Probability of 7 6s before a 7

Could someone tell me what the probability of rolling 7 or more 6s before a 7 in craps would be, and the expected number of rolls before rolling 7 or more 6s before a 7? If I was unclear above, the particular system in question would define a losing game by 7 or more 6s being rolled before a 7, so if more than 7 6s are rolled before a 7 it is still just as much a losing game as 7 6s. I am unsure if the "or more" part counts or how to incorporate it. Please show the calculation if you would be so kind so I can use it for the expectation of the other points as well.

Thanks,

Derek

The probability of at least 7 sixes before a seven is (5/11)^7 or almost exactly 0.4%.

5/11 is the probability of rolling a six before a seven once since there are 5 ways to make a six, and 6 ways to make a seven, so the odds are 6:5 against. We raise this to the power of 7 for the probability that it happens 7 times in a row.

Do you want the average number of rolls for it to happen in the games where it happens, or the average number of rolls it takes before you see at least 6 sixes without a seven in a long continuous series of rolls that span multiple games?

If you want the expected number of rolls until you see 7 sixes without a seven…I’m getting about 1490.61.

I can provide the math later if desired

Quote: PeeMcGee

If you want the expected number of rolls until you see 7 sixes without a seven…I’m getting about 1490.61.

I can provide the math later if desired

I agree, and that's probably what he wanted.

(11/5)^7 * [1 - (5/11)^7] / (1-5/11) * 36/11 ≈ 1490.615

First focus only on the rolls that produce a six or seven. Since the probability of 7 sixes before a seven is (5/11)^7, the average number of TRIALS is 1 over this or (11/5)^7 where we define a trial as a sequence of rolls of six or seven which terminate in either 7 sixes or a seven. Now we must multiply that by the average number six or seven rolls in a trial. We can get that from the geometric series

1 + 5/11 + (5/11)^2 + (5/11)^3 + ... + (5/11)^6

= [1 - (5/6)^7] / (1 - 5/6).

That is, it always takes at least 1, then with probability 5/11 it takes another one when we get a six, then with probability (5/11)^2 it takes another one when we get 2 consecutive sixes, etc. The series stops when we get 6 sixes in a row since the trial ends on the next six or seven roll regardless. Finally, we need to multiply all this by the average number of rolls to get a six or seven. The probability of this is 11/36, so the average number of rolls to get one is 36/11.

I had said something earlier that I took down regarding a different question "What is the average number of rolls for the game in which we roll 7 sixes before a seven". It was kind of an instructive subtle error. You can't just ignore the sevens and say each six has probability 5/30 instead of the normal 5/36. We know more than the fact that we got no sevens. We know that we got 7 sixes before a seven. This makes the sixes relative to the other numbers more likely than usual. Just changing the size of the sample space to 30 doesn't reflect that. I noticed it when I tried to use that type of assumption to get the average number of rolls to get a seven before 7 sixes. I broke it into a weighted sum of cases where we get 0 to 6 sixes, and that produced something greater than 9 which obviously can't be right since 6 is the unconditional expectation, and the answer must be slightly less than that.

Quote: BruceZ

The probability of at least 7 sixes before a seven is (5/11)^7 or almost exactly 0.4%.

5/11 is the probability of rolling a six before a seven once since there are 5 ways to make a six, and 6 ways to make a seven, so the odds are 6:5 against. We raise this to the power of 7 for the probability that it happens 7 times in a row.

Do you want the average number of rolls for it to happen in the games where it happens, or the average number of rolls it takes before you see at least 6 sixes without a seven in a long continuous series of rolls that span multiple games?

Just to clarify, your calculation, (5/11)^7, is for 7 sixes before a 7 in any order, and not for the probability of having 7 sixes all in a row? To address your second question, I would be interested in seeing both now that you mention it.

Thanks,

Derek

Quote: BruceZ

1 + 5/11 + (5/11)^2 + (5/11)^3 + ... + (5/11)^6

= [1 - (5/6)^7] / (1 - 5/6).

