Chances Of Winning At Craps

1. Chances Of Winning At Craps
2. Chances Of Winning At Blackjack

Craps odds of winning for this wager type depends on the number the dice shows. For 2 and 12 it’s 2.77%, while the payouts are 2 to 1 and 3 to 1. For 3, 4, 9, 10 and 11 the probability is 3.88%. In this case you’ll get even money. Here is an overview of craps bets with medium odds that you can make. Have a 16.7% chance of winning an Any 7 bet, with a payout of 4/1. This gives the house a substantial edge of 16.9%. This is quite high compared to some of the bets we’ve seen above, yet still a solid choice if you’re chasing more slightly more daring bets. The allure of the craps table has been the downfall of legendary gamblers from Nick 'The Greek' Dandalos to poker pro T.J. Cloutier, but you can be a winning craps player if you follow some very important rules, like learning the odds at craps. Learn craps rules and you'll be thankful. If you want to be sure to get the best chances to win at craps, you need to pick the Pass / Don’t Pass or the Come / Don’t Come bets. Picking any of the four bets indicated above reduces the house advantage to 1.40%.

pass line win probability = 244/495
(244/495)^3 = 0.119771609 exactly 3
you want 3 or more
formula for the sum of a geometric series is a/1-r
where a is the first term
r = the ratio
a = (244/495)^3
r = 244/495
so (244/495)^3 / (1-(244/495) = 0.236202973
But now you want this to end. Multiply the above result by 1-(244/495)
0.119771609
((244/495)^3 / (1-(244/495)) * (1-(244/495))
added my table

at least in a row then lose	Prob	1 in
2	0.242979288	4.1
3	0.119771609	8.3
4	0.059038934	16.9
5	0.02910202	34.4
6	0.014345238	69.7
7	0.007071188	141.4
8	0.003485596	286.9
9	0.001718152	582.0
10	0.000846928	1,180.7
11	0.000417475	2,395.4
12	0.000205786	4,859.4
13	0.000101438	9,858.3
14	5.00017E-05	19,999.3
15	2.46473E-05	40,572.4
16	1.21494E-05	82,308.7
17	5.98878E-06	166,978.8
18	2.95205E-06	338,748.0
19	1.45515E-06	687,214.1
20	7.17286E-07	1,394,143.4

pass line win probability = 244/495
(244/495)^3 = 0.119771609 exactly 3

Mustangsally: Maybe I'm missing something, but I get different results from what you show. Suppose (for simplicity) that it was a coin flip with p=0.5 instead of 244/495. To get at least one win in a row would be P=0.5. To get at least two in a row is P=0.5^2, etc. To get the answer for exactly n in a row, you need to multiply the 'at least' by the probability of losing on the n+1 try.
For the pass line problem, I think the 0.119771609 figure is for 3 or more, not for exactly 3.
Isn't that correct, or what did I miss?

For the pass line problem, I think the 0.119771609 figure is for 3 or more, not for exactly 3.
Isn't that correct, or what did I miss?

rudeboyoi and Sally both arrived at the same value.
Sally did it differently by starting with 3 pass line wins in a row in 3 trials.
The OP asked a unique question.
Most ask the probability of winning 3 pass line bets in a row. And for 3 trials it is simply p^3
OP wanted to add the probability of 3 *or more* and *followed by a loss*.
Sally's math shows 3 in a row in 3 trials and the OPs Q arrives at the same value. It should.
IF the OP had asked what is the probability of winning 3 pass line bets in a row then losing, we would have p^3 * q or 0.06073
Let us see if OP is happy and replies.
added
average number of trials to see a run of 3 or more pass line wins: 16.466
4 or more: 33.404
5 or more: 67.765
6 or more: 137.475
Multiple streaks of pass line winners.
15 trials about 30 minutes of play at 100 rolls per hour
30 trials about 1 hour of play
Example: 30 pass line trials
about a 90% chance of at least 3 pass line wins in a row at least one time
about a 58% chance of at least 3 pass line wins in a row at least two times
about a 23% chance of at least 3 pass line wins in a row at least three times
here is the losing streaks (miss) for the pass line per N trials

added
average number of trials to see a run of 3 or more pass line wins: 16.466
4 or more: 33.404
5 or more: 67.765
6 or more: 137.475

You guys are great. Thanks for the detailed responses.
How did you calculate the average number of trials to see a run of 3, 4, 5, 6 or more pass line wins?
How do you define a trial? Would each shooter be a new trial? Or does a new trial begin after any losing pass line bet, in which case a single shooter could have multiple trials that end and start over with a losing pass line bet from throwing 2, 3, or 12 on a come out roll?
If a bettor were to power press the pass line bet with a $100 wager:

at least in a row then lose	Prob	1 in	Bet	Win	Lose
1	0.492929293	2.0	$100.00	$100.00	$100.00
2	0.242979288	4.1	$200.00	$300.00	$100.00
3	0.119771609	8.3	$400.00	$700.00	$100.00
4	0.059038934	16.9	$800.00	$1,500.00	$100.00
5	0.02910202	34.4	$1,600.00	$3,100.00	$100.00
6	0.014345238	69.7	$3,200.00	$6,300.00	$100.00

