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Gialmere
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July 4th, 2019 at 1:34:59 AM permalink
I tried asking this question several months ago but got carried away with the concept and no one ended up answering it. So let me try again by distilling it down to its essence.

You've been placed in charge of table games at a medium size casino under new (private) ownership. There is only one rectangular pit and the owners want a craps table at one end and a roulette table at the other. In between there is room for eight blackjack size tables (four on each side). The owners say you can fill it with whatever games you want with the following provisos...

1) At least two of the tables must offer some sort of blackjack. There can be more but see #2.

2) They would like a wide assortment of games that would appeal to a wide variety of players but will yield to your discretion because of #3.

3) The pit must make a reasonable profit. One or two low HE games are fine for promotional reasons but high HE games and side bets are a good thing too.

Which eight games would you choose?


[Hmm ... I should make a logic puzzle out of this.]
Have you tried 22 tonight? I said 22.
ChumpChange
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July 4th, 2019 at 5:34:11 AM permalink
3:2 Blackjack ($5 table) ($25 table)
Spanish 21 ($5/$10 table)
Baccarat ($20 table)
War ($5 table)
Pai Gow Poker ($10 table)
Mississippi Stud ($5 table)
Sic Bo (High HA!)
charliepatrick
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July 4th, 2019 at 7:42:24 AM permalink
In the UK most small casinos would have Blackjacks (perhaps two low limit (CSM) and one high limit (shoe based) ). Next a poker based game: 3CP is in most UK casinos and occasionally other poker games appear. I can imagine other games such as Pai Gow Poker, Blackjack variant and carnival games for busy times.

If you're one of several casinos in the area then it might be worth including a unique (possibly carnival) game as a "loss leader" to attract people into your casino rather than the one next door. Also, if cheap and enjoyable, it introduces people to casinos and hopefully they go away having had a good evening and will return.

Personally I think some places find it hard to accept that a "not very profitable table" is actually good for business.
Zcore13
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July 4th, 2019 at 10:03:30 AM permalink
5 blackjack tables with side bets
1 texas hold'em poker style table
1 High Card Flush table
1 Three Card Poker variant table


ZCore13
I am an employee of a Casino. Former Table Games Director,, current Pit Supervisor. All the personal opinions I post are my own and do not represent the opinions of the Casino or Tribe that I work for.
FCBLComish
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July 4th, 2019 at 10:52:05 AM permalink
Quote: Zcore13

5 blackjack tables with side bets
1 texas hold'em poker style table
1 High Card Flush table
1 Three Card Poker variant table


ZCore13




You would not have some sort of Pai Gow Poker?
Beware, I work for the dark side.... We have cookies
TigerWu
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July 4th, 2019 at 10:56:29 AM permalink
8 Pai Gow Tile tables.

Seriously, though:

2 Blackjack
1 Tiles
1 PGP
1 midi-Baccarat
1 mini-Baccarat
1 Sic Bo
1 something carny like Caribbean Stud or Let it Ride
TinMan
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July 4th, 2019 at 11:02:18 AM permalink
Depends a bit on clientele and whether the market will support a 6/5 game. If this casino is in Macau, it’s 2 BJ tables and 6 Bacc. If there were no requirement for 2 BJ tables, I’d have 1 BJ and 7 Bacc. If this is a US casino not on the LV strip, I’d prob go with 1 $5 6-5 BJ, 3 $15 or $25 3/2 BJ, 1 Bacc, 1 3 card poker, 1 Spanish 21, 1 UTH. If I could have another roulette wheel, I’d sacrifice the Spanish 21 table.
If anyone gives you 10,000 to 1 on anything, you take it. If John Mellencamp ever wins an Oscar, I am going to be a very rich dude.
FCBLComish
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July 4th, 2019 at 11:38:10 AM permalink
It absolutely depends on the location of the casino, and the demographic of the clientele.
Beware, I work for the dark side.... We have cookies
Zcore13
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July 4th, 2019 at 11:54:14 AM permalink
Quote: FCBLComish

You would not have some sort of Pai Gow Poker?



