Which is more likely after a 7? Need you to show me the math

In a series of 1000, rolls about 2 percent will be double 7s or 7 then immediately another 7. Which is more likely? A seven followed by another 7 or a 7 followed immediately by a 6 or 8, and then no other 6 or 8 until another 7 hits?

So probability of a 7-7 vs a 7-(6or8)-XXXXX-7 where X are the other rolls in the series but can't be a 6 or an 8.

Thanks for any and all help.

Keith

The most likely roll will be a 7 (1-in-6). It doesn't matter what has come before. The dice do not know that they are following a 7, 6, 8 or even three 7's in a row.

But what's the most like roll after five 7s in a row?!?! And after six 7s in a row!?!? What about seven 7s in a row!?!?

Like wise, what is the most likely roll after the dice fly off the table? Or after someone orders a "seven and seven," while wearing a dark green "seven-up" t-shirt and yelling "Let's go darkside!"

Isn't it funny that the answer is still the same.

And that is why you should *always* bet the Big Red Seven on every roll... To bad it pays so poorly...

Quote: Ayecarumba

The most likely roll will be a 7 (1-in-6). It doesn't matter what has come before. The dice do not know that they are following a 7, 6, 8 or even three 7's in a row.

True, but that doesn't quite answer the question.

the question, reworded, is: Which is more likely: a) 7-7, or b) 7 - 6 or 8 - (indeterminate number of non 6/8 numbers) - 7

The answer is a), but I also would appreciate seeing the actual math.

Quote: rdw4potus

True, but that doesn't quite answer the question.

the question, reworded, is: Which is more likely: a) 7-7, or b) 7 - 6 or 8 - (indeterminate number of non 6/8 numbers) - 7

The answer is a), but I also would appreciate seeing the actual math.

Expounding:

7 - 6/8 is more likely than 7 - 7. But 7 - 7 is more likely than 7 - 6/8 - 7. Does the string of non-7, non-6/8 numbers even matter in b)?

7 followed immediately by 7 = (1/6)(1/6) = 1/36, which is closer to 3% of all possible pairs of rolls than 2%.

In the second case, I think you mean that after establishing the 6 or 8, that particular number doesn't appear before another 7, and you don't care whether the "other" number (that wasn't the roll immediately after the initial 7) appears or not.

In that case the probability is (1/6)(10/36)(6/11) = (5/18)(1/11) = 5/198 or very close to 2.5%.

If you mean that the intervening period has NEITHER 6 NOR 8 regardless of what the point is, the probability is (1/6)(10/36)(6/16) = (5/18)(1/16) = 5/288 or slightly under 2%.

I left these as fractions, without calculating a closer decimal equivalent, on purpose. One reason why is to better show all my work. Another is that I do math like this all the time at the table, where the ability to convert from fractions to decimals in your head is helpful.

Exactly my question RDW!! Everyone knows the most likely number to come up is a seven...that wasn't my question. I'm trying to settle an argument with my buddy...Scenario is craps...First roll is a seven or even a seven out...he says its better to bet the 6/8 place bet immediately (ON), the come out roll than to double up the seven bet (if a natural) or place a pass line bet. His thinking is it's more likely to be a 6 or 8 right after the 7 and win that bet than to wait for a 6 or 8 point...

So, if a seven follows a seven, he loses his place bet, but if a 6 or 8 comes out on the very next roll, he wins his one of his two placed bets and takes them down.

Look at it this way...

Seven out...new shooter.

Scenario 1a. Place $10 on pass line...7 hits on the come out. win $10.

Scenario 1b. Place $10 on pass line...6 or 8 hit...now waiting for another 6 or 8 to win, doesn't happen, then another 7 happens loses.

Scenario 2a. Place bet on 6/8 $6 each. BET IS WORKING. 7 hits. Lose $12

Scenario 2b. Place bet on 6/8 $6 each. BET IS WORKING. 6 hits. win $14 take place bets off.

So which one of us is more likely to win any money at all?

So again which sequence of numbers is more likely?

A) 7-7
B) 7-6or8-n....n-7 where n cannot be a 6 or 8.

And what is the exact probability...show math if possible?

Keith

Quote: kwalter827

... Scenario 2b. Place bet on 6/8 $6 each. BET IS WORKING. 6 hits. win $14 take place bets off. ...

I think you need to reanalyze this part -- you would only win $7 in this step, not $14.

Here is my guess for the math of which sequence is more likely. Of the 36 possible outcomes for 2 dice, there are six possibilities for a 7 and 5 possibilities each for 6 or 8. The probability of 7 on one roll, followed immediately by a 6/8, then followed by another 7 before another 6/8 is:

p=(6/36)*(10/36)*(6/16)=0.01736111. This all assumes that you do not care whether a 6/8 is a 6 or an 8.

