Odds on 6,7,8

Hey Mr. Wizard, I have been corrected on my etiquette and I am thankful for the advice. My question is... in a series of rolls, what are the odds of seeing seven 7's in that series of rolls before 1 six or 1 eight pops. These sevens are not necessarily consecutive. Any help you can provide would be much appreciative as is all of your other information!

Thanks again!

The math would evade me, so can't help.

did this happen to you?

Quote: wrragsdale

Hey Mr. Wizard, I have been corrected on my etiquette and I am thankful for the advice. My question is... in a series of rolls, what are the odds of seeing seven 7's in that series of rolls before 1 six or 1 eight pops. These sevens are not necessarily consecutive. Any help you can provide would be much appreciative as is all of your other information!

Thanks again!

How many faces are on each die? How many are you rolling at once? I'll assume this relates to craps and will use standard 6 sided dice and rolling 2 at the same time.
There are six ways to roll a 7 and five ways to roll a 6 and five ways to roll a 8. The odds that you will roll at least one 7 before a 6 or 8 is (6/16) , at least two sevens before a 6 or 8 (6/16)*(6/16), 3 ...... (6/16)*(6/16)*(6/16) For seven 7's before a 6 or 8 is (6/16)⁷ = 0.00104284286 or 1 in about 959.

It's something I stumbled onto my last Vegas trip...I thought it interesting. I know, I know...another so-called "system". No just something I began using last month and in the casino, I've had 8 sessions, if you will, and all is well. My curiosity peaked and I had no idea how to calculate the math...
BTW, thanks for the advice earlier. I am on the internet alot for business but I hardly ever pay attention to my etiquette. Good info to have!

"You truly have a dazzling intellect".....Wesley to Vizzini after ingesting iocaine powder in the movie The Princess Bride.
Thank you so much!!!
How do like those odds 0.00104284286?

The odds favor the house, be cautious.

Quote: wrragsdale

"You truly have a dazzling intellect".....Wesley to Vizzini after ingesting iocaine powder in the movie The Princess Bride.
Thank you so much!!!
How do like those odds 0.00104284286?

I stand corrected... It was "dizzying" not "dazzling"....sorry!

Quote: miplet

How many faces are on each die? How many are you rolling at once? I'll assume this relates to craps and will use standard 6 sided dice and rolling 2 at the same time.
There are six ways to roll a 7 and five ways to roll a 6 and five ways to roll a 8. The odds that you will roll at least one 7 before a 6 or 8 is (6/16) , at least two sevens before a 6 or 8 (6/16)*(6/16), 3 ...... (6/16)*(6/16)*(6/16) For seven 7's before a 6 or 8 is (6/16)⁷ = 0.00104284286 or 1 in about 959.

The answer miplet gave you is exactly correct, being an elementary school answer.
The easy and fast answer.
Meaning it does not tell the complete story and fails to define the results as apples to apples and oranges to oranges.
At first glance, there appears to be no errors in his calculations and there are none and I am not picking on him.
Time to show the world here how to deal with streaks so they are better understood.
I am no expert, but I understand the terms "mean time" and "waiting time" as it applies to streaks.

For a good discussion about this, view this pdf.
Gambler’s fallacy, hot hand belief, and the time of patterns http://www.decisionsciencenews.com/sjdm/journal.sjdm.org/10/91117/jdm91117.pdf

When dealing with winning and /or losing streaks, the math can get confusing and we can easily mis-understand the results.

Quote: wrragsdale

Hey Mr. Wizard, I have been corrected on my etiquette and I am thankful for the advice. My question is... in a series of rolls, what are the odds of seeing seven 7's in that series of rolls before 1 six or 1 eight pops. These sevens are not necessarily consecutive. Any help you can provide would be much appreciative as is all of your other information!

Thanks again!

First, Let me re-phrase your question as to pertaining to the game of craps.
Q: What is the probability that either 1 or both place6 or place8 would lose 7 times in a row before a win?
Most gamblers want to know about losing streaks. Me too.

