What is the probability of losing 11 coin flips in a row?

Question

If I choose HEADS for a coin flip, what is the probability that it will land on heads 11 times in a row?

Answer 1

its easy to figure out. simply take the chances of it happening (1/2) and multiply it by itself as many times as you plan to try.

so it becomes (1/2)^11

or 1/2048

Answer 2

The probability of WINNING 11 in a row is 0.0004882%

0.5x0.5x0.5.......................

The probability of LOSING after 11 times is 99.99952%

♣

Answer 3

Let X be the number of heads tossed.  X has the binomial distribution with n = 11 trials and success probability p = 0.5



In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n = 11 , p = 0.5 )

the mean of the binomial distribution is n * p = 5.5

the variance of the binomial distribution is n * p * (1 - p) = 2.75

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.658312

The Probability Mass Function, PMF,

f(X) = P(X = x) is:

P( X =  0 ) = (0.50)^11 =  0.0004882812 <<<<<< ANSWER

P( X =  1 ) =  0.005371094

P( X =  2 ) =  0.02685547

P( X =  3 ) =  0.0805664

P( X =  4 ) =  0.1611328

P( X =  5 ) =  0.2255859

P( X =  6 ) =  0.2255859

P( X =  7 ) =  0.1611328

P( X =  8 ) =  0.0805664

P( X =  9 ) =  0.02685547

P( X =  10 ) =  0.005371094

P( X =  11 ) =  0.0004882812