The probability of 7 on the one roll, followed immediately by another 7 is:

p=(6/36)*(6/36)=0.02777; i.e., more likely, by a factor of 1.6.

I do not guarantee that this analysis is appropriate for the playing strategies that you describe, since I think there may be some errors in other parts of your analysis.

Quote: kwalter827

...Scenario is craps...First roll is a seven or even a seven out...he says its better to bet the 6/8 place bet immediately (ON), the come out roll than to double up the seven bet (if a natural) or place a pass line bet. His thinking is it's more likely to be a 6 or 8 right after the 7 and win that bet than to wait for a 6 or 8 point...

Well, now that you explain it...


When you combine the 6 and 8 as one option, then it is more likely that a 6 OR 8 will come out. But that's AT ANY TIME. The prior roll is still irrelevant.

I.E. If you are trying to decide if, ay any particular moment, if: A-6, B-7 or C-8 is more likely to appear, the answer is 7. But if the choice is: A-either 6 or 8, or B-7, THEN the most likely roll is either 6 or 8.

Get it?

You are correct that 6 or 8 is more likely than a 7 on any roll. But the $7 you sometimes win does not make up for the $12 you sometimes loose.

You'll roll 6 five times to win $35.
You'll roll 8 five times to win $35.
You'll roll 7 six times to loose $72.

Overall you loose $2 but you can have a lot of fun in the process.

Doc

Thanks so much for that help. (And yes, $7 is the win. My bad, I usually play $10 tables.)

Your analysis makes sense. And yes, I don't care if it's a six or eight right after the seven and then can't be a six or eight until the next seven.

Keith

I think you posted this just as I was clarify. Thanks to you as well. Same answer as DOC. Much appreciated.

Keith

Your friend is impatient. Since the seven is the most likely number to come up, he should best the pass line for the best chance to win. Betting the 6 / 8 for $6 each won't give you the best chance. On the pass line, you have 244 ways to win, and 251 ways to lose. On the 6/8, you have 5 ways to win, and 6 ways to lose. The pass line is better. Splitting it between the 6 and 8 won't make a difference on the overall bet, but will lower variance.

It's just close enough for it to make a minuscule difference in the long run. But in the real world -- not so much. I would have fun with it. He bets his way, you bet the other way, and see who comes out ahead more often.

Quote: kwalter827

Exactly my question RDW!! Everyone knows the most likely number to come up is a seven...that wasn't my question. I'm trying to settle an argument with my buddy...Scenario is craps...First roll is a seven or even a seven out...he says its better to bet the 6/8 place bet immediately (ON), the come out roll than to double up the seven bet (if a natural) or place a pass line bet. His thinking is it's more likely to be a 6 or 8 right after the 7 and win that bet than to wait for a 6 or 8 point...

So, if a seven follows a seven, he loses his place bet, but if a 6 or 8 comes out on the very next roll, he wins his one of his two placed bets and takes them down.

Look at it this way...

Seven out...new shooter.

Scenario 1a. Place $10 on pass line...7 hits on the come out. win $10.

Scenario 1b. Place $10 on pass line...6 or 8 hit...now waiting for another 6 or 8 to win, doesn't happen, then another 7 happens loses.

Scenario 2a. Place bet on 6/8 $6 each. BET IS WORKING. 7 hits. Lose $12

Scenario 2b. Place bet on 6/8 $6 each. BET IS WORKING. 6 hits. win $14 take place bets off.

So which one of us is more likely to win any money at all?

So again which sequence of numbers is more likely?

A) 7-7
B) 7-6or8-n....n-7 where n cannot be a 6 or 8.

And what is the exact probability...show math if possible?

Keith

The timing of placing a -EV wager is inconsequential. In the long run, you're going to lose money regardless. Essentially, what you're asking is whether a place bet of the 6 or 8 is better than a pass line bet. The house advantage of a 6 or 8 place bet is 1.52%. For a pass line wager (without odds), it's 1.41%.

The pass line bet is the better wager, especially when you back it up with odds after the point is established.

kwalter, an easy way to look at what is happening is to analyze each bet individually.
The come out roll, before you roll it, has a house edge of 1.4% or so.
A $ 6 bet on 6 or 8 should pay $7.20, but only pays $7, so it has a house edge of 1.5% or so.
Any combination of bets that over time will tend to lose, will tend to lose.
If you can understand the math behind why each single bet favors the house, then you should also be able to figure out that combining a bunch of bets that favor the house will never turn the odds in your favor.

It's actually 10 ways to win vs 6 to lose. If he wins either the 6 or 8 and then takes both down. He loses if two sevens come up with no 6 or 8 between. He can take his place bets down after a win. I have to leave my pass line bet out and hope the same 6 or 8 comes out again...
Still house always wins...

True, but he is only betting $6 on six and eight, which is half of what you're betting (approx.), so really he'd have to hit twice to match what you make (more or less).