February 20th, 2013 at 3:21:52 PM
permalink

To get the measure of the ups and downs of passline play, I ran ten 1,000,000-round simulations in a spreadsheet program. All ten simulations produced streaks of at least 18 wins in a row and there was one instance of 22 wins in a row. I’m fairly certain the output of the random number generator was programmed correctly. Although none of the win/loss ratios conformed exactly to the 0.01414 casino edge, all ten were not far from it. Still, even 18 wins in a row seems a bit improbable even for 1,000,000 rounds. The number of required rolls before going 7-out would likely be considerable even with a high frequency of naturals on the come out roll. Some statisticians have claimed the odds against the 154-roll run that happened at Borgota in 2009 were more than 5 billion to 1. That would seem to put the veracity of my simulation data in doubt. Is my simulation data all wet?

February 20th, 2013 at 3:41:39 PM
permalink

All you really have to do is take the odds of getting one point and take it to the exponent 22. I am assuming 7,11 are passline decisions?

If that's the case, the odds of having a single positive passline decision is: 8/36 + 6/36*1/3 + 8/36*2/5 + 10/36 *5/11 = .4929293. The odds of having 22 decisions in a row is 5,737,704.75:1 but that is the odds if you start at a certain point. The odds of getting 22 passes in a row is not 5.737:1 (though I think it's pretty close) because the trial starts at the 1st roll.

I think the odds are 1- (1-(.4929293^22)^999979 = 1- .8400601 = .1599399

Someone will correct me I am sure.

If that's the case, the odds of having a single positive passline decision is: 8/36 + 6/36*1/3 + 8/36*2/5 + 10/36 *5/11 = .4929293. The odds of having 22 decisions in a row is 5,737,704.75:1 but that is the odds if you start at a certain point. The odds of getting 22 passes in a row is not 5.737:1 (though I think it's pretty close) because the trial starts at the 1st roll.

I think the odds are 1- (1-(.4929293^22)^999979 = 1- .8400601 = .1599399

Someone will correct me I am sure.

-----
You want the truth! You can't handle the truth!

February 20th, 2013 at 3:59:58 PM
permalink

NoQuote:mathdoltIs my simulation data all wet?

From Sally's Streak Calculator (and VBA Excel sheet linked to),

two different programs returned the same results.

http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html

For at least 18 in a row pass line wins: 0.776180221094645

1.5 is the average number of runs 18+ in 1 million trials

The average number of trials to see 18 in a row: 668,047

Event Run Probability 1 in

0 runs of length 18 or more 0.223819779 4.47

at least 1 run of length 18 or more 0.776180221094645 1.29

could even be more

at least 2 runs of length 18 or more 0.441131957559661 2.27

at least 3 runs of length 18 or more 0.190365055189856 5.25

at least 4 runs of length 18 or more 0.065245344895849 15.33

at least 5 runs of length 18 or more 0.018425880846359 54.27

at least 6 runs of length 18 or more 0.004410623975975 226.73

at least 7 runs of length 18 or more 0.000914568073469 1,093.41

at least 8 runs of length 18 or more 0.000167099713893 5,984.45

at least 9 runs of length 18 or more 0.000027270073853 36,670.23

at least 10 runs of length 18 or more 0.000004019318281 248,798.41

at least 11 runs of length 18 or more 0.000000539942486 1,852,049.11

For at least 22 in a row pass line wins: 0.084581177224990

0.088373413 is the average number of runs 22+ in 1 million trials

The average number of trials to see 22 in a row: 11,315,392

`Event Run Probability 1 in`

0 runs of length 22 or more 0.915418823 1.09

at least 1 run of length 22 or more 0.084581177224990 11.82

at least 2 runs of length 22 or more 0.003682169900757 271.58

at least 3 runs of length 22 or more 0.000107655858051 9,288.86

at least 4 runs of length 22 or more 0.000002367580999 422,372.03

at least 5 runs of length 22 or more 0.000000041714686 23,972,372.54

added: The probability of a 22 length streak and the average number of streaks in 1 million trials are very close.

winsome johnny (not Win some johnny)

February 20th, 2013 at 4:00:42 PM
permalink

Here's how I figured it.

On average, it takes 3.376 rolls to get a pass-line decision. That means you'll have roughly 296,208 pass-line decisions for every 1,000,000 rolls.

The pass line wins 49.3% of the time.

