Thread Rating:
Before I get to the side-bet let me say a little about the base game. This was an install at Ballys Wild West in AC. Max odds were 2X on BOTH do and dont
Do odds
4 and 10 = 3 for 1
5 anf 9 = 5 for 2.
6 and 8 = 11 for 5
Dont odds
4 and 10 = 3 for 2.
5 and 9 = 5 for 3.
6 and 8 = 11 for 6
For those APs looking to get to Diamond with Doey/Dont the game was rated at $200 per point
Lucky Shooter
Here is the sidebet and paytables at Ballys
Per the other thread there is a higher paytable offering $150 and $1000 for the top 2 wins respectively
Wagers are $1-5 and must be made in conjunction with the pass/dont pass base bet. The sidebet is only available on the comeout
If the initial roll is any Craps (2,3,12) wager is lost
If the initial roll is 7/11 the wager remains but does not win. I.E. you can keep the sidebet wager while removing or not replacing the base bet. However you cannot raise the sidebet if you did not have the max placed without an additional base bet
You may also remove the sidebet at this time
If a point is thrown the side bet is active and locked until a resolution which must occur within the next 6 throws based on the rules
The point of the side bet is to get all the points thrown within the next 6 throws with certain orders established as follows
Succesive throws
1st successive throw:
If any point number OTHER THAN the original point is thrown the wager locks that point and the wager continues.
If a 2,3,7,11,12 or the initiating point is thrown the wager is lost
No payout is made
2nd successive throw:
If any point number OTHER THAN the original point AND the point number thrown in the first successive throw that additional point number is locked and the wager continues
If a 2,3,7,11,12 or the initiating point or the point thrown in the first successive throw is thrown then player loses wager
No payout is made
3rd successive throw:
If any point OTHER THAN the original point AND the 2 locked points from the previous successive throws are thrown that point number is locked and the wager continues
If a 2,3,7,11,12 or the initiating point OR any of the 2 successive locked points are thrown wager is ended
Payout of 2 for one is made
4th successive throw:
If any point OTHER THAN the original point AND either of the 3 locked successive points are thrown that point number is locked and the wager continues
If a 2,3,7,11,12 or the initiating point OR any of the 3 locked successive points are thrown the wager ends
Payout is made for 4 for 1
5th successive throw:
If the remaining point number not thrown appears the wager continues
If a 2,3,7,11,12 or the initiating point OR any of the 4 locked successive points are thrown the wager is lost
Payout is made for 10 for 1
6th successive throw:
If the ORIGINAL POINT NUMBER is thrown again on this throw then wager is successfully concluded and player wins jackpot of 500 for 1 (1000 for 1 on some paytables)
If a 2,3,7,11,12 OR any of the 5 locked successive points are thrown the wager is lost
Payout is made for 100 for 1 (150 for 1 on some paytables)
This seems like a horrible bet to me
Im certain the Wizard will enjoy the math breakdown on this one
there is one hereQuote: darkozLucky Shooter side bet is on E-Craps installs from Organic.
This seems like a horrible bet to me
Im certain the Wizard will enjoy the math breakdown on this one
https://wizardofvegas.com/forum/gambling/craps/28447-interblock-lucky-shooter-craps-sidebet/
let us see who agrees on the results
I also agree with the results when a push is considered as
NOT a bet resolved
event | prob | net | return |
---|---|---|---|
push | 0 | 0 | 0 |
craps | 0.142857143 | -1 | -0.142857143 |
1pt-0 | 0.384920635 | -1 | -0.384920635 |
1pt-1 | 0.265873016 | -1 | -0.265873016 |
1pt-2 | 0.139311434 | -1 | -0.139311434 |
1pt-3 | 0.052646727 | 4 | 0.21058691 |
1pt-4 | 0.012860082 | 14 | 0.180041152 |
1pt-5 | 0.001360855 | 149 | 0.202767435 |
Lucky6 | 0.000170107 | 999 | 0.169936802 |
total | 1 | edge | -0.169629929 |
consider the push as a bet resolved (as in Baccarat)
I show for the pay table -0.131934389 edge
event | prob | net | return |
---|---|---|---|
push | 0.222222222 | 0 | 0 |
craps | 0.111111111 | -1 | -0.111111111 |
1pt-0 | 0.299382716 | -1 | -0.299382716 |
1pt-1 | 0.206790123 | -1 | -0.206790123 |
1pt-2 | 0.108353338 | -1 | -0.108353338 |
1pt-3 | 0.040947455 | 4 | 0.163789819 |
1pt-4 | 0.010002286 | 14 | 0.140032007 |
1pt-5 | 0.001058443 | 149 | 0.157708005 |
Lucky6 | 0.000132305 | 999 | 0.132173068 |
total | 1 | edge | -0.131934389 |
I have played this a few times and think I did ok on it. never hit them all!