Did you mean,

1 + 5/11 + (5/11)^2 + (5/11)^3 + ... + (5/11)^7

= [1 - (5/11)^7] / (1 - 5/11)?

Quote: CasinoCrasher

Could someone tell me what the probability of rolling 7 or more 6s before a 7 in craps would be,

To answer just this question, the probability is 0.004009048 (corrected).

Next, you're suspended for three days for an inappropriate avatar. Please change it so something G rated.

Quote: CasinoCrasher

Just to clarify, your calculation, (5/11)^7, is for 7 sixes before a 7 in any order, and not for the probability of having 7 sixes all in a row?

Yes, 7 or more sixes before a seven, and all other numbers are ignored.

Quote: CasinoCrasher

Quote: BruceZ

1 + 5/11 + (5/11)^2 + (5/11)^3 + ... + (5/11)^6

= [1 - (5/6)^7] / (1 - 5/6).

Did you mean,

1 + 5/11 + (5/11)^2 + (5/11)^3 + ... + (5/11)^7

= [1 - (5/11)^7] / (1 - 5/11)?

No, since that would be incorrect.

Quote: Wizard

Quote: CasinoCrasher

Could someone tell me what the probability of rolling 7 or more 6s before a 7 in craps would be,

To answer just this question, the probability is 1 in 167,178.

What? (5/11)^7 is about 1 in 249.4

Quote: BruceZ

What? (5/11)^7 is about 1 in 249.4

You're right. I was thinking of consecutive sixes.

Quote: Wizard

You're right. I was thinking of consecutive sixes.

Sorry, 1 in 167178 isn't the probability of 7 consecutive sixes before a seven either. That result is about 1 in 194142. You are apparently computing

(5/36)^7 / [(5/36)^7 + 1/6] ≈ 1 in 167178.

That would be the probability of a game in which you lose if you roll a 7 on the first roll, win if you roll 7 consecutive sixes, and any other result you start over. What we have is different from that. You will lose if you roll a 7 on any of those rolls, and you can still win if you finish with some number of consecutive sixes which would carry over to the next game.

So how do we compute the probability that we get 7 consecutive sixes before a seven?

Quote: Wizard

To answer just this question, the probability is 0.004009048 (corrected).

Next, you're suspended for three days for an inappropriate avatar. Please change it so something G rated.

CasinoCrasher has been unsuspended pending his avatar change (which he couldn't do while suspended). I'm giving him 24 hours to change it, starting now.

Quote: BruceZ

Do you want the average number of rolls for it to happen in the games where it happens, or the average number of rolls it takes before you see at least 6 sixes without a seven in a long continuous series of rolls that span multiple games?

Quote: Wizard

To answer just this question, the probability is 0.004009048 (corrected).

I would like to see the average number of rolls before you see 7 sixes before a 7 in both the games where it happens and both the average number of rolls it takes in a long continuous series of rolls? Just to clarify when you say in the games where it happens, are you defining a "game" as the average number of rolls between a 7 out when it happens, or are you defining a "game" as the time I play when I walk up to a craps table? I would assume you're referring to a game as the average number of rolls between a 7 out when it happens, and the latter being the same as the previously agreed upon probability of 0.004009048 or 249.4358 rolls which would also be the same as a long continuous series of rolls; each time I played, the number of rolls where I was betting would be a representative fraction of this 249.4358 rolls that would eventually add up to the 249.4358 rolls, if I didn't play for 249.4358 rolls in one game.

Thanks,

Derek

Quote: CasinoCrasher

I would like to see the average number of rolls before you see 7 sixes before a 7 in both the games where it happens and both the average number of rolls it takes in a long continuous series of rolls? Just to clarify when you say in the games where it happens, are you defining a "game" as the average number of rolls between a 7 out when it happens, or are you defining a "game" as the time I play when I walk up to a craps table? I would assume you're referring to a game as the average number of rolls between a 7 out when it happens, and the latter being the same as the previously agreed upon probability of 0.004009048 or 249.4358 rolls which would also be the same as a long continuous series of rolls; each time I played, the number of rolls where I was betting would be a representative fraction of this 249.4358 rolls that would eventually add up to the 249.4358 rolls, if I didn't play for 249.4358 rolls in one game.