Put the following numbers into a streak calculator (http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html):

Trials: Series Length: (n) = 296208

Streak Length (or more): (k) = 22

Probability of Success: (p) = 0.493

...and we see that in 1,000,000 rolls you'll get 22 or more pass-line wins in a row about 2.6% of the time.

On average, it takes 3.376 rolls to get a pass-line decision. That means you'll have roughly 296,208 pass-line decisions for every 1,000,000 rolls.

The pass line wins 49.3% of the time.

Put the following numbers into a streak calculator (http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html):

Trials: Series Length: (n) = 296208

Streak Length (or more): (k) = 22

Probability of Success: (p) = 0.493

...and we see that in 1,000,000 rolls you'll get 22 or more pass-line wins in a row about 2.6% of the time.

YouCanBetOnThat.com, a podcast for the recreational gambler

February 20th, 2013 at 4:04:01 PM
permalink

OP also used the word "rounds" instead of rolls.

Maybe he will confirm

added: your numbers look good for 1 million rolls

Maybe he will confirm

added: your numbers look good for 1 million rolls

winsome johnny (not Win some johnny)

February 20th, 2013 at 4:04:37 PM
permalink

Just re-reading the whole post. My calculations were based on 1,000,000 rolls (as indicated in the subject line), not 1,000,000 decisions.

YouCanBetOnThat.com, a podcast for the recreational gambler

February 20th, 2013 at 4:07:27 PM
permalink

It is closer if useQuote:boymimboI think the odds are 1- (1-(.4929293^22)^999979 = 1- .8400601 = .1599399.

.4929293^23 and change the other exponent 1.

(I also think you are just calculating the average number of such runs in 1 million trials)

I saw the Wizard use this method one time and got real close.

added:

The Wizard's method is here

http://wizardofvegas.com/forum/questions-and-answers/math/4582-policing-the-media/2/#post58529

1. A run of exactly 20 or more heads has probability of (1/2)^21 for any given starting point. The 21 in the exponent is because a tail must proceed the run of at least 20 heads.

2. The expected number of such runs is 1000000*(1/2)^21 = 0.476837.

3. The probability of 0 such runs is exp(-0.476837) = 0.620744.

4. The probability of at least 1 such run is 1-0.620744=0.379256.

Using the Wizards method I get 0.082323674

But in my old worksheet I have a different formula that returns 0.084581016 for the probability... even closer

Alternate Formula:

=1-EXP(-A)

A=(p^run)*(1+((trials-run)*q))

-------------------------------------------

p= probability of success

run = length of the run (streak)

q = 1-p

-------------------------------------------

A is "the expected number of runs for length X" formula.

(EXP is an Excel function

Returns e raised to the power of number. The constant e equals 2.71828182845904, the base of the natural logarithm)

This method works well when the length of the run is quite high (like over 15)

and the probability of a run and the average number of runs are close

or errors start creeping in and can get quite large for shorter run lengths

You never know when you can not get to a streak calculator

and are driving in your car and you just need to calculate this in your head,

the alternate formula does the job so you can get back to enjoying your ride and drive.

winsome johnny (not Win some johnny)

February 20th, 2013 at 6:34:59 PM
permalink

My intended meaning was that a round requires a passline decision to be completed. Thus a round can be composed of 1 or many rolls.

February 20th, 2013 at 7:18:19 PM
permalink

A few of us got that.Quote:mathdoltMy intended meaning was that a round requires a passline decision to be completed. Thus a round can be composed of 1 or many rolls.

You have now both answers for rolls and rounds

as the title of the thread had "rolls" and your first post had "rounds"

to summarize since we also showed how this could be calculated,

1 million rounds (pass line decisions)

For at least 18 in a row pass line wins: 0.776180221094645

For at least 22 in a row pass line wins: 0.084581177224990 or about 1 in 12

and "1,000,000 rolls you'll get 22 or more pass-line wins in a row about 2.6% of the time" or about 1 in 39

your sims and results look just fine

winsome johnny (not Win some johnny)

February 20th, 2013 at 7:42:30 PM
permalink

Thank you all for responding and answering with such clarity. Like the blind man said, I now see. A good number of hours were put into that friggin simulation database. It's a relief to know that it’s at least reasonably sound. I still wonder though about how many back-to-back passline wins the run at the Borgata actually consisted of. The information should have been retrievable from the casino video tapes.