When 7 rears it's ugly head...

I (poorly) asked this question earlier so I'll try to refine it.

Is there a way to calculate or to estimate when the 7 will wipe the board?

To say it another way - is it more common to see a Point-7-out or a Point, Roll, 7-out, or a Point, roll, roll, 7-out, etc.?

In 100 sessions after a point has been set, what would the "expected" number of P-S-Os be?

Is this even a reasonable question to ask?

~N

It is more common to see a point-7 out than any of the other combinations. It is the most common roll sequence.

Quote: foolshope

I (poorly) asked this question earlier so I'll try to refine it.

Is there a way to calculate or to estimate when the 7 will wipe the board?

To say it another way - is it more common to see a Point-7-out or a Point, Roll, 7-out, or a Point, roll, roll, 7-out, etc.?

In 100 sessions after a point has been set, what would the "expected" number of P-S-Os be?

Is this even a reasonable question to ask?

~N

A reasonable question to ask. After a point has been established, 16.67 times out of a hundred the next roll will be a 'seven out'. Since only a 'seven' will seven out you, and seven happens one out of 6 rolls, the math is ONE divided by SIX = 16.67%

Quote: foolshope

Is there a way to calculate or to estimate when the 7 will wipe the board?

To say it another way - is it more common to see a Point-7-out or a Point, Roll, 7-out, or a Point, roll, roll, 7-out, etc.?

These aren't the same question. You've already gotten the answer to "how often should a 7 appear", but there's a difference between "how often" and "when". There is no way to calculate, estimate, or predict when the 7 will appear. "When" reflects a point in time, so asking "when will the 7 appear" is not an answerable question. What you can properly ask is a question related to the likelihood of the 7 appearing, or of the 7 appearing before some other event, etc. For example:
1) How likely is it that a seven will appear in the next N rolls?
2) How many sevens should I see before X happens (rolling a point number, hard way, craps, yo, etc.)?
3) Your final question, how likely is it to seven out immediately after a point has been established?

All of those are answerable. But "when will the 7 appear" isn't.

Consider perhaps that newbie at the Borgata who held the dice for over four hours or something like that. All during that roll Sevening Out was the most likely event to occur. Had this 16.67 percent chance... but it didn't happen.

The average number of rolls needed to get a seven is six. However, after you roll five non-sevens you are no closer to a seven than you were when you started. The chances of a seven are always 1 in 6. The dice do not have a memory and a seven is never overdue. This memoryless property is true for many other casino games, such as roulette, slots, keno, and video poker.

Quote: Wizard

The dice do not have a memory and a seven is never overdue. This memoryless property is true for many other casino games, such as roulette, slots, keno, and video poker.

Perhaps its part of The Buzzz or just part of human nature that players forget that. Its always the irrational feeling that somehow this will be the night when the dice do something to his advantage because of some special relationship with fate.

After a few rolls, I switch to the dark side because surely Seven is "due" or "approaching due". After a few Reds, surely Black is due. After all, if its something like 6.6 rolls or the like that is average then surely it will be that "next" roll that is going to be "the one" because although it can be four hours its most likely to be 6.6 rolls or something like that. Those four hour rolls are rare. They give you champagne after the four hour roll. They don't give champagne after a two minute roll.

Ok, so let's say I place 150 across on any given point #... for every 100 of those rolls, 16.666 will be 7 an cause me to lose 150*16.666 = 2500.

But every 83.443 * expected win (call it $40) = 3372 (not sure what the expected payout on the average place bet is relative to the place bet itself) this *could* be a reasonable strategy... the curse, of course being, dice have no memory and we've all seen PSO > 7x in a row... and at a cost of $1K + any pass line bets... the night could be short, correct?

Hmm... appears I forgot about craps and yo - that'd change my expected win if I did not have money there... ick.

~N

Quote: foolshope

Ok, so let's say I place 150 across on any given point #... for every 100 of those rolls, 16.666 will be 7 an cause me to lose 150*16.666 = 2500.

But every 83.443 * expected win (call it $40) = 3372 (not sure what the expected payout on the average place bet is relative to the place bet itself) this *could* be a reasonable strategy... the curse, of course being, dice have no memory and we've all seen PSO > 7x in a row... and at a cost of $1K + any pass line bets... the night could be short, correct?

Hmm... appears I forgot about craps and yo - that'd change my expected win if I did not have money there... ick.

~N

You're definitely missing a little bit of the math to account for the come-out rolls and whether or not you had your place numbers working on the come-out or not. I'll leave that math to someone a little more proficient than me, but I think approximately 30% of the rolls would be come-out rolls. If you ignore the pass line (assuming someone else shooting), the 2/3/12 wouldn't hurt you (obviously). You'd have to decide all of those things in order to get closer to the "real" numbers.

I think a lot of people play other shooters by letting them set a point and then throwing out a place spread on the 5-6-8 + field bet or just 5-6-8-9. With the 5-6-8-field, you win on the next roll unless a 7 is rolled. The players will then turn their bets off or reduce their exposure somehow. This sounds like a great strategy other than the notorious "PSO" and the fact that you're risking hundreds to win maybe 40 or 50 bucks on the roll. If you get nailed by a couple PSOs, you better have a sufficient bankroll if you want to have a chance to make a comeback.