Thanks,

Derek

I provided (and Bruce verified and explained) the expected number of rolls until you see 7 sixes without a seven...which was 1490.61.

The expected number of rolls in games where it happens is difficult to math out. So by simulation, I’m getting about 22.9.

A “game” is defined as the set of rolls until you get 7 sixes OR you roll a 7. If either of those two conditions occurs, the current game is terminated and a new one begins.

Quote: PeeMcGee

The expected number of rolls in games where it happens is difficult to math out. So by simulation, I’m getting about 22.9

Correct, it's exactly 252/11, and here's the math. The average number of rolls to get 7 sixes given that we get no sevens first will be 7 times the average number of rolls to get 1 six given that we get no sevens first. Letting S be the number of rolls to get a six and V be the event that we get no sevens before the first six, we have

E(rolls) = 7*sum{k = 0 to infinity) P(S > k | V).

By Bayes' theorem this is

E(rolls) = 7*sum{k = 0 to infinity} P(V | S > k)*P(S > k) / P(V)

= 7*sum{k = 0 to infinity} [(25/31)^k*5/11] * (31/36)^k / (5/11)

= 7*sum{k = 0 to infinity} (25/36)^k.

By the geometric series

E(rolls) = 7 * 1/(1 - 25/36)

= 7*36/11

= 252/11.

Quote: PeeMcGee

I provided (and Bruce verified and explained) the expected number of rolls until you see 7 sixes without a seven...which was 1490.61.

The expected number of rolls in games where it happens is difficult to math out. So by simulation, I’m getting about 22.9.

A “game” is defined as the set of rolls until you get 7 sixes OR you roll a 7. If either of those two conditions occurs, the current game is terminated and a new one begins.

Okay, makes sense, thanks!

Quote: BruceZ

Quote: PeeMcGee

The expected number of rolls in games where it happens is difficult to math out. So by simulation, I’m getting about 22.9

Correct, it's exactly 252/11, and here's the math. The average number of rolls to get 7 sixes given that we get no sevens first will be 7 times the average number of rolls to get 1 six given that we get no sevens first. Letting S be the number of rolls to get a six and V be the event that we get no sevens before the first six, we have

E(rolls) = 7*sum{k = 0 to infinity) P(S > k | V).

By Bayes' theorem this is

E(rolls) = 7*sum{k = 0 to infinity} P(V | S > k)*P(S > k) / P(V)

= 7*sum{k = 0 to infinity} [(25/31)^k*5/11] * (31/36)^k / (5/11)

= 7*sum{k = 0 to infinity} (25/36)^k.

By the geometric series

E(rolls) = 7 * 1/(1 - 25/36)

= 7*36/11

= 252/11.

Once again, well done Bruce.

The math wasn’t as difficult as I assumed…but that was clever as hell. Kudos.

A quick thought though…you didn’t really have to use Bayes’ right? As P(S>k | V) is the probability of not rolling a 6 or 7 for k rolls. Which is just (25/36)^k?

Quote: BruceZ

I agree, and that's probably what he wanted.

(11/5)^7 * [1 - (5/11)^7] / (1-5/11) * 36/11 ≈ 1490.615

First focus only on the rolls that produce a six or seven. Since the probability of 7 sixes before a seven is (5/11)^7, the average number of TRIALS is 1 over this or (11/5)^7 where we define a trial as a sequence of rolls of six or seven which terminate in either 7 sixes or a seven. Now we must multiply that by the average number six or seven rolls in a trial. We can get that from the geometric series

Can you explain the significance of (5/11)^7 in your equation? I understand, and you state here that "the probability of 7 sixes before a seven is (5/11)^7" but, the entire equation is supposed to represent the average number of rolls before seeing 7 sixes before a 7. If (5/11)^7 is the probability of seeing 7 sixes before a 7, then I would think that 1 over this, "1/(5/11)^7=(11/5)^7" would represent the average number or rolls before the event, which is actually the first term in the equation.