they have have a funny way of rolling, not like shoot to win
Sally
added: this pay table (to 1)
-1
0
1
3
9
99
499 (500 for 1)
returns -0.125261435
I used a Markov chain solution in Excel
event | prob | net | return |
---|---|---|---|
push | 0 | 0 | 0 |
craps | 0.142857143 | -1 | -0.142857143 |
1pt-0 | 0.384920635 | -1 | -0.384920635 |
1pt-1 | 0.265873016 | -1 | -0.265873016 |
1pt-2 | 0.139311434 | 1 | 0.139311434 |
1pt-3 | 0.052646727 | 3 | 0.157940182 |
1pt-4 | 0.012860082 | 9 | 0.115740741 |
1pt-5 | 0.001360855 | 99 | 0.134724672 |
Lucky6 | 0.000170107 | 499 | 0.084883347 |
total | 1 | edge | -0.161050417 |
push included
event | prob | net | return |
---|---|---|---|
push | 0.222222222 | 0 | 0 |
craps | 0.111111111 | -1 | -0.111111111 |
1pt-0 | 0.299382716 | -1 | -0.299382716 |
1pt-1 | 0.206790123 | -1 | -0.206790123 |
1pt-2 | 0.108353338 | 1 | 0.108353338 |
1pt-3 | 0.040947455 | 3 | 0.122842364 |
1pt-4 | 0.010002286 | 9 | 0.090020576 |
1pt-5 | 0.001058443 | 99 | 0.104785856 |
Lucky6 | 0.000132305 | 499 | 0.066020381 |
total | 1 | edge | -0.125261435 |
Quote: gordonm888I get a house edge of 7.6% (7.5992 %) on this side bet with the 500, 100,10,4,2 paytable.
That cant be right. The link mustangsally gave worked out to 17% with the higher paytable
Its a complicated bet resolution scheme. You certain you got it all?
at first the rules seems confusingQuote: darkozIts a complicated bet resolution scheme. You certain you got it all?
but the rules are very basic (especially when one plays it)
In CA I saw this machine only at Fantasy Springs Casino a couple years ago
a bit more data
the average number of rolls for any resolution (including a push 7,11)
I get 2.258968717
let us say 2.26
Sally
Event | Pays | Probability | Return |
---|---|---|---|
Five numbers | 500 | 0.000184 | 0.091879 |
Four numbers | 100 | 0.001632 | 0.163177 |
Three numbers | 10 | 0.007174 | 0.071743 |
Two numbers | 4 | 0.020838 | 0.083352 |
One number | 2 | 0.045010 | 0.090021 |
Zero numbres | 0 | 0.077160 | 0.000000 |
Come out roll 7 or 11 | 1 | 0.222222 | 0.222222 |
Come out roll 2, 3, or 12 | 0 | 0.111111 | 0.000000 |
Point and loss | 0 | 0.514668 | 0.000000 |
Total | 1.000000 | 0.722393 |
Quote: mustangsally
push included
event prob net return push 0.222222222 0 0 craps 0.111111111 -1 -0.111111111 1pt-0 0.299382716 -1 -0.299382716 1pt-1 0.206790123 -1 -0.206790123 1pt-2 0.108353338 1 0.108353338 1pt-3 0.040947455 3 0.122842364 1pt-4 0.010002286 9 0.090020576 1pt-5 0.001058443 99 0.104785856 Lucky6 0.000132305 499 0.066020381 total 1 edge -0.125261435
Maybe I misunderstand the rules, but let me challenge your probability of "1pt-0." I assume this to be when the player makes a point and achieves it the very next roll. I see no other way this event can occur. That said, wouldn't the probability be:
pr(4)^2 + pr(5)^2 + pr(6)^2 + pr(8)^2 + pr(9)^2 + pr(10)^2 = 0.077160494?
