PBguy
Joined: Sep 4, 2013
• Posts: 278
March 14th, 2015 at 3:19:30 AM permalink
I once saw a guy lose well over 10 grand when the 7 roll. But he had gone from about 2 grand to over 50 grand in the last two shooters. He continuously pressed and never pulled his bets down.

If you are up 2K in your rail then why not leave that \$540 across (or whatever it is) up and hope for even bigger profit? The worst that happens is the next roll is a 7 and you walk away up 2 grand.

It's not like a good long roll doesn't necessarily have any 7s rolled. It's all about when they're rolled that counts.
goatcabin
Joined: Feb 13, 2010
• Posts: 665
March 14th, 2015 at 10:22:01 AM permalink
Quote: mikes0805

Lets say we have the same type of scenario; I walk up with \$1500 and I am up 2k, and there is a lot of money still on the table. If I were to call my bets "off" every other roll, what does that do to my percentage of losing all of my bets after 20-30 some odd rolls?

No lets say I am up 2k (in my rail) and I have 540+ across plus hard ways. I'm sure they (dealers) won't like it but for the purpose of this equation they don't care lol. The shooter has rolled approx. 20-30 times, with an occasional seven in between points.

It doesn't matter how many times he's rolled, or how many sevens have occurred on comeout rolls. It sounds like you have 24 ways to win a bet (losing the hardway if a number rolls soft) and 6 ways to lose it all. So, for every roll, you have a .67 probability of winning some bet, a .167 probability of losing it all, and a .167 probability of a non-resolving roll; ignoring the non-resolving rolls, you have a .8 probability of a win and a .2 probability of losing the \$540+.

By calling your bets off every other roll, you avoid the risk on those rolls, but you also waste one of the 24 if it rolls. You might as well just guess. OTOH, presumably the reason you haven't taken all your bets down is that you want to win some more money. Let's say you have these bets (ignoring the hardways):

80+4 90 +4 96 96 90 +4 80 +4 (buy on 5/9, 4/10)
Your wins will be 156, 131, 112, the weighted average of which is 129.3, so 4 wins before a loss will almost break even, assuming you replace each win rather than pressing further.

Here are the probabilities of different numbers of hits before the seven:
0 .2 lose it all
1 .16 (cum .36) win one, average amount \$129, then lose \$540 ~= -411
2 .128 (cum .488) etc.
3 .1024 (cum .5904)
4 .08192 (cum .67232)
5 .065536 (cum .737856)

So, 59% of the time you won't get 4 hits before the seven. The average number of hits is four. (BTW, notice all the powers of two in there?)
NOTE: those numbers only apply before the first decision. IOW, if you hit a number, the .16 probability is now for two hits, since you've already hit one. (The underlying probabilities are always .8 and .2.)

Here's another approach: each number that hits, take that bet down, so you reduce the amount lost when the seven comes. So, let's say you hit a 4, win \$156 and take that bet down; now you only have \$456 out there to lose.

Another approach would be to take everything down if you win one, or two.

It's all a matter of how much is the money out there worth to you, compared to what you might win if you keep betting.

However, skipping rolls has no effect whatsoever on your chances, unless you are clairvoyant.
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
mikes0805
Joined: Jan 2, 2015