what are the odds?
October 2nd, 2012 at 10:21:26 AM
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For those that want to learn more
added: More from WinCraps
Excellent reading.
I wonder why Steen does not post here at WoV.
Maybe all the years at other forums took away the excitement.
"This is arguably the most useful expression of EV because it takes into account the probabilities of winning and losing, the amounts won or lost, and the amounts of money directly responsible for each outcome (the action.)
Since the amount of action received per decision in a solitary bet is the same as the amount wagered, this value can also be called the EV per decision (which is how it appears in the Advantage tab) as long as you remember that it's the amount wagered in each decision that matters.
This may sound the same as the EV per dollar wagered but it's not.
The distinction is that action is the amount wagered when the bet resolves.
In other words, it's the EV per dollar resolved.
This concept is vital when considering multiple, simultaneous bets.
For instance, occasionally you'll come across someone who'll try to convince you that the expected value of some combination of bets is actually lower than any of the included bets by themselves (a synergistic effect, so to speak).
One such argument claims that a combined bet on Place 5, Place 6, Place 8, & Place 9 loses only 1.136% per decision."
Steen continues
"One such argument claims that a combined bet on Place 5, Place 6, Place 8, & Place 9 loses only 1.136% per decision. Using an example of $5 Place 5, $6 Place 6, $6 Place 8, & $5 Place 9 (total of $22 in wagers) the reasoning is this:
There are 18 ways to win $7 and 6 ways to lose $22 for a total of 24 decisions.
Hence, the EV is [18/24 * $7] + [6/24 * -$22] = -$0.25 per decision.
Divide by the amount wagered and the EV% becomes -$0.25/$22 = -1.136% per decision.
That's lower than the individual EV of the any of the included bets!"
"Well, guess what?
Craps bet are not synergistic.
The erroneous conclusion of the the above argument comes from not expressing the EV as a function of the action.
When one of the included bets wins, the amount won is solely a function of the amount that was wagered on the winning bet and not on the other three bets.
For instance, if a 9 hits then the $7 win is a function of the $5 that was wagered on Place 9 and not the $17 on the other three bets.
Yet the argument treats the full amount wagered on all four bets as if it had a bearing on each winning outcome (i.e. by dividing the average loss per decision by the full amount wagered.)
This is wrong.
Intuitively, since the Place 5 & 9 each have EV's of -4.0% per decision and the Place 6 & 8 each have EV's of -1.52% per decision, we know that the combined wagers should have a weighted EV which lies somewhere between -1.52% and -4.0%.
To arrive at this figure we simply need to divide our average loss per decision by the average action per decision.
The average action is figured as follows: There's an 8/24 chance of rolling a 5 or 9 and producing $5 worth of action, a 10/24 chance of rolling a 6 or 8 and producing $6 worth of action, and a 6/24 chance of rolling a 7 and producing $22 worth of action. Hence, action = [8/24 *$5] + [10/24 * $6] + [6/24 * $22] = $9.67 per decision. Now divide -$0.25 by $9.67 and the average loss per dollar of action is 2.59%."
Those that push the per roll HE values to use in all cases, always, I ask,
where are their values for the variance?
I see none.
Not that hard to calculate
IMO, Not the same as per decision.
Now try to use the standard normal distribution using per roll ev and sd and see what a mess it makes.
Do not believe me.
Stewart Ethier in a famous world gambling conference showed why including pushes in calculations just leads to very big messes.
Donald Catlin also agrees.
But these guys are just hacks!
stone them all with rose petals
This is why most just use HE and EV, simple math, the KISS principle.
screw the variance, it is useless.
added: More from WinCraps
Excellent reading.
I wonder why Steen does not post here at WoV.
Maybe all the years at other forums took away the excitement.
"This is arguably the most useful expression of EV because it takes into account the probabilities of winning and losing, the amounts won or lost, and the amounts of money directly responsible for each outcome (the action.)
Since the amount of action received per decision in a solitary bet is the same as the amount wagered, this value can also be called the EV per decision (which is how it appears in the Advantage tab) as long as you remember that it's the amount wagered in each decision that matters.
This may sound the same as the EV per dollar wagered but it's not.
The distinction is that action is the amount wagered when the bet resolves.