Quote: darkozThat cant be right. The link mustangsally gave worked out to 17% with the higher paytable
Its a complicated bet resolution scheme. You certain you got it all?
Sorry, I, too, was going out the door so I basically used Chuckleberry's table of outcome probabilities. I didn't pick up on the fine points of the bet resolution.
"1st successive throw:Quote: WizardMaybe I misunderstand the rules,
If any point number OTHER THAN the original point is thrown the wager locks that point and the wager continues.
If a 2,3,7,11,12 or the initiating point is thrown the wager is lost"
a 7 or a horn number causes the bet to lose
I played this bet about the time I did the math, so I guess I understood the rules better than those who may have just read them a few times
My transition matrix shows this for Point-0=
3/36*15/36+
4/36*16/36+
5/36*17/36+
5/36*17/36+
4/36*16/36+
3/36*15/36 = 388/1296
Sally
Quote: mustangsallyI played this bet about the time I did the math, so I guess I understood the rules better than those who may have just read them a few times
My transition matrix shows this for Point-0=
3/36*15/36+
4/36*16/36+
5/36*17/36+
5/36*17/36+
4/36*16/36+
3/36*15/36 = 388/1296
Maybe we are just disagreeing on how to label losing events. I would file a sequence of 4 and then 7 as a "point and loss." I don't think there is a disagreement on the rules.
Still I think I have a minor error somewhere.
Let me think.
There is an easier way to populate the transition matrix that I used in Excel. I may re-do it later.
Here is my Excel in Google for those that want to see how I did it
https://goo.gl/b8HCGJ
come to think of it, I do not even see a simulation I ran
Hmmm
Sally
btw, the bet IS addictive, imo. I tried not to make it but it kept winning (the Point2 version)
Chuckleberry Pay Table
Event | Pays | Probability | Return |
---|---|---|---|
Five hits plus point | 1000 | 0.000132 | 0.132305 |
Five hits | 150 | 0.001058 | 0.158766 |
Four hits | 15 | 0.010002 | 0.150034 |
Three hits | 5 | 0.040947 | 0.204737 |
Two hits | 0 | 0.108353 | 0.000000 |
One hit | 0 | 0.206790 | 0.000000 |
Zero hits | 0 | 0.299383 | 0.000000 |
Come out roll 7 or 11 | 1 | 0.222222 | 0.222222 |
Come out roll 2, 3, or 12 | 0 | 0.111111 | 0.000000 |
Total | 1.000000 | 0.868066 |
darkoz Pay Table
Event | Pays | Probability | Return |
---|---|---|---|
Five hits plus point | 500 | 0.000132 | 0.066153 |
Five hits | 100 | 0.001058 | 0.105844 |
Four hits | 10 | 0.010002 | 0.100023 |
Three hits | 4 | 0.040947 | 0.163790 |
Two hits | 2 | 0.108353 | 0.216707 |
One hit | 0 | 0.206790 | 0.000000 |
Zero hits | 0 | 0.299383 | 0.000000 |
Come out roll 7 or 11 | 1 | 0.222222 | 0.222222 |
Come out roll 2, 3, or 12 | 0 | 0.111111 | 0.000000 |
Total | 1.000000 | 0.874739 |
Thanks, Sally, for keeping me on my toes.
One would assume the $1000 paytable had the better return
Apparently the higher payouts are balanced by the first payout being one extra combination being made before reaching a winning resolution
I do not understand, but ok to you.Quote: WizardThanks, Sally, for keeping me on my toes.
standing flat-footed is ok too.
(not saying you are flat-footed, I doubt you are)
as combinations (nCk) are my weakest link in most cases, care to show (and share) how you did your math?
Chuckleberry looks to have done it the same way too in the first link I provided.
