This is a rather famous probability problem.
The easiest way to solve this is to calculate the probability that you WON'T roll a "double 6" (or a twelve) each time you roll the dice. There are 36 ways in which dice rols can appear and only one is a twelve. So, for one roll, the probability that you will NOT get a twelve is (35/36)^n where 35/36 is about .97222222 and n would equal 1 for the first trial. So for your first roll the odds that you WON'T get a 12 is .97222222.
For the second roll we calculate (35/36) to the second power or (35/36)^2 which equals about .945216.
When we get to the 24th roll we calculate (.97222222)^24 which equals 0.508596.
For the 25th roll, we calculate (.97222222)^25 which equals 0.494468. For the first time we have reached a probability which is lower than 50 per cent. That is to say, after 25 rolls, we have reached a point in which the probability is less than 50 per cent that we will NOT roll a twelve.
To phrase this more clearly, after 25 rolls we reach a point where the probability is greater then 50 per cent that you will roll a 12 at least once.
Please go to this page 1728.com/puzzle3.htm and look at puzzle 48. (The last puzzle on the page). An intersting story associated with this probability problem is that in 1952, a gambler named Fat the Butch bet someone $1,000 that he could roll a 12 after 21 throws. (He miscalculated the odds [as we know you need 25 throws] and after several HOURS, he lost $49,000!!!)
Please go that page and it has a link to the Fat the Butch story.
6=2m-2= 8=2m = 4=m. 9m=36. m=. Yes, it is the same solution.
Answer:
the answer to a is $75.14
the answer for B is $5,773
Step-by-step explanation:
It's 21 on the left and 34 on the right
9:11, 9/11 are the only ones because girls are first
