Comparing Full Odds and No Odds
April 26th, 2014 at 7:20:20 PM
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Say I have a bankroll of $1000 and I would like to try to make $100. $10 table with 3-4-5X odds.
How likely am I to get up by $100 by betting only pass line ($10) with no odds versus betting pass line ($10) with full odds.
How likely am I to double my money?
From my perspective, this is the most fundamental question to be asked about craps. Should I bet conservatively (no odds) in order to stretch out my bank roll or should I bet full odds and increase my variance, and possibly my winnings.
How likely am I to get up by $100 by betting only pass line ($10) with no odds versus betting pass line ($10) with full odds.
How likely am I to double my money?
From my perspective, this is the most fundamental question to be asked about craps. Should I bet conservatively (no odds) in order to stretch out my bank roll or should I bet full odds and increase my variance, and possibly my winnings.
April 26th, 2014 at 7:28:40 PM
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You are much more likely to get up $100 by betting $10 with full odds than $10 flat with no odds.
You are much more likely to double your $1000 by betting $10 with full odds than $10 flat with no odds.
You would be even more likely to get up $100 by playing a negative progression, starting at $100, flat with no odds, than to use odds.
You would be even more likely to get up $1000 by betting all $1000 one time on the line, than doing anything else.
You are much more likely to double your $1000 by betting $10 with full odds than $10 flat with no odds.
You would be even more likely to get up $100 by playing a negative progression, starting at $100, flat with no odds, than to use odds.
You would be even more likely to get up $1000 by betting all $1000 one time on the line, than doing anything else.
April 26th, 2014 at 7:30:36 PM
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If your goal is to increase your bankroll then you should always take full odds. The odds have a house edge of 0 while you passline has a HE of 1.41%. If your goal is to play for as long as possible then bet no odds.
April 26th, 2014 at 8:28:13 PM
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Yeah, that would be great fun... but only if I happened to win.Quote: sodawaterYou would be even more likely to get up $1000 by betting all $1000 one time on the line, than doing anything else.
April 27th, 2014 at 7:34:47 AM
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Quote: sodawaterYou are much more likely to get up $100 by betting $10 with full odds than $10 flat with no odds.
If you bet $10 flat trying to win $100 with no odds, you will double your money 40.349% of the time with an average of 377 rolls.
Same strategy will lose the $100 buy-in 59.651% of the time with an average number of 361 rolls.
If you bet $10 with 345x odds trying to win $100, you then have a 43.495% chance of success with an average of 21.22 rolls.
You have a 56.505% chance of failure with an average of 21.32 rolls.
So it appears from what I know that you are 7.779% more likely (43.496/40.349 - 1)*100% to win.
Betting odds shortens your session more than it improves your chances of winning when you bet 345x odds in my view, especially when NOT given explicit advice about how to use odds to have the best chance of doubling your money (like what you do when you don't have enough money for 345x odds, or betting slightly less to make it more likely to exactly meet your goal instead of haphazardly winning more because you always are betting 345x).
In the real world, on average a $100 buy-in at a $10 table will last you about four hours without playing odds and about 20 minutes when you play 345x. And this is just pass line with odds and no come bets by the way.
Quote: sodawaterYou are much more likely to double your $1000 by betting $10 with full odds than $10 flat with no odds.
You would be even more likely to get up $100 by playing a negative progression, starting at $100, flat with no odds, than to use odds.
You would be even more likely to get up $1000 by betting all $1000 one time on the line, than doing anything else.
When you get to 345x odds with $1000 bankroll, your chance of doubling your money actually improves compared to trying to double $100 with $10 and 345x odds. Similarly, flat-betting $10 trying to win $1000 is very nearly impossible. The house edge will overtake even the players with the highest endurance levels for boredom. You have about a 4.6% chance of making $1000 before losing $1000 with $10 on the pass line and no other bets. And it takes over 20,000 rolls on AVERAGE to do it, which is an AVERAGE of 200 playing hours or about five weeks at eight hours a day.
You have a 46.408% chance of winning. That's actually better than betting it all in the field (which has a 44% chance of at least doubling your money which includes a 2.78% chance of tripling your money and a 2.78% chance of quadrupling your money). The field is still a better approach than 345x if you want the biggest average win amount when you do win.