In other words, it's the EV per dollar resolved.
This concept is vital when considering multiple, simultaneous bets.
For instance, occasionally you'll come across someone who'll try to convince you that the expected value of some combination of bets is actually lower than any of the included bets by themselves (a synergistic effect, so to speak).
One such argument claims that a combined bet on Place 5, Place 6, Place 8, & Place 9 loses only 1.136% per decision."
Steen continues
"One such argument claims that a combined bet on Place 5, Place 6, Place 8, & Place 9 loses only 1.136% per decision. Using an example of $5 Place 5, $6 Place 6, $6 Place 8, & $5 Place 9 (total of $22 in wagers) the reasoning is this:
There are 18 ways to win $7 and 6 ways to lose $22 for a total of 24 decisions.
Hence, the EV is [18/24 * $7] + [6/24 * -$22] = -$0.25 per decision.
Divide by the amount wagered and the EV% becomes -$0.25/$22 = -1.136% per decision.
That's lower than the individual EV of the any of the included bets!"
"Well, guess what?
Craps bet are not synergistic.
The erroneous conclusion of the the above argument comes from not expressing the EV as a function of the action.
When one of the included bets wins, the amount won is solely a function of the amount that was wagered on the winning bet and not on the other three bets.
For instance, if a 9 hits then the $7 win is a function of the $5 that was wagered on Place 9 and not the $17 on the other three bets.
Yet the argument treats the full amount wagered on all four bets as if it had a bearing on each winning outcome (i.e. by dividing the average loss per decision by the full amount wagered.)
This is wrong.
Intuitively, since the Place 5 & 9 each have EV's of -4.0% per decision and the Place 6 & 8 each have EV's of -1.52% per decision, we know that the combined wagers should have a weighted EV which lies somewhere between -1.52% and -4.0%.
To arrive at this figure we simply need to divide our average loss per decision by the average action per decision.
The average action is figured as follows: There's an 8/24 chance of rolling a 5 or 9 and producing $5 worth of action, a 10/24 chance of rolling a 6 or 8 and producing $6 worth of action, and a 6/24 chance of rolling a 7 and producing $22 worth of action. Hence, action = [8/24 *$5] + [10/24 * $6] + [6/24 * $22] = $9.67 per decision. Now divide -$0.25 by $9.67 and the average loss per dollar of action is 2.59%."
Those that push the per roll HE values to use in all cases, always, I ask,
where are their values for the variance?
I see none.
Not that hard to calculate
IMO, Not the same as per decision.
Now try to use the standard normal distribution using per roll ev and sd and see what a mess it makes.
Do not believe me.
Stewart Ethier in a famous world gambling conference showed why including pushes in calculations just leads to very big messes.
Donald Catlin also agrees.
But these guys are just hacks!
stone them all with rose petals
This is why most just use HE and EV, simple math, the KISS principle.
screw the variance, it is useless.
winsome johnny (not Win some johnny)
October 2nd, 2012 at 11:38:51 AM
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Let me see if I, and others, can grasp your simple math.
Why? not lunch time yet.
I am really into Laura Nyro (?) these days but always like a good math question. (CrystalMath)
agree
or about 1 in 110.1996
If we do not just look at the very next 8 rolls and keep rolling until 8 Field losses in a row happen,
from
https://wizardofvegas.com/forum/questions-and-answers/math/8141-on-average-how-many-trials-will-it-take-to-see-a-streak-of-8-qs-for-fun/
shows the math and formulas for the average number of trials until a streak of X?
and the answer looks to be 1 in 245.70
So, I went even 2 steps further and simulated it.
1 million such streaks. The average wait time shows below.
Your answer of
1 in 25?
care to show the simple math for this?
last Q
And why did you not just roll a 2 or 12 for this player and tell him he won because of your DI skills?
Have a Nice Day
Why? not lunch time yet.
I am really into Laura Nyro (?) these days but always like a good math question. (CrystalMath)
probability of losing the very next 8 fair coin flips in a row is 0.50^8 = 0.00390625 or 1 in 256 (1/answer)Quote: AhighThey had a friend there and he was explaining to the friend that the chances of that happening were rare. Even at 50/50 chance of winning, it happens every 256th time.
agree
To lose the very next 8 Field bets in a row should be (20/36)^8 = 0.009074 (rounded)Quote: AhighBut with a 44.4% chance of winning like the field, it happens ten times as often at 1 in 25. So a 5.55% reduction in the likelihood of the event happening makes the losing streak TEN TIMES AS LIKELY.
And this is SIMPLE math.
If I had quickly come up and told the guy yesterday that you will see 8 non-fielding numbers in a row every 25 rolls, he probably wouldn't have believed me,
or about 1 in 110.1996
If we do not just look at the very next 8 rolls and keep rolling until 8 Field losses in a row happen,
from
https://wizardofvegas.com/forum/questions-and-answers/math/8141-on-average-how-many-trials-will-it-take-to-see-a-streak-of-8-qs-for-fun/
shows the math and formulas for the average number of trials until a streak of X?
and the answer looks to be 1 in 245.70
So, I went even 2 steps further and simulated it.
1 million such streaks. The average wait time shows below.
grouped data
items: 1000000
minimum value: 8.00
first quartile: 75.00
median: 172.00
third quartile: 338.00
maximum value: 3183.00
mean value: 245.79 <<<<<<<<<<
midrange: 1595.50
range: 3175.00
interquartile range: 263.00
mean abs deviation: 176.30
sample variance (n): 57610.79
sample variance (n-1): 57610.85
sample std dev (n): 240.02
sample std dev (n-1): 240.02
Your answer of
1 in 25?
care to show the simple math for this?
last Q
And why did you not just roll a 2 or 12 for this player and tell him he won because of your DI skills?
Have a Nice Day
winsome johnny (not Win some johnny)
October 7th, 2012 at 12:30:52 PM
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Yep. I always try to confuse the confused.Quote: AhighDefinitely some good stuff in there. But when it comes to right and wrong, seven craps still has a lot of rights to go to get his average to be a little more respectable from my view.
And as far as this as a reply to my comment, I think that 7 craps thrives on confusion rather than seeking to reduce it.
Great entertainment and some even do learn something.
You and Mission 146 can read my above two posts.
Especially you and your simple math example.
See who is confused and why
after your reading.
I'll give you 30 days.
You still want everyone to calculate HE, and bet on the lowest HE bets per roll,
as the Wizard also suggests,
based from per roll for every situation. Won't work all the time.
Every situation is not the same and can never be.
A one roll resolved bet vs a 2 stage per decision resolved bet can never be compared by unless the common denominator is total action.
I showed with a very simple (actually WinCraps did) $5 Field bet vs. a $5 place5 bet why this is not true.
A Place 5 wager wins $7 for a $5 bet because of the $5 bet on it, NOT the $5 that is also bet on the Place 9.
The function junction.
The HE per roll does not even show what actually happens when we compare actual wins and losses, the true measure of a bet over time.
Review and comment.
Got to give credit,
You do way better where it actually counts.
On video and at home, no bets made, just proof of your DI skills throwing those dice.
See you at the Craps table,
I will know who you are, but you will not know who I am.
I might be the guy that spills my Bud all over your pants as I buy-in.
or that could be buzz
winsome johnny (not Win some johnny)
October 7th, 2012 at 1:09:04 PM
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" See you at the Craps table,
I will know who you are, but you will not know who I am.
I might be the guy that spills my Bud all over your pants as I buy-in.
or that could be buzz "
Just make sure you apologize for being in my way. And to buy me a fresh beer !
I will know who you are, but you will not know who I am.
I might be the guy that spills my Bud all over your pants as I buy-in.
or that could be buzz "
Just make sure you apologize for being in my way. And to buy me a fresh beer !
October 7th, 2012 at 1:14:27 PM
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MISSION 146
"This will be your only Warning on this matter, and further blanket statements made as to an individual's character or personality of this nature will result in a Suspension. If you wish to belabor this point, I insist that you do so via PM or by creating a post in the Rules Forum. I have no intention of further discussing this matter in this thread. "
Yes, we must behave like gentlemen. I would never dare speaks for others, but I abhor rowdyism.
