Steen
Joined: Apr 7, 2014
• Posts: 126
May 6th, 2014 at 6:45:33 PM permalink
Quote: Ahigh

Steen, I'm going to humor you for a second. If I made a SPECIAL FELT. And I put a little square on it. And inside the square it said "FOUR OR TEN." You were able to put green chips in there and it paid \$49 for \$50. It requires denominations of \$50, and the max bet is \$100. It wins on a four or a ten, and it loses on a seven.

Tell me, this new bet, what would be the edge per bet resolved?

How is it different from having to put one green chip on the four and the other on the ten?

It's possible you don't know how to do your own math at all for new bets that don't exist already.

I create games for a living. I actually think about bets that don't exist already.

Maybe that's what slowing you down?

Ahigh, this is an old argument that I've been through before.

First off, I think you meant to say it pays \$49 TO \$50 (not FOR).

Second, if a casino were to create a special bet called "4 or 10" and it paid 49 to 50 on 4 or 10 and lost on 7 then it would have an EV of -1%.

However, unfortunately for you, they don't offer such a bet. In order to construct such a bet, you have to utilize bets that are defined in different terms and more importantly, you have to pretend that the lack of action on one bet has an affect on the other bet.

Steen
Ahigh
Joined: May 19, 2010
• Posts: 5168
May 6th, 2014 at 6:45:57 PM permalink
Quote: mustangsally

He wants the HE calculation to ONLY be -.5 / \$50 = 0.01 * 100 = 1%
(I see that this can make for easy EV calcs over any N rolls.)

You're wrong.

Quote: mustangsally

He does not want to see this:
-\$0.50/37.50 = -1.33%
as he does not believe at all in house edge per bet resolved

Sally

\$1 / \$100 = 1.00% Sally.

\$50 risked. \$49 won, \$99 returned exactly one half the time.

\$1 / \$100 = 1.00% Sally.

It is absolutely incredible that you don't get this.
http://dumbass.website
Ahigh
Joined: May 19, 2010
• Posts: 5168
May 6th, 2014 at 6:46:48 PM permalink
Quote: Steen

However, unfortunately for you, they don't offer such a bet. In order to construct such a bet, you have to utilize bets that are defined in different terms and more importantly, you have to pretend that the lack of action on one bet has an affect on the other bet.

No I don't. If you argue I have two bets, I only have to resolve one bet, not two. So you're wrong.

You can argue about this all day long and you'll still be wrong just like Sally and just like Math Extremist.

It's humorous though.
http://dumbass.website
Ahigh
Joined: May 19, 2010
• Posts: 5168
May 6th, 2014 at 6:53:54 PM permalink
Quote: Steen

However, unfortunately for you, they don't offer such a bet. In order to construct such a bet, you have to utilize bets that are defined in different terms and more importantly, you have to pretend that the lack of action on one bet has an affect on the other bet.

If such a bet is not "offered," describe the difference between the bet that I describe and putting one chip on the four and one chip on the ten and taking both down any time I take one chip down.

Can you describe to me what, exactly, is the difference between the bet that you acknowledge is a 1.00% HE and putting one chip on the four and one chip on the ten?

Can you?

The location of chips? I will give you that.

The fact that you have to verbally ask for both chips to come down?

Okay!

Anything else?!?
http://dumbass.website
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 6th, 2014 at 6:54:20 PM permalink
Quote: Ahigh

You're wrong.

I am always wrong because everything I post is just an opinion
just like every thing you post is just an opinion

House edge when considering all \$50 at risk (both bets)=
-.5 / \$50 = 0.01
* 100 = 1%

yes, my 1% is not equal to your 1% when considering \$50 at risk.
as we only gave our opinions

what did you say again was the expected value of your bet per roll?

hehe
Sally
I Heart Vi Hart
Ahigh
Joined: May 19, 2010
• Posts: 5168
May 6th, 2014 at 6:58:46 PM permalink
Quote: mustangsally

I am always wrong because everything I post is just an opinion
just like every thing you post is just an opinion

House edge when considering all \$50 at risk (both bets)=
-.5 / \$50 = 0.01
* 100 = 1%

yes, my 1% is not equal to your 1% when considering \$50 at risk.
as we only gave our opinions

what did you say again was the expected value of your bet per roll?

hehe
Sally

Sally, I will create a video and I will post it on you tube.

I will mention your name and ME's name and Steel's name as being the intended recipients of this video and I will explain very clearly what you guys are failing to understand.

Maybe that will help.

I will use actual chips on an actual craps felt and I will go REALLY SLOWLY so it's super clear.
http://dumbass.website
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5369
May 6th, 2014 at 7:01:46 PM permalink
Quote: mustangsally

Quote: ThatDonGuy

Pardon me for asking, but how do you calculate the mean number of rolls?

seems easy to find. I even linked to it in Don Catlin's article

No - that's what I was asking. I didn't notice it at the end of Catlin's article. I wasn't expecting a method to count rolls to be headed "House Edge."
Steen
Joined: Apr 7, 2014
• Posts: 126
May 6th, 2014 at 7:02:06 PM permalink
Quote: Ahigh

Does this mean that \$5 on the pass and \$5 on the don't pass is not a bet because I'm willing to give the casino \$5 if I roll a 12, but I have no way to collect from them? I have never heard of this not being a bet. Only that I have no way to win the bet, only a way to lose.

I am completely unaware of how a bet ceases to become a bet when one party has no opportunity to win.

Where is your source for this information?

There is more than one definition of a bet. One definition is, "an act of risking a sum of money on the outcome of a future event." I think my bet qualifies. If you think otherwise, let's just consider that your opinion.

Your responses are becoming quite animated. Do you feel it helps your argument?

Gee, a \$40 Lay 4 and simultaneous \$20 Buy 4 bet would also never win - oh my!

Oh but wait, it's not "A" bet, it's TWO bets. Each bet in it's own right offers each party a chance to win. The fact that you can simultaneously make bets which oppose one another doesn't change this fact. Do you think for one moment that any gambling commission anywhere would approve a casino to offer bets wherein players have no opportunity to win?

Steen
Steen
Joined: Apr 7, 2014
• Posts: 126
May 6th, 2014 at 7:09:30 PM permalink
Quote: Ahigh

No I don't. If you argue I have two bets, I only have to resolve one bet, not two. So you're wrong.

You can argue about this all day long and you'll still be wrong just like Sally and just like Math Extremist.

It's humorous though.

Yes, you only have to resolve one but you had to make TWO.

Steen
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 6th, 2014 at 7:28:27 PM permalink
Quote: Ahigh

Sally, I will create a video and I will post it on you tube.

YES!