CharlesMousseau
CharlesMousseau
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September 9th, 2019 at 8:35:02 PM permalink
Hello all,

doing some analysis on Lunar Poker, a game covered in detail on Stephen How's Discount Gambling website.

One of the features in the game involves paying a 1x Ante cost when the dealer does not qualify, to make them discard their highest card and replace it with a random card from the deck. Note that if you do this, and the dealer still does not qualify, you will no longer receive an Ante payment; if the dealer DOES qualify, you are paid on your Ante and Raise wagers as before.

Stephen gives two example hands. The first is:

player: Qd Qs Qh Ac 6h
dealer: Ad 9c 7d 5s 4c
forceEV: +0.7143

But when I work through this by hand, I get a force EV of exactly 1.0

If you force the dealer, they will discard their highest card (the ace), and draw one of the 42 remaining cards in the deck.

There are three 4s, three 5s, three 7s, and three 9s left, for a total of 12 cards that will give the dealer a qualifying hand of one pair. If they get one pair, you will win seven units (2x raise bet @ 3:1 + 1:1 Ante). On the remaining 30 cards, you will push the hand.

This gives a total expected value (before factoring in the 1-unit cost to force) of:

12 * 7 / 42 = 84 / 42 = 2.0000
and then subtracting the one-unit flat cost, you get an EV of 1.0000

It's worth noting that I am off by Stephen's figure by 0.2857 which equals 12/42, or 1 unit per winning replacement card

Similarly, on the second example:

player: 4d 4s 4c Qc 2h
dealer: Tc 8s 6s 5c 2s
forceEV: +0.5714

here there are now only eleven cards that make the dealer a qualifying hand when they throw away their top card (the ten), namely: two 2s, three 5s, three 6s, and three 8s.

I get an EV of 11 * 7 / 42 = 1.8333, and after subtracting the one-unit cost to force, I get 0.8333

This differs from Stephen's figure by 0.2619 which equals 11/42, or again, one unit per winning replacement card.

This suggests to me that my calculations are right and that the figures on Stephen's site are either not paying the Ante when the forced dealer now qualifies, or it's double-counting the cost on the winning hands only.

HOWEVER there is a very good chance that I'm missing something as this game has given me a migrane and a half; I've never played one hand and I'm already sick of it :P so if there's something I'm missing here PLEASE I'm all ears.

Thanks,

Charles.
MichaelBluejay
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September 9th, 2019 at 11:23:52 PM permalink
This is over my head, but is there some reason you haven't asked Stephen How himself? (Or maybe you did, and he didn't reply? He might not reply to Joe Blow, but I'm sure he'd reply to you.) Or maybe you prefer to have an independent third party give their take?
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CharlesMousseau
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September 9th, 2019 at 11:29:03 PM permalink
Stephen participates on this forum, plus any third party. Basically just looking for a vote of confidence is all.
kubikulann
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September 10th, 2019 at 2:19:59 AM permalink
Don’t know the game, so it’s just from your examples.
It appears like How’s figures are based on a x6 rather than x7 payment.
So essentially it would mean that when forcing the dealer you forfeit your (original) ante?

OR
You already got this ante paid when the dealer did not qualify, and this calculation is only for the *net* value of buying a forced redraw.
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charliepatrick
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September 10th, 2019 at 4:29:49 AM permalink
If I understand it, if you force then you don't get the Ante and also have to pay 1 unit. However you will be paid out (for Trips) 3xRaise = 6 units. There are 12 winning outs so...
(i) 12 wins (win 6, paid 1 to force) = 5x12 =+60.
(ii) 30 losses (no win, paid 1 to force) = -1x30 = -30.
So net EV = 30/42.
When there are only 11 ways to win, the figures are 5x11 (+55) - 1x31 (-31) -> EV = 24/42.

I did stumble over this game in Prague many years ago (driving back from Hungary), but had no idea of the correct strategy, so didn't do too well.
gordonm888
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charliepatrick
September 10th, 2019 at 6:10:55 AM permalink
Quote: charliepatrick

If I understand it, if you force then you don't get the Ante and also have to pay 1 unit. However you will be paid out (for Trips) 3xRaise = 6 units. There are 12 winning outs so...
(i) 12 wins (win 6, paid 1 to force) = 5x12 =+60.
(ii) 30 losses (no win, paid 1 to force) = -1x30 = -30.
So net EV = 30/42.
When there are only 11 ways to win, the figures are 5x11 (+55) - 1x31 (-31) -> EV = 24/42.

