Rolling a 12

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December 15th, 2010 at 8:31:50 AM permalink

Quote: Ayecarumba
Ask the Wizard #265 included a q/a regarding the number of rolls of 2 dice required to have a 50/50 chance of rolling a 12. The Wizard's precise answer was between 24 and 25.

If the above is true, is it possible to have an advantage by waiting for a streak of non-12 rolls, then flat betting it until it hits? If so, how long would this streak need to be in order to have a 55% chance of rolling a 12 within the next 30 rolls?

Of course the dice don't know that they have not come up 12, so the true odds on any single roll are always 36-1. I am looking for an explaination for how these two, apparently contradictory truths can co-exist.

Here's why there's no contradiction.

Proposition 1) The chances of a 12 in a single roll of the dice are 1/36.
Proposition 2) The chances of at least one 12 in the next 25 rolls are just over 1/2.

If you wait for 30 non-12 rolls and then make a bet, what are you betting on?

If that question is unclear, re-read proposition 2 and focus on the word "next".

Hint: the house doesn't offer bets on proposition 2. But even if they did, would it be on previous rolls or next rolls?

"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

superrick

superrick

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December 15th, 2010 at 9:52:52 AM permalink

The math of the game gives you just the probability and nothing more, just because the math tells you that the 12 should roll 1/36 times it doesn't mean that it will happen!

I have seen players that thought the math of the game was infallible only to learn the truth the hard way!
There are days that you might not see a 12 for hours on the craps table, and you would be losing if you went by the math of the game, nothing is written in stone, and you can't use the math of the game to make you a winner, as it's all based on the probabilities of the math and that is all there is to it!

Note, all my post start with this is just my opinion...! You do good brada ..! superrick Winning comes from knowledge and skill when your betting and not reading fiction http://procraps4u2.myfanforum.org/index.php ...

kauboj

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December 19th, 2010 at 8:12:03 PM permalink

what are the odd that a 12 will not be rolled in 100 rolls. i am just trying to determine how the rng compares to real world. i mean realistically you would expect to see at least 3 12s in 100 rolls and with true randomness probably more.

MathExtremist

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December 19th, 2010 at 8:54:17 PM permalink

The odds that a 12 will not be rolled in 100 rolls is (35/36)^100. This has nothing to do with whether you're using an algorithm to generate two numbers between one and six, or two dice to generate numbers between one and six. It is not true that "real world" is meaningfully different than using a computer that models the real world.

"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

kauboj

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December 19th, 2010 at 9:59:20 PM permalink

please answer that in laymans terms i mean by this is what is the likelyhood that you will not roll a 12 in 100 rolls of the dice

aahigh

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December 19th, 2010 at 10:32:30 PM permalink

you have a 6% chance or 1 in 16 chance that you will hit no 12's in 100 rolls given the data that was so well put together in the previous post.

94% of the time you will hit at least one 12 in 100 rolls. Yay.

MathExtremist

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December 19th, 2010 at 10:36:00 PM permalink

Quote: kauboj
please answer that in laymans terms i mean by this is what is the likelyhood that you will not roll a 12 in 100 rolls of the dice

You can take the formula I gave and just put it into Google. It returns:

(35 / 36)^100 = 0.0597797863

"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

superrick

superrick

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December 19th, 2010 at 10:36:47 PM permalink

kauboj

I am no math guy, but I do know one thing as far as rolling any point in 100 rolls there is no way of telling, if a 12 is going to roll in the 100 rolls of the dice!

Today there were two 12's rolled in 106 rolls of the dice on the table I was playing on, yesterday there were 4 out of 222 rolls.

Two days ago there were 2 out of 74 rolls!
Three days ago there were 5 out of 106 rolls!

The math of the game does not always add up in real play, unless you are going to be at the table for 10,000 rolls of the dice!

I have seen plenty of days that you will not see a 12, even rolled in two or three hours of play!

Take it for what it's worth, just because you haven't seen a 12 in 36 rolls of the dice it does not mean that it's about to come up!

There is a big difference between the math of the game and what really happens on the tables!

...

Note, all my post start with this is just my opinion...! You do good brada ..! superrick Winning comes from knowledge and skill when your betting and not reading fiction http://procraps4u2.myfanforum.org/index.php ...

kauboj

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December 19th, 2010 at 10:38:19 PM permalink

now that is stastical probabilities. in my time at the casino i have never seen that many rolls go by with out having several 12's show

kauboj

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December 19th, 2010 at 10:44:30 PM permalink

please understand i am not asking these questions of ignorance.. i am just trying to improve the longevity of play time i have at the tables.. i love craps i love the atsmosphere around the time. i have fun playing. i go several times a week and only take what i am willing to loose at the given time. however i do get aggrevated when i am there for 5 minutes and my bankroll is gone.. id just like to be able to extend my play time to get the time to enjoy it before my bankroll goes to the wayside..

and sometimes i get a little ansy and make wagers i know better than to make and i wind of cursing myself right after i do it.

thecesspit

thecesspit

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December 19th, 2010 at 10:55:03 PM permalink

There is no difference between the math of the game and what happens at the tables.

Simple variance, which is part of the game of rolling two dice, models this perfectly.

"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829

kauboj

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December 19th, 2010 at 11:04:22 PM permalink

ehh i guess i continue to play as i play... sometimes i win sometimes i lose... but most of the time i have fun...

DeMango

DeMango

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December 20th, 2010 at 4:45:38 AM permalink

Nothing like buying the 12 in a Northern Mississippi casino for say $35 which is a $209 payout after paying $1 vig! (yea I'm braggin!)

When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.

FleaStiff

FleaStiff

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December 20th, 2010 at 5:20:32 AM permalink

Quote: Ayecarumba
is it possible to have an advantage by waiting for a streak of non-12 rolls, then flat betting it until it hits? If so, how long would this streak need to be in order to have a 55% chance of rolling a 12 within the next 30 rolls?
Of course the dice don't know that they have not come up 12, so the true odds on any single roll are always 36-1. I am looking for an explaination for how these two, apparently contradictory truths can co-exist.

The essential point in your question is that irrespective of your obvious realization that the dice do not know they have not come up 12 for awhile, you still think that the non-occurrence of the event will affect that next roll. If you want to make money betting on twelve, you should feel free to do it immediately after there have been a noticeable number of rolls that were 12. The dice don't know about the occurrence of the event any more than they know about the non-occurrence of the event. It is you, and a great many other players, who somehow imbue the dice or Nature or The World with some viewpoint that the Next Roll will obey some ultimate "evening out" desire that smooths the Real World results into something that is compatible with moral expectations. Whether there have been lots of 12s or none at all, that Next Roll is most likely to be a 7. It could also be 12. There may have been a dry spell of 12s. There may also have been a dry spell of 7s. That next roll won't smooth out nothing! It will simply be a roll like any other roll.

ElectricDreams

ElectricDreams

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December 20th, 2010 at 7:53:23 AM permalink

Honestly, the best way to lengthen your playing time is to just bet less. I try to not have more than one pass line bet + one other low-HE bet at a time (something like a come bet or place 6 or 8). Sure, you don't win huge gobs of money, but I can stick around and actually enjoy craps for hours at a time, usually.

I have a friend who plays it much differently. He'll blow through like half his bankroll in no time, and then wanders off to play some table game because he gets pissed he lost so much so fast. It's kind of funny.

aahigh

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December 20th, 2010 at 9:29:32 AM permalink

Yes, if you want to make your money last the longest, it's easy .. bet the pass line for the minimum bet until you either double your money or lose it all.

Chances are good that you will do neither unless you have a lot of time on your hands, or else you are betting more than 5% of your bankroll on the passline for the first bet. (EG: start with $100 at a $5 minimum table is the most common way to do this).

Don't place odds or any other bet at all. The dealers will let you stay there and you can get free drinks.

I have been accused of "playing for free drinks" and thus, I assume I have demonstrated to the dealers that I know how. LOL.

Those with simulation models should give an answer to this question more completely. The true answer is bet the don't pass, but you will look awkward if you do this, when others lose a bunch of money and you win $5, and I don't recommend it.

Anyone with a simulator got an answer for the average number of rolls if you bet a $5 bet for a $100 bankroll on the passline until it's all gone .. or doubled .. whichever comes first?

