Help to work out the Odds

Could someone please post a guide how to calculate the odds of an outcome.

I understand small calculation like the chance of hitting craps seven 6/36 .... but how do you calculate the following example


1. chance of rolling 7 after 10 rolls ( i am guessing the more roll the higher the chance of hitting)
2. guessing a single number from sic bo ( 3 dice) if that number hasn't come up after 10 rolls
3. chance of black/red coming up 10 times and the 11th is also black

Sic Bo
91 way to guess correctly that 1 will come upon any of the 3 dice
125 ways to lose - no 1 will come up
216 way total

so if I play 3 times, what is the chance that a 1 will come up once or more in a single roll ( 1,2,3, or 1,1,2 or 1,1,1)

is it ? 91/216 + 91/216 + 91/216 + 91/216 = 1.263888889

Is that correct?

So if you use a martingale strategy ...
if I waited 10 times for 1 not to come up ... and play the next 5 games ( 15 games ) so the first bet of game 11 .... would be 4.634259259 chance of hitting of not hitting 1 on any of 3 dice


this example could also be added to roulette - like out of 1000 spin ... what the chance of it been black more than 10 times in a row

I guess if there 216 way , and 91 way to win

that means after 10 games there 2160 total combination , and 910 way to win so it still 42% chance on the 11 game

Can someone confirm this


What the correct way to work out the odds that it repeats or change

Chance of win/total ^how many games = chance


91/216^10 = 0.0001761474708 or 1 out of 5677. that the next game will have at least 1 dice with 1#


or

Chance of Loss/total ^how many games = chance
125/216^10 = 0.004212720233 or 1 out 237., that the next game will have at least 1 dice with 1#

As a craps player, I just do too much tsk-tsking looking at the house edge of sick,bro to want to delve into it. Though it looks like Australia is the place to play it if you must. But as far as I can tell the past does not matter in this game, so "the more roll the higher the chance of hitting" theory is shot to hell.

See if this helps

https://wizardofodds.com/games/sic-bo/appendix/1/

not really look for theory and ideas, just want to learn how to do the math correctly,

I want to correct user that say .... "i just wait for 10 black in a row and then bet" , but I want correct them at let them know the true odds ( but i don't know how to do the maths )

Quote: Billyboy402

not really look for theory and ideas, just want to learn how to do the math correctly,

I want to correct user that say .... "i just wait for 10 black in a row and then bet" , but I want correct them at let them know the true odds ( but i don't know how to do the maths )

Well, the Gamblers Fallacy might be a good place to start... that because something has happened x times in a row y is "due" to hit when they are independent trials, such as your dice example. Ever single time the dice are thrown it's an independent trial. The dice have no memory. Similar to each spin on a roulette wheel. Just because X black in a row doesn't mean red is any more likely. Similarly to craps X points in a row don't mean a 7-out is any more likely.

Lot of good craps probability/maths: https://wizardofodds.com/ask-the-wizard/craps/probability/

Quote: Billyboy402

<snip>
this example could also be added to roulette - like out of 1000 spin ... what the chance of it been black more than 10 times in a row

need ham and eggs for more math
you want to do lots of waiting around!
good 4u

ok
11 in a row black in 1000 spins (you said more than 10)
00Roul

takes a calculator or spreadsheet
try Wolfram Alpha
(forgot-WA does exact so will time out)
back to excel
at least 1 run of length 11 or more
0.1312684938716
about 1 in 7.62

or here
http://maxgriffin.net/CalcStreaks.shtml

hope this helps for now
Happy New Year!

Sally

Quote: Billyboy402

Could someone please post a guide how to calculate the odds of an outcome.

I understand small calculation like the chance of hitting craps seven 6/36 .... but how do you calculate the following example

3. chance of black/red coming up 10 times and the 11th is also black

On a double-zero wheel, the chance that the next 11 spins will be 10 of the same color (besides green) followed by black is about 1 / 3712.

However, the chance that they are 10 of the same color followed by red are the same, and the chance of 10 of the same color followed by green is 1/18 of the black (and red) probability, just as they would be if any of the other color permutations appeared in those first 10 spins.

In other words, the results of the previous 10 spins have no effect on what the next spin will be. The same applies to Sic Bo (and craps, for that matter).
Besides - even if they did, and the previous 10 spins were red, what if the 20 before that were black? Wouldn't that affect the next spin?

Quote: ThatDonGuy

On a double-zero wheel, the chance that the next 11 spins will be 10 of the same color (besides green) followed by black is about 1 / 3712.

adding an example for the OP

As was stated, the odds of the NEXT 11 spins being 10 red followed by 1 black are 1 / 3712

However, AFTER 10 reds have been rolled, the odds of the NEXT 1 spin being black are 18 / 38

Further more, AFTER ANY 10 rolls, the odds of the NEXT 1 spin being black are 18 / 38

So, as ThatDonGuy stated:

Quote: ThatDonGuy

the results of the previous 10 spins have no effect on what the next spin will be.

I just wanted to highlight that.

Quote: Billyboy402

.... but how do you calculate the following example

<snip>
Sic Bo
91 way to guess correctly that 1 will come upon any of the 3 dice
125 ways to lose - no 1 will come up
216 way total

so if I play 3 times, what is the chance that a 1 will come up once or more in a single roll ( 1,2,3, or 1,1,2 or 1,1,1)

is it ? 91/216 + 91/216 + 91/216 + 91/216 = 1.263888889

Is that correct?