I did stumble over this game in Prague many years ago (driving back from Hungary), but had no idea of the correct strategy, so didn't do too well.



charliepatrick is 5 or 6 time zones ahead of me, so he almost always beats me to it. And, dang it, he always gets it right. Which I believe he has done here.

My only contribution is to point out that you (Charles Mousseau) also mis-spelled 'migraine.'
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
CharlesMousseau
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September 10th, 2019 at 7:18:52 AM permalink
Quote: charliepatrick

If I understand it, if you force then you don't get the Ante and also have to pay 1 unit. However you will be paid out (for Trips) 3xRaise = 6 units. There are 12 winning outs so...
(i) 12 wins (win 6, paid 1 to force) = 5x12 =+60.
(ii) 30 losses (no win, paid 1 to force) = -1x30 = -30.
So net EV = 30/42.
When there are only 11 ways to win, the figures are 5x11 (+55) - 1x31 (-31) -> EV = 24/42.

I did stumble over this game in Prague many years ago (driving back from Hungary), but had no idea of the correct strategy, so didn't do too well.



See, that's the thing, both rule sheets on Stephen's site and on the Wizard's state that, if the dealer qualifies after a force, the Ante is paid.

"With a cost of 1x ante, the player may, if dealer doesn't qualify, exchange one card (always the highest one) from the dealers hand to make the dealer to qualify. Player hand will push (no ante will be paid) if the dealer still doesn't qualify after this." - WoO

"If the dealer qualifies after the draw, then the player’s Ante and Raise resolve as before. If the dealer doesn’t qualify, then the Ante and Raise push." - DG

So the net win should be 6 units when a forced dealer qualifies, not 5.
charliepatrick
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September 10th, 2019 at 8:06:10 AM permalink
I think there are various rules (one I found some that don't allow you to force the dealer but just pays the Ante at 4-1 for good hands).
- https://www.ukcasinotablegames.info/pokerlunar.html - I'm not sure this is correct as it still pays the Ante before you can force the dealer with a payment and isn't clear what the subsequent payout is on a win.
- https://www.skycityauckland.co.nz/casino/table-games/ - if you click the game it will open a PDF, which says (p4) you don't get paid on the Ante etc.
MichaelBluejay
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September 10th, 2019 at 10:20:47 AM permalink
Quote: CharlesMousseau

Stephen participates on this forum, plus any third party.

Okay, but he might not see your post, though. I can't count how many times someone started a thread here addressed to me, just because I'm a forum member, and I didn't see it until years later when I was searching for something else.

Then again, maybe he *will* see it.
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CharlesMousseau
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September 10th, 2019 at 10:46:33 AM permalink
Quote: charliepatrick

I think there are various rules (one I found some that don't allow you to force the dealer but just pays the Ante at 4-1 for good hands).
- https://www.ukcasinotablegames.info/pokerlunar.html - I'm not sure this is correct as it still pays the Ante before you can force the dealer with a payment and isn't clear what the subsequent payout is on a win.
- https://www.skycityauckland.co.nz/casino/table-games/ - if you click the game it will open a PDF, which says (p4) you don't get paid on the Ante etc.



Would you agree then that, assuming the rules are as I reported (i.e. if you pay to switch you now get paid on Raise AND Ante if dealer qualifies) that my EV calculations are good? Like I should be confident it's super simple and the difference in results is so obvious but still just insecure :P
charliepatrick
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September 10th, 2019 at 11:00:42 AM permalink
^Yes as you add 12/42 and 11/42 to my figures. Note all these ignore any discards you may have made as I think you should draw to dealt Trips.
kubikulann
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September 10th, 2019 at 12:55:36 PM permalink
Quote: CharlesMousseau

Would you agree then that, assuming the rules are as I reported (i.e. if you pay to switch you now get paid on Raise AND Ante if dealer qualifies) that my EV calculations are good? Like I should be confident it's super simple and the difference in results is so obvious but still just insecure :P

(again, just from reading the posts here)
Two possible rules appear from what has been written.
1) if you do not force, you are paid 1x ante. If you force and the dealer still not qualifies, you are NOT paid the ante (nor the raise): -1. If s/he qualifies, you are paid normally, ante 1 and raise 3: +6
2) if you do not force OR if the dealer doesn’t qualify after a force draw, you get paid your ante. If qualify, you receive in addition the raise x3: +6

Your formula would imply a third ruling, where you are not paid one ante even in the case you don’t force.// or you receive a second pay on your ante! +7


