Roulette Math

Can anyone tell me the probabilities of getting a number that ends in 7, and a number that ends in 8, before a number ending in 9 hits on a single zero roulette table? Thanks

Quote: allinriverking

Can anyone tell me the probabilities of getting a number that ends in 7, and a number that ends in 8, before a number ending in 9 hits on a single zero roulette table? Thanks

If the seven and eight have to be in that order:

(3/6 * 3/6) = .25 or 25%

If the seven and eight can appear in any order:

(6/9 * 3/6) = 0.33333333333 or 33.33333%

The fact that it is a single zero, rather than a double zero is irrelevant, because there are still three numbers ending in seven, three in eight and three in nine. All other numbers are irrelevant because they don't resolve anything.

Quote: Mission146

If the seven and eight have to be in that order:

(3/6 * 3/6) = .25 or 25%

I think this should be 1/3 * 1/2 = .1667 = 16.67%

Quote: CrystalMath

I think this should be 1/3 * 1/2 = .1667 = 16.67%

I disagree, but my interpretation could be wrong of the initial question. My assumption is:

1.) Initially, only the sevens and nines matter and eights do nothing.

2.) Finally, only eights and nines matter.

I agree with you, though, if it has to be a seven BEFORE and eight OR nine, and then an eight before a nine, but that's not how I interpreted it.

For betting roulette Even Chance I subscribe to the Barstow principle:

" Diminishing probabilities , " as any repeated pattern of chance events continues , its reversal becomes progressively imminent.

Quote: Mission146

I disagree, but my interpretation could be wrong of the initial question. My assumption is:

1.) Initially, only the sevens and nines matter and eights do nothing.

2.) Finally, only eights and nines matter.

I agree with you, though, if it has to be a seven BEFORE and eight OR nine, and then an eight before a nine, but that's not how I interpreted it.

In that case, I agree.

any number ending in 7 and any number ending in 8, in any order, before a number ending in 9.

Quote: allinriverking

any number ending in 7 and any number ending in 8, in any order, before a number ending in 9.

If the seven and eight can appear in any order:

(6/9 * 3/6) = 0.33333333333 or 33.33333%

The fact that it is a single zero, rather than a double zero is irrelevant, because there are still three numbers ending in seven, three in eight and three in nine. All other numbers are irrelevant because they don't resolve anything.

Quote: Mission146

1.) Initially, only the sevens and nines matter and eights do nothing.

I say wrong! Eights do matter... Because if you get one, it has absolutely foiled your hope of getting a Seven first, followed by an Eight.

Quote: indignant99

Since getting the 8 first foils you, the first factor has to be 3/9.
Assuming that a repeat 7 doesn't foil you, then yes the second factor is 3/6.
(3/9 * 3/6) = 1/6 = .16667 = 16.667%


This is correct, but they can occur half-and-half (7-8 or 8-7). Cut percent in half 16.667% for each of those two ways.

That was already addressed between myself and CrystalMath several months ago. Further, AllinRiverKing replied in this thread stating that my interpretation of the question, and therefore, my answer is correct.

You wouldn't cut the percent in half for anything, you can either get a seven OR an eight before a nine and then you must get the other one.

Ask the Wizard. You're not right.
16.67% (for 7-8) and 16.67% (for 8-7) sum to 33.33%

Your 25% prob for 7,8 in order before the 9 is incorrect.
And I'm not arguing about RiverKing's clarification of the "rules."

How can you not see that - if 7,8 in order is needed - that an 8 first, spoils it?

Okay, "cut" was a poor word choice. Apportion the 33.33% equally to the two ways you can achieve the desired result (7,8 and 8,7).

Quote: allinriverking

any number ending in 7 and any number ending in 8, in any order, before a number ending in 9.

Quote: CrystalMath

I think this should be 1/3 * 1/2 = .1667 = 16.67%

Quote: Mission146

I disagree, but my interpretation could be wrong of the initial question. My assumption is:

1.) Initially, only the sevens and nines matter and eights do nothing.

2.) Finally, only eights and nines matter.

I agree with you, though, if it has to be a seven BEFORE and eight OR nine, and then an eight before a nine, but that's not how I interpreted it.

Okay, on this point that was resolved months ago, as you can see from the quotes above:

1.) My initial interpretation of the question was that you had to get a seven before a nine and the eight didn't matter at first, and then you had to get an eight before a nine.

If that's the case, then you would have a 1/2 followed by a 1/2 which is .25 which is what CrystalMath agreed with under my interpretation of the question.

2.) As it turns out, that's not even what the OP was asking, the OP later stated that the seven and eight can appear in any order. You wouldn't cut anything in half for this:

3-Sevens
3-Eights
3-Nines

The first of these results must be a seven OR eight before a nine, therefore, that is 6/9. The second result must be the opposite of the seven or eight (whichever hit first) before a nine and is therefore 3/6. (6/9 * 3/6) = .333333

3.) In the above post, I agreed with CrystalMath's probability of .1667 if a Seven and Eight had to come, in that order before a nine. That's where the 3/9 would come in, because you don't want an eight OR nine before the seven.

