One outcome eliminated

I'm guessing it would be a player advantage, but someone with better math skills can probably figure it out pretty essy...

What if in craps, every time the dice were rolled, you knew that the number on each die that just hit, would not repeat on the next roll? For example, the dice end up 6 4 on the comeout roll. On the next roll, the first die could be anything, except a 6. The second die could be anything except a 4. So, I bet no 10 on the next roll, but then change it to no whateverr the next roll is after that.


ZCore13

5,5 Hardway

Quote: Zcore13

I'm guessing it would be a player advantage, but someone with better math skills can probably figure it out pretty essy...

What if in craps, every time the dice were rolled, you knew that the number on each die that just hit, would not repeat on the next roll? For example, the dice end up 6 4 on the comeout roll. On the next roll, the first die could be anything, except a 6. The second die could be anything except a 4. So, I bet no 10 on the next roll, but then change it to no whateverr the next roll is after that.


ZCore13
link to original post

I think you should say you know the idea is a dead end, I know you know that.

the chances of rolling a 10 would be the ways 5+5 and 4+6, while 6+4 can't repeat due to magical influence. 2 ways. The ways to win would be 6 ways to roll a seven... 8 total ways. Fair odds should pay 2:6 while the unsuspecting house will pay 3:6 [1:2] minus a commission I guess [I never bet this].

let's say you bet $6 on 'no 10', you'd win 6 times out of 8 and lose 2 times out of 8 ...

[6/8*$3]-[2/8*$6]............ or [3/4*3]-[1/4*6] = 75 cents , 12.5% player advantage , but minus commission

Quote: odiousgambit

I think you should say you know the idea is a dead end, I know you know that.

Thanks for your input. It's not a dead end if it's an electronic version and the big dice aren't flipping very strong.


ZCore13

Quote: odiousgambit

Quote: Zcore13

I'm guessing it would be a player advantage, but someone with better math skills can probably figure it out pretty essy...

What if in craps, every time the dice were rolled, you knew that the number on each die that just hit, would not repeat on the next roll? For example, the dice end up 6 4 on the comeout roll. On the next roll, the first die could be anything, except a 6. The second die could be anything except a 4. So, I bet no 10 on the next roll, but then change it to no whateverr the next roll is after that.


ZCore13
link to original post

I think you should say you know the idea is a dead end, I know you know that.

the chances of rolling a 10 would be the ways 5+5 and 4+6, while 6+4 can't repeat due to magical influence. 2 ways. The ways to win would be 6 ways to roll a seven... 8 total ways. Fair odds should pay 2:6 while the unsuspecting house will pay 3:6 [1:2] minus a commission I guess [I never bet this].

let's say you bet $6 on 'no 10', you'd win 6 times out of 8 and lose 2 times out of 8 ...

[6/8*$3]-[2/8*$6]............ or [3/4*3]-[1/4*6] = 75 cents , 12.5% player advantage , but minus commission
link to original post

This isn’t right. There are no longer six ways to make a 7. There are 4. One die won’t show a 6 next roll eliminating a 6,1 combo. One die won’t show a 4, eliminating a 3,4 combo.

I'll let someone else factor that in!

Quote: unJon

Quote: odiousgambit

Quote: Zcore13

I'm guessing it would be a player advantage, but someone with better math skills can probably figure it out pretty essy...

What if in craps, every time the dice were rolled, you knew that the number on each die that just hit, would not repeat on the next roll? For example, the dice end up 6 4 on the comeout roll. On the next roll, the first die could be anything, except a 6. The second die could be anything except a 4. So, I bet no 10 on the next roll, but then change it to no whateverr the next roll is after that.


ZCore13
link to original post

I think you should say you know the idea is a dead end, I know you know that.

the chances of rolling a 10 would be the ways 5+5 and 4+6, while 6+4 can't repeat due to magical influence. 2 ways. The ways to win would be 6 ways to roll a seven... 8 total ways. Fair odds should pay 2:6 while the unsuspecting house will pay 3:6 [1:2] minus a commission I guess [I never bet this].

let's say you bet $6 on 'no 10', you'd win 6 times out of 8 and lose 2 times out of 8 ...