Not with only 8 games in what I assume would be a small town.


ZCore13
I am an employee of a Casino. Former Table Games Director,, current Pit Supervisor. All the personal opinions I post are my own and do not represent the opinions of the Casino or Tribe that I work for.
DJTeddyBear
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July 5th, 2019 at 6:03:01 AM permalink
I assume this will be somewhere in the US, where BJ is king. Therefore, emphasis on BJ varieties.

BJ with Lucky Ladies
BJ with 21+3
FreeBet BJ
Down Under BJ
Face up PGP
3CP
Criss Cross Poker or Let It Ride
Mini Bac


I’m kinda surprised you didn’t ask, and nobody volunteered, specs about the craps and roulette tables.

Craps:
5x odds
Double/triple field
25¢ chips - only for buy bets less than $20
Fire bet

Roulette:
Double zero
Poker For Rouletre (Did you expect me to say anything else???)
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
mtcards
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July 5th, 2019 at 12:24:50 PM permalink
Spanish 21 and one 21+3 blackjack
Kriss Kross Poker
Ultimate Texas Holdem
High Card Flush
Three Card Poker
Pai Gow Poker
Mississippi Stud

I personally love the variety. Seems like so many casinos are going to the same games now.
DRich
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July 5th, 2019 at 1:52:55 PM permalink
6 BJ tables with side bets available
1 Ultimate Texas Holdem
1 Three Card poker
At my age, a "Life In Prison" sentence is not much of a deterrent.
Gialmere
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July 5th, 2019 at 2:15:08 PM permalink
Quote: DRich

6 BJ tables with side bets available


3/2 or 6/5?
Have you tried 22 tonight? I said 22.
TheOKCPanda
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July 5th, 2019 at 2:39:53 PM permalink
4 blackjack, two are 25 dollar min. The two lower limits are continuous shuffle, 25+s are double deck pitch. One pai gow poker, two three card, one UTH.

I'd also never bar any player but probably have a 1000 limit. Skilled APs are few and far between and itll get the word out. The player edge even in a high count is not enough to sweat. It doesn't happen that often and itll be taken care of by the higher than normal traffic
Zcore13
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July 5th, 2019 at 5:44:00 PM permalink
Quote: mtcards

Spanish 21 and one 21+3 blackjack
Kriss Kross Poker
Ultimate Texas Holdem
High Card Flush
Three Card Poker
Pai Gow Poker
Mississippi Stud

I personally love the variety. Seems like so many casinos are going to the same games now.



You would get fired pretty quickly with only 2 BJ tables.


ZCore13
I am an employee of a Casino. Former Table Games Director,, current Pit Supervisor. All the personal opinions I post are my own and do not represent the opinions of the Casino or Tribe that I work for.
Gialmere
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July 5th, 2019 at 6:44:26 PM permalink
Logic Puzzle [I've never actually built a logic puzzle before so this one is pretty easy.]

You're in charge of setting up a new casino. The owners have purchased and delivered 8 table games: Big 6 Wheel, Craps, Roulette, Spanish 21, Three Card Poker, Sic Bo, Super Fun 21 and Ultimate Texas Hold'em. The live game pit is to be in the shape of an octagon with each game making a side. The side facing the front door is table #1 and the rest are numbered in a clockwise order. The owners, however, like puzzles and give you these cryptic instructions on how to set the tables up...

1) Place the craps table one position clockwise from 3CP but not next to B6.

2) The two poker type games shall neither be placed next to each other nor next to a blackjack type game.

3) Spanish 21 and Roulette shall be placed at opposite ends of the pit.

4) Place the Sic Bo table as far from the front as possible.


Can you set up the pit as they requested?
Table #1=
Table #2=
Table #3=
Table #4=
Table #5=
Table #6=
Table #7=
Table #8=

Have you tried 22 tonight? I said 22.
beachbumbabs
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July 5th, 2019 at 11:12:37 PM permalink
Quote: Gialmere

Logic Puzzle [I've never actually built a logic puzzle before so this one is pretty easy.]

You're in charge of setting up a new casino. The owners have purchased and delivered 8 table games: Big 6 Wheel, Craps, Roulette, Spanish 21, Three Card Poker, Sic Bo, Super Fun 21 and Ultimate Texas Hold'em. The live game pit is to be in the shape of an octagon with each game making a side. The side facing the front door is table #1 and the rest are numbered in a clockwise order. The owners, however, like puzzles and give you these cryptic instructions on how to set the tables up...

1) Place the craps table one position clockwise from 3CP but not next to B6.

2) The two poker type games shall neither be placed next to each other nor next to a blackjack type game.

3) Spanish 21 and Roulette shall be placed at opposite ends of the pit.

4) Place the Sic Bo table as far from the front as possible.


Can you set up the pit as they requested?



Think there are at least 2 correct answers, but here's one:

Table #1=Craps
Table #2=superfun 21
Table #3=BIG 6 Wheel
Table #4=SPANISH 21
Table #5= Sic bo
Table #6=UTH
Table #7=roulette
Table #8=3CP


If the House lost every hand, they wouldn't deal the game.
Gialmere
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July 5th, 2019 at 11:22:22 PM permalink
Sorry. No. Note instruction #3. I suppose I should have clearly stated "directly oppisite". Apologies. As I said, I'm new at this.
Have you tried 22 tonight? I said 22.
charliepatrick
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July 8th, 2019 at 11:59:58 AM permalink
I have assumed position 1 is not considered one round from position 8.

The logic I used started with looking at poker games and BJ games, called P and B here...
R2: could be (starting any where)
-P-B-P-B (have P and B alternate)
-P-P-BB- (have PPBB, PP can't be adjacent, the B's can be in any two of last four positions)
-P-P-B-B
-P-P--BB

R3: cannot be -P-B-P-B or -P-P-B-B as R needs to be opposite one of the B's (Sp21)
R3: gives R between the poker tables,
-PRP-?B? where one of the ? is a Blackjack

R1: gives Craps one posn clockwise from 3CP.
This gives - UTH R 3CP Craps ? Sp21 ? where either ? is a BJ.

R1: Craps ? Sp21, the ? cannot be Big6.

R4: merely says that Sic Bo shall be in position 5, with everything else shuffled round. Thus I don't know whether this says that 3CP can't be in poisition 8 with Craps in position 1. If so this gives a unique solution...
- UTH R 3CP Craps ? Sp21 SicBo (where 3CP is at position1 and Craps at position 2)

R2:
Big6 UTH R 3CP Craps SF21 Sp21 SicBo shuffled so SicBo is in position 5.

Assuming 8 and 1 isn't allowed, I got
1 = 3CP (Poker)
2 = Craps
3 = Super Fun 21 (BJ)
4 = Spanish 21 (BJ)
5 = Sic Bo
6 = Big 6
7 = UTH (Poker)
8 = Roulette
I got the positions of Sic Bo and Big 6 could be interchanged and everything shuffled one position if Craps could be in position 1 with 3CP in position 8. Hence the above question and assumption to get a unique answer.
beachbumbabs
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July 8th, 2019 at 12:31:56 PM permalink
Quote: Gialmere

Sorry. No. Note instruction #3. I suppose I should have clearly stated "directly oppisite". Apologies. As I said, I'm new at this.



No apology necessary. I understood the instructions - had those 2 lined up correctly at one point, and moved them. My error, not yours.

I thought instructions 2 and 3 were mutually exclusive, but they're not: there's an out. Revised solution below.

Table #1=Craps
Table #2=superfun 21
Table #3=Spanish 21
Table #4=Big 6 wheel
Table #5= Sic bo
Table #6=UTH
Table #7=roulette
Table #8=3CP
If the House lost every hand, they wouldn't deal the game.
charliepatrick
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July 8th, 2019 at 12:37:52 PM permalink
Quote: beachbumbabs

...Revised solution below...

Hi Babs, I agree it works, but did you see my assumption above?
beachbumbabs
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July 8th, 2019 at 12:47:07 PM permalink
Quote: charliepatrick

Hi Babs, I agree it works, but did you see my assumption above?



I did see that, but I'm not sure that assumption is inherent in Gialmere's wording. So perhaps that language should be added to refine his puzzle into a single, unique correct answer. As it stands, I believe you and I are both correct.

But of course, I could be wrong. :)
If the House lost every hand, they wouldn't deal the game.
Gialmere
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July 8th, 2019 at 5:26:48 PM permalink
Quote: charliepatrick

I have assumed position 1 is not considered one round from position 8.

The logic I used started with looking at poker games and BJ games, called P and B here...
R2: could be (starting any where)
-P-B-P-B (have P and B alternate)
-P-P-BB- (have PPBB, PP can't be adjacent, the B's can be in any two of last four positions)
-P-P-B-B
-P-P--BB

R3: cannot be -P-B-P-B or -P-P-B-B as R needs to be opposite one of the B's (Sp21)
R3: gives R between the poker tables,
-PRP-?B? where one of the ? is a Blackjack

R1: gives Craps one posn clockwise from 3CP.
This gives - UTH R 3CP Craps ? Sp21 ? where either ? is a BJ.

R1: Craps ? Sp21, the ? cannot be Big6.

R4: merely says that Sic Bo shall be in position 5, with everything else shuffled round. Thus I don't know whether this says that 3CP can't be in poisition 8 with Craps in position 1. If so this gives a unique solution...
- UTH R 3CP Craps ? Sp21 SicBo (where 3CP is at position1 and Craps at position 2)

R2:
Big6 UTH R 3CP Craps SF21 Sp21 SicBo shuffled so SicBo is in position 5.

Assuming 8 and 1 isn't allowed, I got
1 = 3CP (Poker)
2 = Craps
3 = Super Fun 21 (BJ)
4 = Spanish 21 (BJ)
5 = Sic Bo
6 = Big 6
7 = UTH (Poker)
8 = Roulette
I got the positions of Sic Bo and Big 6 could be interchanged and everything shuffled one position if Craps could be in position 1 with 3CP in position 8. Hence the above question and assumption to get a unique answer.


Wow CP! Impressive. I think you're a few levels above me in this arena and would encourage you to make a puzzle of your own. (Just don't expect me to solve it.) Anyways, I considered spots 8 and 1 to be adjacent and will have to work my wording. So, along with BBB...


Quote: beachbumbabs

Table #1=Craps
Table #2=superfun 21
Table #3=Spanish 21
Table #4=Big 6 wheel
Table #5= Sic bo
Table #6=UTH
Table #7=roulette
Table #8=3CP


Ding!! Ding!! We have two winners!!
Have you tried 22 tonight? I said 22.
Gialmere
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July 9th, 2019 at 7:25:57 PM permalink
Still messing around. This one is slightly more difficult...

At the WOV Spring Fling, GWAE, Pokergrinder, Drich, Axlewolf and Darkoz all agreed to a contest. Each of them would gamble for an hour with each on a different game: Craps, Blackjack, Roulette, Video Poker and Slots. The player accumulating the most money during the allotted time wins the contest. You're trying to listen to the end results on a live podcast but, unfortunately, the audio is terrible. You can hear that the final cash totals were $1000, $800, $600, $400 and $200 but the audio is mostly garbled. Can you determine who won what amount playing which game from what you could make out?

1) Neither Axelwolf nor Pokergrinder won the top amount, but one of them played slots.

2) Pokergrinder, Drich and the craps player are now going for drinks with the blackjack player who came in 2nd place.

3) The video poker player, along with the last place player both have names that start with the same letter.

4) GWAE did not play roulette, but the player who did didn't come in 3rd or last.

5) Axelwolf won $600.



The Order: Name/Amount Won/ Game Played
1st:
2nd:
3rd:
4th:
5th:



Hopefully this one works out.
Have you tried 22 tonight? I said 22.
Gialmere
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July 13th, 2019 at 11:47:09 AM permalink
White mates in three...





What's the next card?
Have you tried 22 tonight? I said 22.
rawtuff
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July 13th, 2019 at 12:05:40 PM permalink
Quote: Gialmere

White mates in three...



This is a well known mate: 1. Ng6++ Kg8; 2. Qf8+ Rxf8; 3. Ne7#


Quote: Gialmere

What's the next card?



5 of clubs...maybe
Tits are good, but the most important thing is the soul.
Gialmere
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July 13th, 2019 at 12:10:51 PM permalink
Quote: rawtuff

Quote: Gialmere

White mates in three...



This is a well known mate: 1. Ng6++ Kg8; 2. Qf8+ Rxf8; 3. Ne7#

Winner Winner Chicken Dinner! (yep, a classic)

Quote: Gialmere

What's the next card?



5 of clubs...maybe

Sorry no.

Have you tried 22 tonight? I said 22.
beachbumbabs
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July 13th, 2019 at 1:23:00 PM permalink

The Order: Name/Amount Won/ Game Played
1st: Drich/1000/Video Poker
2nd: GWAE/800/Blackjack
3rd: Axelwolf/600/Slots
4th: PokerGrinder/400/Roulette
5th: DarkOz/200/Craps



That was tricky, Gialmere. Good problem. I thought at first it was unsolvable, but I think I found the path.
If the House lost every hand, they wouldn't deal the game.
rdw4potus
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July 14th, 2019 at 10:04:50 AM permalink
Quote: Zcore13

You would get fired pretty quickly with only 2 BJ tables.


ZCore13



You're right, but I'm not sure why that is. Every time I'm in a casino in vegas, there are a dozen empty bj tables, one full uth table, one full pgp table, and people milling around watching those games and waiting to play.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
Gialmere
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July 14th, 2019 at 12:34:36 PM permalink
Quote: beachbumbabs


The Order: Name/Amount Won/ Game Played
1st: Drich/1000/Video Poker
2nd: GWAE/800/Blackjack
3rd: Axelwolf/600/Slots
4th: PokerGrinder/400/Roulette
5th: DarkOz/200/Craps



That was tricky, Gialmere. Good problem. I thought at first it was unsolvable, but I think I found the path.


Ding! Ding! Ding! You are correct!



Quote: Zcore13

You would get fired pretty quickly with only 2 BJ tables.

Quote: rdw4potus

You're right, but I'm not sure why that is. Every time I'm in a casino in vegas, there are a dozen empty bj tables, one full uth table, one full pgp table, and people milling around watching those games and waiting to play.


Now that you mention it, I was one of the people milling around the full PGP table while staring at all the empty blackjack tables.
Have you tried 22 tonight? I said 22.
Gialmere
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September 5th, 2019 at 5:34:42 PM permalink
Pai Gow Riddle

Since coming to the wizard sites I've been slowly learning the game of Pai Gow. I've finally reached a level where I've memorized all the tile names and ranks and can set a hand in the traditional way in 10 seconds or less. So it's time to learn the Wizard Way and complete my masters in what I call "casino game skills for recreational players".

While I found the WOO PG pages to be invaluable once reaching an intermediate level of understanding, I found even them to be intimidating at first. I ended up at PaiGow.com which is a nice site for players who need to be taken by the hand and slow walked through the game from the beginning. (It also has a good practice area that helps you memorize things although you have to pay to get its advanced startegy.)

Anyways, there's a fun facts section still under construction that presents the following riddle: Which of these two hands would you rather be dealt? Here are the hands...



Hand #1 is the obvious, knee-jerk choice but is it the correct one? I've pondered on this, moving the available remaining tiles around in my mind, but have decided to present it here cold. Are there any PG masters here that can elucidate? (I confess my brain is stuck in "uhhhhh" mode.) I should note that the site does not specify if you're banking, nor if you're playing to win or not to lose. It does, however say that all you assets are on the line. Also, since the section is under construction, the answer hasn't been posted yet.
Have you tried 22 tonight? I said 22.
charliepatrick
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September 5th, 2019 at 10:31:58 PM permalink
I was surprised when I worked it out but you would prefer to have hand #2! I suspect part of the logic is you have one of the tiles that could create a higher pair for the dealer.
Hand #1 - chances of winning 96.601% - chances of losing 0.015% - EV = .917560
Hand #2 - chances of winning 96.918% - chances of losing 0.181% - EV = .918916
Gialmere
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September 6th, 2019 at 5:16:11 PM permalink
Thank you CP. The math is certainly nonintuitive to say the least. I think you're right about the banker high hand getting choked off. It's still weird though.

For the first deal you're offered paired Yun and Gor. Only paired Gee Joon, Teen and Day can beat that and the bank would need two of those three to win. How could a hand with a mere (by this riddle's standards) Gong possibly improve things?

For the second offering you still have a paired Gor now with said Gong. The Gong, however, contains Teen and Yun tiles which eliminate two of the four pairs above the paired Gor leaving the bank only two pairs (Gee Joon and Day) to beat it. The Gong is obviously more vulnerable. The bank could tie it with the other Teen and a Chop Bot. A Wong could be made from either the remaining Teen tile or by using a Day Tile (although using a Day would mean only a paired Gee Joon would be left to beat the paired Gor). There are also, of course, a dozen pairs still available above the Gong.

But the math doesn't lie. Hand #2 is superior due to some freak mathematical wizardry for this special case. The riddle really does boggle the mind.
Have you tried 22 tonight? I said 22.
Gialmere
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October 23rd, 2019 at 7:39:53 AM permalink
Dice Riddle
A dealer slams a shaker containing two dice down on the table, peeks under the lid and tells you that one of the dice is showing a 1. You then place a $1 bet that snake eyes has been rolled. The dealer opens the shaker lid and reveals two 1s. Assuming true odds, how much money should you be paid?

Have you tried 22 tonight? I said 22.
charliepatrick
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October 23rd, 2019 at 8:52:31 AM permalink
This one, albeit in a different guise, was asked on another thread a few years ago.
Think one die is red and the other is green.
Essentially there are 11 ways out of 36 of throwing at least one "1", think 1-1 1-2 1-3 1-4 1-5 1-6 2-1 3-1 4-1 5-1 6-1. Thus true odds would be 10/1.
unJon
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October 23rd, 2019 at 1:48:50 PM permalink
Quote: Gialmere

Dice Riddle
A dealer slams a shaker containing two dice down on the table, peeks under the lid and tells you that one of the dice is showing a 1. You then place a $1 bet that snake eyes has been rolled. The dealer opens the shaker lid and reveals two 1s. Assuming true odds, how much money should you be paid?



Does “one of the dice is showing a 1” really mean “AT LEAST one of the dice is showing a 1”?

Also need to know if the dealer would always say the same sentence if at least one dice showed a 1. Otherwise no way to get to true odds, because you don’t have an the conditional probability to run a Bayesian.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
charliepatrick
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October 23rd, 2019 at 3:25:20 PM permalink
Quote: unJon

...Does “one of the dice is showing a 1” really mean “AT LEAST one of the dice is showing a 1”?...

I think that's a reasonable assumption - at least one of the dice has a "1".
Gialmere
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October 29th, 2019 at 9:18:39 AM permalink
Quick Logic (easy)
At a casino, Andrew is closely watching Susan while Susan is closely watching David. Andrew works for the casino but David does not. Is a casino employee closely watching a non casino employee?

Yes?
No?
Insufficient data?
Have you tried 22 tonight? I said 22.
unJon
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October 29th, 2019 at 9:47:09 AM permalink
Quote: Gialmere

Quick Logic (easy)
At a casino, Andrew is closely watching Susan while Susan is closely watching David. Andrew works for the casino but David does not. Is a casino employee closely watching a non casino employee?

Yes?
No?
Insufficient data?



Nice one.

Yes. If Susan is a casino employee, then her watching David makes the question a yes. If Susan is not a casino employee, then Andrew watching her makes it a yes.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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October 29th, 2019 at 9:53:54 AM permalink
Quote: Gialmere

Dice Riddle
A dealer slams a shaker containing two dice down on the table, peeks under the lid and tells you that one of the dice is showing a 1. You then place a $1 bet that snake eyes has been rolled. The dealer opens the shaker lid and reveals two 1s. Assuming true odds, how much money should you be paid?



