Try some simple math, it's nice to be able to look at different betting schemes, as how much money do I need to reach some level of "bust". In this instance of betting a hard way for $5, and adding a $1 more each time it misses, try this. Just pick one hard way, and work the numbers, assuming a hard 6-8 pays 9 to 1, and a hard 4-10 pays 7 to 1. Example: Start $5 hard four. 1st. bet........$5.....expected win $35 2nd bet........$6.....expected win $37.......$42 minus $5 1st bet 3rd bet.........$7.....expected win $38 ......$49 minus $11 invested 4th bet.........$8.....expected win $38.......$56 minus $18 invested 5th bet.........$9.....expected win $37.......$63 minus $26 invested 6th bet.........$10...expected win $35.......$70 minus $35 invested 7th bet (you do the math) 8th bet (you do the math) and etc. Note: you will get to a negative win expectation where you will have to increase your bet by $2, then $3, etc. to get your investment back. You can add up the investments to get a bankroll required to reach a level that you consider a "bust". Get in the habit of doing these kind of exercises....you can use a spread sheet if you find that easier, than a quick mental exercise. Disclaimer: math done in my head, check it.

If not for established table limits for specific bet types, the Martingale would always produce a winner. Keep betting larger and larger until you finally win to recapture all of the losses. With a "semi" or "fractional" martingale system of play, applied to a wager or series of wagers which pay off at favorable odds (that is, much better than flat wins), there might be a way to consistently (but obviously not always) come out ahead. Likely this would not often result in big wins, but a winner is a winner. As with any wager on a craps table which loses with the appearance of the seven, the fewer bets made the better. In this sense, it might be better to choose just one of the four hard way bets to play this game. If you choose all four, more money is to be involved in the chase. There will always be more losing bets than winners, and the winners are NOT paid off fairly, 10 for 1 or 8 for 1 instead of 10 to 1 or 8 to 1. Maybe this sort of play would work better hopping the sevens at 15:1 on every roll, until you see two or three winners in a row??

I look at these types of betting patterns often, never considered going up to $2 or $3 increases to keep it going. Interesting idea if a bit expensive.... I always just figure from first bet to a point my increases would equal at best a break even point, then call that the last bet for bust.....

Absolutely agree! You must decide on a "bust" limit on these schemes. I use an "odds" of success (risk-reward) scale to determine my level of bust....along with available funds at the time.

The problem with any of these progressive bet schemes is that you start with a cumulative probability of success. That usually sounds really good, then as you lose each bet, that cumulative probability changes....for the worse!

According to your math then, the probability of a 7 appearing is 97.39 % if it hasn't appeared in 20 tosses ; would that be an indicator that it is due.......or do you want to use your 1 in 6 math. I know you like to pull out the math gun that most supports your position....there is so much math to pick from. 777

Cumulative probability is a "before" the fact look at the likelihood of something happening. It looks at future expectation. That has nothing to "due" with anything that has happened!

Off subject a bit on the 7s but I hit 4 in a row and 3 in a row on two seperate come out rolls the other day and still refused to bet red

If you feel it is due, then bet accordingly. I certainly do not have a problem with anyone playing a "due" type strategy. Luck is luck, no matter that the bet was placed using a less than rational reason. In fact, I don't think we can use "rational" for making any bet in a negative expectation game....just don't try and sell "due" theory with any hint of math attached to that sell.

Because the game is random and we know numbers clump and then hide we must play a game in which we follow the trend and then figure out when the trend is at the end and then bet the 'due'. It's enough to drive you a little crazy, and it's important to have a good 6th sense for these things. If the game is random, you have to be random too and not always play one way. The math can only help you so much....after the math the voo-doo must take over. Of the people here, The Comeback Kid recognizes this the most. All the aforementioned is in my humble opinion. 777

Of course! It is called "In search of LUCK". How one searches for that elusive LADY, is an individual decision! TCK, just suggests that you do it with a positive view, and with a little self help Guru guidance.

While my own negative results with this advice are likely due to limited bankroll, I have NEVER changed a losing session into a winning one by changing style of play. I guess the problem being that I have already lost half the buy-in when concluding that a change is necessary. The handful of times when willing to lose it all with the change have always succeeded, every time. As a result, I'd rather accept the loss of half of the small buy-in and look to do better next time in.

With the "regularity" of the seven, it appears that hopping them for let's say $2 each as a starting point, adding $1 to every loser, would keep this inexpensive, unless you happen to see the now dreaded 50 roll hand. tercol, would have been nice had you been either parlaying the line bet or hopping the reds for a few bucks at the time of your good fortune. But you know how it is with the would haves, the could haves, and should haves....

? Has the math been done on this? Starting at $2 and increasing $1 each loss, how much on average is needed to bet do you assume before a 7 shows or you reach a point that $1 increases won't recover all prior losses? Cumulative betting can add up fast, $2 + $1, $1, $1, etc..... total bet before a win subtracted from the win gives how much of a payday? Sometimes it is not enough to end up, but to end up enough that all the hard work and patience has paid off.... will this accomplish a satisfactory end result when it is working? It sounds simple but you must be aware of the point your increased bet only equals a break even win and understand that all future bets can only equal a loss..... You are not just betting $1 on each new bet but $1 more than the bet before: $2 $3 $4 $5 $6 $20 down in just 5 bets that lose and how much will you win if your $7 bet wins? $7 $8 $9 $10 $11 $45 more in the next 5 rolls for a total loss of $65 on 10 rolls, will your $12 bet recover it? A strategy is only as good as the player understands the concept of Break Even Analysis. If you don't know when to call the bet a loss, and calculate accordingly, the strategy is doomed to fail.

Can't and won't argue with your logic. It all depends on how they tumble. As far as the increased wagers go, Martingale (or our incremental variation thereof) says to increase with the loser and stand pat with the winner. If you hit a couple in a row, which is not all too uncommon with the seven, you'll do OK with the 15 to 1 minus your two losing sevens pay out.

- $ Everything does good IF you can hit a couple in a row...... At 15 to 1 minus 2 unit payout structure, what is the break even point of this bet? Bet $2 each or $6 total if you win $30 - $4 = $26 if lose next bet Bet $3 each or $9 total if you win $45 - $6 -$6 prior loss = $33 if lose next bet Bet $4 each or $12 total if you win $60 - $8 -$15 prior loss= $37 if lose next bet Bet $5 each or $15 total if you win $75 -$10 -$27 prior loss = $38 if lose next bet Bet $6 each or $18 total if you win $90 -$12 - $42 prior loss = $36 if lose next bet Bet $7 each or $21 total if you win $105 -$14 -$60 prior loss = $31 if lose next bet Bet $8 each or $24 total if you win $120 -$16 -$81 prior loss = $23 if lose next bet Bet $9 each or $27 total if you win $135 -$18 -105 prior loss = $12 if lose next bet Bet $10 each or $30 total if you win $150 -$20 -$132 prior loss = (-$2) if win or lose you lose. Starting at $8 level your total bet is more than your expected net win on the strategy. Starting at $10 level your are negative even on a win. $9 is the last positive expectation level. This means that you have an 8 bet range to catch a seven before you can't end up. Counting only bets $2 - $9 the total strategy has a cost of $132 to either win or lose. Max win is at level $5 for $38 and max lose is after level $9 for negative $132. 8 Bet strategy using $132 to win between $12 and $38 depending on level won on or losing it all.