Confused by the Game

So I've been playing Craps for 5-6 years now... I go every now and then and tend to play fairly low limit ($1-10) minimums ($1, local tribal casino, $10, Vegas).

The more I watch the game, the more I believe it is incredibly random and that even "variance" in craps is incredibly difficult to define (yet I seem to find the massive run on 7's variance a lot!).

I'm re-thinking my whole strategy - Normally I'll play 26/7 (52/54) across and at least 3x on the pass - sometimes I will play the come instead of the odds. What I'm thinking of doing is playing only the pass with max odds.

What I can't figure out is this:
From what I understand, out of 100 opportunities to point, roughly 41 will and 59 won't. Is it possible to figure out the expected win/loss if I were to go through 100 come-outs, place the table minimum (call it $5) and back it up with max odds for that point?

I'm also curious if anyone could estimate (or predict) or whatever, after a point is set, how many rolls, on average happen until a 7-out? My gut tells me that can't be figured out.

Anyone?

Quote: foolshope

What I'm thinking of doing is playing only the pass with max odds.

that is the Wizard recommended way, with the precaution that you should not strain your roll just to do max odds, thinking that is how to win more.

Quote:

Is it possible to figure out the expected win/loss if I were to go through 100 come-outs, place the table minimum (call it $5) and back it up with max odds for that point?

Easy, you risk $500 overall on the line, at 1.41% house edge ... you are expected to lose $7.05. Here's the rub: it will be a rare session that will resemble that outcome. The free odds do not affect the expected loss, but if you play them the session results will vary like crazy.

Quote:

I'm also curious if anyone could estimate (or predict) or whatever, after a point is set, how many rolls, on average happen until a 7-out? My gut tells me that can't be figured out.

Bear in mind Craps math is actually totally crunchable. Many people have figured this one out; don't have time to search but I think it is about 3 rolls till 7-out or point won.

The Wizard has answered:

Quote: https://wizardofodds.com/ask-the-wizard/craps-probability/, bottom of page

Q: What is the average number of rolls (throws?) till a shooter "sevens out"? I know that a 7 will appear every 6 rolls, but with come-out 7-11s and craps, plus the possibility of shooters making multiple points, I think the average number of rolls may be higher than expected. Is there any mathematical reference material on this? - Grshooter from Kansas City, Missouri

First, if the probability of an event is p then the expected number of trials for it to occur is 1/p. Let's call x the expected number of rolls per shooter. The probability that any given round will end in one roll (with a 2, 3, 7, 11, or 12) is 1/3. If the player rolls a 4 or 10 on the come out roll the expected number of additional rolls is 4, because the probability of rolling a 4 or 7 is (6+3)/36 = ¼. . Likewise If the player rolls a 5 or 9 on the come out roll the expected number of additional rolls is 3.6 and for a 6 or 8 is 36/11. Assuming a point was thrown the probability of it being a 4 or 10 is 3/12, a 5 or 9 is 4/12, and a 6 or 8 is 5/12. So the expected number of throws per round is 1+(2/3)*((3/12)*4 + (4/12)*3.6 + (5/12)*(36/11)) = 3.375758. Next, the probability that the player will seven out is (2/3)*((3/12)*(2/3) + (4/12)*(3/5) + (5/12)*(6/11)) = 0.39596. The probability that player will not seven out is 1 - 0.39596 = 0.60404. So...

x = 3.375758 + 0.60404*x
0.39596*x = 3.375758
x = 8.52551.

The wizard states that the expected number of throws PER ROUND is 3.375758. After the point is established however, the number of rolls is 3/12 * 36/9 + 4/12 * 36/10 + 5/12 * 36/11 = 3.56363.

Quote: foolshope

I'm re-thinking my whole strategy -
Normally I'll play 26/7 (52/54) across and at least 3x on the pass - sometimes I will play the come instead of the odds.

What I'm thinking of doing is playing only the pass with max odds.

Pass with max odds can give you a good shot at a profit in your 100 PL bets. Also requires a smaller bankroll than going across with pass/full odds.
results from 1 million session simulation:
added parameters (forgot first time around)
100 pass line resolved wagers. $1 pass, 345x odds

Bankroll decreased 50.876% of the time
Bankroll increased 48.304% of the time
Avg (mean) -$1.36
Std-dev ending bankroll $49.18
edge -0.359%

Quote: foolshope

What I can't figure out is this:
From what I understand, out of 100 opportunities to point, roughly 41 will and 59 won't.

Is it possible to figure out the expected win/loss if I were to go through 100 come-outs, place the table minimum (call it $5) and back it up with max odds for that point?

41 and 59 looks to be correct on the point ratios.
Yes it is possible to figure out the expected win/loss from your above example.

I'm too lazy to do that, maybe someone else will, since I already have simulation results that match your example parameters.

100 pass line resolved wagers. $5 pass, 345x odds
Place across $26 or $27 not including the point.

Results
mean: -$73.38
sd: $340.51
high: $1285
low: -$1290
edge: -2.2%
41% of sessions resulted in a net profit

Quote: foolshope

I'm also curious if anyone could estimate (or predict) or whatever, after a point is set, how many rolls, on average happen until a 7-out? My gut tells me that can't be figured out.
Anyone?

boymimbo also shows how to create the distribution with a few examples in his post
Pass-Odds only Scheme Thread

My last reply didn't make it... I'll try to repeat and condense...

In your simulation, if only the pass line was played, then it was about $50/50 with 3 Std-devs +/- $150. I presume this would double if I was playing $10 minimums and be 20% if I was playing $1?

In simulation #2, profit happened about 41% of the time and 3 Std-devs were ~ +/- $1200.

Correct?

So if I want to win big money, be prepared to lose big money, yes?

Quote: foolshope

My last reply didn't make it... I'll try to repeat and condense...

In your simulation, if only the pass line was played, then it was about $50/50 with 3 Std-devs +/- $150. I presume this would double if I was playing $10 minimums and be 20% if I was playing $1?

Nice question. I forgot to add the parameters for sim#1.
100 trials, pass line $1 with 345x odds.
So you can just multiply the results by the pass line bet factor. 5x for a $5 bet, 10x for a $10 wager etc.

Here are the theoretical numbers for $1 pass line with 345x odds and the standard deviation (if I can locate a thread with the math for this I will post the link)
$1 pass
345X odds
avg bet: 3.78
expected loss: -$1.41 (for 100 trials)
sd for 1: $4.91563184

Remember to take the square root of the number of trials and multiply that by the SD and you will have the sd for the number of trials.

Example:
$5 pass line
sd: $24.5781592
square root of 100 = 10.
10*$24.5781592 = $245.78

Quote: foolshope

In simulation #2, profit happened about 41% of the time and 3 Std-devs were ~ +/- $1200.
Correct?
So if I want to win big money, be prepared to lose big money, yes?

Yes or have Lady Luck smile down on you more than expectation. It does happen.

Now I did run these sims with a very high bankroll so I would not bust out.
The pass line/odds high was ~$900 and low ~$900 with $5 pass flat bet, more than 3SDs.

Foolshope: You are on the right track with your thinking. Just bet the pass-odds with the max odds you can afford with at least a 10 shooter session bankroll, and you will do quite well overall. $5 pass with $50 odds with $600 session bankroll does quite well overall, most sessions.

I like the following just a little better though: Similar to the the 3x4x5x odds but doubling it to 10x8x6x odds. Meaning, $5 pass and if the point is 6 or 8 then $50 odds, if the point is 5 or 9 then $40 odds, if the point is 4 or 10 then $30 odds. Any point win is $65. For some reason, this does a little better than level odds on all points.