dwheatley
Joined: Nov 16, 2009
• Posts: 1246
October 8th, 2012 at 8:05:13 PM permalink
You have a ~6% chance of throwing 148 or less 7s in 1000 throws. That's somewhat convincing, but not really enough.

You still have a 21% chance of throwing 159 or less 7s in 1000 throws. Not convincing.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
CrapsForever
Joined: Mar 6, 2012
• Posts: 517
October 8th, 2012 at 8:11:10 PM permalink
Good Times.....
Craps is the most "Jekyll and Hyde" casino game ever invented!
buzzpaff
Joined: Mar 8, 2011
• Posts: 5328
October 8th, 2012 at 8:13:01 PM permalink
I once saw a drunk pick a 15 team parlay. He was not the world's greatest handicapper, just a statistical anomoly.
dwheatley
Joined: Nov 16, 2009
• Posts: 1246
October 8th, 2012 at 8:14:14 PM permalink
I'd be convinced around a 1% chance, so 2100 throws for 14.8%, and 13'000 throws for 15.9%.

Or I'd have to see what 7craps just posted about. That would convince me.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
7craps
Joined: Jan 23, 2010
• Posts: 1977
October 8th, 2012 at 8:18:11 PM permalink
from
http://forumserver.twoplustwo.com/47/science-math-philosophy/statistics-question-calculating-required-sample-size-271828/

Do not put WongBo down.

It comes down to the degree of certainty and margin of error wanted.

If it is really all about just rolling less 7s than a certain acceptable range of outcomes,
and we know it is not...

For the range of a random shooter (DI would have to be outside that range more times)

The standard deviation for the binomial distribution is sqrt(Npq) = sqrt(N*1/6*5/6).

Say we want N such that 0.01*N = 3.09025 standard deviations.

This would be a CL of 99.9% with a error of 0.01(1%)

0.01*N = 3.09025*sqrt(N*1/6*5/6)

N = (3.09025/0.01)^2*(1/6*5/6)

N= 572,978.70
The error rate is about 1/sqrt(N)

The SD range and margin of error can be lowered. (lower the overall certainty)
a CL of 95% with an error of 0.01
1.959964 SD
N=230,487.53

Still too high?
a CL of 90% with an error of 0.01
1.644854 SD
N=230,487.53

Maybe just a .05 (5%) error rate instead
N=6,493.31

===================================
OK
How about a 50% CL and a 10% error rate
N=272.96
Nice!!!!

Last offer
An 80% CL with a 13.5% error rate
N=540.70
added: (cashed on the Jets \$\$\$... got to love NFL home dogs!)

IMO,
Any DI that claims to roll less 7s should beat any random range consistently.

Not really sorry for all the math.
The numbers can be agreed upon, but are still only meaningful to a certain degree of confidence.

No one really wants to be confident in any DI claim.

Just have faith and bet with the DI.

Especially Ahigh, he IS 1 in a million.
Special talents with the dice.

Even if you have an edge as a DI, you better make the right bets,
or you are just as -EV as the rest.
winsome johnny (not Win some johnny)
7craps
Joined: Jan 23, 2010
• Posts: 1977
October 8th, 2012 at 8:24:11 PM permalink
Quote: dwheatley

You have a ~6% chance of throwing 148 or less 7s in 1000 throws. That's somewhat convincing, but not really enough.

You still have a 21% chance of throwing 159 or less 7s in 1000 throws. Not convincing.

Even blind Gorillas throwing the dice can do this.
Nothing special we can all agree
winsome johnny (not Win some johnny)
7craps
Joined: Jan 23, 2010
• Posts: 1977
October 8th, 2012 at 9:06:35 PM permalink
Quote: dwheatley

I'd be convinced around a 1% chance, so 2100 throws for 14.8%, and 13'000 throws for 15.9%.

Or I'd have to see what 7craps just posted about. That would convince me.

No convincing needed.

My idea since large sample sizes are too large,
set the number of dice rolls to 36 per session.

Since 1 in 21 random shooters would roll 0,1 or 2 - 7s (EV=6)
The DI would have to do better.
or
Since 1 in 86 random shooters would roll 0 or1 - 7s (EV=6)
The DI would have to do better more times on average to win some confidence.

Maybe just back to back
winsome johnny (not Win some johnny)
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
October 8th, 2012 at 9:26:23 PM permalink
Quote: 7craps

No convincing needed.

My idea since large sample sizes are too large,
set the number of dice rolls to 36 per session.

Since 1 in 21 random shooters would roll 0,1 or 2 - 7s (EV=6)
The DI would have to do better.
or
Since 1 in 86 random shooters would roll 0 or1 - 7s (EV=6)
The DI would have to do better more times on average to win some confidence.

How can someone do better than rolling no 7s?
"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
7craps
Joined: Jan 23, 2010
• Posts: 1977
October 8th, 2012 at 9:33:28 PM permalink
Maybe just back to back
0,1 or 2 per 36 rolls

1 in 21 is an average as you know.
They would just have to do better than average per 36 roll session

I can't do it
winsome johnny (not Win some johnny)
SOOPOO
Joined: Aug 8, 2010
• Posts: 6907
October 8th, 2012 at 9:38:37 PM permalink
Quote: tupp

How many rolls would satisfy you if someone claims that they can roll 15.9% (or less) 7s?

... How about if they claim 14.8% (or less) 7s?

I am not a 'math guy' anymore, but I'll bet any sum you want that 148 or fewer out of 1000 won't be acheived.
A 'math guy' here will probably say that 148 or lower would happen a few percent of the time by normal variance, but I'll take that chance.
The 15.9% would require too many rolls to be practical for me to get involved in, likely 10000 rolls...