 Joined: Feb 10, 2023
• Posts: 1
February 10th, 2023 at 9:32:45 AM permalink
What are the odds of shooting 7 7's before shooting a 6 or an 8?
DJTeddyBear Joined: Nov 2, 2009
• Posts: 10973
February 10th, 2023 at 10:33:13 AM permalink
This is very similar to a thread from earlier this week.

Read the question and my reply (I�m the first to respond). Then just change a couple variables because the odds on 6/8 are different.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ������������������������������������� Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5956
February 10th, 2023 at 1:22:29 PM permalink

What are the odds of shooting 7 7's before shooting a 6 or an 8?

If you mean, "rolling 7 7s without rolling either a 6 or an 8":
The only numbers that matter are 6, 7, and 8; of those, a 7 comes up 6 times, while a 6 or 8 comes up 10, so the probability of rolling a 7 before a 6/8 is 3/8.
You want 7 of these, so the probability is (3/8)^7 = 2187 / 2,097,152, or about 1 / 959.

Pardon me for asking, but what is your strategy? As far as I know, there is no bet where you win if you roll a 6 or 8 before a 7 (and if there was, it would pay less than 3-5.)
Based on your title, is it something like this:
1. Start with, say, \$12 place bets on the 6 and 8
2. If either wins, you are up \$14; you take the other place bet down and walk (or run, if it's "hit and run")
3. If the 7 comes up, you are down \$24, so you bet \$24 on the 6 and 8
4. Repeat until either one of your bets wins, or that 1 in 959 event happens that 7 comes up 7 times without a 6 or 8.
There are two problems.
First, since the bet is actually paying only 7-12 odds (if you bet \$6 on the 6 and another \$6 on the 8, you are risking \$12 to win \$7), you will have to more than double your bet after each loss in order to guarantee a profit if you win.
Second, if you do lose all 7 bets, the loss is far more than 959 times the amount you expect to win.

Here is a table that assumes that each bet (a) is a multiple of \$6 (so you can get true 7-6 odds when placing the 6 and 8), and (b) is high enough so you will profit at least \$10 if you win. The four columns are (a) the bet number in the series, (b) the total bet on 6/8 (so, for the first bet, it is 12 on each number; for the second, it is 30), (c) the total profit from the entire run if you win, and (d) the total loss in the run up to that point if a 7 comes up.
Bet #Total BetProfit
with Win
Total Loss
with Loss
1241424
2601184
316814252
445614708
51236131944
63360165304
791201614424

In this case, you are risking \$14,424 to win anywhere from \$11 to \$16.

Now, if you are willing to risk \$14,424 every time, then the "black box" house edge (where the house edge is the expected return divided by the amount you are willing to bet) is only 0.0122%, but remember, in that case, you are playing a game that pays at most 1-to-900 odds.
Ace2 Joined: Oct 2, 2017
• Posts: 2671
February 10th, 2023 at 2:13:54 PM permalink
A slightly more difficult question:

What�s the probability of rolling a 6 and an 8 before seven 7s

So at least one 6 and at least one 8 before seven 7s
It�s all about making that GTA
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5956
February 10th, 2023 at 6:05:38 PM permalink

I'm a little surprised this worked, but it matched both my simulation and my Markov calculation

P(6 and 8 before 7 7s) = 1 - P(7 7s before 6 and 8)
= 1 - P(7 7s before 6) - P(7 7s before 8) + P(7 7s before either a 6 or 8)
= 1 - (6/11)^7 - (6/11)^7 + (3/8)^7
= 39,736,041,395,425 / 40,867,559,636,992

Ace2 Joined: Oct 2, 2017
• Posts: 2671
February 10th, 2023 at 7:03:17 PM permalink
I agree. It�s perfect for the inclusion-exclusion principle

You can also get there via the complement of the integral from zero to infinity of :

(x/6)^6 / (6! * e^(x/6)) * (1 - (1 - 1/e^(5x/36))^2) * 1/6 dx

Which is: 1 - 1131518241567 / 40867559636992
It�s all about making that GTA
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5956
February 10th, 2023 at 7:16:28 PM permalink
Quote: Ace2

I agree. It�s perfect for the inclusion-exclusion principle

You can also get there via the complement of the integral from zero to infinity of :

(x/6)^6 / (6! * e^(x/6)) * (1 - (1 - 1/e^(5x/36))^2) * 1/6 dx

Which is: 1 - 1131518241567 / 40867559636992

Let me see if I have this figured out:
1/e^(5x/36) represents the probability of not rolling, say, a 6
1 - 1/e^(5x/36) represents the probability of rolling one or more 6s
(1 - 1/e^(5x/36))^2 represents the probability of rolling one or more 6s AND one or more 8s
1 - (1 - 1/e^(5x/36))^2 represents the probability of rolling no 6s or 8s at all
(x/6)^6 / (6! * e^(x/6)) represents the probability of rolling exactly 6 7s
The product represents the probability of rolling exactly 6 7s and no 6s or 8s
Multiply by 1/6 to represent the seventh 7

