dice probability question

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

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.

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%

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.

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

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.

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%.)

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.

+1 on the 4/10 buy... and I'll add twice as nice when the commission is on a winning bet only!

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.

Alright, I tried to show this with as little verbage as possible to make it simple.

"do,1p,0x" is $5 passline with no odds ($0.02 per roll expected loss)
The other lines show the grind on placing the four at $5, $10, $20, and $25.

At $5, you can see the line of the edge for the $5 which costs a buck for every time you get paid. I modelled $10 pays you $19 instead of $18 (automatic buy).

But what you're visualizing here is that you can get some volatility for that buck at $25 that can compete with the edge.



Now this chart is the exact same data with pressure. (schedules shown but note all pressure stops at $500 max bet as indicated in the charts by "l500")

IE: "buy4,s25,4,l500" means buy the four starting at $25, press schedule 4 and don't press beyond $500.

@buy_press_schedule6 = ( 5, 15, 25, 75, 225, 675, 2025 );
@buy_press_schedule5 = ( 5, 15, 30, 80, 260, 750, 2000 );
@buy_press_schedule4 = ( 5, 10, 25, 60, 160, 460, 1350 );
@buy_press_schedule3 = ( 5, 10, 20, 55, 160, 420, 1200 );
@buy_press_schedule2 = ( 5, 10, 25, 50, 75, 100, 150, 250, 500 );
@buy_press_schedule1 = ( 5, 10, 15, 20, 25, 40, 80, 100, 150, 250, 300, 350, 400, 450, 500 );



When you press, the combined parlayed edge stays low when you press it hard (unlike hardways).

This gives you a betting strategy that can survive longer by keeping your edge low.

Alright, hopefully that makes sense!

Quote: EdCollins

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.)

I ran 100 million, and I got around 51.16%.

Our figures are so close to each other; I'm trying to think of odd circumstances where I might be getting a false positive or where you might be getting a false negative.

Again, we're looking at groups of 30 come out rolls and looking for 7s back-to-back at least once.

We might have to post something over in the Math forum and have someone figure it out for us. (I thought about it for a while and couldn't come up with a formula.)

Quote: Ahigh

When you press, the combined parlayed edge stays low when you press it hard (unlike hardways).

This gives you a betting strategy that can survive longer by keeping your edge low.

Thanks for the info, Ahigh. Pressing (or parlaying) buys on the 4 and 10 are some of my favorite bets.

I would caution that there is a pretty significant difference in house edge between "vig only on wins" versus "vig on all bets" when you parlay these buys. For example, take the case where you start with $25 on the 4 with the intent of parlaying to $75. "Vig only on wins" has a house edge of 2.22% on that parlay. "Vig on all bets" has a house edge of 8.97%. Ouch.

Quote: YouCanBetOnThat

Thanks for the info, Ahigh. Pressing (or parlaying) buys on the 4 and 10 are some of my favorite bets.

I would caution that there is a pretty significant difference in house edge between "vig only on wins" versus "vig on all bets" when you parlay these buys. For example, take the case where you start with $25 on the 4 with the intent of parlaying to $75. "Vig only on wins" has a house edge of 2.22% on that parlay. "Vig on all bets" has a house edge of 8.97%. Ouch.

This is a really good point. I only am contemplating vig on the win. I'm also assuming that they round down from $1.25 vig to $1.00 vig for a $25 buy. You need to make sure the place where you're playing allows this. But this is the Wizard of Vegas forum. The only place I ever went here that wouldn't give me a $1.00 vig on the win for a $25 buy on the four was the Riviera. They claimed it was vig up front because they were offering 1000x odds at the time. But generally, you will find $1.00 vig on the win for $25 buy's all across Las Vegas. Your mileage may vary elsewhere.

Money needed	Total money required	Money on felt	Description	Compounded Edge
$25	$25	$25	$25 buy on the 4/10	N/A
$1	$26	$75	First hit throw in a buck say "go to $75."	1/75 = 1.33%
$3	$29	$225	Second hit throw in three bucks say "go to $225."	4/225 = 1.78%
$11	$40	$675	Third hit throw in eleven bucks say "go to $675."	15/675 = 2.22%
$33	$73	$2025	Fourth hit throw in thirty three bucks say "go to $2025."	48/2025 = 2.37%
$101	$174	$6075	Fifth hit throw in $101 and say "go to 6075."	149/6075 = 2.45%

This outlines the BEST case scenario (all partial dollars are rounded down). If they round up for vig on the $75, or later bets, the edge is a little higher. But the best case scenario shows the compounded edge stays relatively contained.

Quote: YouCanBetOnThat

I ran 100 million, and I got around 51.16%.

Our figures are so close to each other; I'm trying to think of odd circumstances where I might be getting a false positive or where you might be getting a false negative.

Again, we're looking at groups of 30 come out rolls and looking for 7s back-to-back at least once.

We might have to post something over in the Math forum and have someone figure it out for us. (I thought about it for a while and couldn't come up with a formula.)

30 = 51.1780330210008
40 = 61.8314558175545
50 = 70.1601992882844

Ah.

I figured out my coding error.

After another 100 million run, I now get 51.178315 for 30 rolls. Thanks.

Quote: tomshark30

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?

Numbers look good.
Try this streak calculator.
No simulation required.
Even shows the expected number of attempts needed.

http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html

Seems very good and has links to WoV thread to do the math yourself

Ultimately you have to be at the right place at the right time to hit 2 consecutive 7s... regardless if it is a come out roll or not... the probabilities are always the same . However, I wouldnt suggest playing the any 7 bet in the middle of a roll. Someone my hit you in the head with the dice! .lol

Quote: EdCollins

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.

The streak calculator at http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html

gives .0816 for the 11's and .5119 for the 7's.

As someone pointed out, the more comeouts, the more times you are likely to lose before this happens.

Edge * action, no way around it. Get lucky!
Cheers,
Alan Shank