redtop43

I was playing craps last week when I saw someone playing an odd system. I eventually figured out what he was doing.

After the come-out roll he would play all the point numbers - $5 on the 4/5/9/10 and $6 on the 6/8.

I think that when he lost he would increase his bet. I forget the exact amounts, It might have been 25/30/36 on the 4&10, 5&9, 6&8.

If a number hit, he would take down his bets and immediately go back to $32. If he lost on the first roll, his next play was (I think) $125/$150/$202.

If he hit a number he'd go back to $32, or if this lost on the first roll, he'd go to $650/$850/$1002.

It was pretty unusual to see him buy into a $5 table for $6200. He said the system would lose on average once every 8 hours.

I make his odds of a losing sequence 1/625. He has 24 winning numbers (4/5/6/8/9/10) and 6 losers (all 7's) so the first roll after the come-out that's not 2/3/11/12 has to be a 7, 4 times in a row.

Eventually I figured out that it was just a variation on the doubling-up system. When he hits, if he hits before going broke, he makes back all his prior losses plus a small profit.

What I don't think he knew is that there is some variation in the payout odds. I don't usually play the numbers (just the pass line and odds) but I know at this casino if you get $25 or more on the 4/10, they pay $49 rather than $45. I haven't really noticed what they pay for large bets on 5/9. 6/8 are already only about 1.52% house advantage.

I had just never seen that before. The table limits were $5-$2000 so he was fine that way. I was only there for about 15 minutes while he was doing that, and he did get up to the highest level but he hit a number. A quick 7 and he'd have lost almost his entire stake.

RevJim

Sounds like a variation of the Martingale system, which has been proven like every other system to lose money over time at approximately the rate of the house advantage. You can read about it here. As you already noted, a player can win small gains for a while but eventually they lose a large stake that makes up for the small gains they've been making.

redtop43

It kind of was. He acknowledged that the system would fail once every 8 hours and he'd lose his stake. You're still going to lose the house edge times the amount on the table.

gengar

Just to note, there aren't different odds based on the bet amount. In craps, buying numbers pays even odds (no house advantage) minus a "vig", short for vigorish (commission). The vig may vary on bet amount depending on how the casino handles rounding.

wds17

To further clarify, the minimum commission is $1. So only for bets above $25 it is more favorable to buy the 4/10 vs simply to place the number. In your example we assume the person is buying those numbers.

mileage junkie

redtop43

I'm not sure you're right. I occasionally play 6 and 8, and never 4/5/9/10. I believe the normal odds are 7/6, 7/5, and 9/5, but I'm pretty certain I've seen them pay $49 back for a $25 bet on the 4 and 10. When you lay a number (bet it to lose) I think they give you fair odds less a 5% commission on the amount won. I saw a guy who would win about $500 on a side bet if another 5 was rolled, so he hedged and bet the 5 to lose, and I'm pretty sure he bet $155 to win $100.

gengar

Not sure where anything you've said here at all contradicts anything I said in my post.

redtop43

You said the odds don't vary with the amount of the bet. I said that at least when you play the 4 or 10, they do vary, as you get better odds for a bet of $25 and up.

mileage junkie

gengar

Yes, I said that because it's correct. Again, what varies is how the casino handles rounding of the vig, or commission. The odds do not vary. These are different things and not to be conflated.

mileage junkie

Although I have no doubt that the gambler in question was indeed betting the way you state he was, I think this post is virtually moot. I believe, with all due respect, that your "number" of 625-1 is WAY off. I calculate the odds of a 7 coming up before a single 4/5/6/8/9/10 as.....6-1 in one "round" (a round meaning the bet is resolved after any 4-10 is rolled, with any number but 7 being a winner. A losing 7 results in a significantly increased bet on each number, and, subsequently, the next "round." Any winning bet results in a downgrade to the original bet and you're back to "round one"); 18-1 in round two, 53-1 in round three, and finally, 167-1 (not 625-1), in round 4. Hypothetically, if you could get to a round five before reaching the table limit (highly doubtful), and/or the end of your bankroll(!), your odds are still only 500-1.

Somebody please check my math and prove me wrong, as I would like to try this. Albeit, on "machine craps," (ie..."Shoot to Win"), with much lower minimum bets (some start at $1), and exact payouts to the penny.