Question about Michael's Probability Table: Seven-Out and Survival

First, thanks to everyone who maintains this website, and also thanks to those who participate and try to help.

I'm not new to the game, but I am newer at studying it. Thirty years ago I was taught the Pass w/ Odds method, and the Don't etc. I haven't really played much because I haven't lived near a casino until recently, and my preferred vacation is to the beach and my boat versus elsewhere.

Probability Table...this table is published as an answer to the question, "what is the probability of a shooter lasting x rolls in craps?"

Given that the average shooter lasts between 8-9 rolls, is this table saying that somewhere between rolls 1 and 9 the average shooter will roll a seven following an unresolved point? A possible scenario then might be: (1) rolls 1 and 2 are 5's, rolls 3 and 4 are 6's, rolls 5 and 6 are 8's, roll 7 is a 7, roll 8 is a 10, and roll 9 is a 7. In other words, the table is not saying that the average shooter will roll 8 times before he rolls his first seven, rather it's saying the average shooter will roll 8 times before disqualifying his turn. Is this a correct interpretation of the table?

Quote: nowakezone

Is this a correct interpretation of the table?

Yes. The expected rolls until any seven is 6.

Thanks Wizard.

I'm going to apologize in advance for starting my next thread because I know you've answered it here and on your other website before, but it still isn't making complete sense to me. The topic is betting the do and don't on the come out and taking the odds. If you'd prefer to zap my keyboard and ban me now I'll understand...lol!!! I just really like that bet so maybe hear me out, then zap me.

I was shown a similar chart with very different figures.. I have to find it and post it

Quote: Wizard

Yes. The expected rolls until any seven is 6.

Isn't it 3.8018 rolls?

(5/6)^x = .5
x * ln(5/6) = ln(.5)
x ~ 3.8018

Quote: wudged

Isn't it 3.8018 rolls?

(5/6)^x = .5
x * ln(5/6) = ln(.5)
x ~ 3.8018

You simply calculated the median (the 50/50 point)
number of rolls to see a 7

rolls	Prob
1	16.6667%
2	30.5556%
3	42.1296%
4	51.7747%
5	59.8122%
6	66.5102%
7	72.0918%
8	76.7432%
9	80.6193%
10	83.8494%
11	86.5412%
12	88.7843%
13	90.6536%
14	92.2113%
15	93.5095%
16	94.5912%
17	95.4927%

The length of a shooter's hand is a bit different.
Why the Wizard refuses to call this, the length of a shooter's hand,
by what other gaming math writers call it, is beyond me.

like calling a ROSE a 'thorny stemmed pretty petal flower'
or maybe a 'Crap Out' when the shooter rolls a 7 with a point established?
close

How about this book
The Doctrine of Chances
Probabilistic Aspects of Gambling
Stewart N. Ethier (yes, he can easily be my newest boyfriend)

15.2 The Shooter’s Hand
"A shooter rolls the dice until he sevens out, that is, until he misses a point.

When this happens, all bets are resolved and a new shooter comes out. The
sequence of rolls by a shooter, from the initial come out to the seven out,
is known as the shooter’s hand.

so the expected length of the shooter’s hand follows easily...

Table 15.2 The distribution of the length L of the shooter’s hand."

This is easily (Sally says so) calculated in a spreadsheet using either a Markov chain
or a simple recursive formula.
I thought the Wizard showed how he did this, not the Markov chain, but that seems to be very difficult to find.


Remember
23 May 2009, at the Borgata Hotel Casino & Spa in Atlantic City, Patricia DeMauro
finally sevening out at the 154th roll?

The media loves this kind of story.
The Wizard was quoted in the news about it.
Had only a simulation completed for the answer that was close.

The wizardofodds website is silent on this event today. (searching Google)
must have got lost in the shuffle.

I do not play Craps these days.
Too many hands lasting at most 6 rolls. (not the average but the median. here they are not the same)
Takes the fun out of playing (for me) if you want to keep rolling the dice.
Sally

These were the numbers I remember seeing

Quote: mustangsally

You simply calculated the median (the 50/50 point)
number of rolls to see a 7

How is this different than "expected rolls until any 7" ?

Quote: wudged

How is this different than "expected rolls until any 7" ?

average (mean) verses the median
The mean is just the average of the numbers.
It is easy to calculate: add up all the numbers, then divide by how many numbers there are

The Median is the "middle number" (in a sorted list of numbers).
http://www.mathsisfun.com/mean.html
http://www.mathsisfun.com/median.html

took me some extra time to understand the differences
Sally

I know the difference between mean and median. I just misinterpreted "expected rolls until any 7" as meaning the median number of rolls instead of mean.

Quote: wudged

I know the difference between mean and median. I just misinterpreted "expected rolls until any 7" as meaning the median number of rolls instead of mean.

yes

for others

flip a fair coin 10 times.

The average (mean or expected) number of heads = 5
the median = 5
because here is the list of the # of heads
0
1
2
3
4
5<< right in the middle (5 higher and 5 lower)
6
7
8
9
10

for the 7
the mean or average = 6 (p=1/6 so 1/p)
the median = 4 (that was shown - calculated nicely) as close as one can get
Sally

Let's say you have four kids ages 5, 7, 7, and 13.

The mean age is (5+7+7+13)/4 = 8.
The median age is 7, because it is in the middle of the distribution.
The mode, not that anyone asked, is also 7, because it is the more frequent age.

mustansally




I have met Pat and that is the whole part of playing the game, it could
happen to you or me.

But at my age i did not want to lose my house trying so i started with
dice controll.

The past 3-4 years i have many long rolls over 40-60 , but none higher
and maybe i wont ever get one higher, but 30-60 minute rolls are great
fun. To be sure, i have had many more very short rolls, and they
are not fun.

So i guess you have to take the good with the bad.

dicesetter