thanks!



Nice! I don't have a poker eye, just math.
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