JyBrd0403
Joined: Jan 25, 2010
• Posts: 548
December 12th, 2013 at 8:43:03 PM permalink
Quote: MathExtremist

However, there is no information that you can use to conditionally inform the probability of winning a passline bet prior to it starting. You don't know what you're going to roll on the comeout roll, and the past wager has no impact on the next one. The unconditional probability of winning a passline bet is 244/495, period. It's certainly true that the conditional probability of winning the passline is far less if your first roll is 4, but it's just as true that the conditional probability of winning the passline is (or was) 100% if your first roll is 11, and 0% if your first roll is 3.

Here's an exercise: break down the passline into its component parts, comeout and post-comeout. Determine the probability of a streak of 3 passline wins for each possible permutation of component outcomes. Then weight them according to their probabilities and add them up. Here's a start: the probability of 3 come-out losses is 1/729, which doesn't count as a successful 3-streak. The probability of 3 come-out wins is 8/729, which does. You do the rest.

It turns out the result will equal 19927/166375, or exactly the same as just taking the overall probability of a win (244/495) and cubing it.

I understand what your saying here. I don't know how else to explain that this is an average of all possible outcomes. Here's what I would say. I would say you have a 27% chance of winning 1/3 of your trials, and 60% chance of winning 2/3 of your trials, as a Don't Pass better. 1/3 of the trials will be settled on the come out roll, and 2/3 will be decided on an established point.

My guess is you're going to say each trial is .493 chance of winning. I'm offering the reality of what will happen, and your giving an average chance of winning. Somebody's streaks are going to be off. That's all I'm saying.
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
December 12th, 2013 at 9:54:22 PM permalink
Quote: JyBrd0403

I understand what your saying here. I don't know how else to explain that this is an average of all possible outcomes. Here's what I would say. I would say you have a 27% chance of winning 1/3 of your trials, and 60% chance of winning 2/3 of your trials, as a Don't Pass better. 1/3 of the trials will be settled on the come out roll, and 2/3 will be decided on an established point.

My guess is you're going to say each trial is .493 chance of winning. I'm offering the reality of what will happen, and your giving an average chance of winning. Somebody's streaks are going to be off. That's all I'm saying.

I think you're agreeing with me, you just don't know it. If you actually compute the probability of the chances of a 3-streak in the way you propose, where you break it down into your 1/3 vs. 2/3 trials, you'll see that the final probability is exactly the same as just doing it using the overall average. Go ahead and try.
"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
January 3rd, 2014 at 6:52:08 PM permalink
Quote: JyBrd0403

My guess is you're going to say each trial is .493 chance of winning.
I'm offering the reality of what will happen, and your giving an average chance of winning.
Somebody's streaks are going to be off. That's all I'm saying.

I go with nobody's streaks will be off from expectation.
If they are, I do not see it in actual data.

Streak distribution is based from winning and losing probabilities.
For bets that can win at different steps in different rounds of play, the total is the value to use.
both the pass line and don't pass streaks are as calculated using total win probability for both rounds of play.

the don't pass win streaks
the Zumma actual 35,097 dice rolls from 2 Las Vegas casinos
I see a distribution based from the winning probabilities from both rounds.

Here are the DPass win streaks with the expected number from the total win probability of 949/1925
a push does not break or stop a win streak
# of decisions: 10338
wins: 5112
win percentage: 49.448636%
proba of a win: 49.298701%
standard deviation of win percentage over 10,338 bets: 0.4917%

the lose streaks are also very close as are both from the pass line bets.
(maybe I post those later once I get them on the same page and the percentages for each length also)
win streak lengthfreqor lessexpected numberor lessdiffor less
1135913591310.3637231310.36372348.6362774548.63627745
26412000645.92982211956.293545-4.92982209643.70645535
33282328318.4042142274.6977599.59578597253.30224132
41622490156.95395862431.6517175.0460413958.34828271
570256077.368777822509.020495-7.36877782250.97950489
628258838.138112472547.158608-10.1381124740.84139242
718260618.799774922565.958382-0.79977492140.0416175
81326199.2671480292575.2255313.73285197143.77446947
9526244.5681414882579.7936720.43185851244.20632799
10126252.2518164592582.045488-1.25181645942.95451153
11126261.1100088142583.155497-0.11000881442.84450271
12126270.5471669562583.7026640.45283304443.29733576
13126280.2697200882583.9723840.73027991244.02761567

Maybe you can run some other sets of dice rolls to show that your point of view is the correct one.
I would be interested in seeing that data.
I can run larger simulations also if needed.

If the streaks were different as you say they should be, how could that information be used?
That sounds interesting too.
Craps is just so damn interesting.
winsome johnny (not Win some johnny)