[6/8*$3]-[2/8*$6]............ or [3/4*3]-[1/4*6] = 75 cents , 12.5% player advantage , but minus commission
link to original post

This isn’t right. There are no longer six ways to make a 7. There are 4. One die won’t show a 6 next roll eliminating a 6,1 combo. One die won’t show a 4, eliminating a 3,4 combo.
link to original post

For me, it's easier just to create new dice and go from there, so here you go:

Die 1 (Blue): 1, 2, 3, 4, 5

Die 2 (Red): 1, 2, 3, 5, 6

Possible combinations:

BR
1 1
1 2
1 3
1 5
1 6
2 1
2 2
2 3
2 5
2 6
3 1
3 2
3 3
3 5
3 6
4 1
4 2
4 3
4 5
4 6
5 1
5 2
5 3
5 5
5 6

All 25 possible combinations are covered. We know that there are 25 combinations because combinations for any two dice are simply (# of Faces) * (# of Faces), for comparison, if you had a game that involved rolling a six-sided die and flipping a coin, then there would be 12 possible combinations as follows:

1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T

With that, we will now look at the probabilities for these combinations:

Snake Eyes: 1/25
Three: 2/25
Hard Four: 1/25
Soft Four: 2/25
Five: 3/25
Hard Six: 1/25
Soft Six: 3/25
Seven: 4/25
Hard Eight: Impossible 0/25
Soft Eight: 3/25
Nine: 2/25
Hard Ten: 1/25
Soft Ten: 1/25
Yo: 1/25
Midnight: Impossible 0/25

Sum: 25

Okay, really, you can do any bet you want just by multiplying the new probability (found above) by the payouts.

Snake Eyes: 1/25*****4% compared to 2.778% (!)*
Three: 2/25*****8% compared to 5.556% (!)
Hard Four: 1/25*****4% compared to 2.778% (!)
Soft Four: 2/25*****8% compared to 5.556% (!)
Five: 3/25*****12% compared to 11.1111% (!)
Hard Six: 1/25*****4% compared to 2.778% (!)
Soft Six: 3/25*****12% compared to 11.1111% (!)
Seven: 4/25***** 16% compared to 16.667%
Hard Eight: Impossible 0/25*****0% compared to 2.778%
Soft Eight: 3/25*****12% compared to 11.1111% (!)
Nine: 2/25:*****8% compared to 8.333%
Hard Ten: 1/25*****4% compared to 2.778%
Soft Ten: 1/25*****4% compared to 5.556%
Yo: 1/25*****4% compared to 5.556%
Midnight: Impossible 0/25*****0% compared to 2.778%

*(!) denotes more likely than before.

With that out of the way, before we do our Place Bets on a one roll basis, we are going to go ahead and look at our single roll combination bets to determine whether they are more likely (warrants further examination) or less likely.

Crap Check: Okay, so our any Any Craps bet would normally be 11.1111% to occur, but it is now 12% to occur despite the fact that Midnight is impossible.

Field: Um...are we tripling the two as opposed to 12? Are we in Reno? Either way, Field is 44% to win compared to 56% to lose. This would normally be 44.444% to win compared to 55.556% to lose, so the Field has become a little worse on the face of it.

On the other hand, if Snake Eyes pays triple on the Field, then the Field has become better by virtue of eliminating a double and keeping a triple with the probabilities otherwise being the same. Single ways to make the Field have also been eliminated, so it's a good trade off for us.

Oh, screw it, let's do the Crap Check and Field first:

Crap Check

What sounds good? 7:1? I think so. All bets are $5:

(7 * 5 * .12) - (5 * .88) = -0.2

Okay, so this bet loses twenty cents every time and is, therefore, bad; it's just not as bad as it would normally be.

Field

Triple Snake Eyes:

(5 * 3 * .04) + (5 * .4) - (5 * .56) = -0.2

Once again, this loses twenty cents every time and is, therefore, bad. It's actually worse than usual.

However, suppose the previous roll had been 3-3, we are still left with 25 combinations that are possible, though all combinations that involve a three are now gone. In this scenario, we lose three possible Field rolls, but, perhaps crucially, we keep our doubler and our triple.

Also, the previous scenario only left us with 11 possible Fields, but now we have twelve.

( 5 * 3 * .04) + (5 * 2 * .04) + (5 * .4) - (5 * .52) = $0.40

Under this scenario, we are winning $0.40 on every $5.00 bet and are, therefore, at an 8% advantage.

With that, the questions are simply:

1.) How many Fields become disqualified?

AND:

2.) Which Fields become disqualified?

In the above, even if 2 and 12 only doubled, you would still be winning $0.20 on every $5 bet and would be, therefore, at a 4% advantage.

Hard Hops

That bring us to Hard Hops, of which we have four that are possible and all four of which that are equally likely.

