Calculating the probability of a sequence of events

Simultaneously throw 3 dices N times. It is collected 1 point for each figure 6 appear.
When we collect >= 10 points the event A is happened.
1. What is the probability of the event A after N attempts (for example N = 10 and N = 100)?
2. What is the probability of the event A , if during N attempts it can be happened several times? (will be collected several times >= 10 figures 6)

"It is collected 1 point for each figure 6 appear"

Can you explain this with an example ? (1,6,6) means 2 points in one throw ? (6,2,5) means 1 point ?

Quote: ssho88

"It is collected 1 point for each figure 6 appear"

Can you explain this with an example ? (1,6,6) means 2 points in one throw ? (6,2,5) means 1 point ?

Yes, exactly.
And if each 10 points give 1$, how to estimate whis value?
I tryed to use Bernulli formula to estimate the probability of appearing 6 more than 10 times in N trials. But it gives the probability which became close to 1 with growing N. And with big number of N the event A can be happened several times.

Quote: Myshell

1. What is the probability of the event A after N attempts (for example N = 10 and N = 100)?

I'll give it a try, For N = 10, total no of dice outcomes = 30,

For exact 9 points, the probability = P(x=9) = (1/6)^9 * (5/6)^21 * 30! / 9! / 21! = 0.03085927379

For exact 8 points, the probability = P(x=8) = (1/6)^8 * (5/6)^22 * 30! / 8! / 22! = 0.06312124215

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For exact 0 points, the probability = P(x=0) = (1/6)^0 * (5/6)^30 * 30! / 0! / 30! = 0.00421272033

P(x>=10) = 1 - P(x=9) - P(x=8) - P(x=7) - P(x=6) - P(x=5) - P(x=4) - P(x=3) - P(x=2) - P(x=1) - P(x=0) = 0.01970628325

So probability of the event A after 10 attempts = 1.97% ? Please help to verify it.

Quote: Myshell

And if each 10 points give 1$, how to estimate whis value?

P(x<=19) - P(x<=9) = a
P(x<=29) - P(x<=19) = b
P(x=30) = c

So the estimated value =a*($1) +b*($2) +c*($3) ?

Quote: ssho88

"So probability of the event A after 10 attempts = 1.97% ? Please help to verify it."

Yes, i got the same result = 0,0197062838
But how to estimate the final result? How many times 'A' wil be occurs? May be it can be estimated with diferent levels of certainties?

Quote: ssho88

P(x<=19) - P(x<=9) = a
P(x<=29) - P(x<=19) = b
P(x=30) = c

So the estimated value =a*($1) +b*($2) +c*($3) ?

It seems like a right way)! Thanks a lot. I will try to check it.

Quote: Myshell

It seems like a right way)! Thanks a lot. I will try to check it.

c = b = 0.0000000000000000000001

Why ?

Quote: ssho88

c = b = 0.0000000000000000000001

I get:
a = 0,019706282
b = 1,4647E-09
c = 4,52337E-24

Quote: Myshell

Why ?


I get:
a = 0,019706282
b = 1,4647E-09
c = 4,52337E-24

You are right, I didn't calculate the value of b and c. LOL