https://wizardofvegas.com/forum/questions-and-answers/math/37934-multiple-4s-10s-before-big-red/#post880731

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.

Quote:Gamblehead49What are the odds of shooting 7 7's before shooting a 6 or an 8?

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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 Bet | Profit with Win | Total Loss with Loss |
---|---|---|---|

1 | 24 | 14 | 24 |

2 | 60 | 11 | 84 |

3 | 168 | 14 | 252 |

4 | 456 | 14 | 708 |

5 | 1236 | 13 | 1944 |

6 | 3360 | 16 | 5304 |

7 | 9120 | 16 | 14424 |

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.

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

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

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

Quote:Ace2I 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

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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

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 !

Quote:Ace2Logic 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

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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.

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

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).