This actually implies that the 7-7s ARE consecutive 7s. Your statement "These sevens are not necessarily consecutive." is not correct. They must be consecutive.
Please continue reading.
Let x = 2 or 3 or 4 or 5 or 9 or 10 or 11 or 12.
any consecutive 16 dice rolls could look like...
x6x6xx78xxx6x87x
(I took this string from an actual game I played last week)

so you are looking for something like this:
7x7xxx7x77x7x7xx8
7 consecutive 7s, only counting the 6,7,and 8s.

Remember... We are only dealing with 3 numbers in our formula. The 6,7 and 8.

(6/16)^7 = 0.00104284286 or 1 in about 959.(the answer miplet gave) is the probability for only when n(number of trials of 6,7s and 8s) = 7, and only 7.

What?

Let me continue.
----------------------------------------------
Point #1
1 in about 959 I call that number the mean time or average rolls. But the 959 is only rolls that are 6 or 7 or 8s. Let us convert those rolls into Total Dice Rolls.
Expected number of rolls to see one (1) 6,7 or 8 is 1/(16/36) = 2.25. A nice number.
959 (6,7,8s) * 2.25 = 2,157 expected rolls to see 959 6,7 and 8s.
Now we all can chew on that number.
2,157 total dice rolls. 100 dice rolls per hour. You can take it from there.
Both answers, 959 and 2,157 are correct.
959 are total rolls only counting the 6,7,and 8s that have rolled. (apples)
2,157 are total dice rolls counting all the numbers that have rolled.(oranges)
My answer shows how rare it can be as it relates to the total number of dice rolls.
-------------------------------------------------------
Point #2
I also said above, before Point #1, the answer of 959 is for only when the number of trials, n, is equal to 7, and only 7.
The more trials, the probability of at least 1 run of 7 losses, or more, in a row increases as the number of trials also increase and at the same time the average or mean decreases.
Computer simplations prove my statement is correct and anyone would see that if they run a simulation. I use WinStats for some of my simulations. You can Google WinStats and download it for free. It is worth more than free in my opinion.
When p=.375 and n=7 in my simulations, actually I did not use WinStats, I arrive very close to 959 for the average number of groups of 7 rolls to see 7 in a row.

I would also be concerned about this sequence.
Another dice roll sequence.
x6xxx8xx6x66xxxx7x7xxx7x77x7x7xx8xx68xx
A streak of 7 losses in a row where n = 15. How about when n = 100, 200, 500. Most craps players see those rolls easily in one day of play.
Worse yet, how about 2 runs of 7 losses or more in n trials?
one can use a calculator HERE

Remember to multiply n by 2.25 to get total dice rolls.
One can also DIVIDE the total number of dice rolls by 2.25 to get the expected number of 6,7 and 8s.
How about these larger values of n. (remember n = only 6,7 and 8s.)
n = 445(1001 total dice rolls) 25.02% chance of at least 1 streak of 7 or more.
n = 1064(2394 total dice rolls) 50.01%
n = 7,037(15833 total dice rolls) 99.00%
------------------------------------------------------------------
Point #3: Waitng time (expected number of trials to see a success)
Let me use the #7 in craps as the example, then I will return back to your question's example.

Let us look at the probability of the 7.

We all know, or at least should know, it is 6/36 or 1/6.
Expected number of trials to see 1 seven? (I call the mean recurrence time)
Most math folks know the formula as 1/p. OK. 1/(1/6)=6 or 1 in 6 rolls.

How about 2 in a row.
Easy: (1/6)^2 = 1/36. Expected number of trials to see 2 in a row?
Most math folks know the formula as 1/p. OK. 1/(1/36)=36 or 1 in 36 rolls.

How about 3 7s in a row? I have seen it many times.
Easy: (1/6)^3 = 1/216. Expected number of trials to see 3 in a row?
Most math folks know the formula as 1/p. OK. 1/(1/216)=216 or 1 in 216 rolls.

Looks good so far? The probability of an event(the frequency)is realted to the expected number of trials in order to see the event, on average.
Wrong! When it comes to runs or streaks.
Expected number of trials (wait time) for 1 seven is correct.
The others are wrong.
Why? I was using the wrong formula.
Most math folks know the formula as 1/p for expected number of trials. That formula is for independent trials. A run or streak is not an independent event.