"This will be your only Warning on this matter, and further blanket statements made as to an individual's character or personality of this nature will result in a Suspension. If you wish to belabor this point, I insist that you do so via PM or by creating a post in the Rules Forum. I have no intention of further discussing this matter in this thread. "
Yes, we must behave like gentlemen. I would never dare speaks for others, but I abhor rowdyism.
October 7th, 2012 at 1:26:27 PM
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No problem.Quote: buzzpaff" See you at the Craps table,
I will know who you are, but you will not know who I am.
I might be the guy that spills my Bud all over your pants as I buy-in.
or that could be buzz "
Just make sure you apologize for being in my way. And to buy me a fresh beer !
Those drinking drunks!
Happens ALL the time.
They gots their FREE Bud in a cup and
know about the small shelf at the Craps table,
but the effort to place it there while lowering their head makes them dizzy,
the effect of too much alcohol, and they spill it everywhere, even on the pants of the player next to them.
I do find it good comedy.
winsome johnny (not Win some johnny)
October 7th, 2012 at 3:27:19 PM
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I was wrong on the 10x more likely and the math. You got me!
Congratulations.
My simulator told me something like 2.067 times as likely to have a losing streak of 8 or more for a 44% chance of winning compared to a 50% chance of winning. Does that jive with what you got?
That is a far cry from 10x.
4014 / 1942 = 2.0674
Congratulations.
My simulator told me something like 2.067 times as likely to have a losing streak of 8 or more for a 44% chance of winning compared to a 50% chance of winning. Does that jive with what you got?
That is a far cry from 10x.
#!/usr/bin/perl
&teststreak( 0.5 );
&teststreak( 0.44444 );
sub teststreak
{
my( $chancewin ) = @_;
my ( $i ) = 0;
my( $totalwins ) = 0;
my( $totallosses ) = 0;
my( %losses );
my( %wins );
printf( "\nStreak test for %.3f chance of winning\n", $chancewin );
for( $i=0; $i<1000000; $i++ )
{
my ( $r ) = rand( 1.0 );
if( $r < $chancewin )
{
# print "Won!\n";
$won++;
$losses{$lost}++ if( $lost > 0 );
$lost = 0;
$totalwins++;
}
else
{
# print "Lost!\n";
$lost++;
$wins{$won}++ if( $won > 0 );
$won = 0;
$totallosses++;
}
}
$losses{$lost}++ if( $lost > 0 );
$wins{$won}++ if( $won > 0 );
printf( "%d totalwins %d totallosses for %.3f observed chance of winning\n", $totalwins, $totallosses, $totalwins / ( $totalwins + $totallosses ) );
my( $count ) = 0;
printf( "Wins streaks at %.3f chance of winning:\n", $chancewin );
foreach( sort { $a <=> $b } keys %wins )
{
printf( "%d:%d ", $_, $wins{$_} );
$count++;
if( $count > 15 )
{
print "\n";
$count = 0;
}
}
print "\n";
$count = 0;
printf( "Losing streaks at %.3f chance of losing\n", 1 - $chancewin );
foreach( sort { $a <=> $b }keys %losses )
{
printf( "%d:%d ", $_, $losses{$_} );
$count++;
if( $count > 15 )
{
print "\n";
$count = 0;
}
}
print "\n";
printf( "Losing streaks of n or greater:\n" );
foreach( sort { $a <=> $b }keys %losses )
{
my( $streak ) = $_;
my( $total ) = 0;
foreach( sort { $a <=> $b }keys %losses )
{
if( $_ >= $streak ) { $total += $losses{$_}; }
}
printf( "%d:%d ", $streak, $total );
}
print "\n";
}
Streak test for 0.500 chance of winning
499832 totalwins 500168 totallosses for 0.500 observed chance of winning
Wins streaks at 0.500 chance of winning:
1:125055 2:62698 3:31375 4:15475 5:7890 6:3893 7:1909 8:936 9:471 10:258 11:117 12:59 13:28 14:21 15:6 16:3
17:3 18:2
Losing streaks at 0.500 chance of losing
1:124934 2:62725 3:31412 4:15639 5:7732 6:3903 7:1911 8:963 9:475 10:236 11:143 12:64 13:31 14:15 15:7 16:1
17:4 18:2 19:1
Losing streaks of n or greater:
1:250198 2:125264 3:62539 4:31127 5:15488 6:7756 7:3853 8:1942 9:979 10:504 11:268 12:125 13:61 14:30 15:15 16:8 17:7 18:3 19:1
Streak test for 0.444 chance of winning
444318 totalwins 555682 totallosses for 0.444 observed chance of winning
Wins streaks at 0.444 chance of winning:
1:137370 2:61616 3:26913 4:12012 5:5316 6:2411 7:1006 8:426 9:186 10:100 11:44 12:10 13:4 14:4 15:2 20:1
Losing streaks at 0.556 chance of losing
1:110326 2:61028 3:33787 4:18746 5:10473 6:5807 7:3238 8:1811 9:983 10:550 11:295 12:164 13:81 14:67 15:25 16:19
17:8 18:7 19:3 21:1 24:1
Losing streaks of n or greater:
1:247420 2:137094 3:76066 4:42279 5:23533 6:13060 7:7253 8:4015 9:2204 10:1221 11:671 12:376 13:212 14:131 15:64 16:39 17:20 18:12 19:5 21:2 24:1
4014 / 1942 = 2.0674
aahigh.com
October 7th, 2012 at 3:55:46 PM
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Quote: 7crapsSee you at the Craps table,
I will know who you are, but you will not know who I am.
I might be the guy that spills my Bud all over your pants as I buy-in.
or that could be buzz
I'll ask again: why do you gotta be like this? Is this just how you are? Do you think this is funny? I'll tell you that I don't think it's funny and I think it reflects poorly on you.
aahigh.com
October 7th, 2012 at 4:14:31 PM
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Quote: 7crapsAnd why did you not just roll a 2 or 12 for this player and tell him he won because of your DI skills?
Have a Nice Day
I was betting on the 6 and 8 and that guy hadn't said anything at all to me and was doing his own thing on the other end of the table. I won my bet when he lost his. He could have known he was betting against me and I was aiming for strings of 6's and 8's, but he was assuming that the math was going to protect him from a string of losses and he was wrong.
Here's the field streaks for my last 1429 rolls:
Quote:
Field streaks for 1429 rolls
Field win streaks (655 total field wins 0.458%)
1:168 2:87 3:41 4:24 5:11 6:4 7:1 8:1
Field loss streaks (774 total field losses 0.542%)
1:158 2:74 3:36 4:23 5:25 6:12 7:4 8:3 9:1 10:1
Compared to random results:
Quote:
Streak test for 0.444 chance of winning
620 totalwins 809 totallosses for 0.434 observed chance of winning
Wins streaks at 0.444 chance of winning:
1:202 2:85 3:45 4:13 5:6 6:3 7:1 8:1
Losing streaks at 0.556 chance of losing
1:159 2:81 3:46 4:37 5:14 6:8 7:3 8:3 9:2 10:1 11:1
Looks like I have fewer 2 and 3 losers in a row on the field and much more 5, 6, in a row to make up for the fewer 2 and 3 in a row on the losing streaks.
On the winning side of streaks, it looks like I have fewer single-field wins and more 4, 5, and 6 in a row field win streaks.
Guys who expect random distributions on my throws can be surprised by streaks in my outcomes that are relatively unlikely but still happen a little more often than random.
aahigh.com
October 7th, 2012 at 6:33:11 PM
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" a little more often than random" roflmao
October 7th, 2012 at 8:54:10 PM
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Quote: buzzpaff" a little more often than random" roflmao
What are you laughing about?
aahigh.com
October 7th, 2012 at 9:40:09 PM
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Quote: AhighWhat are you laughing about?
" This concept of randomness suggests a non-order or non-coherence in a sequence of symbols or steps, such that there is no intelligible pattern or combination. "
October 7th, 2012 at 10:14:37 PM
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Dude. I ran two simulations, one with random 1400 random samples and one with my recorded 1400 rolls. Then I looked for streaks of losses and streaks of wins in the field. If you read my message you would understand that.
I know that you like to just make one-liner quips, but it's damn annoying when you are laughing without even knowing what I'm talking about.
Go back and read the message before the one-liner that you responded to and tell me that you understand it and how it's funny.
What's really funny is that you comment on so much crap that you don't even bother to understand.
Having more losing or winning streaks in the field than what one would expect with random data is absolutely meaningful.
And I invite anyone to correct me on this. Obviously it's more meaningful with more data, but it means something to me even at 1400 samples.
I know that you like to just make one-liner quips, but it's damn annoying when you are laughing without even knowing what I'm talking about.
Go back and read the message before the one-liner that you responded to and tell me that you understand it and how it's funny.
What's really funny is that you comment on so much crap that you don't even bother to understand.
Having more losing or winning streaks in the field than what one would expect with random data is absolutely meaningful.
And I invite anyone to correct me on this. Obviously it's more meaningful with more data, but it means something to me even at 1400 samples.
aahigh.com
October 9th, 2012 at 1:48:34 PM
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Quote: AhighHere's the field streaks for my last 1429 rolls:
Quote:
Field streaks for 1429 rolls
Field win streaks (655 total field wins 0.458%)
1:168 2:87 3:41 4:24 5:11 6:4 7:1 8:1
Field loss streaks (774 total field losses 0.542%)
1:158 2:74 3:36 4:23 5:25 6:12 7:4 8:3 9:1 10:1
Compared to random results:Quote:
Streak test for 0.444 chance of winning
620 totalwins 809 totallosses for 0.434 observed chance of winning
Wins streaks at 0.444 chance of winning:
1:202 2:85 3:45 4:13 5:6 6:3 7:1 8:1
Losing streaks at 0.556 chance of losing
1:159 2:81 3:46 4:37 5:14 6:8 7:3 8:3 9:2 10:1 11:1
Thanks for the data. I think it looks better in a table.
Field Win streaks
| Run Length | Freq | Total | Expected # | Exp# Ahigh sim | Difference | SD | Actual #SD | or more | Freq | Run Length |
|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 168 | 168 | 196.2414 | 202 | -28.241 | 12.54 | -2.2521 | 353.037037 | 337 | 1 |
| 2 | 87 | 174 | 87.1574 | 85 | -0.157 | 8.32 | -0.0189 | 156.7956104 | 169 | 2 |
| 3 | 41 | 123 | 38.7095 | 45 | 2.290 | 5.73 | 0.3997 | 69.63816491 | 82 | 3 |
| 4 | 24 | 96 | 17.1922 | 13 | 6.808 | 3.96 | 1.7191 | 30.9286186 | 41 | 4 |
| 5 | 11 | 55 | 7.6356 | 6 | 3.364 | 2.69 | 1.2507 | 13.73641853 | 17 | 5 |
| 6 | 4 | 24 | 3.3912 | 3 | 0.609 | 1.81 | 0.3363 | 6.100793043 | 6 | 6 |
| 7 | 1 | 7 | 1.5062 | 1 | -0.506 | 1.22 | -0.4149 | 2.709560526 | 2 | 7 |
| 8 | 1 | 8 | 0.6689 | 1 | 0.331 | 0.82 | 0.4037 | 1.203403323 | 1 | 8 |
| 337 | 655 |
5678
Field Lose streaks
| Run Length | Freq | Total | Expected # | Exp# Ahigh sim | Difference | SD | Actual #SD | or more | Freq | Run Length |
|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 158 | 158 | 157.0919 | 159 | 0.908 | 12.17 | 0.0746 | 353.1481481 | 337 | 1 |
| 2 | 74 | 148 | 87.2123 | 81 | -13.212 | 8.62 | -1.5328 | 196.0562414 | 179 | 2 |
| 3 | 36 | 108 | 48.4174 | 46 | -12.417 | 6.39 | -1.9433 | 108.8439262 | 105 | 3 |
| 4 | 23 | 92 | 26.8797 | 37 | -3.880 | 4.87 | -0.7967 | 60.42651019 | 69 | 4 |
| 5 | 25 | 125 | 14.9227 | 14 | 10.077 | 3.68 | 2.7384 | 33.54676248 | 46 | 5 |
| 6 | 12 | 72 | 8.2846 | 8 | 3.715 | 2.79 | 1.3317 | 18.62402307 | 21 | 6 |
| 7 | 4 | 28 | 4.5993 | 3 | -0.599 | 2.1 | -0.2854 | 10.33941993 | 9 | 7 |
| 8 | 3 | 24 | 2.5534 | 3 | 0.447 | 1.58 | 0.2827 | 5.740089097 | 5 | 8 |
| 9 | 1 | 9 | 1.4176 | 2 | -0.418 | 1.18 | -0.3539 | 3.186697784 | 2 | 9 |
| 10 | 1 | 10 | 0.7870 | 1 | 0.213 | 0.88 | 0.2421 | 1.769142878 | 1 | 10 |
| 337 | 774 |
You are within 2SD on all run lengthsQuote: AhighGuys who expect random distributions on my throws can be surprised by streaks in my outcomes that are relatively unlikely but still happen a little more often than random.
except 2 that are within 3SD of every mean for each run length.
Nothing special in this group of 1429 dice rolls.
No Headlines, no appearance on Oprah.
You are not producing relatively unlikely events as you claim.
You may have just convinced yourself you have.
You (me and others)
need to learn more about and how to apply variance (standard deviation)
to all your averages and calculate those averages with the proper formulas.
Sims are OK, but others have already paved the way using formulas like Feller to name a good one.
(q = 1-p and R = Run length)
Expected number of total runs in N trials.
p^2+(p*q*trials)
Expected number of runs of length X or more in N trials.(Excel)
=IF(R>trials,0,(p^R)*(1+((trials-R)*q)))
Expected number of runs of length X in N trials.
=((P_Win)^Run)*(1+((N_Trials-Run)*P_Lose)) - ((P_Win)^(Run+1))*(1+((N_Trials-(Run+1))*P_Lose))
Sally has her Excel sheet here. It is Live.
Expected Number of Runs per N Trials
For those that are bored and just do not care one way or another
because we just want to see Ahigh
call his shots and roll the hardways in a casino
with our bets winning on almost every decision.
See ya at the Craps table
winsome johnny (not Win some johnny)
October 9th, 2012 at 1:59:44 PM
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Thanks and I appreciate the time you put into your response. I do need to learn more. Thanks in advance for the help on the math as I continue.
aahigh.com
October 9th, 2012 at 3:02:35 PM
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" For those that are bored and just do not care one way or another
because we just want to see Ahigh
call his shots and roll the hardways in a casino
with our bets winning on almost every decision. "
Nothing succeeds like success. Or proving you believe in your delusions, by backing them up with cash.
Do the words " lucid interval " seem familiar ?
because we just want to see Ahigh
call his shots and roll the hardways in a casino
with our bets winning on almost every decision. "
Nothing succeeds like success. Or proving you believe in your delusions, by backing them up with cash.
Do the words " lucid interval " seem familiar ?
March 11th, 2013 at 2:24:06 AM
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"i've been successful recently with a simple craps betting method, and i'm here because i can't seem to find anywhere on the net that can tell me how to figure the odds to see if i'm good or lucky. i just make equal bets pass line and field(come out only) and take full odds after the point. yes, there are a lot of break-even rolls with the 3 and 7, but the 2, 11(x2), 12(x2) are pure wins. and yes, the 5,6,8 are initial losses with a chance to break even, but the 4,9,10 are initial wins with a chance to double up. can anyone do the math for me on this?"
There is a better variation for this system:
We could maximize profits by minimizing loses,for examble on the outcomes 4,9 and 10 we are winning the field bet,so we are already + initial bet (it could be 10)
So let's say plus 10 then instead of living the outcome to chance,we place a lay bet on the respective numbers (4,9,10) of 9 dollars/euros/pounds...etc
There are two possible outcomes: 1st we'll win the pass line bet 10 (field bet) + 1 (pass line-lay bet) = 11 net profit
2nd possible outcome is to come out 7,in this case you are winning the lay bet instead of the pass line which gains approximately 13.3 accumulated profit or 4.3 net thus 10 (field bet net earnings) minus 5.7 (19-13.3) = 4.3 of net profit
This way you win either way,furthermore if you raise your initial bet on every winning streak,your winnings would be increased significantly,take a look on the following progression:
First bet won net profit 20,second bet will be half of the net profit,thus 10 (5 field + 5 pass) even if you lose both,you would still be in profit because you just used the half of the previous bet's winnings.
If you win one or both of the bets you keep raising your bet accordingly by increasing your previous bet from 10% up to 50% of your net profit from the winning streak.