Thanks,

Derek

Quote: CasinoCrasher

Quote: BruceZ

I agree, and that's probably what he wanted.

(11/5)^7 * [1 - (5/11)^7] / (1-5/11) * 36/11 ≈ 1490.615

First focus only on the rolls that produce a six or seven. Since the probability of 7 sixes before a seven is (5/11)^7, the average number of TRIALS is 1 over this or (11/5)^7 where we define a trial as a sequence of rolls of six or seven which terminate in either 7 sixes or a seven. Now we must multiply that by the average number six or seven rolls in a trial. We can get that from the geometric series

Can you explain the significance of (5/11)^7 in your equation? I understand, and you state here that "the probability of 7 sixes before a seven is (5/11)^7" but, the entire equation is supposed to represent the average number of rolls before seeing 7 sixes before a 7. If (5/11)^7 is the probability of seeing 7 sixes before a 7, then I would think that 1 over this, "1/(5/11)^7=(11/5)^7" would represent the average number or rolls before the event, which is actually the first term in the equation.


Thanks,

Derek

Keep in mind that a “game” continues until you roll 7 sixes OR you roll a 7.

Then, (5/11)⁷ is the probability that a game ends with the 7 sixes.

Therefore, (11/5)⁷ is expected number of games needed to get the 7 sixes.

So then we multiply the expected number of games by the expected number of rolls per a game….to get the expected number of rolls (1490.61).

Quote: PeeMcGee

A quick thought though…you didn’t really have to use Bayes’ right? As P(S>k | V) is the probability of not rolling a 6 or 7 for k rolls. Which is just (25/36)^k?

It's the probability of not rolling a 6 for k rolls AND not rolling a 7 before the first 6 (including after k rolls) divided by the overall probability of not rolling a 7 before the first 6 by the definition of conditional probability. That's

(25/36)^k * (5/11) / (5/11)

= (25/36)^k.

The 5/11 cancels because the probability of not getting a 7 before a 6 starting after k is the same as it was at the beginning, so it ends up just being the probability of not rolling a 6 or 7 for k rolls as you say. I felt needed to make it explicit why it works out that way, but just using the definition of conditional probability like this would have been simpler than turning it around with Bayes' theorem.

Quote: BruceZ

I agree, and that's probably what he wanted.

(11/5)^7 * [1 - (5/11)^7] / (1-5/11) * 36/11 ≈ 1490.615

First focus only on the rolls that produce a six or seven. Since the probability of 7 sixes before a seven is (5/11)^7, the average number of TRIALS is 1 over this or (11/5)^7 where we define a trial as a sequence of rolls of six or seven which terminate in either 7 sixes or a seven. Now we must multiply that by the average number six or seven rolls in a trial. We can get that from the geometric series

1 + 5/11 + (5/11)^2 + (5/11)^3 + ... + (5/11)^6

= [1 - (5/6)^7] / (1 - 5/6).

That is, it always takes at least 1, then with probability 5/11 it takes another one when we get a six, then with probability (5/11)^2 it takes another one when we get 2 consecutive sixes, etc. The series stops when we get 6 sixes in a row since the trial ends on the next six or seven roll regardless. Finally, we need to multiply all this by the average number of rolls to get a six or seven. The probability of this is 11/36, so the average number of rolls to get one is 36/11.

I used this equation for finding the expected number of rolls before seeing 7 6s before a 7 and extended it to obtain the expected number of rolls before seeing 6 6s before a 7, 5 6s before a 7, 4 6s before a 7, 3 6s before a 7, 2 6s before a 7, and 1 6 before a 7. I calculated 674, 303, 135, 58, 23, and 7 respectfully, rounded of course. Since I am defining 7 6s before a 7 as a losing game, I then wanted to find how many times I would expect to see each combination before a losing game, so how many times did I see 6 6s before a 7, 5 6s before a 7, ect. in 1491 rolls. To find this I divided 1491, the number of rolls before seeing 7 6s before a 7 by 674, the number of rolls before seeing 6 6s before a 7. This gave me approximately 2, again rounded. So, I would expect to see 6 6s before a 7 twice in 1491 rolls, or before seeing 7 6s before a 7.