The average number of rolls to double your $1000 buy-in with $10 and 345x odds is 1449 rolls.
The average number of rolls to lose all $1000 with $10 and 345x odds is 1444 rolls.
At 100 rolls per hour, that's a 14-hour+ session.
The advantage play is ordering a $5 drink every fifteen minutes and tipping $1, which will net you $16/hour compared to the average loss of $4.00 per hour on the $10 passline @ 100 rolls per hour.
I simulated 100,000 sessions to come up with each of these calculations.
Just some sanity for those who actually believe things like "You are much more likely to get up $100 by betting $10 with full odds than $10 flat with no odds."
The generalization of this statement is true as the initial amount of money being wished to double increases. As the ratio of the flat bet on the passline decrease to the amount you wish to win, it becomes nearly impossible to double your money. But betting odds requires a bankroll. A bankroll that most people following this advice fail to understand with the other complexities going on related to the game.
Here's the truth: you're much more likely to have a quick and bad experience doing this believing that you have a significantly higher chance of accomplishing your goal because of the odds when in fact, you only have at most a 7.779% higher chance of accomplishing your goal and it will be at a eighteen times to the pace of play in terms of how long your buy-in lasts you.
#!/usr/bin/perl
$verbose = 0;
$initial_br = 1000;
$min = 10;
if( 1 )
{
$odds[4] = 3;
$odds[5] = 4;
$odds[6] = 5;
$odds[8] = 5;
$odds[9] = 4;
$odds[10] = 3;
}
else
{
$odds[4] = 0;
$odds[5] = 0;
$odds[6] = 0;
$odds[8] = 0;
$odds[9] = 0;
$odds[10] = 0;
}
sub roll
{
$rollcount++;
$d1 = int( rand( 6 ) + 1 );
$d2 = int( rand( 6 ) + 1 );
$sum = $d1 + $d2;
print "Roll was $d1-$d2 = $sum\n" if( $verbose );
}
sub resolve
{
if( $point == 0 )
{
if( $sum == 2 || $sum == 3 || $sum == 12 )
{
print "$sum craps $sum -- reel 'em in\n" if( $verbose );
$pass = 0;
}
elsif( $sum == 7 )
{
print "Seven winner seven pay the line\n" if( $verbose );
$br += $pass * 2;
$pass = 0;
}
elsif( $sum == 11 )
{
print "Yo eleven pay the line\n" if( $verbose );
$br += $pass * 2;
$pass = 0;
}
else
{
$point = $sum;
print "Point is $sum mark it up\n" if( $verbose );
}
}
elsif( $sum == $point )
{
print "$sum winner $sum pay the line\n" if( $verbose );
my( $save ) = $br;
$br += $odds + $pay[$sum] * $odds;
$br += $pass * 2;
$pass = 0;
$odds = 0;
print "Put \$" . ( $br - $save ) . " in rack ($br total) from hitting the $point\n" if( $verbose );
$point = 0;
}
elsif( $sum == 7 && $point > 0 )
{
print "Seven out!\n" if( $verbose );
$pass = 0;
$odds = 0;
$point = 0;
}
}
sub bet
{
if( $pass == 0 && $point == 0 && $br >= $min )
{
$br -= $min;
$pass = $min;
print "Betting $min on the pass with $br in the rail\n" if( $verbose );
}
elsif( $point > 0 && $odds == 0 )
{
my( $maxodds ) = $odds[$point] * $pass;
if( $br >= $maxodds )
{
$odds = $maxodds;
$br -= $odds;
}
elsif( $br >= $min )
{
$odds = $br;
$br = 0;
}
print "Betting \$$odds odds on the $point with $br in the rail\n" if( $verbose );
}
}
sub start_session
{
$br = $initial_br;
$pass = 0;
$odds = 0;
$rollcount = 0;
$point = 0;
}
sub totalbets
{
return $pass + $odds;
}
&start_session;
$pay[4] = 2;
$pay[5] = 3/2;
$pay[6] = 6/5;
$pay[8] = 6/5;
$pay[9] = 3/2;
$pay[10] = 2;
$point = 0;
&bet;
while( $success + $fail < 100000 )
{
&bet;
&roll;
&resolve;
if( $br >= $initial_br * 2 )
{
print "+++ SUCCESS in $rollcount rolls\n\n" if( $verbose );
$success++;
$nrs += $rollcount;
&start_session;
}
elsif( $br < $min && &totalbets == 0 )
{
print "+++ FAILURE in $rollcount rolls\n\n" if( $verbose );
$fail++;
$nrf += $rollcount;
&start_session;
}
}
print "$success SUCCESSES and $fail FAILURES\n";
print "Chance of success was " . 100 * $success / ( $success + $fail ) . "%\n";
print "Chance of failure was " . 100 * $fail / ( $success + $fail ) . "%\n";
print "Average number of rolls for success was " . ( $nrs / $success ) . "\n";
print "Average number of rolls for failure was " . ( $nrf / $fail ) . "\n";
aahigh.com
April 27th, 2014 at 7:40:23 AM
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Quote: geoffIf your goal is to increase your bankroll then you should always take full odds. The odds have a house edge of 0 while you passline has a HE of 1.41%. If your goal is to play for as long as possible then bet no odds.
If your goal is to increase your bankroll, you should either work more and gamble less, or learn how to advantage play. IN THE LONG RUN increasing your odds does not improve your chance of increasing your bankroll more than it increases your chance of reducing your bankroll. So I would suggest that your statement is incorrect.
IN THE SHORT RUN you only make your point 40% of the time. The remaining 60% of the time you will lose your odds bet on average. So in the SHORT run, if your goal is increasing your bankroll, you want to lay odds, not take them as you're more likely to win that way.
The funny thing about dice is that they don't care what you are trying to do. And in the LONG RUN the most practical thing that odds do for a player is to temporarily give them money in the rail that is an opportunity for a dealer to beg for a tip. That's a house edge too, you know. Or maybe you could call it a "dealer edge." There is a reason dealers remind you about betting more odds.
aahigh.com
April 27th, 2014 at 9:26:49 AM
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Quote: AhighIf your goal is to increase your bankroll, you should either work more and gamble less, or learn how to advantage play. IN THE LONG RUN increasing your odds does not improve your chance of increasing your bankroll more than it increases your chance of reducing your bankroll. So I would suggest that your statement is incorrect.
IN THE SHORT RUN you only make your point 40% of the time. The remaining 60% of the time you will lose your odds bet on average. So in the SHORT run, if your goal is increasing your bankroll, you want to lay odds, not take them as you're more likely to win that way.
The funny thing about dice is that they don't care what you are trying to do. And in the LONG RUN the most practical thing that odds do for a player is to temporarily give them money in the rail that is an opportunity for a dealer to beg for a tip. That's a house edge too, you know. Or maybe you could call it a "dealer edge." There is a reason dealers remind you about betting more odds.
This has got to be your most acutely poignant post that I've read.
April 27th, 2014 at 9:57:16 AM
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First off,
Bankroll $1000 and $10 flat bets (bankroll units = 100. So it does not now matter what your $ unit is. bankroll of $100 and flat bet of $1 is equal to a Bankroll of $2500 and $25 flat bets and on and on)
and this post will finally set me as my syndicates' official number cruncher.
Super!
I mean I do have the legs for it, as I am also told.
many ask these type of questions all the time
goal: $100 win (10 units) or BUST trying (for simulations, bust also includes the sessions where a full bet is not able to be made)
Using the Gambler's Ruin formula that has been around and derived by many over the last 350 years
no odds
success = 74.2149768% with an average number of trials = 1,298.56
simulation by 1 million shooters
Avg. No. games played . = 1295.79
Avg. No. dice rolls . . = 4374.35
Bankroll was busted . . = 25.741% of the time ( 257414)
Win goal was met . . . = 74.259% of the time ( 742586)
added:
as pointed out, it can take a lot of time to hit your win goal.
74.2149768% is really a limit
here are some probabilities and trials
200: 0.41382205
500: 0.560445529
1000: 0.638792258 ruin: 0.005341015
2000: 0.692413962 ruin: 0.063885309
Now using the odds at 345x
Using a transition matrix
(matrix algebra. I was not exposed to this in High School, but in college for 2 weeks meaning I really had to learn this on my own. Actually, everyone should be taught this by High School IMO. Most use it but are not aware that they do)
345x odds = 88.406245%
(at first I had only 80% but I noticed I had the win goal set for $200 or 20 units. My first error found)
simulation by 1 million shooters
Avg. No. games played . = 53.23
Avg. No. dice rolls . . = 179.74
Bankroll was busted . . = 11.725% of the time ( 117248)
Win goal was met . . . = 88.275% of the time ( 882752)
summary
$1000 into $1100
ONE attempt probability
88.406245% with 345X odds
74.2149768% with 0 odds
and do not forget this
"You are much more likely to get up $100 by betting..."