The point is to ride the wave of the winning streaks and doing so by raising when you win and not the opposite.
There are winning and losing streaks on every game session,you should have noticed this,don't you?
When you eventualy lose,you should return back to your initial bet,in our examble 10,also by betting this way it aims to protect your bankroll by the loses because the loses after winning streaks are not actually loses but just minimizing your net profits.
A winning streak could be anything from 2 up to 10 consecutive wins,when you experience a losing streak,you just keep betting exactly the same initial bet,not increasing,neither decreasing.
Besides after a good winning streak you could say enough for today by stopping when you are still ahead.
We don't have to change any odds,neither to predict any outcome,we just increasing our wins and decrease our loses.
You don't even had to be always winner,if you have stop win/loss limits the same amount,for examble 200 units minus/plus then you only need to win 6 out of 10 times: 6 x 200 = 1200 - 800 (4 x 200) = 400 units net profit after 10 game sessions with a very moderate estimation!
All these are great but I have created an even better roulette system which isn't out there...
There is a better variation for this system:
We could maximize profits by minimizing loses,for examble on the outcomes 4,9 and 10 we are winning the field bet,so we are already + initial bet (it could be 10)
So let's say plus 10 then instead of living the outcome to chance,we place a lay bet on the respective numbers (4,9,10) of 9 dollars/euros/pounds...etc
There are two possible outcomes: 1st we'll win the pass line bet 10 (field bet) + 1 (pass line-lay bet) = 11 net profit
2nd possible outcome is to come out 7,in this case you are winning the lay bet instead of the pass line which gains approximately 13.3 accumulated profit or 4.3 net thus 10 (field bet net earnings) minus 5.7 (19-13.3) = 4.3 of net profit
This way you win either way,furthermore if you raise your initial bet on every winning streak,your winnings would be increased significantly,take a look on the following progression:
First bet won net profit 20,second bet will be half of the net profit,thus 10 (5 field + 5 pass) even if you lose both,you would still be in profit because you just used the half of the previous bet's winnings.
If you win one or both of the bets you keep raising your bet accordingly by increasing your previous bet from 10% up to 50% of your net profit from the winning streak.
The point is to ride the wave of the winning streaks and doing so by raising when you win and not the opposite.
There are winning and losing streaks on every game session,you should have noticed this,don't you?
When you eventualy lose,you should return back to your initial bet,in our examble 10,also by betting this way it aims to protect your bankroll by the loses because the loses after winning streaks are not actually loses but just minimizing your net profits.
A winning streak could be anything from 2 up to 10 consecutive wins,when you experience a losing streak,you just keep betting exactly the same initial bet,not increasing,neither decreasing.
Besides after a good winning streak you could say enough for today by stopping when you are still ahead.
We don't have to change any odds,neither to predict any outcome,we just increasing our wins and decrease our loses.
You don't even had to be always winner,if you have stop win/loss limits the same amount,for examble 200 units minus/plus then you only need to win 6 out of 10 times: 6 x 200 = 1200 - 800 (4 x 200) = 400 units net profit after 10 game sessions with a very moderate estimation!
All these are great but I have created an even better roulette system which isn't out there...
Author of "BlueAngel system" and "RNG & ROULETTE WHEELS" available at Amazon and at www.theultimategamblingcollection.com
March 11th, 2013 at 7:43:29 AM
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" Having more losing or winning streaks in the field than what one would expect with random data is absolutely meaningful."
It is only meaningful if you can repeat it. Otherwise it is just RANDOM. What don't you understand about that ?
It is only meaningful if you can repeat it. Otherwise it is just RANDOM. What don't you understand about that ?
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
March 11th, 2013 at 9:03:32 AM
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Are you talking to me Buzzard? I didn't mentioned that I don't understand something. The odds for such a game are slightly on the house favor,but since you bet more on the winning streaks rather than the losing,then yes you could be a winner and yes you could repeat the same procedure every time.This doesn't have to do with predicting or luck,has to do with statistics analysis and math calculations.So you could say that can be proved scientificaly,by Maths and Stats science.
Author of "BlueAngel system" and "RNG & ROULETTE WHEELS" available at Amazon and at www.theultimategamblingcollection.com
