If two fair dice are tossed, what is the smallest number of throws, n, for which the probability of getting at least one double 6 exceeds 0.5? (note: this was one of the first problems that de méré communicated to pascal in 1654.)

You can use the binomial distribution to model such problems.

The smallest number of throws, n, for which the probability of getting at least one double 6 exceeding 0.5 is 25

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

The probability that out of n trials, there'd be x successes is given by

Let the success be defined as getting a double six.

And let the failure be defined as not getting a success in a toss of two fair dice (that is, not getting a double six)

Let we have
E = getting a double six in a throw of two dice

Then,

(as there are 36 different outputs available in toss of two dice(by rule of product) and that getting a double six is possible in one way only)

Thus, probability of success is 1/36

and probability of failure is

Let the random variable X tracks number of successes in n trials.

Then, we have

Using the probability function, we have probability of at least 1 success as

This is the probability of getting at least one double 6,

As per the question, we need to find value of n, for which the probability found above exceeds 0.5, thus,

(as n is integer since it is number of times dice will be tossed)

Thus,

The smallest number of throws, n, for which the probability of getting at least one double 6 exceeding 0.5 is 25

Learn more about binomial distribution here:

link

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.

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

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step-by-step explanation: