jeffwarren75
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June 18th, 2012 at 6:36:04 AM permalink
just a comment - i love the site and am always on here- went back and was reading old ask the wizard articles and noticed this year has the least activity of any year since the site started, I loved that section and hope you bring it back more frequently
DJTeddyBear
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June 18th, 2012 at 6:54:45 AM permalink
Have you noticed that, lately, nearly all the questions have originated here?

Maybe Mike has decided to end the redundancy....
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? 😁
jeffwarren75
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June 18th, 2012 at 6:59:51 AM permalink
its a shame, i loved that
Wizard
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June 18th, 2012 at 8:20:12 AM permalink
I would love to write another column, but I have a hard time getting worthy questions that haven't been asked before. I'm just one question short of having enough material for the next one.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
teliot
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June 18th, 2012 at 9:59:54 AM permalink
Quote: Wizard

I would love to write another column, but I have a hard time getting worthy questions that haven't been asked before. I'm just one question short of having enough material for the next one.

I thought of a question while out on my walk this morning -- based on a progression thread. This one should be attractive to the markov-chain-o-philes.

Assume a game played with a fair coin. On the first toss, the player wagers 1 unit. Going forward, if the coin is tails, the players next wager is twice his previous wager. If the coin is heads, the players wager is either 1/2 his previous wager, if his previous wager was > 1, or if his previous wager was 1, his wager remains 1. That is:

w(1) = 1.
For n > 1:
w(n) = 2*w(n-1), if coin is tails.
w(n) = w(n-1), if coin is heads and w(n-1) = 1.
w(n) = (1/2)*w(n-1), if coin is heads and w(n-1) > 1.

Question, what is the expected value of w(1000)? Give all digits for the integer part.

I don't know the answer. Just thought of the question.
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teliot
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June 18th, 2012 at 10:08:35 AM permalink
Quote: Wizard

I would love to write another column, but I have a hard time getting worthy questions that haven't been asked before. I'm just one question short of having enough material for the next one.

I thought of a question while out on my walk this morning -- based on a betting progression thread. This one should be attractive to the markov-chain-o-philes.

Assume a game played with a fair coin. On the first toss, the player wagers 1 unit. Going forward, if the coin is tails, the players next wager is twice his previous wager. If the coin is heads, the players wager is either 1/2 his previous wager, if his previous wager was > 1, or if his previous wager was 1, his wager remains 1. That is:

w(1) = 1.
For n > 1:
w(n) = 2*w(n-1), if coin is tails.
w(n) = w(n-1), if coin is heads and w(n-1) = 1.
w(n) = (1/2)*w(n-1), if coin is heads and w(n-1) > 1.

Question, what is the expected value of w(1000)? Give all digits for the integer part.

I don't know the answer right off.
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DJTeddyBear
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June 18th, 2012 at 11:07:31 AM permalink
Are you trolling for questions? Here's one that was asked but never answered. Perhaps because I didn't ask at the right time....


In October 2010, you, Miplet and I were playing face-down BlackJack at Terribles. There were times when I wasn't sure what to do, so rather than whip out my trusty Wizard's Strategy Card, I just showed you the cards and asked. As I recall, there was at least one time where you and Miplet showed cards as well.

After a while, I asked, "Aren't the cards dealt face down to prevent other players, some of whom might be counting, from getting extra information prior to making their own decision? As such, why aren't they stopping us from showing each other our cards?"

Your response was, "That's a good question for 'Ask The Wizard'." So now I'm formally asking.

I suppose the answer at the moment was that since we were playing low stakes, or that we were losing, that they didn't care. But do they allow such sharing at higher stakes?

---

By the way, I later regretted talking about counting while we were plaing. It hadn't occurred to me that you might actually BE counting and wanted to keep a low profile.
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? 😁
mustangsally
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June 18th, 2012 at 11:29:46 AM permalink
removed
silly
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Wizard
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June 18th, 2012 at 11:57:11 AM permalink
Thanks for the ideas.

I normally stay away from betting system questions. However, I'm open to making an exception if there is some interesting mathematical property at play. If there is a formulaic solution to the system you mentioned I would be happy to consider it. However, if the answer is found by simulation or brute force recursion only, it isn't going to interest me very much.

Thanks for the correction on my terminology on the point rolls question.

I proudly have a copy of Stuart's book. I will go through it for questions, but tend to think that most too academic to interest the average reader. As much as I enjoy math, I'm trying to help players lose less money in the casinos, so prefer more down to earth questions. If there are any you wish to specifically bring to my attention, please just quote me a page and problem number.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
guido111
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June 18th, 2012 at 12:12:32 PM permalink
Quote: mustangsally


That opens up the question.
What is the expected number of point rolls for a point decision?
It does not have to include the come out roll that establishes the point.
This is a good math question and I have never seen this asked or even answered.
I simulated it then solved it.

Sally

This has been answered here at WoV.
can't seem to find it at the moment.
I think miplet solved it.

Here is the formula for cumulative probability distribution from my notes.
n=roll number
6/24*(1-(27/36)^n)+8/24*(1-(26/36)^n)+10/24*(1-(25/36)^n)
you should get a table like this
rolls	probability
1 0.282407407
2 0.484567901
3 0.629418867
4 0.733304517
5 0.80788109
6 0.861468782
7 0.900011824
8 0.92776081
9 0.947758041
10 0.962183009
11 0.972598561
12 0.980126424
13 0.985572479
14 0.989516248
15 0.992374878
16 0.994448924
17 0.995955148
18 0.997050031
19 0.997846645
20 0.998426776
mustangsally
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June 18th, 2012 at 12:26:14 PM permalink
removed
silly
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Pokeraddict
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June 18th, 2012 at 2:03:58 PM permalink
I have one that may fit. The UTH machine at Vegas Club is not set up like a legit table game. Two or more players can receive the same card. Are there any different strategies that come into play when another player has the same card as you, theoretically creating a 53 or 54 card deck? Does it change the payback? It seems to me that there is an advantage to the player here as it makes you less likely to have your pair cards in other hands if your hole cards can duplicate.

Also, Sec. 14.042(b) of the NGR states this:

Quote:

(b) For gaming devices that are representative of live gambling games, the mathematical
probability of a symbol or other element appearing in a game outcome must be equal to the
mathematical probability of that symbol or element occurring in the live gambling game. For other
gaming devices, the mathematical probability of a symbol appearing in a position in any game
outcome must be constant.



Since it is impossible for me to have the same card as my neighbor in a live UTH game, how can it be legal for it to happen on a machine? The machine is licensed and approved and this issue is known.

At first I was irritated by this issue, but I suspect it is actually player friendly since it makes it less likely for another player to have your pair/set/quad cards. It makes me wonder why they would set it up this way. These machines were all over town and disappeared on the same day last year. When they came back, they had this "feature".

I am really just curious about why it is this way. Maybe I should have made this its own thread.
DJTeddyBear
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June 18th, 2012 at 2:38:36 PM permalink
Regarding the UTH machine, I'll venture a guess...

In the real table game, you're not supposed to show other players your cards because you can then figure out if the dealer has less chances to beat you.

My guess is that the machine internally reserves the dealer's 2 cards, as well as the 5 card board. Then it selects 2 cards for each player, from their own deck of the 45 remaining cards.

As a result, you can't tell if the dealer has cards that could beat you, but you can tell if you're less likely to hit your draw.

Bottom line: Assuming you're not mistaken, I find it hard to believe that this game got approval.
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? 😁
Wizard
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June 18th, 2012 at 3:09:57 PM permalink
Quote: Pokeraddict

Two or more players can receive the same card. Are there any different strategies that come into play when another player has the same card as you, theoretically creating a 53 or 54 card deck? ...



Good question. I think the way the game works is the community cards are dealt first from a dummy deck. Then each player, and the dealer, is given his own deck with those five cards removed. By sharing information one could narrow down the remaining cards by eliminating any card seen in any player hand.

Without collussion, the math would be the same as against a human dealer with a single deck of cards. With collusion, it is possible with and without this separate decks rule. It would seem less effective with separate decks, because you'll be able to eliminate fewer cards due to some of them appearing more than once.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ayecarumba
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June 18th, 2012 at 3:51:48 PM permalink
Has the existence of this feature been independently confirmed? If the machine is operating as described, with the dealer playing a single hand against opponents who may or may not have the same cards, I think it is in violation of the NGC regulation that the machine operate with the same probabilities as the live game.
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Pokeraddict
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June 18th, 2012 at 4:07:05 PM permalink
I was playing it last night at LVC and three of us got the 7 of hearts. Here is a screenshot of one I took at Binions a year ago when they had the machine. I talked to a slot supervisor and tech that both stated NGCB had approved the new version. All of these machines were removed from the floors weeks before and had just returned with this new "feature". I called gaming and could not get a hold of anyone that understood my complaint. Figuring it did not really affect the outcome of the game, I gave up. I believe the game dealing this way violates the gaming law I quoted in my other post. OTOH, now that there is only one left in the city that I know about, I don't want to kill it either.

Here is proof:



Note that seat 3 and 5 have the 10d. It has happened probably 100+ times in my play and have seen three players with the same card a handful of times. This was at Binions where there is no longer this machine. The Vegas Club machine will deal the same card to multiple players just as this did.
Pokeraddict
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June 18th, 2012 at 4:10:16 PM permalink
Quote: Wizard

Good question. I think the way the game works is the community cards are dealt first from a dummy deck. Then each player, and the dealer, is given his own deck with those five cards removed. By sharing information one could narrow down the remaining cards by eliminating any card seen in any player hand.

Without collussion, the math would be the same as against a human dealer with a single deck of cards. With collusion, it is possible with and without this separate decks rule. It would seem less effective with separate decks, because you'll be able to eliminate fewer cards due to some of them appearing more than once.



Although at first players told me they had been dealt the same card as the dealer, I have never seen this. The dealer is getting cards that cannot duplicate among the player AFAIK. If it were any other way I would have pursued this already as that would considerably change the outcome of the game.

This machine operates like the Pennsylvania machines would have to operate (unless the law changed, I do not follow that state closely). If they have these machines in Pennsylvania, this machine would have a setting or software setup that would make this possible. I have wondered if the Shufflemaster video BJ games do this too. I cannot think of any way to test them. I don't think the 3CP, Mississippi Stud, or Paigow versions of these games exist in the city anymore. If they are, I would be interested in knowing where so I could check them out. I actually enjoyed playing the video Paigow over the table game.

Edit: At Binions, part of the instructions stated the game played with one 52 card deck. The next time I went there after complaining that was removed. LVC's machine has no such disclaimer.
Nareed
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June 18th, 2012 at 4:50:12 PM permalink
I played the UTH machine at Vegas Club, twice, last May. I played alone both times. I did not notice getting the same cards as the dealer, but then I did not pay much attention to the dealers' cards anyway. Just enough to see if it beat me.

Have you tried contacting the manufacturer? perhaps they can explain how it works, and whether the house edge, hit frequencies, etc differ from the live game.
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Wizard
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June 18th, 2012 at 5:03:26 PM permalink
As much as I like to keep the casinos complaint on Gaming regulations, I don't think this is worth making a stink about, and I don't think it is obviously incompliant with the regulation quoted above.

If a players does not concern himself with the cards of other players then his odds are the same as if the game were dealt by a live dealer, assuming the player and dealer cannot get the same cards.

In other words, if the player is minding his own business, the odds are the same as in a live game. Only colluders should care about the difference, and players aren't supposed to collude in the first place.

Then again, it does not seem unreasonable that the rules should at least footnote that each player is given a separate deck, with the predestined community cards removed.

If Gaming is having a hard time understanding the issue, take a picture of the game with the same card appearing more than once at the same time, and mail it to them with a letter of explanation. That should at least make it clear what the issue is.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
RogerKint
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June 19th, 2012 at 3:45:56 PM permalink
Quote: Pokeraddict


I have wondered if the Shufflemaster video BJ games do this too.



The instructions on the Shufflemaster video BJ game at Binions indicate each player is dealt from their own 6 deck shoe, shuffled every hand.
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teddys
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June 19th, 2012 at 4:09:13 PM permalink
Question, regarding the Shufflemaster games:

For the ones that use a six deck shoe, shuffled after 2/3rds of the deck, I've heard it said that you can "decay" the count and essentially count the game.

Is this possible and can you speak about it?
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Wizard
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June 22nd, 2012 at 11:51:53 AM permalink
Quote: teddys

Question, regarding the Shufflemaster games:

For the ones that use a six deck shoe, shuffled after 2/3rds of the deck, I've heard it said that you can "decay" the count and essentially count the game.

Is this possible and can you speak about it?



Any game that does not always shuffle between hands is, in theory, countable. Most Shufflemaster games are poker variants, and thus shuffle between hands. For the rest it is just a question of how effective it is. For example, baccarat is in theory countable, but without a computer you could play all day waiting for one positive EV bet.

Anyway, per the original post, I have a draft of my first Ask the Wizard column since January. Before announcing it on my Odds site, I welcome all comments and corrections.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
JB
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June 22nd, 2012 at 12:11:30 PM permalink
Hmm. In the answer to the question I offered, I disagree about Joker Poker having a higher chance of a royal. Running the 17/7 paytable through the calculator indicates that perfect strategy yields a natural royal once every 40,466.58 hands.

Also, while it's probably not around any more, 9/6 Royal Aces offers a prize of 800 times the bet every 3,729.24 hands on average.
Wizard
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June 22nd, 2012 at 12:35:02 PM permalink
Quote: JB

Hmm. In the answer to the question I offered, I disagree about Joker Poker having a higher chance of a royal. Running the 17/7 paytable through the calculator indicates that perfect strategy yields a natural royal once every 40,466.58 hands.



I just double check it and stand by my figure. Perhaps you put in a different pay table. I put in the 98.44% Kings pay table.

Quote:

Also, while it's probably not around any more, 9/6 Royal Aces offers a prize of 800 times the bet every 3,729.24 hands on average.



Good point. I just added mention of that game.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
JB
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June 22nd, 2012 at 12:57:08 PM permalink
Quote: Wizard

I just double check it and stand by my figure. Perhaps you put in a different pay table. I put in the 98.44% Kings pay table.


Yes, that's where the difference is. I was using a prize of 800 for the royal. Using 940 does yield the figure you quoted.
AcesAndEights
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June 22nd, 2012 at 5:22:19 PM permalink
Quote: Wizard

Anyway, per the original post, I have a draft of my first Ask the Wizard column since January. Before announcing it on my Odds site, I welcome all comments and corrections.


Slight typo on the last question,
Quote:

What we see is that by the third generation the proportion of the population with red hair will converge to 3.85%.


The table shows 3.89% (and technically 3.90% after the 3rd generation if you're rounding to 2 decimal points).
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JJ
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June 25th, 2012 at 10:20:21 AM permalink
At our local bingo (Centennial Bingo) there is a jackpot called "Strive for Five". A 3-up book costs $5 and basically the goal is to get a standard start (any line, 4 outside corners, or a corner postage stamp) in 5 numbers or less during any of the 8 regular games. There is no limit to the amount of books you can play although you must have at least a regular 6-up door book. You can use the free for a line.

It is important to note that if a bingo is called on the 4th number, all bingos on the 5th number become void. It goes up $50 per regular game and starts at $500. If there is more than one winner on the eligible books, the pot is shared.

What is the chance of hitting it (in 4 and 5 numbers)?

Is there a break even point?
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