I see a difference
goal: to DOUBLE the starting bankroll or bust out trying.
This assumes that the player does not remove money from the Bankroll if it increases during play.
That would result in a different question and result
success = 5.5804887% with an average number of trials = 6,282.19
=1/((q/p)^b_x+1) where b_x = bankroll units
simulation by 1 million shooters
Avg. No. games played . = 6278.16
Avg. No. dice rolls . . = 21193.64
Bankroll was busted . . = 94.397% of the time ( 943966)
Win goal was met . . . = 5.603% of the time ( 56034)
transition matrix
345x odds = 46.885%
simulation by 1 million shooters
Avg. No. games played . = 425.40
Avg. No. dice rolls . . = 1436.06
Bankroll was busted . . = 53.498% of the time ( 534979)
Win goal was met . . . = 46.502% of the time ( 465021)
summary
$1000 into $2000 (100 units into 200 units)
ONE attempt probability
46.885% with 345X odds
5.5804887% with 0 odds
Bet $100, lose bet $200, lose bet $400
sounds like a Marty!
But a loss here means next bet is all-in for $300.
I set this up in Excel and BruceZ was kind enough to set the code up for the free program R
success = 0.904106868
only 0.901300869 by and including the 5th bet
my Excel
in R
[1] 0.9041068
R code
Less than 2 bets on average for success and failure
Maybe also also because you can lose more times, on average, than you win.
final summary (100 units into 110 units)
$1000 into $1100
90.410687% from Bold Play
88.406245% with 345X odds
74.2149768% with 0 odds
$1000 into $2000 (100 units into 200 units)
49.2929% from Bold Play
46.885% with 345X odds
5.5804887% with 0 odds
Sally Oh
Bankroll $1000 and $10 flat bets (bankroll units = 100. So it does not now matter what your $ unit is. bankroll of $100 and flat bet of $1 is equal to a Bankroll of $2500 and $25 flat bets and on and on)
and this post will finally set me as my syndicates' official number cruncher.
Super!
I mean I do have the legs for it, as I am also told.
Nice questions I see.Quote: longtimelancerSay I have a bankroll of $1000 and I would like to try to make $100. $10 table with 3-4-5X odds.
How likely am I to get up by $100 by betting only pass line ($10) with no odds versus betting pass line ($10) with full odds.
How likely am I to double my money?
many ask these type of questions all the time
sometimes many answers given trying to help out are also also wrong for many reasons.
Ahigh's post above looks to have many results that differ from mine.
We both can not be 100% correct here.
Ahigh has a past history of posting result errors, why? Don't know and don't care.
I have a past and current history of copy/paste problems (wrong data copied) I say because of my fat fingers
But not all care to be gifted IMO.
I do like to dot my eyes and cross my tees.
Ahigh's post above looks to have many results that differ from mine.
We both can not be 100% correct here.
Ahigh has a past history of posting result errors, why? Don't know and don't care.
I have a past and current history of copy/paste problems (wrong data copied) I say because of my fat fingers
But not all care to be gifted IMO.
I do like to dot my eyes and cross my tees.
How much more likely?Quote: sodawaterYou are much more likely to get up $100 by betting $10 with full odds than $10 flat with no odds.
goal: $100 win (10 units) or BUST trying (for simulations, bust also includes the sessions where a full bet is not able to be made)
Using the Gambler's Ruin formula that has been around and derived by many over the last 350 years
no odds
success = 74.2149768% with an average number of trials = 1,298.56
simulation by 1 million shooters
Avg. No. games played . = 1295.79
Avg. No. dice rolls . . = 4374.35
Bankroll was busted . . = 25.741% of the time ( 257414)
Win goal was met . . . = 74.259% of the time ( 742586)
added:
as pointed out, it can take a lot of time to hit your win goal.
