Steen
Joined: Apr 7, 2014
• Posts: 126
May 6th, 2014 at 5:51:08 PM permalink
Quote: endermike

I will preface my thoughts with an admission, I looked at the names on the posts and immediately assumed Ahigh was wrong. I started looking to make my own counterargument and found...he has a point. Specifically regarding the question of a 1% HE bet via the 4 and 10.

Quite simply, his "hybrid buy" has 1% HE upon resolution, .33% HE per roll, and average of 3 rolls per resolution (assuming always on and vig only on win). First, on a per roll basis:

<snip tables>

That his "hybrid buy" bet takes 3 rolls for which it is on to resolve is trivial (geometric RV with p=1/3).

I think this is best explained by not saying he is averaging 1.33 to 1.33 and getting 1.00. Instead he is averaging 1.33 with .67 and getting 1.00. This is better because he is averaging the HE of two buy bets (resolving) half the time (when he loses) and the HE of one buy bet (resolving) half the time (when he wins).

Allow me to tailor an excerpt from WinCraps' help file:

Craps bet are not synergistic. The erroneous conclusion of Ahigh's argument comes from not expressing the EV as a function of the action. When one of his Buy 4 or Buy 10 bets wins, the amount won is solely a function of the amount that was wagered on the winning bet and not on the other bet. For instance, if a 10 hits then the \$49 win is a function of the \$25 that was wagered on Buy 10 and not the \$50 that was wagered on both. Yet the argument treats the full amount wagered on both bets as if it had a bearing on each winning outcome. This is wrong.

Intuitively, since both Buy bets have EV's of -1.33%, we know that a combined wager should also have an EV of 1.33%. To arrive at this (per decision) figure we simply need to divide our average loss by our average action.

The average loss is figured as follows:

6/12 chance to win \$49 = \$24.50
6/12 chance to lose \$50 = -\$25.00
----------------------------------------
Average loss per decision = -\$0.50

The average action is figured as follows:

3/12 chance of rolling a 4 and producing \$25 worth of action
3/12 chance of rolling a 10 and producing \$25 worth of action
6/12 chance of rolling a 7 and producing \$50 worth of action
-----------------------------------------------------------------------
Average action per decision = \$37.50

Hence, the EV per dollar of action = -\$0.50/37.50 = -1.33%.

By the way, MathExtremeist is correct that the generic Buy 4 or 10 bet EV is -1.67%. This is figured on the correct vig of 5%. As we know, the casinos often round down fractional vigs, so rather than collect the correct \$1.25 vig for a \$25 Buy 4 bet, some casinos bend a little and only collect \$1 vig thus reducing the house advantage (in this case) to -1.33%

Steen
Ahigh
Joined: May 19, 2010
• Posts: 5197
May 6th, 2014 at 6:10:10 PM permalink
Quote: Steen

When one of his Buy 4 or Buy 10 bets wins, the amount won is solely a function of the amount that was wagered on the winning bet and not on the other bet.

This is impossible if you consider that I have only one bet. You fail to understand that you can have a bet that's not on the felt.

END.
aahigh.com
Ahigh
Joined: May 19, 2010
• Posts: 5197
May 6th, 2014 at 6:17:00 PM permalink
Here's another bet I give dealers. I did this on Sunday while at the Rio.

Two hard fours rolled, and I told the stick man, "STICK MAN IF YOU CALL ANOTHER HARD FOUR, \$100 DEALER HAND-IN -- THIS ROLL ONLY."

That's a bet too! And you better bet your ass I would hand in the \$100 if the stick man called the hard four for the third time in a row. I would have offered \$1,000 for the next one too.

If you want to say that's not a bet because it's not on the felt, I would wager even more money that you have NEVER received a bet from me like this (that you can call something specific on the next roll).

I have offered MUCH more. My thinking is that it's cheaper to do this than to let the casino take an edge off a wager for the dealers. And much like betting the four and the ten for \$25 each, if I want a 50/50 chance to win \$49 for \$50 risked, I am going to do it with a 1.00% house edge ..

BECAUSE I KNOW THINGS THAT OTHER PEOPLE WHO CONSIDER THEMSELVES VERY SMART DO NOT KNOW.

Like you can give a dealer a bet that's not marked on the felt by acting as the banker for the bet, and have a ZERO EDGE BET OF ANY F\$&%^G KIND YOU WANT for the dealer.

I find it hilarious that people don't understand bets that aren't on the felt.
aahigh.com
Steen
Joined: Apr 7, 2014
• Posts: 126
May 6th, 2014 at 6:21:22 PM permalink
Quote: Ahigh

This is impossible if you consider that I have only one bet. You fail to understand that you can have a bet that's not on the felt.

Really? Wow! You know, I had the same thing once. I had \$20 in my pocket when a hard 8 rolled. I asked the dealer to pay me but he refused! Go figure.

It doesn't matter if you call it one bet or ten bets, the fact remains that anytime you win, the casino figures your amount won solely as a function of the amount that was wagered on the winning bet and not on ANY other bets.