There are 2 formulas I use to calulate the Expected number of trials (wait time or mean recurrence time)
here is one. (1-(p^r))/(q*(p^r))
Here is the other(p^-n)-1)/1-p
You can find the proof of formula #2 in a free pdf.
http://people.ccmr.cornell.edu/~ginsparg/INFO295/mh.pdf
p = probability of success. In my example that would be (1/6)
q = 1-p
r = run length
Correct answer for 2 in a row 7s is: 42 (not 36. 36 is the mean time for exactly 2 rolls)
Correct answer for 3 in a row 7s is: 248 (not 216. 216 is the mean time for exactly 3 rolls) I can see a pattern developing.
And bringing back point #2, 1 in 36 for 2 in a row is the mean time where n = 2 and only 2 and not the expected number of trials.
same for 3 in a row.
Again, computer simulations have proven to me that my statements are accurate and correct.
----------------------------------------------------------------------------
So, the waiting time for 7 losses of 6 and 8s is not 959, which is the mean for when n=7.
1532.66 is my result and verified by simulations. And * 2.25 = 3448 total dice rolls.
-----------------------------------------------------------------------------
Oh, yes, I swore at it the first time until my math friend showed me why, the correct formula and simulation results, where I finally accepted the truth about the math of runs (streaks).
I also have run the simulations and have seen with my own eyes the difference between mean time and waiting time as it applies to runs or streak sequences. If you disagree after reading the pdfs I pointed to earlier and my examples, that is too bad. I am here not to prove or convince anyone. Proving to myself only mattered.
I share this so we can understand that those easy, elementary school answeres as they pertain to runs or streaks just do not tell the whole truth and nothing but the truth, whan apples to apples and oranges to oranges matters.

This was also not to be a text book chapter. One can learn by searching and reading also.
I could also mention the formula for variance, which is almost equal to the mean recurrence time. It also can be found.

Back to your regularly scheduled program.
Not bad English for a Spaniard.
My Mother passed away last month, my Father this past Saturday, looks like it is my turn.
Turn, turn, turn

Unbeatable without a slanted table or crooked dice, unless one is an expert thrower.

Quote: nope27

This was also not to be a text book chapter. One can learn by searching and reading also.
I could also mention the formula for variance, which is almost equal to the mean recurrence time. It also can be found.

Back to your regularly scheduled program.
Not bad English for a Spaniard.
My Mother passed away last month, my Father this past Saturday, looks like it is my turn.
Turn, turn, turn

Well, It all reads like a text book.

Thanks for some links and the info.
strange ending, good luck.

Sorry for your loss!
I guess what I'm not understanding is the sequence. May I phrase it another way?.....
Come Out Roll.................7
Come Out Roll.................4 Point
Next Roll.........................9
Next Roll.........................5
Next Roll.........................7 Out
Come Out Roll.................12
Come Out Roll.................10 Point
Next roll...........................7 Out
Come Out Roll..................9 Point
Next Roll..........................3
Next Roll..........................2
Next Roll..........................9
Next Roll..........................7 Out
Come Out Roll..................5
Next Roll..........................7 Out
Come Out Roll..................7
Come Out Roll..................9 Point
Next roll...........................7 Out...............That's SEVEN 7's popping without a 6 or an 8 hitting in that sequence. What are those odds?

Quote: wrragsdale

Sorry for your loss!
I guess what I'm not understanding is the sequence. May I phrase it another way?.....
Come Out Roll.................7
Come Out Roll.................4 Point
Next Roll.........................9
Next Roll.........................5
Next Roll.........................7 Out
Come Out Roll.................12
Come Out Roll.................10 Point
Next roll...........................7 Out
Come Out Roll..................9 Point
Next Roll..........................3
Next Roll..........................2
Next Roll..........................9
Next Roll..........................7 Out
Come Out Roll..................5
Next Roll..........................7 Out
Come Out Roll..................7
Come Out Roll..................9 Point
Next roll...........................7 Out...............That's SEVEN 7's popping without a 6 or an 8 hitting in that sequence. What are those odds?