I followed this methodology for the remaining number of sixes before a 7, so for 5 6s before a 7, 4 6s before a 7, 3 6s before a 7 ect. and I calculated 5, 11, 26, 65, and 207 respectfully, again rounded. So, now I know how many times I should expect to see each number of 6s before a 7, before seeing 7 6s before a 7, or before a losing game, 1491 rolls. This will allow me to observe how many times I would be able to win on each "sequence" of sixes before a 7, before a losing game. All is well until I get down to the number of times I should expect to see 1 6 before a 7 in 1491 rolls, which is 207. I noticed that 207 is in fact the "exact" number of sixes that you would expect in 1491 rolls, found by multiplying 1491 by the probability of rolling a six on any given roll, 5/36. If I am expected to see 207 6s in 1491 rolls but, also expected to see 1 6 before a 7 207 times in 1491 rolls, then I thought the formula you provided must be giving me the number of times I should expect "at least" 1 6 before a 7. Undoubtedly 207 times you expect "at least" 1 6 before a 7 in 1491 rolls.

Then this is where my dilemma started, how would I know how many times I would be expected to bet against a 6 1 time before winning (seeing a 7), bet against a 6 2 times before winning (seeing a 7), ect. Currently I only know that I am expected to see 207 6s before a 7 in 1491 rolls, not how many times I am expected to see only 1 6 before a 7 in 1491 rolls, only 2 6s before a 7 in 1491 rolls. How do I find the number of times I would be expected see "only" 1 6 before a 7, "only" 2 6s before a 7, ect. in 1491 rolls?

Quote: PeeMcGee

If you want the expected number of rolls until you see 7 sixes without a seven…I’m getting about 1490.61.

I can provide the math later if desired

How would you find the number of times you would expect to see only 1 6 before a 7, and only 2 6s before a 7 ect. before seeing 7 6s before a 7 (or in 1490.61 rolls)?

Thanks,

Derek

Quote: PeeMcGee

If you want the expected number of rolls until you see 7 sixes without a seven…I’m getting about 1490.61.

I can provide the math later if desired

So I guess to simplify my last two posts even further; I am asking if you have some number of rolls, lets say in this case 1491, how many times would you expect to see only 1 6 before seeing a 7, only 2 6s before seeing a 7 ect.

Thanks,

Derek

Quote: CasinoCrasher

Quote: PeeMcGee

If you want the expected number of rolls until you see 7 sixes without a seven…I’m getting about 1490.61.

I can provide the math later if desired

So I guess to simplify my last two posts even further; I am asking if you have some number of rolls, lets say in this case 1491, how many times would you expect to see only 1 6 before seeing a 7, only 2 6s before seeing a 7 ect.

Thanks,

Derek

So lets’ assume the case that we are rolling until we see 7 sixes before a 7. We know the probability for this is (5/11)⁷.

Now what is the probability that we see 6 sixes before a 7? We must remember in such a case the game will continue on when the sixth 6 is rolled (it only terminates on a roll of 7). So therefore, the probability is (5/11)⁶(6/11). Do some algebra; you’ll see that this probability is 6/5 times greater than seeing 7 sixes. Therefore, we will see this event 6/5 times more. Since we are seeing 7 sixes once, we are then expected to see 6 sixes 1.2 times.

Follow the same idea for the other x number of sixes to get the following:
0 | 136.06
1 | 61.84
2 | 28.11
3 | 12.78
4 | 5.81
5 | 2.64
6 | 1.2

The case where we just roll 1491 times will result in slightly smaller numbers (since one of the conditions to terminate a game is removed). I’ll leave the math out for now, here are the numbers:
0 | 135.55
1 | 61.61
2 | 28.01
3 | 12.73
4 | 5.79
5 | 2.63
6 | 1.20
7 | 0.54
8 | 0.25
9 | 0.11
…and so on…

Quote: PeeMcGee

So lets’ assume the case that we are rolling until we see 7 sixes before a 7. We know the probability for this is (5/11)⁷.