Does this mean that on average you would be betting about $1600 to win $700 on a press of 3 pass line wins for a net loss of $900?
$3,300 to win $1,500 on 4 pass line wins losing $1,800?
$6,700 to win $3,100 on 5 pass line wins losing $3,600?
$13,700 to win $6,300 on 6 pass line wins losing $7,400?
What's the best way to interpret this expected value for power pressing a pass line bet?

Thread Rating:

Probably similar to 18.... what nevermind.
So this is basically a firebet. Are you saying if the point is a 6 and there is a 3 then the bonus bet is over, even though it was not a seven out? If so then this has got to be the worst bet in the house.

Yes, it pays for hitting fewer numbers. 5 numbers would pay 150 to 1, even if the bet ended. 4 paid less, and 3 barely paid anything.
To win the main prize, 1,000 to 1, yes, you'd hit the other 5 numbers then the point.
The bet is placed before the come-out roll. I can't remember the name of the bet - something like 'six-shooter bonus' or the like.
I'm still trying to figure out what the chances are of hitting those 5 numbers and the point again all in a row.

Chances Of Winning At Craps

Even if you dodge the 2-3-7-11-12 all those times, it seems more likely than not that the point would change in the course of throwing all those other numbers. You have to hit the current point after you've got all 6 numbers marked, then, no matter how many times it might have changed during the hand?Babs, I'm pretty sure that you'd lose if you hit the point early... tantamount to hitting the same number twice.
Nobody has answered your question because you did not give us a complete set of rules and it probably requires a lot of time commitment. People get paid for this stuff. With that said, I think that this would require a simulation or a lot of math by hand. There are a limited number of win scenarios (200 maybe?) with any other sequence of six rolls (after setting the point) losing. I mean, if I understand this correctly, you have to hit all 5 other 'numbers' then the point... any other situation and the bet loses.
Hopefully someone gets interested in this. I'll keep thinking about it. This has got to be worse than the firebet, right?

Hi. Last month I went into a smaller casino and found their craps table was closed. They did, however, happen to have a large 'craps machine' that was basically 6 seats with screens for betting, and a large dome that contained 2 large dice that would roll when the platform they were on started vibrating.

This sounds like the interblock 'organic craps' I have seen and played a few times.

Chances Of Winning At Blackjack

There were 3 dice in the shaker I do remember.
so with 2 dice maybe it is another game totally...
2 dice slot is the shoot-to-win craps and that is all over Vegas but I have yet to see the side bet.
The bet is called Lucky Shooter (Great name for a XXX rated flick) from Interblock's site
This is what I found last year when it was warmer outside
'The Lucky Shooter is a side bet for Craps that will improve the Hold on the game.
The Side Bet can be made before each new “Come Out” roll.
The object
of the Lucky Shooter Side Bet is to first establish a Point,
then roll the 5 other Box Numbers (4, 5, 6, 8, 9, 10 excluding the Initial Point) in any order
during the next 5 rolls,
and then roll the “Initial Point” on the 6th roll.
Once one of the “5 other Box Numbers” is rolled it cannot be rolled again.
If a 7 or 11 is rolled on the Come Out Roll, the game ends and the bet remains working for the next Come Out roll.
If a 2, 3 or 12 is rolled on the Come Out Roll,
the game ends and the bet loses.
If a point is established on the Come Out Roll, the game ends when Any Number other than one of the remaining (not already rolled) “5 other Box Numbers” is rolled.
If all “5 other Box Numbers” are rolled, then the game ends after the 6th and final roll. The Player has a chance to win up to 1000 to 1.'

My question is, what are the odds or chances of actually winning the bonus bet like that?
Hitting all 6 required numbers within 6 rolls after the come-out.

I get (with simple multiply and adding in a 207x207 transition matrix in old Excel 2007) about 1 in 7,558.27
for what I call the Lucky6
for some to see
I B in Palm Desert (just next to Palm Springs)
for the long weekend and I plan on visiting Fantasy Springs Casino in Indio (plug) as they told me they have the bet there.
Hope it is the 1000 for 1 as I have seen the 500 for 1 paytable too.
I B very lucky this year (so far so goooooooood)
and I should hit that Lucky6 after dinner and dancing and B4 sex.
sounds like you had fun getting lucky!
SallyI get winning the 6-number bet 1 time in 7558.272