Answer: average of 3,448 total dice rolls to see a run of at least 7-7s before a 6 or 8.
(average is 1533 dice rolls only counting 6,7 and 8s.)
Three reasons why the basic math formula is so far off is we are always thinking in total dice rolls, the basic math only considers rolls of 6,7 and 8s since every roll does not resolve in a 6,7 or 8 rolling and there is more to runs ( or streaks) than just (6/16)^7

Read on only if you want to understand the math behind the answer.
Let me give this a try.
I will use some of nope27 comments since he did answer your question but did not present them in an easy to read post. He emailed me about his math but I have not heard back from him since. It also took me a few readings of his post and a few simulations to see what he really was saying.

Your sequence above still has 7-7s in a row before any 6 and/or 8s. ANY OTHER NUMBER DOES NOT MATTER when it comes to calculating the probability.
This actually implies that the 7-7s ARE consecutive 7s. Your statement "These sevens are not necessarily consecutive." is not correct. They must be consecutive since we are only interested in 6,7, and 8s.

Remember... We are only dealing with 3 numbers in our formula. The 6,7 and 8.

(6/16)^7 = 0.00104284286 or 1 in about 959.(the answer miplet gave) is the probability for only when n(number of trials of 6,7s and 8s) = 7, and only 7.
The 959 is NOT 959 consecutive dice rolls. I think most, if not all, would think that. That is totally wrong in looking at the statistic that way.
The 959 is only about 6,7 and 8s that are rolled, and we all know they do not roll every dice roll.
And they can overlap.
rolls 1-7, 2-8, 3-9. If every one of these 9 rolls was a 6,7 or 8 the count would be at 3. A run of 9 counts as 1 run of 9, 2 runs of 8 and 3 runs of 7.

How about these larger values of n.
n = 1001 total dice rolls 25.02% chance of at least 1 streak of 7 or more.
n = 2394 total dice rolls 50.01%... the median, chance of at least 1 streak of 7 or more.
n = 15,833 total dice rolls 99.00% chance of at least 1 streak of 7 or more.

If this all makes sense, re-read nope27 post and his statements and the math will make more sense.
the 1 in 959 is NOT 959 total dice rolls.
Even my computer simulations (quick 1,000 sessions) show the average to be around 3450 total dice rolls, sd 3500, median 2395, min 17 and max 22,136.
This goes to show that there is way more to streaks than meets the eye.

Yes, honey it is my turn to walk the dog.

Quote: wrragsdale

That's SEVEN 7's popping without a 6 or an 8 hitting in that sequence. What are those odds?

Possibly what is confusing people is that you are asking this question in the context of a casino, which implies a craps game. But then you say you don't care if the seven's are coming out, or 7-outs. That would imply you are asking a simple dice question.

What are the odds of rolling 7 sevens before I roll a 6 or an 8 with a pair of dice, (regardless of what game is being played). I could just be sitting in a corner rolling dice.

It's a funny way to ask the question.
The answer is 1 in 959, or (16/6)^7. And yes I agree that calculating the odds, does not tell you how long it will take.

Quote: pacomartin

Possibly what is confusing people is that you are asking this question in the context of a craps game. But then you say you don't care if the seven's are coming out, or 7-outs. That would imply you are asking a simple dice question.

What are the odds of rolling 7 sevens before I roll a 6 or an 8 with a pair of dice, (regardless of what game is being played). I could just be sitting in a corner rolling dice

It's a funny way to ask the question. Is that your question?

That is exactly how I looked at the question.
Now I will run a sim, seeing how many rolls on average does it take to roll 7 7s B4 a 6 or 8-not counting come out rolls, since most place bettors never have their place bets working on the come out roll. A sim would be fast. The math?
Since 29.62% of all rolls happen on a come out roll the average should be that much higher than 3448.
stay tuned...OK a real quick 500 session sim, I am tired now
Not counting any 7s that roll on a come out roll

To see 7-7s B4 a 6 or 8 not counting the come out roll
4645.6 average rolls
4863.5 SD
3219.5 median
44 min
28925 max
Looks close enough to me.

moral to the story...
When someone says the odds of winning or losing or happening "X times in a row" is p^x, it is partly true when n=X and most times the complete story is not told.
Otherwise it is a totally different formula to arrive at the correct answer.

Quote: pacomartin

Possibly what is confusing people is that you are asking this question in the context of a casino, which implies a craps game. But then you say you don't care if the seven's are coming out, or 7-outs. That would imply you are asking a simple dice question.