Now what is the probability that we see 6 sixes before a 7? We must remember in such a case the game will continue on when the sixth 6 is rolled (it only terminates on a roll of 7). So therefore, the probability is (5/11)⁶(6/11). Do some algebra; you’ll see that this probability is 6/5 times greater than seeing 7 sixes. Therefore, we will see this event 6/5 times more. Since we are seeing 7 sixes once, we are then expected to see 6 sixes 1.2 times.

Follow the same idea for the other x number of sixes to get the following:
0 | 136.06
1 | 61.84
2 | 28.11
3 | 12.78
4 | 5.81
5 | 2.64
6 | 1.2

The case where we just roll 1491 times will result in slightly smaller numbers (since one of the conditions to terminate a game is removed). I’ll leave the math out for now, here are the numbers:
0 | 135.55
1 | 61.61
2 | 28.01
3 | 12.73
4 | 5.79
5 | 2.63
6 | 1.20
7 | 0.54
8 | 0.25
9 | 0.11
…and so on…

Okay, thanks! Can you further explain the difference between your first and second set of numbers above? I believe you are saying that your first set of numbers is based on 7 sixes before a 7 and your second set is based on 1491 rolls but, 1491 rolls was the number or rolls before seeing 7 sixes before a 7.

Also, can you explain why your second set of numbers doesn't indicate that 7 sixes before a 7 would show 1 time vs 0.54 times in 1491 rolls?

Quote: CasinoCrasher

Quote: PeeMcGee

So lets’ assume the case that we are rolling until we see 7 sixes before a 7. We know the probability for this is (5/11)⁷.

Now what is the probability that we see 6 sixes before a 7? We must remember in such a case the game will continue on when the sixth 6 is rolled (it only terminates on a roll of 7). So therefore, the probability is (5/11)⁶(6/11). Do some algebra; you’ll see that this probability is 6/5 times greater than seeing 7 sixes. Therefore, we will see this event 6/5 times more. Since we are seeing 7 sixes once, we are then expected to see 6 sixes 1.2 times.

Follow the same idea for the other x number of sixes to get the following:
0 | 136.06
1 | 61.84
2 | 28.11
3 | 12.78
4 | 5.81
5 | 2.64
6 | 1.2

The case where we just roll 1491 times will result in slightly smaller numbers (since one of the conditions to terminate a game is removed). I’ll leave the math out for now, here are the numbers:
0 | 135.55
1 | 61.61
2 | 28.01
3 | 12.73
4 | 5.79
5 | 2.63
6 | 1.20
7 | 0.54
8 | 0.25
9 | 0.11
…and so on…

Okay, thanks! Can you further explain the difference between your first and second set of numbers above? I believe you are saying that your first set of numbers is based on 7 sixes before a 7 and your second set is based on 1491 rolls but, 1491 rolls was the number or rolls before seeing 7 sixes before a 7.

Also, can you explain why your second set of numbers doesn't indicate that 7 sixes before a 7 would show 1 time vs 0.54 times in 1491 rolls?

Correct. The first set of numbers is based on the fact that rolling ends when 7 sixes before a 7 occurs. The second set of numbers is based on the fact that rolling ends after a constant 1491 rolls.

The reason you are expected to see 7 sixes before a 7 only 0.54 times in 1491 rolls is because rolling doesn’t terminate when you roll a 7th six. It’s possible to get more sixes…That is why the list goes on to 8 sixes, 9 sixes and so on. It’s essentially asking 7 or more sixes before a 7. And since rolling is allowed to continue there’ll be slightly more rolls per a game—and therefore, less games—hence why the second set of numbers is slightly smaller.

Okay, thanks so much for the help! I applied the same logic to the other point numbers and it came out nicely. Does the probability expressions used for finding the expected number of rolls before seeing 7 sixes before a 7 ((11/5)^7 * [1 - (5/11)^7] / (1-5/11) * 36/11 ≈ 1490.615) shown by BruceZ, and the number of times each number of sixes will be seen before a 7 that you showed here have a name, or are they just fundamentally derived from "basic" probability?