Ahigh
Joined: May 19, 2010
• Posts: 5167
May 4th, 2014 at 10:01:32 PM permalink
Quote: mustangsally

funny it still is.
actually very sad that he thinks all of his results is the truth.

I mean, he starts his first post in this thread with some sim data that was not even asked about.
Got half of his results quite accurate. 50% is better than 0%
and when presented by two others that his results could be very accurate using a simple formula,
no simulation required,
he says his work is close enough to show his simulation is working. hehe

stop it
hehe

He continues on and on and I really can not read much more of his first post without hurting myself from out-of-controlled laughing.
at least the laughing is much cheaper (+EV) than just smoking pot in Colorado.

I still
hehe

he says right away too for the OP
"In the real world, on average a \$100 buy-in at a \$10 table will last you about four hours without playing odds"
more what? conclusions (about) hehe... *real world*

I get, hehe, the average number of rolls to ruin a 10 unit bankroll = 2,300 (only a 5k sim)
median = 540
hehehe

So I found 100,000 craps players, real ones, to play until ruin and count the number of bets it took to be totally broke.
They had free food and drinks from Sally's Casino.
(for rolls multiply by 3.375 I did hehe)

median: 168.00 (567 rolls)
maximum value: 58,042 (195,891 rolls) ****WOW!
this player must have thought she (it was a she) IS the greatest craps player of all time.****
hehe

stop it
stop it

we all have our opinions
Sally

If there is a bug in the sim find it. Sounds like an awful lot of defense mechanisms at work to defend that you couldn't program in Perl and want to prop up some excuses along the lines that math can solve the problem so you don't need to understand the simulator that I wrote.

Keep using other people's software and other known formulas and provide the answers you can obtain.

And when something is outside your domain of expertise, continue to attack based on your ignorance of things you don't know how to do.

Plenty of people believe that you have "put me in my place" even though you apparently have no knowledge how to write code at all.

You are an excellent source of entertainment for yourself at least.
http://dumbass.website
Ahigh
Joined: May 19, 2010
• Posts: 5167
May 4th, 2014 at 10:34:46 PM permalink
Running the sim to see if you can double a \$1000 doing \$10 pass and no other bets takes a long time.

ahigh~/craps% ./roll
7 SUCCESSES and 93 FAILURES
Chance of success was 7%
Chance of failure was 93%
Average number of rolls for success was 28009
Average number of rolls for failure was 19005.7956989247
ahigh~/craps% vi roll
ahigh~/craps% ./roll
51 SUCCESSES and 949 FAILURES
Chance of success was 5.1%
Chance of failure was 94.9%
Average number of rolls for success was 20974.6666666667
Average number of rolls for failure was 22216.1264488936
ahigh~/craps% vi roll
ahigh~/craps% ./roll
530 SUCCESSES and 9470 FAILURES
Chance of success was 5.3%
Chance of failure was 94.7%
Average number of rolls for success was 21595.7811320755
Average number of rolls for failure was 21179.8907074974

I'm not sure this is the answer you were getting, but that's what the sim tells me. I figured I would run it with more sessions since you hadn't yet figured out how to do it yourself. If you get a different answer, maybe there's a bug in my sim. But I would expect that I was right in the beginning when I suggested I didn't get enough samples to have an answer precise enough for your requirements of being correct, yet the sim seems to be working from my perspective in spite of the fact you don't know how to run it yourself.

I'm sure I could get four or five decimal places of accuracy if anybody besides you cared about that. But it would take an hour or so with a simulation to get that accurate.

The fact that you took the opportunity to characterize my answer as incorrect rather than inaccurate due to a limited number of session simulations is very illuminating, however. Especially given that you had plenty of time to figure this out on your own if you were skilled in the ability to read other people's code and understand it.

Of course you didn't even know I provided a sim at first from what I understood. Just wanted to jump straight to "he doesn't know what he's talking about" and other discounting efforts aimed in my general direction.
http://dumbass.website
Ahigh
Joined: May 19, 2010
• Posts: 5167
May 4th, 2014 at 10:44:07 PM permalink
Maybe we should move on to the other things that you were wrong about and still don't acknowledge, huh?

By the way, what do you do for a living that gives you so much time to fart around answering these questions on this forum and making false claims about people like me?
http://dumbass.website
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5356
May 4th, 2014 at 10:50:44 PM permalink
Quote: Ahigh

If there is a bug in the sim find it.