If I may reword the question as "A dealer will slam a shaker containing two dice down on the table, peek under the lid until at least one of the dice is showing a 1. You then place a $1 bet that snake eyes has been rolled. The dealer opens the shaker lid and reveals two 1s. Assuming true odds, how much money should you be paid?

Given that wording my answer is

$10 + your original $1 wager


As was noted, the same problem was debated here for months already. It is still a sore subject with me.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ayecarumba
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October 29th, 2019 at 10:37:56 AM permalink
Quote: Wizard

Quote: Gialmere

Dice Riddle
A dealer slams a shaker containing two dice down on the table, peeks under the lid and tells you that one of the dice is showing a 1. You then place a $1 bet that snake eyes has been rolled. The dealer opens the shaker lid and reveals two 1s. Assuming true odds, how much money should you be paid?



If I may reword the question as "A dealer will slam a shaker containing two dice down on the table, peek under the lid until at least one of the dice is showing a 1. You then place a $1 bet that snake eyes has been rolled. The dealer opens the shaker lid and reveals two 1s. Assuming true odds, how much money should you be paid?

Given that wording my answer is

$10 + your original $1 wager


As was noted, the same problem was debated here for months already. It is still a sore subject with me.



Hehe.... Happy Halloween Wizard! The undead are coming back...
Simplicity is the ultimate sophistication - Leonardo da Vinci
Gialmere
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October 30th, 2019 at 9:40:04 AM permalink
Card Conundrum
While sitting at a card table you agree to an unusual wager. First, you are blindfolded so you cannot see. Next, eleven playing cards are placed in a pile before you. Finally, you are (correctly) told that seven of the cards scattered in the pile are faceup while the remaining four are facedown.

Using all eleven cards while blindfolded, the challenge is to make two groups of cards and each group must contain an equal number of facedown cards. Can you win this bet without literally resorting to blind luck?

Yes (How?)
No (Why?)
Insufficient Data
Have you tried 22 tonight? I said 22.
charliepatrick
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October 30th, 2019 at 9:58:12 AM permalink
Let's call the cards XXXXXXX YYYY, you need and equal number of Y's in both piles.

(i) Divide the cards into two piles, one with 7 and the other with 4.
This can result in N=0-4 Y's in the 4-pile (and 4-N X's in the 4-pile). Thus the remainder, 4-N Y's, are in the 7-pile.

(ii) Turn over all the cards in the 4-pile.
All Y's turn into X's, and all X's turn into Y's.
So now there are 4-N Y-cards in the 4-pile.

So both piles have equal number of Y's.
beachbumbabs
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October 30th, 2019 at 10:06:48 AM permalink
Quote: charliepatrick

Let's call the cards XXXXXXX YYYY, you need and equal number of Y's in both piles.

(i) Divide the cards into two piles, one with 7 and the other with 4.
This can result in N=0-4 Y's in the 4-pile (and 4-N X's in the 4-pile). Thus the remainder, 4-N Y's, are in the 7-pile.

(ii) Turn over all the cards in the 4-pile.
All Y's turn into X's, and all X's turn into Y's.
So now there are 4-N Y-cards in the 4-pile.

So both piles have equal number of Y's.



That's really well figured, CharliePatrick! Impressive.
If the House lost every hand, they wouldn't deal the game.
Gialmere
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October 31st, 2019 at 5:17:24 PM permalink
Note that if you turn over the 7-card stack you end up with an equal number of faceup cards. Also note that this will work with any number of cards in any up/down combination. Knowing this, the trick can be weaponized into an entertaining bar bet. Using a standard 52 card deck, have the mark cut the cards, turn one stack up and shuffle the stacks together. If you're told that there are 20 facedown cards, simply count the top 20 cards off the deck, turn them over and spread both stacks. You'll win every time.

------------------------

New Game (?)
At G2E you talk to some designer friends at their booth and they show you a new 2-player game they are developing. Nine cards numbered 1-9 are placed faceup on the table. Each card is worth its face value. To begin, the first player takes any card on the table. The second player then takes any one of the remaining cards. Play alternates back and forth with each player taking a remaining card. The winner is the first player to collect a combination of three cards whose sum is fifteen (such as 9-4-2).

