$132 Inside Bet: Two-Hit Regressions Explained

Sam Originally published Updated

$132 Inside · Two hits · Regress to $66 · Regress to $22

The $132 Inside strategy uses two-hit stages: collect two $42 wins, regress to $66 inside, collect two $21 wins, then regress to $22 inside. If both stages finish before 7, $104 is locked in the rack with $22 still working.

The payout path and its survival chance

The two regressions reconcile exactly. The other side of the plan is a $132 loss if 7 arrives before any inside hit and a roughly 31.64% chance of completing both two-hit stages before 7 when neutral totals are ignored.

Stage 1: $132 to $66 inside

Stage 1: $132 to $66 inside: Number; Starting bet; Win; After first regression.
NumberStarting betWinAfter first regression
5$30$42$15
6$36$42$18
8$36$42$18
9$30$42$15
  1. Place $132 inside after the point according to the table’s working rule.
  2. Collect the first two inside wins: $42 + $42 = $84.
  3. Regress from $132 to $66, returning another $66.
  4. The rack now holds $150 against the $132 entry: $18 is locked with $66 still working.

Stage 2: $66 to $22 inside

  1. At $66 inside, each 5/6/8/9 hit pays $21.
  2. Collect two hits: $42 more to the rack.
  3. Regress from $66 to $22, returning $44.
  4. Total returned is now $150 + $42 + $44 = $236.
  5. Relative to the original $132, $104 is locked with $22 working.

Taking everything down at a checkpoint

After the first two $42 hits, the rack contains $150 and $66 remains working. If you remove the $66 at that point, the rack becomes $216 against the $132 entry: $84 completed profit. Keeping the $66 working instead means $18 is protected in the rack if the next working result is 7, with the other $66 still exposed until you remove it or it resolves.

After the second pair of $21 hits and the regression, $236 is in the rack and $22 remains working. Taking down that $22 brings the rack to $258, or $126 profit against $132. Leaving it up preserves $104 in the rack if the next working 7 takes the $22. These numbers count only the Place-bet sequence; any line bet and Odds used to establish a point need their own ledger.

Failure-state table

Failure-state table: Sequence before 7; Cycle result if 7 follows.
Sequence before 7Cycle result if 7 follows
No inside hit−$132
One $42 hit−$90
Two hits, then regress to $66+$18
One additional $21 hit+$39
Two additional hits, then regress to $22+$104

Probability of reaching each regression

The inside numbers have 18 combinations; 7 has six. Ignoring totals that affect neither position, the probability the next relevant result is an inside hit is 18/24, or 75%. Two inside hits before 7 therefore occur with probability 0.75² = 56.25%. Completing both two-hit stages before 7 is 0.75⁴ = 31.64%.

The rolls are independent. Completing the first stage does not make the second stage due, and failing one shooter does not improve the next shooter’s chance.

Bankroll and table rules

  • One immediate seven-out costs the full $132.
  • Three complete entries require $396 before line bets or Odds.
  • State whether the point number remains included.
  • Keep returned chips in the rack if they are being counted as locked profit.
  • Do not describe the $66 or $22 still working as free or house money.

Compare the smaller $22 Inside reduction, the point-excluded $135 strategy, and the complete Place-bet guide.

Sam’s table note: The two-hit stages give clear checkpoints. Decide whether to keep the smaller layout working or take it down; returned stake and collected profit are different parts of the rack.

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2 Comments

  1. I like this strategy and am planning to try it out at the casino this week, starting with a $400 bankroll. I’ve been testing it on some various simulators (which all pretty much suck except for WinCraps). I’m a newbie to Craps and still getting to know that program, btw.

    Anyway, the issue with this strategy is my local casino (and I’d imagine many others) has a $10 min bet, so the 3rd regression down to $22 inside would have to be at least $44, which is ok, but does leave you a little more exposed at higher roll counts where the 7 becomes more and more likely to hit. After I regress to $44 inside and hit twice, I start pressing bets until the streak stops. I’ve had streaks up to 35 rolls, but after that one craps out, the 7’s get pretty common. (I’ve seen 4 in a row, which really hurts at $132 a pop!).

    I think if you manage to get through a few shooters with at least 4 hits, you should strongly consider bowing out after your next hot streak craps out. You should have made at least $300 by then.

    Another issue (maybe not at the casino, but most of the programs seem to assume this rule) is the shooter is required to have a pass line bet active, which throws off this strategy a little, as well.

    1. Hello Chris, Thank you for your insight and tips! It is greatly appreciated and I hope others take a few minutes to also contribute here or on other pages! Best Regards!

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