Case(1) is 6 for success, -1 for fail.
Case (2) is 6 for success, 0 for fail. That is what was exhibited as St.How’s result.
Your case is 7 and 0.
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Dobrij
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September 10th, 2019 at 1:06:56 PM permalink
Guys, I don’t understand exactly what the problem is, but if You want the correct calculation will be like this:

Quote:

player: Qd Qs Qh Ac 6h
dealer: Ad 9c 7d 5s 4c
forceEV: +0.7143


Get ante = +1,00
Buy = +0,79

Quote:

player: 4d 4s 4c Qc 2h
dealer: Tc 8s 6s 5c 2s
forceEV: +0.5714


Get ante = +1,00
Buy = +0,68

* what you call "forсe", calling is "buy the game"
kubikulann
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September 10th, 2019 at 1:10:51 PM permalink
Quote: Dobrij

Guys, I don’t understand exactly what the problem is, but if You want the correct calculation will be like this:


Bet = +1,00
Buy = +0,79


Bet = +1,00
Buy = +0,68

* what you call "forсe", calling is "buy the game"

Sorry, but that’s wrong.
Would you please expose your equations ?
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Dobrij
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September 10th, 2019 at 1:16:51 PM permalink
Quote: kubikulann

Sorry, but that’s wrong.
Would you please expose your equations ?



"Get ante", more profitable than trying to "buy a game".

Before saying this wrong, be sure that you are 100% right! And it can be a shame ...
kubikulann
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September 10th, 2019 at 1:45:59 PM permalink
Quote: Dobrij



Before saying this wrong, be sure that you are 100% right! And it can be a shame ...

Neither of us is a native English speaker, but I can make the difference between « THIS is wrong/right » and « YOU are wrong/right ».

And between « please » and « a shame ».

Are you a Russian version of 7craps?
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CharlesMousseau
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September 10th, 2019 at 1:54:12 PM permalink
Enough bickering, you two. I'm trying to have a mathematical and rules conundrum solved. Take it outside. :)
Dobrij
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September 10th, 2019 at 3:14:05 PM permalink
Quote: CharlesMousseau

Enough bickering, you two. I'm trying to have a mathematical and rules conundrum solved. Take it outside. :)



So, calculation for You is not suitable? (is not the answer 4 U?)
¯\_(ツ)_/¯


Quote:


player: Qd Qs Qh Ac 6h
dealer: Ad 9c 7d 5s 4c
forceEV: +0.7143
Get ante = +1,00
Buy(force) = +0,79

player: 4d 4s 4c Qc 2h
dealer: Tc 8s 6s 5c 2s
forceEV: +0.5714
Get ante = +1,00
Buy(force) = +0,68

CharlesMousseau
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MichaelBluejay
September 12th, 2019 at 12:02:38 PM permalink
OKAY I figured it out. Apparently the Ante does NOT pay when the dealer has a qualified hand, it only pays when the dealer does not qualify.

This would explain why my results were off by one unit, and other off-by-one-unit errors that I found.

Thanks to everyone participating. Sometimes even just talking things out is useful, even when it means I need to find my problem ;)
DRich
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September 12th, 2019 at 12:15:07 PM permalink
Quote: CharlesMousseau


Thanks to everyone participating. Sometimes even just talking things out is useful, even when it means I need to find my problem ;)



I used to talk my wife through my programming problems and she had zero clue about any technology. It usually worked because it made me explain the problem in different ways which would help me realize the problem.
At my age, a "Life In Prison" sentence is not much of a deterrent.
MichaelBluejay
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September 12th, 2019 at 12:23:04 PM permalink
Quote: DRich

I used to talk my wife through my programming problems and she had zero clue about any technology. It usually worked because it made me explain the problem in different ways which would help me realize the problem.

I can't count the number of times I've had that same experience! She'll sit there without saying anything while I explain how I'm stuck, then because of the way I'm explaining it I'll realize the answer. I always thank her for helping, but she says, "I didn't do anything!" But really, without her there to listen, I wouldn't have figured it out.

She's also great for troubleshooting interfaces, finding problems that regular users would have that wouldn't have occurred to me.
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Dobrij
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September 12th, 2019 at 2:00:42 PM permalink
Quote: DRich

I used to talk my wife through my programming problems and she had zero clue about any technology. It usually worked because it made me explain the problem in different ways which would help me realize the problem.



Therefore, God created languages so people can communicate

: )
MichaelBluejay
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September 12th, 2019 at 5:06:12 PM permalink
In other news, she just outdid me on a practical math problem. Ugh.

On the one hand, it's handy being married to someone smarter than me, because we wind up making better plans/decisions. On the other hand, it's a bit trying getting constant reminders that my problem-solving ability is poor.
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