You can double it quickly if you get a lot of good natural rolls, which does happen.

aahigh

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December 20th, 2010 at 10:09:28 AM permalink

Lasted 781 rolls with a total of 515 in winnings and 615 in losses resulting in a bankroll of 0
Lasted 2160 rolls with a total of 1570 in winnings and 1665 in losses resulting in a bankroll of 0
Lasted 2374 rolls with a total of 1855 in winnings and 1750 in losses resulting in a bankroll of 200
Lasted 257 rolls with a total of 255 in winnings and 150 in losses resulting in a bankroll of 200
.
.
deleted about 995 lines of output for purposes of brevity....
.
.
Lasted 403 rolls with a total of 255 in winnings and 355 in losses resulting in a bankroll of 0
Average number of rolls over 1000 trials was 1299.09
Walked away a winner 330 times (33%)
Walked away a loser 670 times (67%)

----------- cut here ----- source code -------

#!/usr/bin/perl
$br = 100; # bankroll
$numtrials = 1000;
$minbet = 5;
$point = 0;
&main;
exit 0;

sub roll
{
$d1 = int( rand( 6 ) ) + 1;
$d2 = int( rand( 6 ) ) + 1;
$roll = $d1 + $d2;
$rollcount++;
}

sub resolve
{
if( $point == 0 ) #comeout
{
if( $roll == 12 || $roll == 2 || $roll == 3 )
{
$losses += $passline;
$passline = 0;
}
elsif( $roll == 7 || $roll == 11 )
{
$wins += $passline;
$br += $passline;
}
else
{
$point = $roll;
}

return;
}
else
{
if( $roll == 7 )
{
$losses += $passline;
$passline = 0;
$point = 0;
}
elsif( $roll == $point )
{
$wins += $passline;
$br += $passline;
$point = 0;
}
}
}

sub main
{
$initial_br = $br;

for( $trial=0; $trial < $numtrials; $trial++ )
{
$br = $initial_br;
$wins = 0;
$losses = 0;
$passline = 0;
$rollcount = 0;

while( $br >= $minbet && $br < $initial_br * 2 )
{
if( $passline < $minbet )
{
$passline += $minbet;
$br -= $minbet;
}

&roll;
&resolve;

# $results += sprintf( "$rollcount : br $br roll $roll point $point\n" );
}

$total_rolls += $rollcount;
if( $br > 0 )
{
$total_win_trips++;
}
else
{
$total_lose_trips++;
}

$results .= "Lasted $rollcount rolls with a total of $wins in winnings and $losses in losses resulting in a bankroll of $br\n";
}

print $results;

$avg = $total_rolls / $numtrials;

print "Average number of rolls over $numtrials trials was $avg\n";
printf( "Walked away a winner $total_win_trips times (%d\%)\n", $total_win_trips * 100 / $numtrials );
printf( "Walked away a loser $total_lose_trips times (%d\%)\n", $total_lose_trips * 100 / $numtrials );
}

aahigh

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December 20th, 2010 at 10:48:32 AM permalink

One more line of output I added shows:

Total change in bankroll was -90400 over 8926903 rolls or -0.010127 (-0.2025%) per roll or -0.068422 (-1.3684%) per bet

I think these are already available in the wizard's web pages on house odds and advantages, but it's good to have independent verification. A penny per roll for a $5 passline bet is pretty easy to remember if you are worried about how long you can last or what the long term costs are.

MathExtremist

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December 20th, 2010 at 11:18:48 AM permalink

Something didn't calculate properly there -- the EV/roll of the passline should be twice as high as you're reporting. The rough rule is $5 betting costs 2c/roll, not 1c/roll. That said, if you're playing for free drinks, go to a $10 table at a higher-end casino. Even if you're betting $10 every shooter (which you don't have to do, btw), your hourly EV is in the -$4 range. That's far cheaper than the cost of a good drink.

"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

aahigh

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December 20th, 2010 at 11:48:30 AM permalink

Your probably thinking about place bets per roll. I'm pretty sure $.01/roll is correct on the passline. LMK.

$.02 would be right on a $10 table for the pass.

From:

https://wizardofodds.com/craps

Place Bet Pays Prob. Win House Edge per Bet Resolved House Edge per Roll
6,8 7 to 6 0.454545 0.015152 0.004630
5,9 7 to 5 0.400000 0.040000 0.011111
4,10 9 to 5 0.333333 0.066667 0.016667

MathExtremist

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December 20th, 2010 at 12:39:21 PM permalink

See https://wizardofodds.com/craps/appendix2.html
You got 0.2% per roll; the actual number should be 0.4% or so.

"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

aahigh

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December 20th, 2010 at 1:56:57 PM permalink

Yep, you were right. I had an extra total_rolls += rollcount in my trial loop.

I should have known better than to argue with someone who's name is MathExtremist.