74.2149768% is really a limit
here are some probabilities and trials
200: 0.41382205
500: 0.560445529
1000: 0.638792258 ruin: 0.005341015
2000: 0.692413962 ruin: 0.063885309
Now using the odds at 345x
Using a transition matrix
(matrix algebra. I was not exposed to this in High School, but in college for 2 weeks meaning I really had to learn this on my own. Actually, everyone should be taught this by High School IMO. Most use it but are not aware that they do)
345x odds = 88.406245%
(at first I had only 80% but I noticed I had the win goal set for $200 or 20 units. My first error found)
simulation by 1 million shooters
Avg. No. games played . = 53.23
Avg. No. dice rolls . . = 179.74
Bankroll was busted . . = 11.725% of the time ( 117248)
Win goal was met . . . = 88.275% of the time ( 882752)
summary
$1000 into $1100
ONE attempt probability
88.406245% with 345X odds
74.2149768% with 0 odds
and do not forget this
"You are much more likely to get up $100 by betting..."
I see a difference
SureQuote: sodawaterYou are much more likely to double your $1000 by betting $10 with full odds than $10 flat with no odds.
goal: to DOUBLE the starting bankroll or bust out trying.
This assumes that the player does not remove money from the Bankroll if it increases during play.
That would result in a different question and result
success = 5.5804887% with an average number of trials = 6,282.19
=1/((q/p)^b_x+1) where b_x = bankroll units
simulation by 1 million shooters
Avg. No. games played . = 6278.16
Avg. No. dice rolls . . = 21193.64
Bankroll was busted . . = 94.397% of the time ( 943966)
Win goal was met . . . = 5.603% of the time ( 56034)
transition matrix
345x odds = 46.885%
simulation by 1 million shooters
Avg. No. games played . = 425.40
Avg. No. dice rolls . . = 1436.06
Bankroll was busted . . = 53.498% of the time ( 534979)
Win goal was met . . . = 46.502% of the time ( 465021)
summary
$1000 into $2000 (100 units into 200 units)
ONE attempt probability
46.885% with 345X odds
5.5804887% with 0 odds
How about Bold Play here.Quote: sodawaterYou would be even more likely to get up $100 by playing a negative progression, starting at $100, flat with no odds, than to use odds.
Bet $100, lose bet $200, lose bet $400
sounds like a Marty!
But a loss here means next bet is all-in for $300.
I set this up in Excel and BruceZ was kind enough to set the code up for the free program R
success = 0.904106868
only 0.901300869 by and including the 5th bet
my Excel
in R
[1] 0.9041068
R code
pdoub = function(br,goal,max_bet,pwin,odds,thresh,prob=1) {
if (br < 1 | prob < thresh) return(0)
bet = ceiling((goal-br)/odds) # bet needed to reach goal
limit = min(br,max_bet) # limit = max bet possible
if ( bet < limit )
pwin + (1-pwin)*pdoubb(br-bet,goal,max_bet,pwin,odds,thresh,prob*(1-pwin))
else
pwin*pdoubb(br+odds*limit,goal,max_bet,pwin,odds,thresh,prob*pwin) + (1-pwin)*pdoubb(br-limit,goal,max_bet,pwin,odds,thresh,prob*(1-pwin))
}
require(compiler)
pdoubb = cmpfun(pdoub)
system.time(p <- pdoubb(1000,1100,5000,244/495,1,0.0000001))
pyes, but many would find this type of play not fun at all.Quote: sodawaterYou would be even more likely to get up $1000 by betting all $1000 one time on the line, than doing anything else.
Less than 2 bets on average for success and failure
Maybe also also because you can lose more times, on average, than you win.
overkill
> pdoub = function(br,goal,max_bet,pwin,odds,thresh,prob=1) {
+ if (br < 1 | prob < thresh) return(0)
+ bet = ceiling((goal-br)/odds) # bet needed to reach goal
+ limit = min(br,max_bet) # limit = max bet possible
+ if ( bet < limit )
+ pwin + (1-pwin)*pdoubb(br-bet,goal,max_bet,pwin,odds,thresh,prob*(1-pwin))
+ else
+ pwin*pdoubb(br+odds*limit,goal,max_bet,pwin,odds,thresh,prob*pwin) + (1-pwin)*pdoubb(br-limit,goal,max_bet,pwin,odds,thresh,prob*(1-pwin))
+ }
> require(compiler)
> pdoubb = cmpfun(pdoub)
> system.time(p <- pdoubb(1000,2000,5000,244/495,1,0.0000001))
user system elapsed
0.00 0.00 0.06
> p
[1] 0.4929292
final summary (100 units into 110 units)
$1000 into $1100
90.410687% from Bold Play
88.406245% with 345X odds
74.2149768% with 0 odds
$1000 into $2000 (100 units into 200 units)
49.2929% from Bold Play
46.885% with 345X odds
5.5804887% with 0 odds
Sally Oh
I Heart Vi Hart
April 27th, 2014 at 10:05:49 AM
permalink
FalseQuote: AhighSo in the SHORT run, if your goal is increasing your bankroll, you want to lay odds, not take them as you're more likely to win that way.