Question: you call Bet just before the main loop, and then call it again at the start of the main loop.

Won't that subtract 10 from the initial bankroll before actually counting it as a bet (so, in effect, you are working out whether or not you will reach 2000 or 0 first from a starting point of 990)?

Note that I tried simulating this on my own (just the flat bets), and got somewhere around 330 rolls per win and 339 rolls per loss.
dicesitter
Joined: Jan 17, 2013
• Posts: 1157
May 5th, 2014 at 7:54:46 AM permalink
Mustangsally

In the end you just have to play the game. If you go to the table and bet the 6 & 8
and hit 5 - 9's and a 7 you wont make any money no matter what the math of the
game tells you.

If you are a di and have a decent shot and use throw a 7 and then come back next
time and throw to the exact same place and use the same set and throw another
7 and you do that all night your going to lose, it does not make any difference
what the math of the game is or what bets are the best.

No matter what anyone tells you, there is a pattern to the table most nights and
there is a pattern in your rolls or within a longer roll, you have to see those
and use them to your advantage. You cant just throw the dice the same way and
have them bounce all over the table or back to you and think your a great player
because you threw to hard eights.

thats why i love this game, it is hard and yet it offers so much to be learned that
it is exciting, and each night is different.

We should enjoy talking about it as much as we do playing it.

dicesetter
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 5th, 2014 at 8:56:31 AM permalink
Quote: ThatDonGuy

Won't that subtract 10 from the initial bankroll before actually counting it as a bet (so, in effect, you are working out whether or not you will reach 2000 or 0 first from a starting point of 990)?

Note that I tried simulating this on my own (just the flat bets), and got somewhere around 330 rolls per win and 339 rolls per loss.

Looks like you simulated a 10 unit bankroll (betting 1 unit) instead of the 100 unit bankroll Ahigh is playing with.
your average rolls looks close to mine I showed earlier

My sim for # of games (only 100k) for a 10 unit bankroll (\$100 into \$200 with \$10 flat bets) for duration until ruin or success to double to 20 units
success:
median: 76.00
mean value: 99.49
ruin:
median: 76.00
mean value: 99.51
they look the same to me

here is a table for the average number of games and probabilities (pass line decisions, no odds) to hit a 20 unit bankroll target
calculated using excel and a simple matrix
nice results
unit bankrollsuccess to 20 unitsruinmean # of gamesmean ruinmean success
10.0377148220.96228517817.3747510912.87036809132.3030661
20.0765116280.92348837233.2192696425.09296169131.3030661
30.1164214560.88357854447.4896555236.66587051129.6365772
40.1574762380.84252376260.1407491947.58728136127.3038659
50.1997088210.80029117971.1260955557.85547914124.3053051
60.2431529950.75684700580.3979067767.46884779120.6413738
70.2878435180.71215648287.9070240476.42587123116.3126572
80.3338161470.66618385393.6028782984.72513415111.3198457
90.3811076620.61889233897.4334496692.36532284105.6637352
100.4297559010.57024409999.3452259699.3452259699.34522596
110.4797997860.52020021499.28315977105.663735292.36532284
120.5312793560.46872064497.19062447111.319845784.72513415
130.58423580.415764293.0093689116.312657276.42587123
140.6387114850.36128851586.67947075120.641373867.46884779
150.6947499980.30525000278.13928864124.305305157.85547914
160.7523961730.24760382767.32541278127.303865947.58728136
170.8116961310.18830386954.17261425129.636577236.66587051
180.8726973180.12730268238.61379282131.303066125.09296169
190.9354485390.06455146120.57992323132.303066112.87036809

Sally
I Heart Vi Hart
Ahigh
Joined: May 19, 2010
• Posts: 5167
May 5th, 2014 at 9:39:31 AM permalink
Quote: ThatDonGuy

Question: you call Bet just before the main loop, and then call it again at the start of the main loop.

Won't that subtract 10 from the initial bankroll before actually counting it as a bet (so, in effect, you are working out whether or not you will reach 2000 or 0 first from a starting point of 990)?

Note that I tried simulating this on my own (just the flat bets), and got somewhere around 330 rolls per win and 339 rolls per loss.