All four are now 4% to happen and 96% not to happen. Let's say they pay 30:1:

(30 * 5 * .04) - (5 * .96) = $1.20

In other words, we are expected to win $1.20 on every $5.00 bet and are now at a 24% advantage.

Hard Ways

For Hard Ways, we will DEFINITELY want to potentially pull these back after one roll; the reason why is because, suppose we bet 2-2 under the current conditions, if the next result is 2/5, then our Hardway becomes impossible on the roll after that one.

Since we are talking about the Hard Four, I'm just going to go ahead and do that one.

Hard Four: 4% Easy Four (Loses) 8% Seven (Loses) 16% Nothing Happens (72%)

Okay, so we are getting 7:1 if this hits:

(7 * 5 * .04) - (5 * .24) = $0.20

In other words, we expect to make $0.20 for every $5 bet and are at a 4% advantage accordingly.

Hard Six pays better and we have actually made one Easy Way six impossible in the current scenario, which we did not do to a Easy Four, so let's see what happens:

Hard Six: 4% Easy Six (Loses) 12% Seven (Loses) 16% Nothing Happens (68%)

(9 * 5 * .04) - (5 * .28) = $0.40 for every $5 bet and reflects an 8% advantage.

As we established, a Hard Eight is not possible. Hard Ten has eliminated an Easy Ten, which Hard Four had not, so let's look at it:

Hard Ten: 4% Easy Ten (Loses) 4% Seven (Loses) 16% Nothing Happens (76%)

(7 * 5 * .04) - (5 * .2) = $0.40 which reflects an 8% advantage for one roll on these.

Hard Ways v. Hard Hops

Given the choice, you would want Hard Hops as the advantage is through the roof. I guess if this were an actual certainty you might load the layout with everything possible, but assuming an overall total bet limit with a cap that could, in theory, be reached, you would prefer Hard Hops to Hard Ways easily.

Soft Hops

This depends on the soft hop in question.

Any Soft Hop that is more likely than it was before is now automatically 8% to happen as compared to 5.556% to happen. My thinking is that any Soft Hop that is now more likely immediately becomes good, so let's see how good:

(15 * 5 * .08) - (5 * .92) = $1.40, which reflects an advantage of 28% every time.

An example of a less likely Soft Hop is 10, because B6R4 cannot happen, only B4R6 can. Or, that might be the other way around. I'm sure I mixed the colors up at some point in this description. It doesn't matter. Only one soft ten can happen, the invert of whatever the first one was.

(15 * 5 * .04) - (5 * .96) = -$1.80 or a 36% disadvantage. This is obviously much worse than normal.

Place Bets

Place Bets are effectively going to become one roll propositions because they can become bad bets based on what the dice do on the next roll, so you might end up picking some of them up.

I am not going to bother with calculating all Place Bets because they are going to potentially change with every roll following this one depending on what ways, if any, are made impossible and what ways are made more likely. I will, however, do a best case scenario. If the eliminated dice are a B1R2, then every way to win a Place Ten is possible and the probability of doing so on the next roll goes from 8.333% to 12%. We only care about tens and sevens. Sevens will always be 4/25 (16%) because there is no way to eliminate two faces without eliminating two sevens.

(.12 * 9) - (5 * .16) = $0.28

Okay, this is the best case scenario for a Place Ten bet because the number of sevens eliminated will always be two and every possible ten still exists. This represents a 5.6% advantage and is nothing compared to our advantage of just hopping all possible tens, so you would just want to hop all possible tens and do this only if you did not hit the bet cap yet and there was nothing else (such as other Hops) that are better.

If you want to calculate the other Place Bets, on a one roll basis, just do the correct variation of the above for whatever dice you want to have eliminated.

Come/Don't Come---:Spoiler Not Doing It:

Come bets can be calculated for the next roll as far as the initial bet goes. You would win on the very next roll 20% of the time (7 or 11) as compared to the normal 22.222%, but I'm not going to calculate it beyond that because I don't know if B-6 is always illegal and R-4 is always illegal. In any event, since you now only lose on Snake Eyes and Three, there is a 12% probability of instantly losing, resulting in a +8% probability of winning. Normally, the probability of losing would be 11.11%, for a difference of +11.111% in favor of winning the very next roll, so as far as initial roll Come Bet goes, it is worse.

I assume that Blue die and Red die make illegal whatever happens next, and, yeah, not doing it. The initial Come Bet is worse; good enough for me.

The initial Don't Come bet would normally be 1/36 push, 2/3 is win (3/36) and 7/11 loses (8/36). The difference is 5/36 which means it is 13.8888% more likely to lose as win.

With our new Don't Come bet, pushes are gone, but we are 3/25 to win instantly, which is 12%, though we remain 5/25, or 20% to lose instantly. The difference between the two is 8% more likely to lose instantly as win.

Once again, I don't know the impact going forward of whatever Blue is being illegal and whatever Red is being illegal on the roll after this one. It can be figured out because we know Blue-6 and Red-4 cannot happen, and therefore, will not be illegal the roll after this one, but I'm not even getting into it.

If all else remained equal for the Don't Come bet, and I don't immediately recognize why it wouldn't (though, maybe it does) then we just say that we immediately lose 5.888% less frequently.

Of course, the fact that seven will continue to be less likely that normal is bad for us after we have survived the Come Out roll. Normally, a seven would be 16.667% to appear and now it is 4/25 = 16%. It will always be 16% because every roll disqualifies two possible sevens from the roll after that. Blue 3 Red 3, for example, would still disqualify Blue 3 Red 4 and Blue 4 Red 3 from being possible the roll after this one.

I don't know whether or not it ends up being a wash and I am not going to figure it out. Everything relies upon whatever the next roll after the Don't Come bet is made is, though we do know, if it is a point number, it immediately becomes less likely the following roll as it has immediately eliminated one of the ways to roll itself.

Conclusion

I'm going to say that's all I am doing absent very specific questions that only require a single step/decision. For those, the above should contain enough information that anyone should be able to figure out how to calculate it anyway.

The short answer is that the correct one roll Hop Bets are always going to be the way to go. For example, the following ways to make six are typically possible:

5/1, 4/2, 3/3, 2/4, 1/5

But, we have eliminated 2/4, (or was it 4/2?) whatever the case, one of those is gone, so you would not want to Hop that because:

(15 * 5 * .04) - (.96 * 5) = -1.8

So, that's terrible. You do want to Hop 1/5 because:

(15 * 5 * .08) - (.92 * 5) = $1.40

Is awesome.

I conclude that the best possible bets in this scenario are whatever Hard Hops or Soft Hops bets become more likely to win compared to normal. The worst bets to make are results that have become literally impossible.

ADDED NOTE: All Hard Hops that remain possible are always good and the math is always the same because there will always be one way to roll those and 24 ways to roll anything that is not that. See above for the math on that.

ANOTHER ADDED NOTE:

Okay, the last thing that I am going to do is make every Hop bet that has become better. We are only going to look at the case of one Soft Hop Winning and one Hard Hop winning, so let's do that:

Snake Eyes: 4% ($5)
Threes: 8% ($5)
Hard Four: 4% ($5)
Soft Four: 8% ($5)
Soft Five 3/2: 8% ($5)
Soft Five 1/4: There is only one way to do this, so you would not bet it.
Hard Six: 4% ($5)
Soft 6 1/5: 8% ($5)
Soft 6 2/4: There is only one way to do this, so you would not bet it.
Seven 2/5: 8% ($5)
Hard Eight: Impossible
Soft Eight 3/5: 8% ($5)
Soft Eight 6/2: There is only one way to do this, so you would not bet it.
Nines: There is only one way to roll each Hop Nine, so you would not bet these.
Hard Ten: 4% ($5)
Soft Ten: There is only one way to do this, so you would not bet it.
Eleven: There is only one way to do this, so you would not bet it.
Midnight: Impossible.

Okay, so we would be making $50 in bets in this scenario. 36% of the time, everything just loses.

Hard Ways to Win: 1/1, 2/2, 3/3, 5/5 (4)---4/25

Soft Ways to Win (Being Bet): 1/2, 2/1, 3/1, 1/3, 2/3, 3/2, 1/5, 5/1, 2/5, 5/2, 3/5, 5/3 (12)---12/25

Soft Ways to Lose (NOT Being Bet): 1/4, 4/2, 4/3, 6/1, 6/2, 4/5, 3/6, 4/6, 6/5 (9)---9/25

Okay, so there are three net results possible:

ALL LOSS: There are nine ways to just lose $50.

Win Hard: If we win on one of the four Hard Hops, then we are paid $150 and have lost $45 on the other bets for total gains of $105.

Win Soft: If we win on one of the twelve Soft Hops (six bets), then we are paid $75 and have lost $45 on the other bets for total gains of $30.

With that:

(12/25 * 30) + (4/25 * 105) - (9/25 * 50) = $13.20

Therefore, given the present situation, we will expect to gain $13.20 for every $50 bet, thereby reflecting an advantage of 26.4% on every single roll under the present circumstances.