Quote: MathExtremist
See https://wizardofodds.com/craps/appendix2.html
You got 0.2% per roll; the actual number should be 0.4% or so.

aahigh

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December 20th, 2010 at 3:17:12 PM permalink

Shortest session was 70 rolls
Longest session was 9440 rolls
Average number of rolls over 10000 trials starting with 100 bankroll and stopping at broke or 200 was 1249.36 (3.3789 per $5 bet)
Total change in bankroll was -252785 over 12493613 rolls or -0.020233 (-0.4047%) per roll or -0.068158 (-1.3632%) per bet
Walked away a winner 3597 times (35.970%)
Walked away a loser 6403 times (64.030%)

Is this getting closer? Sorry I am a newb to craps sims.

weaselman

weaselman

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December 20th, 2010 at 3:30:31 PM permalink

Quote: MathExtremist
That said, if you're playing for free drinks, go to a $10 table at a higher-end casino.

Why not 1c slots? $100 should be able to get you really-really drunk ...

aahigh: not that it matters very much in this case, but int( rand( 6 ) ) is a lousy way to generate random numbers. Something like int(rand(1<<31)%6) should be a lot better distributed.

"When two people always agree one of them is unnecessary"

aahigh

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December 20th, 2010 at 4:15:00 PM permalink

From "man perlfunc" .. it seems I'm okay here. 0.0 to 5.999999 equally distributed for rand(6) .. rounded down to 0 to 5 with the int and then add one .. or am I missing something you're trying to explain to me.

rand EXPR
rand Returns a random fractional number greater than or equal to 0 and less than the value of EXPR. (EXPR should be positive.) If
EXPR is omitted, the value 1 is used. Currently EXPR with the value 0 is also special-cased as 1 - this has not been documented
before perl 5.8.0 and is subject to change in future versions of perl. Automatically calls "srand" unless "srand" has already
been called. See also "srand".

Apply "int()" to the value returned by "rand()" if you want random integers instead of random fractional numbers. For example,

int(rand(10))

returns a random integer between 0 and 9, inclusive.

(Note: If your rand function consistently returns numbers that are too large or too small, then your version of Perl was probably
compiled with the wrong number of RANDBITS.)

I'm sure I'm about to get schooled again .. LOL.

weaselman

weaselman

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December 20th, 2010 at 4:28:51 PM permalink

I don't know what I was thinking, your method is actually better than what I suggested (because it uses high order bits).
Today's my "opposite day" :)

"When two people always agree one of them is unnecessary"

nope27

nope27

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December 20th, 2010 at 11:14:05 PM permalink

Quote: superrick
I am no math guy, but I do know one thing as far as rolling any point in 100 rolls there is no way of telling, if a 12 is going to roll in the 100 rolls of the dice!

I think us math guys knows what happens at a craps table relative to the theoretical numbers.
I would like to know the frequencies of each 12 that hit, the math shows how close they are also. Remember half of all 12s that roll will do so within 25 rolls.
=BINOMDIST(B2,A$2,A$4,FALSE) in Excel shows a 5.9780% chance of exactly "0" 12s rolled in 100 rolls of the dice.

Quote: superrick
Today there were two 12's rolled in 106 rolls of the dice on the table I was playing on,

"2.9" 12s was the expected number of 12s in 106 rolls.
Exactly "2" 12s rolled in 106 rolls is the most probable outcome at 22.9338%

Quote: superrick
yesterday there were 4 out of 222 rolls.

"6.1" 12s is the expected number of 12s to roll in 222 rolls. So we have a sample that was 2 off. It still was the 4th most probable outcome.

Quote: superrick
Two days ago there were 2 out of 74 rolls!

"2.1" expected number of 12s. dead on!

Quote: superrick
Three days ago there were 5 out of 106 rolls!

"2.9" expected. Table results 2 higher. The 5th most probable outcome.

Quote: superrick
The math of the game does not always add up in real play, unless you are going to be at the table for 10,000 rolls of the dice!

I know you are wrong with the number of trials needed.
Your example shows 508 rolls and "13" 12s rolled.
"14.1" is the expected number that should have rolled in 508 rolls. Interesting...only 1 off. Standard Deviation is 3.7039.

Quote: superrick
I have seen plenty of days that you will not see a 12, even rolled in two or three hours of play!
Take it for what it's worth, just because you haven't seen a 12 in 36 rolls of the dice it does not mean that it's about to come up!

You are 100% correct there. The 12 can disappear for many many dice rolls. I tracked a game once over 350 rolls before a 12 hit just 1 time. But it showed more often after that happened.

Quote: superrick
There is a big difference between the math of the game and what really happens on the tables!...