First define the short run in terms of how many bets are going to be made, the size of those bets and the starting bankroll and the target bankroll to be used.
this sets the playing field so no one has an advantage.
This should be so much fun!
Then
show your proof for "you're more likely to win that way"
not like
How a pass line bet magically turns into a put bet when the point is established.
funny funny stuff now
Oh, do not forget to address variance because it dominates the short run
and see you after Wednesday
off to the casinos in SoCal
Sally Oh
I Heart Vi Hart
April 27th, 2014 at 10:29:12 AM
permalink
Quote: mustangsallyFalseQuote: AhighSo in the SHORT run, if your goal is increasing your bankroll, you want to lay odds, not take them as you're more likely to win that way.
First define the short run in terms of how many bets are going to be made, the size of those bets and the starting bankroll to be used.
this sets the playing field so no one has an advantage.
This should be so much fun!
Then
show your proof for "you're more likely to win that way"
not like
How a pass line bet magically turns into a put bet when the point is established.
funny funny stuff now
Oh, do not forget to address variance because it dominates the short run
and see you after Wednesday
off to the casinos in SoCal
Sally Oh
I know how it pains you to admit that you are wrong.
An average odds bet wins 1/3rd of the time for a four or a ten, 2/5ths of the time for a five or nine, and 5/11ths of the time for a six or an eight.
I know that you know this.
The chance of establishing a four or ten is 6/24, a five or nine is 8/24, and six or eight is 10/24.
So
6/24ths of points have 1/3 chance of winning
8/24ths of points have 2/5 chance of winning
10/24ths of points have 5/11 chance of winning
(6/24)*(1/3) + (8/24)*(2/5) + (10/24)*(5/11) = 40.60606060606 chance of winning a free odds bet.
Given that there is no house edge, it's easy to know that 100-40.606060 is the chance of winning an average lay odds bets. So that's 59.393939%
My assertion is that if you want to make a single free bet, you are more likely to win the lay odds bet.
The shortest run is a single bet. In the long run (millions of samples that I doubt you have in your play) neither one is better than another.
But there are more folks making a single bet and leaving than folks making a million bets.
If your argument is that a single bet is not a short run, please elaborate. But I am 100% certain that I am right in the short run of a single bet.
Transitively, your assertion that my statement is false is as wrong as other false assertions that you have made about my claims.
I know it hurts to be wrong when you work so hard on your answers. And maybe you can point out an example where you admitted that I was right and you were wrong. The Wizard just did that two days ago, but I can't recall a single instance of you doing that for me.
Here's what I think. I think you feel that you are smarter than I am and it burns you up to have to admit that someone who you consider to be less intelligent that you is correct about something that you disagree with.
So go ahead and tell me how that a single bet is not a short run approach. In fact, if you're really more math orientated than an actual player who actually plays the game, maybe you can calculate the chance of winning on average for a sequence of max odds versus max lay odds bets where n is a small number between 1 and 100. I have already proven that when n is one, you're going to win 60% of the time versus 40% of the time. Given that all sample sizes include at least one bet, I think that until the sample size is greater than one, you have no chance of having a better chance of winning among those who consider "short runs" to include a single bet. Maybe you could define "run" as requiring at least two samples to support your theory that laying odds doesn't increase your chance to win up to 60%!
I personally think you come up with theoretical answers that often don't have any application to the average real-world craps gambler, and that your writing reflect a less pragmatic and more pedagogical (and therefore less practical) approach to the game.
You seem very out of touch with the masses who enjoy the game.
I could be wrong.
There's a statement I have never seen you post.
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