Mechanically, this would also be the easiest thing to do as it would be fast. Eventually, someone would immediately recognize what ways they should bet and what ways they should not, so this is going to be much easier than keeping track of multi-roll bets and/or whether the Field is good this roll or not.

Another thing that would make it quicker is that Hard Hops are really easy. If you didn't see the face on the previous roll, then bet that Hard Hop. If you didn't see either of the faces on the previous roll, then bet those Soft Hops. Anything involving a face you saw on the previous roll you do not want.

Awesome work!

Quote: unJon

Awesome work!
link to original post

Thank you very much! It'll probably either be used in an, "Ask Mission," LCB Editorial or I might write something of a, "What If..." article that looks at other things.

Great idea.

To answer OP’s question (which if Mission directly answered, I missed it) if last roll was a 6 and a 4, the odds on buying or laying a 10 are unchanged. So neither is a good bet.

Normally it’s 3 ways to make a 10 and 6 ways to make a 7. So 2:1 fair odds.

After the roll it’s 2 ways to make a 10 and 4 ways to make a 7. So still 2:1 fair odds.

ETA: Also, Dude, “2/1” is not the preferred nomenclature.

Quote: unJon

Great idea.

To answer OP’s question (which if Mission directly answered, I missed it) if last roll was a 6 and a 4, the odds on buying or laying a 10 are unchanged. So neither is a good bet.

Normally it’s 3 ways to make a 10 and 6 ways to make a 7. So 2/1 fair odds.

After the roll it’s 2 ways to make a 10 and 4 ways to make a 7. So still 2/1 fair odds.
link to original post

That was the question? Oops. These things happen with me. I start with a specific question and my mind wanders off to, "But, what would be the best bet?"

And...then I fail to ever answer the original question.

ha ha.

It never would have occurred to me to start checking out every single conceivable bet, certainly not hop bets. Genius.

I also see that a 2:1 payoff [or 1:2] on the 10 is now unchanged as a fair bet [if no commission]... as per Unjon. But I didn't want to show it and get embarrassed again!! Actually being a good exercise for me is why I did it, but sure enough wrong again!

I'm going to look at a old worn bubble craps machine a little differently now.

Quote: unJon

[snip] So 2/1 fair odds.
link to original post

Please use the convention 2:1 instead of 2/1 for "2 to 1" as I read the latter as "2 divided by one" ... yeah, I know, I'm being anal. Sorry

Quote: odiousgambit

Quote: unJon

[snip] So 2/1 fair odds.
link to original post

Please use the convention 2:1 instead of 2/1 for "2 to 1" as I read the latter as "2 divided by one" ... yeah, I know, I'm being anal. Sorry
link to original post

Have edited my post.

It's been a while since I played, but are all the hop bets available on bubble craps? Also, do you have enough time to place all of the 'good' hop bets between rolls? Oh, and I'd imagine that you will want to take your winning hop bet down after each roll!

As I recall, there is some button that says, "Hop bets open," or some such, so I'd guess you would click on that. I also know that Snake Eyes, Three, Yo and Midnight are on the main layout of these things. Those bets are going to be really easy to make, so just remember what faces you want to have not seen.

There should be a button that removes all bets that can be removed, so that's going to be your first thing to do every roll, if it's even applicable. Aren't Hop Bets one roll and down even if they win? The machine is at least going to remove all of your Hop Bets except for the one that won (I would think), which it also might remove.

As far as speed, I wouldn't know, though I imagine that you would get faster with time. I don't know what machine this is precisely:

https://www.youtube.com/watch?v=KbKWaU2VDKQ

But, if you jump to about 4:20 you will see the Hops open on it, exception to main screen ones, I think.

Out the door, but here's a quick---

SHOWER THOUGHT: Because 4 * 7 = 28, an Any Craps Bet can also be very good, but only if the previous dice faces were 4/3 or 3/4.

If such a machine existed where a number could not repeat itself, I would be able to retire after a day or two. I think my favorite would be to bet on "8" after rolling snake eyes. I think around 17% edge if being paid 7 to 6. The only difficulty would be trying to figure out how hard to hit it before you get 86'ed.

Quote: SOOPOO

If such a machine existed where a number could not repeat itself, I would be able to retire after a day or two. I think my favorite would be to bet on "8" after rolling snake eyes. I think around 17% edge if being paid 7 to 6. The only difficulty would be trying to figure out how hard to hit it before you get 86'ed.
link to original post

yes, but it would only have to "tend to" not repeat itself.

I think the machine is the thing that would get 86'd