This whole thing is a question of interpretation. When I said, "If the seven and eight have to be in that order," I later clarified and said that if the seven must come before the nine and eights do nothing, then that is 1/2 and then if the eight must come before the nine and an additional seven does nothing, that is also 1/2.

Basically, if you look over the posts, everything that you are addressing was fully resolved months ago. The difference between my interpretations of the question and those of yourself and CrystalMath is that I never thought that an eight coming before a seven would cause it to lose. And, in fact, the OP later clarified that such would not cause it to lose. The seven and eight can come in any order, and repeat if they want to, provided they both come before the nine. Therefore, the probability the OP wanted to know is .33.

I understand full well allinriverking's clarification. I'm referring to a different requirement - not allinriverking's clarified problem.

I'm referring to your postulated or surmised requirement that a 7 occur before 8, and of course both before 9. And you calculated its probability WRONG. Please, please, please try again. Using the "7 must precede 8, both preceding 9" proviso.

It's damn well not 25%, unless you bend your own requirement to permissively allow 8's before an eventual 7, and then another eventual 8.

Calculate, please for these requirements (you can call them indignant's requirements):

• Must achieve a 7 before achieving either 8 or 9. (Achieving either 8 or 9 before a 7, is a failure result.)
• After achieving a 7, may achieve more 7's, with no consequence (i.e. no failure nor success, yet).
• After achieving a 7, must achieve an 8, before achieving a 9. (Achieving a 9 before an 8, is a failure result.)
• After achieving both a 7 and an 8 (before a 9), the scenario has been won.

Quote: indignant99

I understand full well allinriverking's clarification. I'm referring to a different requirement - not allinriverking's clarified problem.

I'm referring to your postulated or surmised requirement that a 7 occur before 8, and of course both before 9. And you calculated its probability WRONG. Please, please, please try again. Using the "7 must precede 8, both preceding 9" proviso.

It's damn well not 25%, unless you bend your own requirement to permissively allow 8's before an eventual 7, and then another eventual 8.

Calculate, please for these requirements (you can call them indignant's requirements):

• Must achieve a 7 before achieving either 8 or 9. (Achieving either 8 or 9 before a 7, is a failure result.)
• After achieving a 7, may achieve more 7's, with no consequence (i.e. no failure nor success, yet).
• After achieving a 7, must achieve an 8, before achieving a 9. (Achieving a 9 before an 8, is a failure result.)
• After achieving both a 7 and an 8 (before a 9), the scenario has been won.

The probability of a number ending in 7 coming up before a number ending in 8 or 9 = 3 / 9 = 1/3.
After that, the probability of a number ending in 8 coming up before a number ending in 9 = 3 / 6 = 1/2.
The overall probability = 1/3 x 1/2 = 1/6.

Quote: ThatDonGuy

The overall probability = 1/3 x 1/2 = 1/6.

Correct. Thank you very much, Don.

Quote: indignant99

I understand full well allinriverking's clarification. I'm referring to a different requirement - not allinriverking's clarified problem.

I'm referring to your postulated or surmised requirement that a 7 occur before 8, and of course both before 9. And you calculated its probability WRONG. Please, please, please try again. Using the "7 must precede 8, both preceding 9" proviso.

Try to read this again:

Quote: Mission146

I disagree, but my interpretation could be wrong of the initial question. My assumption is:

1.) Initially, only the sevens and nines matter and eights do nothing.

2.) Finally, only eights and nines matter.

I agree with you, though, if it has to be a seven BEFORE and eight OR nine, and then an eight before a nine, but that's not how I interpreted it.

As you can see, I was allowing an eight to precede seven and then needing another eventual eight:

1.) Initially, only the sevens and nines matter and eights do nothing.

Therefore, I don't know what the Hell you're arguing with me about. For the 117th time, I agree with you just as I agreed with CrystalMath several months ago:

Quote: CrystalMath

I think this should be 1/3 * 1/2 = .1667 = 16.67%

In response:

Quote: Mission146

I agree with you, though, if it has to be a seven BEFORE and eight OR nine, and then an eight before a nine, but that's not how I interpreted it.

That discourse was had in October of last year. Yes, if it has to be a Seven before an eight or nine and then an eight before a nine, you have 1/3 * 1/2 = .1667

You've been arguing with me about something that I agreed to (and already knew) nearly six months ago. The only reason I didn't say that in the first place, six months ago, is because that is not how I interpreted the question.

Anyway, for a change, I didn't calculate anything wrong. Try to read this post, one more time:

Quote: Mission146

I disagree, but my interpretation could be wrong of the initial question. My assumption is:

1.) Initially, only the sevens and nines matter and eights do nothing.

2.) Finally, only eights and nines matter.

I agree with you, though, if it has to be a seven BEFORE and eight OR nine, and then an eight before a nine, but that's not how I interpreted it.

What I was calculating is that only Sevens and Nines matter initially and that an eight before a seven does not affect anything. After that, only eights and nines matter. I clearly stated such was my interpretation several months ago. I then stated that the 1/6 is correct if they had to come in the precise order 7-8 before a nine and that an eight first would cause it to fail.