What are the odds of rolling 7 sevens before I roll a 6 or an 8 with a pair of dice, (regardless of what game is being played). I could just be sitting in a corner rolling dice.

It's a funny way to ask the question.
The answer is 1 in 959, or (16/6)^7. And yes I agree that calculating the odds, does not tell you how long it will take.

Again, your answer is also NOT "THE" correct answer. It answers a different question. That question is what is the probability of 7-7s before a 6 or 8 in exactly and always 7 trials?
It only answers for the rolls that are 6,7 or 8s and when n=7.
1 in 959 dice rolls? Nope, not even close!

nope27 in his post at the end of point#3 has 2 formulas to calculate the correct answer.

Run a sim and you will see.
EDIT:
Or better yet, here is an excellent pdf nope27 emailed me.
Especially the beginning and the ending at page 24.
PenneyAnte.pdf
HERE

Quote: guido111

That is exactly how I looked at the question.

It could be an effective side bet. The casinos offers $800 for $1 for every time this happens. The advantage to the casino is it makes people stay at the blackjack table for long time.

But it would be a real sucker bet since even if you did win you'd have to stay there all day and night. Of course, maybe that is the point, people just walk away. But I can't imagine a gaming control board approving such a game.

Perhaps if it paid $15 to $1 for a streak of 3 seven's before a 6 or an 8 as a second level. That would be effective in keeping people at the table.

Quote: guido111

You mean (6/16)
Again, your answer is also NOT the correct answer.
It only answers for the rolls that are 6,7 or 8s and when n=7.

Run a sim and you will see.

The question is what is the probability of "popping" 7 sevens, before rolling a 6 or an 8.

I think it is the correct answer since the other 20 sums result in a "push". There are only 16 possible states, and 10 of them end the sequence, while 6 of them advance the sequence by 1. The probability of advancing the sequence to seven is (6/16)^7 or alternatively one in (16/6)^7 .

Quote: pacomartin

The question is what is the probability of "popping" 7 sevens, before rolling a 6 or an 8.

I think it is the correct answer since the other 20 sums result in a "push". There are only 16 possible states, and 10 of them end the sequence, while 6 of them advance the sequence by 1. The probability of advancing the sequence to seven is (6/16)^7 or alternatively one in (16/6)^7 .

See my post above. I did an edit to my post.
It is excellent reading.

You are correct that it ((6/16)^7 or alternatively one in (16/6)^7) is a correct answer, but to a different question, only when n=7 and always 7.

In nope27 post, point#3, he gives an excellent example of 1,2 3 7s in a row from 2 dice.
The probability of 1 seven in a row is (1/6)^1 or 1 in 6.
The probability of 2 seven in a rows is (1/6)^2 or 1 in 36. Again, wrong formula for the question asked. The answer is 42 rolls of the dice counting all the numbers rolled.
This question is like the Monte Hall problem. Until you look at the total problem in the 1 correct way, you will never agree with the correct answer.

It took me over a week to understand what nope27 was trying to get across in his post.
Our formula that 99.999% of us uses for the expected number of "runs greater than length of 1" p^r (r=length of run) is not correct for any value of n that is greater than r.
(1-(p^r))/(q*(p^r))
or
(p^-n)-1)/1-p
are both correct to use and both results in the correct answer.
Another formula one can use, for the 7 example above, would be The probability of 2 seven in a rows is (1/6)^-1+(1/6)^-2 or 1 in 42
The average number of rolls to see 3 seven in a rows is (1/6)^-1+(1/6)^-2+(1/6)^-3 or 1 in 258... not (1/6)^-3=216 because 1 in 216 is when n=3.

Added:3-26 9 am
The average number of rolls to see 10 heads in a row flipping a fair coin?
we all know it to be 2^10=1024. 1024 flips?
Again wrong formula to use for the question asked. It gives us a wrong answer.
But, the formula gives the correct answer when n is always and only equal to 1024.