NO, not really. They are close, considering the standard deviation in the short run.

Relative frequencies can tell a better story than just actual hits of any number.
But I am sure you did not track what roll each 12 hit on so the frequencies can be tabulated.

DJ"s table on page 1 is an accurate "cumulative frequency table", meaning the probability of a frequency hit at or below a certain percentage. Again slightly more than 50% of all 12 "hits" will happen within 25 rolls. After that, the table can show how far it can go... and it can disappear for a long time.

The reason for my lengthy reply was to show that the math does show how close actual play can be in the short run.

Some "math numbers" are more valuable than others.
I gamble by frequencies.

HO HO HO
back to craps I go...

MathExtremist

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Posts: 6526

Joined: Aug 31, 2010

December 20th, 2010 at 11:48:25 PM permalink

Quote: weaselman
I don't know what I was thinking, your method is actually better than what I suggested (because it uses high order bits).
Today's my "opposite day" :)

We don't know that unless we know the scaling method used in perl. I think your assumption was based on generating ints, but it appears that the perl RNG generates doubles. I'm not sure how that's done under the covers (but I bet that info is readily available).

"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

weaselman

weaselman

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Posts: 2349

Joined: Jul 11, 2010

December 21st, 2010 at 5:04:11 AM permalink

Quote: MathExtremist
We don't know that unless we know the scaling method used in perl. I think your assumption was based on generating ints, but it appears that the perl RNG generates doubles. I'm not sure how that's done under the covers (but I bet that info is readily available).

Perl doesn't have its own RNG, it just uses whatever the underlying system has. I have never heard of an algorithm, that generates random doubles out of the box (I don't claim a lot of expertise in the area, so may be wrong, but I do know some stuff) - usually it is either individual bits, or groups of bits (like 8 or 32 at once) that get generated. Once you generate enough bits, you can always interpret them as an int, and divide by the maximum possible number to get a [0-1] double.

"When two people always agree one of them is unnecessary"

aahigh

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Posts: 77

Joined: Dec 10, 2010

December 21st, 2010 at 9:10:51 AM permalink

Quote: weaselman
Perl doesn't have its own RNG, it just uses whatever the underlying system has. I have never heard of an algorithm, that generates random doubles out of the box (I don't claim a lot of expertise in the area, so may be wrong, but I do know some stuff) - usually it is either individual bits, or groups of bits (like 8 or 32 at once) that get generated. Once you generate enough bits, you can always interpret them as an int, and divide by the maximum possible number to get a [0-1] double.

It's true that there is no RNG source code built into the base perl.

It can be configured a couple of ways, you can see what the configure script prefers (drand48):

: How can we generate normalized random numbers ?
echo " "
echo "Looking for a random number function..." >&4
case "$randfunc" in
'')
if set drand48 val -f; eval $csym; $val; then
dflt="drand48"
echo "Good, found drand48()." >&4
elif set random val -f; eval $csym; $val; then
dflt="random"
echo "OK, found random()." >&4
else
dflt="rand"
echo "Yick, looks like I have to use rand()." >&4
fi
echo " "
;;
*)
dflt="$randfunc"
;;
esac
cont=true

hightoa/cygdrive/c/perl-5.12.1> grep drand01 *
Configure:drand01=''
Configure: drand01="drand48()"
Configure: drand01="($randfunc() / (double) ((unsigned long)1 << $randbits))"
Configure: drand01="($randfunc() / (double) ((unsigned long)1 << $randbits))"

and the source code for perl's rand() function:

#ifndef HAS_DRAND48_PROTO
extern double drand48 (void);
#endif

PP(pp_rand)
{
dVAR; dSP; dTARGET;
NV value;
if (MAXARG < 1)
value = 1.0;
else
value = POPn;
if (value == 0.0)
value = 1.0;
if (!PL_srand_called) {
(void)seedDrand01((Rand_seed_t)seed());
PL_srand_called = TRUE;
}
value *= Drand01();
XPUSHn(value);
RETURN;
}

PP(pp_srand)
{
dVAR; dSP;
const UV anum = (MAXARG < 1) ? seed() : POPu;
(void)seedDrand01((Rand_seed_t)anum);
PL_srand_called = TRUE;
EXTEND(SP, 1);
RETPUSHYES;
}

There is also a Math::TrulyRandom module available from CPAN for those who want better results than what is available through your operating system.

Boy you guys like to dig deep on this stuff...

MathExtremist