As I already mentioned, I was in a Markov chain mode a few years ago and did it that way
to be different
Sally
Quote: mustangsallycare to show (and share) how you did your math?
I did it in Excel. I modified my Fire Bet spreadsheet, which is also a Markov Chain. Plus, when we didn't agree, I wrote a simulation in C++, which confirmed you were right. I can share the spreadsheet, if you like, on Google Docs.
sure, that would be nice.Quote: WizardI can share the spreadsheet, if you like, on Google Docs.
thank you
I have yet to simulate this one...
since the Astros just won, I will do just that (I won too)
Sally
Quote: mustangsallysure, that would be nice.
Here is a link to the spreadsheet. Sorry it isn't documented well. https://www.dropbox.com/s/whrxq399i4fa83n/lucky%20shooter.xls?dl=0.
Here is some code for my simulation.
void LuckyShooterSimulation(void)
{
int i,end,total,total_array[13],point,nummin,wins;
unsigned int die1,die2;
time_t curtime,endtime;
__int64 count,wins_array[13],tot_no_point,tot_point_loss;
cerr << "Enter number of minutes in simulation: ";
cin >> nummin;
curtime=time(NULL);
endtime=curtime+(60*nummin);
for (i=0; i<=12; i++)
wins_array=0;
count=0;
tot_no_point=0;
tot_point_loss=0;
__int64 tot_jackpot=0;
do
{
count++;
end=0;
wins=0;
for (i=4; i<=10; i++)
total_array=0;
die1=genrand_int32()%6+1;
die2=genrand_int32()%6+1;
point=die1+die2;
if ((point<=3)||(point==7)||(point>=11)) // bet over via repeat, 2, 3, 7, 11, or 12
{
point=0;
end=1;
}
while (end==0)
{
die1=genrand_int32()%6+1;
die2=genrand_int32()%6+1;
total=die1+die2;
if ((total_array[total]==1)||(total<=3)||(total==7)||(total>=11)||(total==point)) // bet over via repeat, point, 2, 3, 7, 11, or 12
end=1;
else
{
wins++;
total_array[total]=1;
}
};
if (point==0)
tot_no_point++;
else if ((wins==5)&&(total==point))
tot_jackpot++;
else
wins_array[wins]++;
if (count%5000000==0)
{
curtime=time(NULL);
cerr << "Minutes remaining: " << (double)(endtime-curtime)/60.0 << "\n";
}
}
while (curtime<endtime);
printf("No point\t%I64i\t%f\n",tot_no_point,(double)tot_no_point/(double)count);
printf("total jackpots\t%I64i\t%f\n",tot_jackpot,(double)tot_jackpot/(double)count);
for (i=0; i<=5; i++)
printf("%i wins\t%I64i\t%f\n",i,wins_array,(double)wins_array/(double)count);
}
I see this pay table is not already listed on my site. Here is my analysis of it:
Event | Pays | Probability | Return |
---|---|---|---|
Five hits plus point | 1000 | 0.000132 | 0.132305 |
Five hits | 150 | 0.001058 | 0.158766 |
Four hits | 20 | 0.010002 | 0.200046 |
Three hits | 5 | 0.040947 | 0.204737 |
Two hits | 0 | 0.108353 | 0.000000 |
One hit | 0 | 0.206790 | 0.000000 |
Zero hits | 0 | 0.299383 | 0.000000 |
Come out roll 7 or 11 | 1 | 0.222222 | 0.222222 |
Come out roll 2, 3, or 12 | 0 | 0.111111 | 0.000000 |
Total | 1.000000 | 0.918077 |
For a multi-roll side bet, an 8.2% house edge ain't bad.
I have heard from a few that have played it, hard to win from it.Quote: WizardFor a multi-roll side bet, an 8.2% house edge ain't bad.
Maybe their data proves it and try to make players win more from playing it.
sucker bets taste sweet until they don't
and then you look for another sucker
ask Telly
Sally
Quote: ChuckleberryAh, yes. The 7 or 11 tie! Good catch. I'm glad my original mistake pushed my calculation a little worse, rather than a little better. Nothing's worse than playing a bad game when you think it's good.
Welcome back! Sorry I never took notice of your original post on this bet.