Simulations show this to be true only when n=10. ie: flips 1-10,2-11,3-12 etc. overlapping flips are counted.
12 heads in a row is actually 1 run of length 12
2 runs of length 11
3 runs of length 10
4 runs of length 9
5 runs of length 8
6 runs of length 7
7 runs of length 6
8 runs of length 5
9 runs of length 4
10 runs of length 3
11 runs of length 2
12 runs of length 1
check out, only if you wish, http://forumserver.twoplustwo.com/25/probability/successes-row-904091/
post#14
BruceZ (a math guru good guy) calculates the average number of tosses to see 10 heads in a row with a fair coin.
His final answer is 2046.(First it was 2048 for runs of 10 or more) But he explains why.

Quote: pacomartin

The answer is 1 in 959, or (16/6)^7. And yes I agree that calculating the odds, does not tell you how long it will take.

pacomartin is real close to the correct reason why. But calculating the odds, with the correct formula, will tell you how long it will take. It is called the "wait time".
(16/6)^7 or (6/16)^7 is NOT the correct formula to use to calculate the wait time.

nope27 and guido111 pointed to 2 of them, actually guido pointed to 3 of them.
(6/16)^-1+(6/16)^-2+(6/16)^-3+(6/16)^-4+(6/16)^-5+(6/16)^-6+(6/16)^-7
or
(16/6)^1+(16/6)^2+(16/6)^3+(16/6)^4+(16/6)^5+(16/6)^6+(16/6)^7=1,532.6676 (then one must multiply that by 2.25 to get the actual dice roll answer of 1 in 3448.5)
edit added: since the above formulas are Summing a Geometric Series the formula would be: ((1/p)^(n+1)-(1/p))/((1/p)-1)

I struggled with this concept until I read the PennyAnte.pdf that guido111 linked to.
As Guido111 pointed out the 1 in 959 is NOT 959 consecutive dice rolls. Far from it as shown by the proper formulas to use and simulations.
The 1 in 959 is overlapping 7-element sequences while only counting or considering the numbers 6,7, and 8.
Rolls 1-7 would be counted as one
rolls 2-8 also would be counted as one.
So would rolls 9-15 be counted as a run of 7.

Then, this brings up and interesting fact.
When some one answers a question about the probability of X happening in a row, the formula 1/p does not work for the expected number of trials for runs or streaks with a length greater than 1.

An Example:
https://wizardofodds.com/ask-the-wizard/blackjack-faq/
Q4 from the bottom:
"According to my blackjack appendix 4 the probability of a win is 43.31%, a loss is 47.89%, and a tie is 8.80%, give or take depending on the exact rules.
Assuming you are ignoring ties the probability of a win is 47.49%."
The Wizard contines...
"So the probability of x wins in a row is 0.4749^x -For example the probability of 5 wins in a row is 0.47495^5=0.0242. 1 in 41."

Now I have always read the answer to mean 1 in 41 hands we can expect on average to see 5 wins in a row at BJ.

That then is not true as I am assuming 1 in 41 means 41 hands. So my assumption is no correct.
What is correct is when "n" (number of hands)=5.

The mean recurrence time from 1 of the formulas by nope27 or guido111 shows: a
1 in 76.90152705 to be the correct answer for 5 wins in a row at BJ (1 in 77 hands on average) and my simulations show that to be also correct.
also a SD of 73.08770458

I leave my post to end here.
I received an email from nope27's (his real name Esteban Nope) son who says his father passed away this week after an unsuccessful heart surgery. I knew he had
heart problems like singer Bobby Darin did.
My father passed this past March 5th. between Jane Russell and Liz Taylor. Nice company to be in line with.

Quote: guido111

1 in 959 dice rolls? Nope, not even close!

I'm surprised you would think that I would make a stupid mistake like that. I never said that it would 959 dice rolls. The original poster's question was not how many dice rolls would it take on average, but his question was what is the probability of being able to roll seven 7s without rolling a 6 or an 8.

In his example he had 5 7-outs, and 2 7's on a come-out role. He was very specific in saying he doesn't care when the 7 occurs.

The answer to his question is 1 chance in 959, which is not the answer to the question of how many rolls of the dice will it take, or what is the probability of surviving seven 7-outs.

I admit, that the question as phrased is somewhat unusual, since most craps players want to know if they can survive a given number of 7-outs, but I am answering his question.