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

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

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#

See if this helps

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

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.Quote:Billyboy402not 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 )

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

need ham and eggs for more mathQuote: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

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:Billyboy402Could 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:ThatDonGuyOn 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:ThatDonGuythe results of the previous 10 spins have no effect on what the next spin will be.

I just wanted to highlight that.

No.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?

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