You think about it and realize the game is flawed. Assuming both players utilize optimal strategy...

a) The player going first will win.
b) The player going second will win.
c) Neither player will win.


Bonus Question
The designers ask your opinion on what to name the game. What should you tell them?
Have you tried 22 tonight? I said 22.
charliepatrick
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October 31st, 2019 at 11:45:49 PM permalink
Quote: Gialmere

You think about it and realize the game is flawed. Assuming both players utilize optimal strategy...
a) The player going first will win.
b) The player going second will win.
c) Neither player will win.

There are only eight ways to have three numbers 1-9 add up to 15. Thus the objective is to prevent your opponent making any of these combinations.

Consider the following where you arrange the cards into a 3x3 perfect square, such as this...
276
951
438

In essence you have to prevent your opponent from making any of the eight possible lines. If you can do this then the game will be a draw. So just use standard Noughts and Crosses strategy to pick your card and this will force a draw.
Gialmere
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November 1st, 2019 at 5:58:31 AM permalink
Correct. Well done again CP!

Although it's possible for either player to win, assuming optimal strategy, neither of them ever will. Essentially the designers have invented Tic Tac Toe.
Have you tried 22 tonight? I said 22.
Gialmere
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November 1st, 2019 at 5:49:40 PM permalink
Amazon Abstruseness
While shooting dice in a craps pit a software engineer standing next to you says that it reminds him of a test question he got while interviewing for a job at Amazon. You're given a pair of dice. One of them has the standard numbers of 1-6 on its faces. The other one is blank. Can you fill in each blank face with a number from 0-6 so that when the dice are rolled together they generate a number from 1-12 with equal probability?
Have you tried 22 tonight? I said 22.
beachbumbabs
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November 1st, 2019 at 10:10:51 PM permalink
Quote: Gialmere

Amazon Abstruseness
While shooting dice in a craps pit a software engineer standing next to you says that it reminds him of a test question he got while interviewing for a job at Amazon. You're given a pair of dice. One of them has the standard numbers of 1-6 on its faces. The other one is blank. Can you fill in each blank face with a number from 0-6 so that when the dice are rolled together they generate a number from 1-12 with equal probability?




The blank die would have to have 3 faces of 0 and 3 faces of 6.

You would then have 3 ways to make each number. A blank face + any number = 1-6, and a 6 + any number = 7-12. Each total has an equal probability.
If the House lost every hand, they wouldn't deal the game.
Gialmere
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November 2nd, 2019 at 4:05:36 PM permalink
Quote: beachbumbabs

The blank die would have to have 3 faces of 0 and 3 faces of 6.

You would then have 3 ways to make each number. A blank face + any number = 1-6, and a 6 + any number = 7-12. Each total has an equal probability.


Winner Winner Chicken Dinner!

An easy way to think of it is that you have 12 numbers and 36 possible outcomes. So each number must have 3 chances to come up. Because the only way to roll a 1 is 1+0, the blank die must have three 0s. To roll midnight, the blank die must also have three 6s. Somewhat surprisingly, doing this makes all the other numbers kosher as well.

This really was an Amazon interview question and it generated some minor buzz around ten years ago. As often happens here at WOV, the debate didn't center on how to solve the riddle but rather on how to correctly word the question. For example, here is one rather vague way I've seen of wording it...

Given a normal dice and a dice with blank faces, fill in the blank dice with numbers from 0-6 so that the probability of each number coming up, when you roll the two dice together is equal.


------------------------------

NFL Nonsense
At a casino sportsbook you examine some futures bets on which NFL teams will make it into the playoffs. After thinking a while, you ask yourself two questions...

1) What is the minimum number of wins an NFL team needs to reach the playoffs?
2) What is the maximum number of losses an NFL team can suffer and still make the playoffs?

What does your brain answer back?
Last edited by: Gialmere on Nov 2, 2019
Have you tried 22 tonight? I said 22.
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