No.
a good try

but your 91/216 is correct.
5/6 is the probability of not getting a 1 on one die
so 5*5*5 / 6*6*6 = 125/216 as you posted

now,
one must multiply the mutually exclusive probabilities to get an answer.
read more here if interested:
https://www.mathsisfun.com/data/probability-events-mutually-exclusive.html
*****
and the probability of an event is between 0 and 1

to win "at least 1 time" in 3 tries

it is way easier to figure out the ways NOT to win in 3 tries.
125/216 * 125/216 *125/216 = 1953125 / 10077696 = A
about 0.193806699
or about 1 in 5.16

so 1 - A = probability of getting "at least 1 time" in 3 tries =
about 0.806193
or 80.6% chance (IF U like % instead)

instead of using a binomial probability distribution formula
or another method that might increase the error factor.
could just use an online calculator too

here is one
http://vassarstats.net/binomialX.html
n=3
k=1
p=91/216
P: 1 or more out of 3
0.806193300532

I doubt one could do this in their head
standing next to someone trying to explain probability

hope this helps some
Sally

Quote: mustangsally

No.


but your 91/216 is correct.
5/6 is the probability of not getting a 1 on one die
so 5*5*5 / 6*6*6 = 125/216 as you posted


to win "at least 1 time" in 3 tries

it is way easier to figure out the ways NOT to win in 3 tries.
125/216 * 125/216 *125/216 = 1953125 / 10077696 = A
about 0.193806699
or about 1 in 5.16

so 1 - A = probability of getting "at least 1 time" in 3 tries =
about 0.806193
or 80.6% chance (IF U like % instead)

instead of using a binomial probability distribution formula
or another method that might increase the error factor.
could just use an online calculator too

here is one
ml
n=3
k=1
p=91/216
P: 1 or more out of 3
0.806193300532

I doubt one could do this in their head
standing next to someone trying to explain probability

hope this helps some
Sally

This was a great heap, I was close, just got it mix up -
So it better to cal the chance of not winning is 5 to 1 .... so if you can afford to play 5 time in a row ... you should win at least once out of 5 games


was a great help, so you can cal the change if there was a new game with 5 Die ... or Roullet game that uses 2 wheels

Quote: mustangsally

No.

here is one
ml
n=3
k=1
p=91/216
P: 1 or more out of 3
0.806193300532


Sally

Is this correct

n=3 ( number games EG 10 )
k=1 ( at least 1 die will have my number ? )
p=91/216 ( chance of winning )
P: 1 or more out of 3

it put Q as 125/216

so that means for 10 it
0.995787279767% that it should happen more than once. + 0.030668603297 it will only happen once ????

but under it say
mean = 4.213
variance = 2.4381
standard deviation = 1.5614

So if I wait for 10 games - and bet once ..... I should win at least Once every 4th time I try it. ( do I use the "mean"? )

Quote: Billyboy402

Is this correct

n=3 ( number games EG 10 )
k=1 ( at least 1 die will have my number ? )
p=91/216 ( chance of winning )
P: 1 or more out of 3

it put Q as 125/216

their example ***my comment

For example: In 100 tosses of a coin, with 60 "heads" outcomes observed or stipulated to occur among the 100 tosses,
n = 100 [the number of opportunities for a head to occur] ***or number of games or number of successful bets
k = 60 [the stipulated number of heads]*** number of successes or wins
p = .5 [the probability that a head will occur on any particular toss]*** successes or win probability
q = .5 [the probability that a head will not occur on any particular toss]*** fail or loss probability

So
n=3 ( number games EG 10 ) yes
k=1 ( at least 1 die will have my number ? ) no. number of wins (game wins)
p=91/216 ( chance of winning ) yes probability of a win 1-(5*5*5/6*6*6)
P: 1 or more out of 3

Quote: Billyboy402

so that means for 10 it
0.995787279767% that it should happen more than once. + 0.030668603297 it will only happen once ????

P: 1 or more out of 10
0.995787279767 or 99.5787279767%

0.030668603297 = not 1 game (or bet) will be a winner.
about 1 in 33 sets of 10 games (on average of course)
bummer

Quote: Billyboy402

but under it say
mean = 4.213
variance = 2.4381
standard deviation = 1.5614

mean = 4.213 = the 'average'
'average' number of successes or game wins (or bets) for 10 games (or bets)
can't win .2 games
(or bets. either win 1 or lose 1. but an 'average' can be between)

formula = p*n

Quote: Billyboy402

So if I wait for 10 games - and bet once .....

your chance of winning the very next bet you make is exactly 91/216

waiting does not change anything about the next bet winning probability.

Quote: Billyboy402

So if I wait for 10 games - and bet once .....I should win at least Once every 4th time I try it. ( do I use the "mean"? )

no. put n=4 and see your probability of winning at least 1 time in 4 tries.
should show this:
P: 1 or more out of 4
0.887843345215

it is not 1
so 1-0.887843345215=0.112156654785 = about 1 in 9
sets of 4 bets will have 0 (zero) wins for you
*****
maybe you need to understand sample space 1st and move forward after
http://www.mathsisfun.com/data/probability.html

math and gambling are (is) fun
IF not
get help
Sally