How often will I see 9 Don't Passes in a row per 100,000 decisions?

I’d like to confirm in Craps that out of every 100,000 decisions, I could expect to see 278 strings of 9 (or longer) Passes or 9 Don’t Passes in a row.

I arrived at that by taking .52 to the ninth power (.52^9 = .00277), and multiplied it by 100,000.
However, if this is correct, does that mean there will be 278 Don’t Pass *and* 278 Pass string of 9 consecutive decisions, or a combined total of 278 Don’t Pass/Pass strings of 9 consecutive decisions *combined*? Meaning about 139 DP and 139 P string of 9+ decisions consecutive (i.e., 9 DPs + 9 P consecutive decisions, which include longer string of 10, 11, 12 etc…)

Or is all of the above wrong and how would you calculate how often I will see 9 DPs in a row (or more) per 100,000 decisions?

In other words, out of 100,000 total decisions (about 1,333 hours of play at 75 decisions per hour) – would I see the DP side have 9 decisions in a row (or more) a total of 139 times or a total of 278 times?

Or, would this number be higher, because string of 10, 11, 12 losses in a row on DP side are in addition to the number of times I would lose 9 times in a row.

Sorry of my explanation was not perfect. I can clarify if needed.

Quote: NewtoTown

I’d like to confirm in Craps that out of every 100,000 decisions, I could expect to see 278 strings of 9 (or longer) Passes or 9 Don’t Passes in a row.

For at least 9 Passes (pass line wins) in a row I get 87.116 per 100,000 decisions.
I think that your 278 average number of times would be way too high for either one

You need to define what you actually want for the don't pass as the 'push' can either end a don't pass run
or
keep it where it is at.
one can not have both

the math for both (or just one, that I use) is more involved than how you calculated it.

There’s a 196/495 chance the next decision will be a DP.

The expected number of rolls to get 9 consecutive DPs is (495/196)^1 + (495/196)^2 ..... + (495/196)^9 = 6,918. So you’d expect 100.000 / 6,918 = 14.5 streaks of 9 DPs. I’m not saying “or more”‘ since, for example, a streak of 27 counts as 3.

If you disregard decisions made on come-out rolls, there’s a 98/165 chance the next decision will be DP. In this case the expected number of rolls to get 9 consecutive DPs is (165/98)^1 + (165/98)^2 ..... + (165/98)^9 = 265. So you’d expect 100.000 / 265 = 377 streaks of 9 DPs

Would that be with odds or just flat bets on the line?

Thank you for the reply. Just to make sure I got it right... not that this is why I am asking, but to illustrate whether I understood the answer, if someone 'Martingaled' using a series of 8 bets...10, 20,40, 80, 160, 320, 640, 1280 etc... one would lose only 14.5 times per 100,000 decisions (assuming all bets were pass line and made including the come out roll). I.e., streaks of DP would only come up 14.5 decisions (not rolls)?

I recorded 7 such streaks in just 8000 actual decisions (meaning P and DP decisions combined).

Example $2550 x 14.5 ‭=$36,975‬ lost Martingaling due to streaks of 9+ ($10+$20+$40...$1280 totaling $2550 for a series)?

The win rate would be much more than the $36,975 lost.

I am not challenging your math, but trying to understand how one could ever lose Martingaling if the actual number were not much higher.

With or without odds, but I think their would be way more streaks of 9+ DPs streaks per 100,000 P/DP decisions.

Would you be able to post how you got to 87.116?

I'd bet on hitting 7-11 on the come-outs for a win streak; and winning the odds bets on the Don'ts for a win streak. I'd try to make it so I win a 10 bet session on 6-7 wins.

Some of my best win days at the tables are done betting the DC with no odds and I win 21 bets.

I get 69.6 runs of 9 consecutive Don't Passes (counting 18 in a row as 2 x 9) per 100,000 comeouts.

As another estimate, assuming a 196/495 chance next decision being DP, your average trial to get 9 consecutive DPs will last 1.65 decisions. So in 100k decisions we expect about 60k trials. Each trial has a (196/495)^9 = 0.000239 chance of success. Apply binomial distribution and the most probable value is 14 at 10.6%

Quote: NewtoTown

Would you be able to post how you got to 87.116?

this was done long ago and the proof of the math was done then also.

expected # of runs = (P^Run)*(1+((Trials-Run)*(1-P)))

for the pass line to lose at least 9 in a row (does not mean the DPass wins 9 in a row)
109.237 in 100k decisions

I would just simulate the DPass to get a close average
or simulate both to get the combined average

turning this into a Marty discussion
is not at all fun
winning $600 in a casino because they called my name is fun
maybe they do it again tomorrow night.

good luck

Quote: ThatDonGuy

I get 69.6 runs of 9 consecutive Don't Passes (counting 18 in a row as 2 x 9) per 100,000 comeouts.

Agreed. In my calculation I had used p=196/495 which is the probability the next decision will be a seven-out (excludes DP wins on come out rolls) instead of 949/1980.

The actual expected number of streaks will be slightly lower than 69.59575 because taking the number of decisions divided by the average number of decisions required to get 9 consecutives includes streaks that are “in progress” at the end of the experiment. Like if the last 5 decisions were P-DP-DP-DP-DP

As a simple example, we know that the average number of coin flips to get one head is 2, two consecutive heads is 6 flips and three consecutive heads is 14 flips. If we flip 3 times, the chance for getting the 3 heads is 1 in 8 permutations. However, dividing the number of flips by the average is 3/14 instead of 1/8 because it includes “in-progress” streaks which are permutations TTH, HTH (both 2/14 of the way to HHH) and THH (6/14 of the way). Aggregate them as (1 + 2/14 + 2/14 + 6/14) * 1/8 weighting = 3/14.

The effect on the Craps example will be tiny...less than a basis point due to 100,000 trials. But worth mentioning.