tomshark30
tomshark30
Joined: Nov 7, 2012
  • Threads: 1
  • Posts: 2
November 7th, 2012 at 6:13:22 PM permalink
So i am interested in the hop bets for the 7 in craps. My question is what are the oods that two 7s will be thrown in a row on the come out roll within 30 come out rolls? When I play craps at the casino it seems like there are always alot of 7s thrown on the come out roll and I want to know if this is just short time occurrence or is it actually that 7's are more likely to come in a row due to it being the number with the highest possible combinations.
tom
s2dbaker
s2dbaker
Joined: Jun 10, 2010
  • Threads: 51
  • Posts: 3259
November 7th, 2012 at 8:32:19 PM permalink
The chances of exactly two sevens in a row is one in 36. Since the come out is still in effect if you roll an eleven, I guess that has to be factored in but I am unfamiliar with these hop bets.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
EdCollins
EdCollins
Joined: Oct 21, 2011
  • Threads: 14
  • Posts: 1036
November 7th, 2012 at 8:45:14 PM permalink
As s2dbaker mentioned, the chances of two consecutive 7s appearing are 1/36.

(1/6 * 1/6)

If I understand your question, (and obviously I THINK I do) I calculate there's a 50.9% chance that two (or more) consecutive 7s will appear on the come out roll, if given a chance to do so over a span of exactly 30 come out rolls.

And yes, two (or more) consecutive 7s are indeed more likely to appear in a row than any other number, for the very reason you mention... because it's easier to obtain.

For example, the chances that two (or more) consecutive 11s appear on the come out roll, in a span of exactly 30 come out rolls, is only about 8.3%
YouCanBetOnThat
YouCanBetOnThat
Joined: Nov 7, 2012
  • Threads: 13
  • Posts: 91
November 8th, 2012 at 11:07:59 AM permalink
There are three hop bets you can make on the 7:
1-6
2-5
3-4

If the table you're playing at pays 15 to 1 on these hop bets, you are better off making all three of these hop bets than making a single bet on Any 7. ($1.00 on each hop bet, for a total $3.00 bet, will pay $13.00. $3.00 bet on Any 7 will only pay $12.00.)

If the table pays 14 to 1 on the hop bets, it doesn't matter whether you bet the hops or the Any 7.

Ed, I'm getting 51.2% using WinCraps simulations (3 million come out rolls). Pretty close to your 50.9%, and I'll defer to actual math.

Anyway, Tom, I'm not sure that any of this can help you. Betting on the 7 with the intention of parlaying it when it hits doesn't change the house edge. And if instead your intent is to wait for a 7 to roll and then bet on the 7, the chances of it appearing on the next roll are 1 in 6, just like every other roll.
YouCanBetOnThat.com, a podcast for the recreational gambler
eagledice
eagledice
Joined: May 27, 2011
  • Threads: 0
  • Posts: 29
November 8th, 2012 at 12:49:33 PM permalink
Forget the math.


Do you want to guarantee that two sevens will show up back to back,


Place $200 in the hardways working on the comeout and I guarantee that you will roll sevens back to back......



EagleDice
tomshark30
tomshark30
Joined: Nov 7, 2012
  • Threads: 1
  • Posts: 2
November 8th, 2012 at 3:45:35 PM permalink
So you figured out my idea. I am banking on the idea that two 7s will come up in a row on the come out roll. You bet all three hop 7s on come out roll and then parlay it up hoping that another 7 hits. If it does, then the payout is much more. I would think this would deminsh the house edge because you are not treating it as a single roll but a combined roll along with 13 bets of the casinos money. Also if it is over 50% for 30 rolls then what if we push it out to 60 come out rolls? Are we at 99% chance of hitting? The profit increase is so much that it would be easy to have a betting system to show profit over 60 come out rolls.
YouCanBetOnThat
YouCanBetOnThat
Joined: Nov 7, 2012
  • Threads: 13
  • Posts: 91
November 8th, 2012 at 4:25:04 PM permalink
Those "13 bets" are not the casino's money. They're your money. When you parlay, it's your money that you've chosen to leave on the table. And just because you've chosen to increase your bet, that doesn't change the house edge.

Also, increasing the number of come out rolls in our discussion won't change anything. Yes, the chances of hitting two 7s in a row increases, but so do the number of losing wagers.

I'm not trying to discourage you from making the bet. A parlay is very exciting when it hits. However, be aware that you will not see a profit in the long run. It will eat away at your bankroll pretty significantly with its huge house edge. (This particular parlay has a house edge over 20%.)
YouCanBetOnThat.com, a podcast for the recreational gambler
Ahigh
Ahigh
Joined: May 19, 2010
  • Threads: 86
  • Posts: 5111
November 8th, 2012 at 4:51:50 PM permalink
The best parlay for a non-free bet is to start with a $25 four or $25 ten and press it all the way until the money seems ridiculous.

The combined edge doesn't EVER get anywhere near a hardway's first hit edge, and you get 9x return after only two hits.
98Clubs
98Clubs
Joined: Jun 3, 2010
  • Threads: 52
  • Posts: 1726
November 8th, 2012 at 5:08:02 PM permalink
+1 on the 4/10 buy... and I'll add twice as nice when the commission is on a winning bet only!
Some people need to reimagine their thinking.
EdCollins
EdCollins
Joined: Oct 21, 2011
  • Threads: 14
  • Posts: 1036
November 8th, 2012 at 5:56:08 PM permalink
Quote: YouCanBetOnThat


Ed, I'm getting 51.2% using WinCraps simulations (3 million come out rolls). Pretty close to your 50.9%, and I'll defer to actual math.


I'm curious... try 100 million come out rolls, instead of just 3 million, and let me know if that brings it down to something closer to 50.9%. (Please note that my 50.9 is not actual math... it's also based upon a simulation, of exactly 100 million rolls. But with that many rolls, 50.9 should be pretty darn close.)

And yes, it's irrelevant anyway... you just can't overcome the house edge with this or similar methods.

  • Jump to: