BDB. A couple math questions

[Penguinfan]

Maybe this is pretty basic, but I have a Couple math questions mostly related to craps. How would I find out how often I could expect to go X number of roll without hitting a certain number?

Example: how often can I expect to go 15 rolls without hitting a 6? How often would I go 20 rolls with no 6. Given the odds that hittng any 6 is always 31:5

I'm interested in how you figure this out so I can do my own math on different numbers.

Thanks

 

[marine]

since each roll is independent of the previous roll, you would just take the odds of not rolling a 6 one time, and multiply out it 15 times.

 

[Penguinfan]

Thank you Marine

 

[BleedDodgerBlue]

Marine is correct....take the probability of rolling or not rolling and multiply for the amount of rolls u are looking for. As stated every roll is independent of the previous. People who look for hot or cold dice are silly..


I'd suggest a good starting point for probability as it relates to gaming .... wizardofodds.com
They have lots of charts and probabilities for most everything.

Gl

 

[Penguinfan]

I'm not looking for hot or cold tables/rolls, I want know how often I will bust a certain size bankroll given a progression of wagers that pay out odds different than one to one.

I'm probay the last casino player to hear of "Oscars grind" and while it eventually leads to busting out, I want to see how far you would have to drag it out based on different bets paying something other than even money.

 

[BleedDodgerBlue]

I know you r not looking for hot cold.. I applaud u for that. i laugh at peoplenot getting the basic premise behind every roll spin whatever being independent of the previous ones.

I don't play table games only poker but if I did I sure would want to know the math...but I know I'm in the minority on that.


Gl

 

[JOSHNAUDI]

if I did I sure would want to know the math...but I know I'm in the minority on that.

The odds of any person wanting to know the math is 50%

The odds of any person not wanting to know the math is 50%

I believe you are neither in the minority nor in the majority.

:0008 :0008 :0008

Just Fuckin With Ya

Another neat read PF
http://homepage.ntlworld.com/dice-play/Odds.htm

"
The Sixes and Double Sixes Bet illustrates an interesting point about odds. The story goes that a French gambler called Antoine Gambaud the Chevaliere de Mere, was making a good living from the Sixes Bet. The bet was that he would roll a 6 within 4 rolls of a die. With 6 possible outcomes of a throw of a single die we can see that an even money game should have been three rolls and that he had a good edge. But after time players refused to play him and so he came up with the Double Sixes Bet. That is that he would roll a double-6 within 24 rolls of the dice. He worked out his game, without modern maths, by multiplying the 4 rolls he originally used by the number of combinations of the second die (6) to get 24 throws. You would have thought that the number of rolls should have been something greater than 18 since there are 36 possible outcomes of two dice and only one double-6 combination. However he soon found that he was losing money and so he approached the French mathematician and scientist Blaise Pascal (1622 - 1663) with the problem.

It was Blaise Pascal with Pierre de Fermat who came up with the basics of probability theory. Pascal concluded that the true odds were the odds against winning in one roll multiplied by the colog of the hyperbolic log of two (0.693). This works out as 35 x 0.693 which equals 24.255 rolls for an even money game. He needed at least 25 rolls of the dice to get an edge and 26 or 27 would have been more like it. The story goes that he took this advice and soon recouped his losses.

Even today many gamblers still calculate the odds the way I would have. 36 combinations and 1 way to make a double-6 would give even odds at 18 rolls. The story goes that in the USA during the 1950s a gambling operator called Fat the Butch was so convinced that this was so, he bet $1,000 a time on Double Sixes and was allowed 21 rolls of the dice, in each game, to achieve it. He gave up 12 hours later having lost $49,000.
"
Conceptual IP and Media.

 

[Mr. Poon]

I know you r not looking for hot cold.. I applaud u for that. i laugh at peoplenot getting the basic premise behind every roll spin whatever being independent of the previous ones.

I don't play table games only poker but if I did I sure would want to know the math...but I know I'm in the minority on that.


Gl

It is amazing how many people don't grasp this premise, and it goes beyond just craps/dice.

 

[Penguinfan]

It is amazing how many people don't grasp this premise, and it goes beyond just craps/dice.

But the last 8 spins were black, odds are the next one will be red. DUDE, bet the house on it!

:facepalm:


My favorite part of that is when they walk away broke one guy is always assuring the other guy that they did the right thing and just got unlucky.

 

[Penguinfan]

Marine/BDB if you are still checking in please check my math, I can't believe this is right.

Let's say I want to see how often I could go 20 rolls without throwing a 9.

The odds of not throwing a 9 are 32/36, right? Simplified that works out to 8/9.

Taking that to the 20th power gives me:
1073741824/3486784401

Divide that out and it comes to .307 rounded.

Am I doing that right? I can expect that 30% of the time I can expect not to see a 9 in 20 rolls?

 

[Penguinfan]

Dealing with "how long can I go without throwing a 7", my math shows that only 6% of the time a player can expect to throw the dice 15 times and not see a 7.

That sounds closer to accurate that the 30% number above for the no 9 in 20 rolls.

 

[marine]

Marine/BDB if you are still checking in please check my math, I can't believe this is right.

Let's say I want to see how often I could go 20 rolls without throwing a 9.

The odds of not throwing a 9 are 32/36, right? Simplified that works out to 8/9.

Taking that to the 20th power gives me:
1073741824/3486784401

Divide that out and it comes to .307 rounded.

Am I doing that right? I can expect that 30% of the time I can expect not to see a 9 in 20 rolls?

maybe this will help you?

First let me say I think your web site is absolutely outstanding. Thanks. I watched a new craps game being played at Grand Casino, Biloxi, MS. called "Four The Money". To win the shooter must throw the dice 4 times without a 7 coming up. What are the odds of throwing the dice:
4 times without throwing a 7?
3 times without throwing a 7?
2 times without throwing a 7?
1 times without throwing a 7?
How does the math work for this? Thanks
? Stan Abadie from Harahan, Louisiana

You?re welcome, thanks for the kind words. The probability of throwing the dice n times without a 7, and then throwing a 7, is (5/6)n*(1/6). The probability of throwing n non-sevens, without specifying the next throw would be (5/6)n. So the probability of throwing the dice at least four times without a seven would be (5/6)4=625/1296=0.4823.

 

[marine]

Marine/BDB if you are still checking in please check my math, I can't believe this is right.

Let's say I want to see how often I could go 20 rolls without throwing a 9.

The odds of not throwing a 9 are 32/36, right? Simplified that works out to 8/9.

Taking that to the 20th power gives me:
1073741824/3486784401

Divide that out and it comes to .307 rounded.

Am I doing that right? I can expect that 30% of the time I can expect not to see a 9 in 20 rolls?

20 rolls without a 9
(8/9)^20 = .09
20 rolls with a 9 on the 21st.
(8/9)^20 * (1/9) = .01

 

[Penguinfan]

20 rolls without a 9
(8/9)^20 = .09
20 rolls with a 9 on the 21st.
(8/9)^20 * (1/9) = .01

OK, 9% of the time sounds much more reasonable.

 

[Penguinfan]

I see where I made my mistake, I only took it to the power of 10 instead of 20.