Betting: ..How many numbers to cover

For me it appears that the better results are covering 3 numbers. Pass-odds and the 6 and 8, and if the point is 6,8 then adding the 5 or 9 to get the three numbers. Not too many numbers but enough to keep it interesting.

This is working better for me compared to inside betting or across betting. For those losing too much with inside and/or across betting as I was, give this one a try.

It's completely a matter of taste. Every bet on the craps table except the free odds bet is negative.

If you're going to place numbers, the 6 and 8 carry lower house edges than the 5,9,4, and 10.


I don't ever place numbers, though. I play pass line and continuous come betting. That way I can cover numbers for the cheapest cost possible. But I don't stop with 2 come bets. I bet come every roll of the dice.

A friend of mine use to encourage me to do the following, and I hope that I'm remembering this correctly (I'm not a craps player)
Place the 6 & 8. Wait for 3 wins, pressing each time and then move everything and buy either a No 4 or a No 9.
He seemed to do fairly well with it......

The number of bets still really depends on why you even play the game.
Most want length of play to be the first choice
Many just want to play to win $X and quit the session no matter how long it takes
Others want it all: play at least a certain length of time and win $Xs.

When the choice comes down to risking the same action say $64, the fewer larger bets,
with the same bankroll at risk,
will hit win goals more often than many smaller bets.
So if your playing goal is mainly to hit win goals and
quit while risking your total starting bankroll, the fewer larger bets the better.
Keep it in mind

4 players and 4 methods (average bets)
1) $64 across (6 bets)
2) $66 across ($18 P6&P8; $10 P5&P9; $5 P4&P5) (6 bets)
3) $30 P6&P8 ($60 total) (2 bets)
4) $60 on P6 or P8 (1 bet)

starting bankroll $640
target bankroll $1280
who has the highest probability of hitting their target?

7craps; So less is better in your examples. Two more examples if you don't mind: 5) $66 inside (4 bets: 15-18-18-15) 6) $60 Pass line only, no odds.

Quote: DeMango

7craps; So less is better in your examples.
Two more examples if you don't mind: 5) $66 inside (4 bets: 15-18-18-15) 6) $60 Pass line only, no odds.

The fewer number of bets for the same action/risk

$66 inside: 33.56%

The $60 pass/no odds is a simple calculation using the Gambler's Ruin formula
or =1/((q/p)^b_x+1)
p=244/495
q=1-p
b_x = # of betting units
40.3% with average number of rolls about 340

$64 pass: 42.98% (exactly 10 betting units)

5 pass/10x odds: 47.92%
avg # rolls: 543
5 pass/11x odds: 47.74%
avg # rolls: 473
1 bet of $640 with pass: 49.29%
(this minimizes the number of games played)