No. The "if( \$pass == 0 ....." part prevents the bug you are suspicious of possibly being present.
http://dumbass.website
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
May 5th, 2014 at 9:55:13 AM permalink
Quote: Ahigh

I respectfully disagree with your conclusions about misleading folks. You and I both agree on the math.

But not on its application. Your reasoning led to the conclusion that the "cost" of the field bet is the same as the "cost" of the place six bet. I subsequently pointed out that you didn't go far enough because under your reasoning, the "cost" of "the place six bet" is actually higher than the "cost" of "the field bet" when you take into account appropriate bet sizing.

Except it isn't. Your comparison between a single \$5 field bet vs. a bunch of \$6 place six bets is not a fair one. From what I've discerned, you have a habit of focusing on the percentage edge (overall, per roll, or per some series of rolls) rather than on the actual expectation. The dollar expectation of a \$5 field bet is -13.9 cents, while the dollar expectation of a \$6 place 6 bet is -9.1 cents. It is obviously true that if you make N place 6 bets, the total expectation is -9.1 * N cents. And if you wanted to, you could solve for -9.1N = -13.9M, factor in the number of rolls to resolve each wager, and determine how often on average you could make a \$5 field bet and have the same dollar expectation as leaving up a \$6 place 6 bet for every roll. (What's the answer?)

Lies, damned lies, and statistics. Especially when you're dealing with percentages, you can make the numbers say anything. A few months ago, I proposed the ELPH as a fair statistic for comparing bets of different resolution times and percentage expectations, as opposed to the Element of Risk:
http://wizardofvegas.com/forum/gambling/other-games/15219-is-the-element-of-risk-a-useful-concept/#post272469
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Ahigh
Joined: May 19, 2010
• Posts: 5167
May 5th, 2014 at 3:39:22 PM permalink
Quote: MathExtremist

But not on its application. Your reasoning led to the conclusion that the "cost" of the field bet is the same as the "cost" of the place six bet. I subsequently pointed out that you didn't go far enough because under your reasoning, the "cost" of "the place six bet" is actually higher than the "cost" of "the field bet" when you take into account appropriate bet sizing.

You're over-generalizing and splitting hairs.

It's a simple point. Most people who place the six or place the eight leave it up for an average of six rolls.

That's a fact that is easily verified.

For those six rolls, the house edge percentage for the money put on the felt is the same as a single bet in the field (yes for one not six rolls).

In addition, the chance that you win as much money as you started with is greater in the field (44.4%).

The whole point, and I expect someone with a name like yours fails to understand, is that there is a pragmatic issue being ignored by folks who think more about the math than about the psychology of having more money in action on every roll rather than taking a risk, winning, and leaving with a high enough probability of coming out ahead.

Keep gambling, and you're going to lose is the pragmatic illustration that I am trying to make. Even a single field bet doesn't suck if you hit it once compared to ANY strategy you apply over and over and over.

And again, it's the beginners that don't understand. Your splitting hairs with me is doing nothing but diverting the discussion.
http://dumbass.website
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5356
May 5th, 2014 at 4:13:54 PM permalink
Quote: Ahigh

No. The "if( \$pass == 0 ....." part prevents the bug you are suspicious of possibly being present.

My bad - you're right. It just seemed strange that you were calling &bet twice in succession. I was wondering how you would not get a glaring discrepancy in your results if this was happening the way I thought it was...

Quote: mustangsally

Looks like you simulated a 10 unit bankroll (betting 1 unit) instead of the 100 unit bankroll Ahigh is playing with.

When using a 100-unit bankroll with 3/4/5x odds, over 4 million runs, I get a mean of of 1444 rolls per success and 1457 rolls per failure. (Note that the simulation allowed the player to bet the full odds even if he doesn't have the bankroll to cover it - e.g. if his bankroll is down to 3 and the point is 8, the odds bet is assumed to be 5.)

Quote: MustangSally

here is a table for the average number of games and probabilities (pass line decisions, no odds) to hit a 20 unit bankroll target calculated using excel and a simple matrix

Pardon me for asking, but how do you calculate the mean number of rolls? I know how to calculate the probability of success/failure (in fact, with a basic probability of success, rather than multiple states (like you would have with 3/4/5x), you don't even need a matrix; there's a straightforward formula), but can't figure out how to count the mean number of comeouts without resorting to simulation.