Steen
Ahigh
Joined: May 19, 2010
• Posts: 5197
May 6th, 2014 at 6:23:32 PM permalink
Quote: Steen

Really? Wow! You know, I had the same thing once. I had \$20 in my pocket when a hard 8 rolled. I asked the dealer to pay me but he refused! Go figure.

It doesn't matter if you call it one bet or ten bets, the fact remains that anytime you win, the casino figures your amount won solely as a function of the amount that was wagered on the winning bet and not on ANY other bets.

Steen

You fail to understand. I hear where your coming from. It's called "I don't understand" land.
aahigh.com
Ahigh
Joined: May 19, 2010
• Posts: 5197
May 6th, 2014 at 6:25:28 PM permalink
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?
aahigh.com
Steen
Joined: Apr 7, 2014
• Posts: 126
May 6th, 2014 at 6:32:56 PM permalink
Quote: Ahigh

Here's another bet I give dealers. I did this on Sunday while at the Rio.

Two hard fours rolled, and I told the stick man, "STICK MAN IF YOU CALL ANOTHER HARD FOUR, \$100 DEALER HAND-IN -- THIS ROLL ONLY."

That's a bet too!

I've no doubt that the dealers appreciate such generosity but to call it a bet is a fundamental misunderstanding of what a bet is.

Now if you had said, "If you call another Hard 4 then I'll give the dealers \$100, and if not then dealers will give me \$20." AND if the dealers would accept your offer THEN you would have a bet. Each party to the wager is risking something to win something.

Steen
Ahigh
Joined: May 19, 2010
• Posts: 5197
May 6th, 2014 at 6:36:12 PM permalink
Quote: Steen

Quote: Ahigh

Here's another bet I give dealers. I did this on Sunday while at the Rio.

Two hard fours rolled, and I told the stick man, "STICK MAN IF YOU CALL ANOTHER HARD FOUR, \$100 DEALER HAND-IN -- THIS ROLL ONLY."

That's a bet too!

I've no doubt that the dealers appreciate such generosity but to call it a bet is a fundamental misunderstanding of what a bet is.

Now if you had said, "If you call another Hard 4 then I'll give the dealers \$100, and if not then dealers will give me \$20." AND if the dealers would accept your offer THEN you would have a bet. Each party to the wager is risking something to win something.

Steen

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.

http://lmgtfy.com/?q=definition+of+bet
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mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 6th, 2014 at 6:40:34 PM permalink
Quote: Steen

Allow me to tailor an excerpt from WinCraps' help file:

Craps bet are not synergistic. The erroneous conclusion of Ahigh's argument comes from not expressing the EV as a function of the action. When one of his Buy 4 or Buy 10 bets wins, the amount won is solely a function of the amount that was wagered on the winning bet and not on the other bet. For instance, if a 10 hits then the \$49 win is a function of the \$25 that was wagered on Buy 10 and not the \$50 that was wagered on both. Yet the argument treats the full amount wagered on both bets as if it had a bearing on each winning outcome. This is wrong.

Intuitively, since both Buy bets have EV's of -1.33%, we know that a combined wager should also have an EV of 1.33%. To arrive at this (per decision) figure we simply need to divide our average loss by our average action.

The average loss is figured as follows:

6/12 chance to win \$49 = \$24.50
6/12 chance to lose \$50 = -\$25.00
----------------------------------------
Average loss per decision = -\$0.50

The average action is figured as follows:

3/12 chance of rolling a 4 and producing \$25 worth of action
3/12 chance of rolling a 10 and producing \$25 worth of action
6/12 chance of rolling a 7 and producing \$50 worth of action
-----------------------------------------------------------------------
Average action per decision = \$37.50

Hence, the EV per dollar of action = -\$0.50/37.50 = -1.33%.

Now, Ahigh is a firm believer in house edge per roll.
and not per bet resolved.

So, I see his per roll method having the exact same expected value as Steen and ME per bet resolved.
Ahigh still refuses to acknowledge this fact.

Ahigh does not see these (expected values) as being equal as he knows he can take his bet down after one roll.
hehe
Not always!
hehehe

It is because as Steen points out, Ahigh also wants to include all the money at risk in his house edge and ev calculations.
(that can easily get tricky (mistakes made) when comparing two very different bets - apples to oranges)

here is what I show per roll
then added the per bet resolved
eventprobnetprob*netdecimal
4 rolls 3/36494 3/364.083333333
10 rolls 3/36494 3/364.083333333
7 rolls 6/36-50-8 12/36-8.333333333
all other rolls 24/3600 0
totalexpected value per roll- 6/36-0.166666667
rolls to resolve (36/12) 3
expected value per decision -0.5

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.)

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
I Heart Vi Hart
Ahigh
Joined: May 19, 2010
• Posts: 5197
May 6th, 2014 at 6:42:36 PM permalink
Quote: mustangsally