Question: why isn't it the integral of this (or is it):
(x/6)^7 / (7! * e^(x/6)) * (1 - (1 - 1/e^(5x/36))^2)
I can see why you would do it the other way in other cases, but since there are no 6s or 8s rolled, you don't have to "single out" the final 7 as being the last roll
Ace2 Joined: Oct 2, 2017
• Posts: 2671
February 10th, 2023 at 7:55:39 PM permalink
Logic looks good except: 1 - (1 - 1/e^(5x/36))^2 represents the complement of both a 6 and 8 having been rolled. Meaning one of three states:

a 6 hasn�t been rolled
an 8 hasn�t been rolled
neither has been rolled

For 7s to win, there can be 6s OR 8s but not 6s AND 8s.

I believe the reason the last integral is incorrect is because it overcounts cases when 7s win. For instance, if the first seven rolls are all 7s (ignoring rolls that aren�t 6-7-8) then that string is correctly counted as a win for 7s. However, if the next roll is a 6, then that string also gets counted as a win because it meets that integral�s criteria (exactly seven 7s and either 6 or 8 hasn�t been rolled). It will get counted again and again as long as 6s are rolled before 7s and 8s. My integral doesn�t overcount those kinds of strings. A win can only be counted once

Try it on https://www.integral-calculator.com/. You�ll see

It�s Friday night and I�ve had a few drinks so caveat emptor on tonight�s math reasoning !
Last edited by: Ace2 on Feb 10, 2023
It�s all about making that GTA
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5956
February 11th, 2023 at 7:54:39 AM permalink
Quote: Ace2

Logic looks good except: 1 - (1 - 1/e^(5x/36))^2 represents the complement of both a 6 and 8 having been rolled. Meaning one of three states:

a 6 hasn�t been rolled
an 8 hasn�t been rolled
neither has been rolled

For 7s to win, there can be 6s OR 8s but not 6s AND 8s.

I believe the reason the last integral is incorrect is because it overcounts cases when 7s win. For instance, if the first seven rolls are all 7s (ignoring rolls that aren�t 6-7-8) then that string is correctly counted as a win for 7s. However, if the next roll is a 6, then that string also gets counted as a win because it meets that integral�s criteria (exactly seven 7s and either 6 or 8 hasn�t been rolled). It will get counted again and again as long as 6s are rolled before 7s and 8s. My integral doesn�t overcount those kinds of strings. A win can only be counted once

Got it. I kept jumping back and forth between "you need to roll both a 6 and an 8" and "you need to roll a 6 or an 8" in there somewhere. Obviously, "there are no 6s or 8s rolled" isn't necessarily true in my reasoning for the second integral.
Ace2 Joined: Oct 2, 2017
• Posts: 2671
February 11th, 2023 at 8:47:15 AM permalink
Another way to do the integral is to evaluate all cases when there are less than seven 7s and either >zero 6s with zero 8s or zero 6s with >zero 8s. Then it�s a 5/36 chance of 6/8 winning from that state. But this is actually a longer integral since you must list the seven possible states of the 7 (0-6)

It�s surprising how often it�s easier to calculate the probability of losing and then take the complement than directly calculating the probability of winning

I�ve also noticed that every Poisson integration I�ve done could also be solved via inclusion-exclusion. For this problem, integration really isn�t necessary (except for practice) since it�s so easy to do with I-E. More complex problems, however, require so many combinations/probabilities to be added/subtracted that it�s much more practical to integrate it�the integral often being quite short if you use the right approach. I-E is also all about the simplest approach�gotta think it through from different angles. Classic problem-solving
It�s all about making that GTA
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5956
February 11th, 2023 at 9:19:01 AM permalink
With my computing background, I usually find Markov easiest for this sort of thing.
In this case, let P(A, B) be the probability of rolling seven 7s before a 6 and an 8, where A is the number of 7s already rolled and B = (1 if you have rolled any 6s) + (1 if you have rolled any 8s)
P(7, 0) = P(7, 1) = 1; P(n, 2) = 0
If you have already rolled a 6 or an 8, the only rolls that matter are a 7 and whichever of 6 or 8 you haven't rolled yet; 6/11 of the time, the 7 will come first
P(n, 1) = 6/11 P(n + 1, 1) + 5/11 P(n, 2) = 6/11 P(n + 1, 1)
If you haven't rolled any 6s or 8s yet, the rolls that matter are 6, 7, and 8; 3/8 of the time, a 7 will come first
P(n, 0) = 3/8 P(n + 1, 0) + 5/8 P(n, 1)
Work backwards, calculating P(6,1), then P(6,0), then P(5,1), and so on; the desired result is P(0,0).
Ace2 Joined: Oct 2, 2017