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

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:AyecarumbaThe 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:rdw4potusTrue, 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)?

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.

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'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.