Paddy's Unfair Coin

Solution 1: Consider what happens when Paddy flips a coin. If it’s a head, he's done. If it’s a tail, he would expect another $E[H]$ flips to get a head. Hence, $E[H] = \frac {1}{4} \times 1 + \frac {3}{4} \times ( 1 + E[H] )$ . Solving this gives us $E[H] = 4$ .

Solution 2: Let $X$ be the number of coin flips needed to make in order to obtain a Head. If $X=n$ , then the first $n-1$ flips are $T$ , and the $n^{th}$ is $H$ , which occurs with probability $\frac {3^{n-1}}{4^{n-1} } \times \frac {1}{4}$ .

Hence, by the definition of expectation, we have $E[H] = \frac {1}{4} \times 1 + \frac {3}{4} \times \frac {1}{4} \times 2 + \frac {3^2}{4^2} \times \frac {1}{4} \times 3 + \ldots$ . Let’s denote this value by $S$ . Consider the value of $\frac {4}{3} S-S$ : $\begin{array}{l l l l l l} \frac {4} {3} S & = \frac {4}{3} \times \frac {1}{4} \times 1 &+ \frac {1}{4} \times 2 & + \frac {3}{4} \times \frac {1}{4} \times 3 & + \frac {3^2}{4^2} \times \frac {1}{4} \times 4 & + \ldots \\ - S & = & - \frac {1} {4} \times 1 & - \frac {3}{4} \times \frac {1}{4} \times 2 & - \frac {3^2}{4^2} \times \frac {1}{4} \times 3 & - \ldots \\ \hline\\ \frac {1}{3} S & = \frac {4}{3} \times \frac {1}{4} & + \frac {1}{4} & + \frac {3}{4} \times \frac {1}{4} & + \frac {3^2}{4^2} \times \frac {1}{4} & + \ldots \\ \end{array}$

The last line is a geometric progression with first term $\frac {4}{3} \times \frac {1}{4}$ and common ratio $\frac {3}{4}$ , hence has sum $\left( \frac {4}{3} \times \frac {1}{4} \right) \left( \frac {1}{1- \frac {3}{4}} \right) = \frac {4}{3}$ . Hence, $E[H] = S = 4$ .

One can imagine that this coin is a dice with four faces. 3 of the faces are tails whereas 1 of the face is a head, which indeed shows that displaying a head is 1/4 as there is one 'desirable' possibility (one head) out of 4 equally likely possibilities. Therefore, if one were to roll this 'coin' 4 times, about 1/4 of the time the coin would display a head.

The probability of an event that will happen is computed as the number of ways that the event can occur is divided by the total number of possible outcomes. Since the probability is 1/4, 4 is the least total number of possible outcome/try.

the formula is nCr= n!/(n-r)!r! this is read as "the combination of n taken r at a time".

Let: n=4 r=1

nCr= 4!/(4-1)!1! = 4.3.2.1/(3)!1 =4.3.2.1/3.2.1.1 =4/1 or 4 use cancellation( cancell 3.2.1 on the right side and 3.2.1 on the other side.

According to the question, the probability of getting a heads is 1 in 4. A normal coin would have a 1 in 2 chance of getting a head. This means that out of two flips, one will be heads and the other tails. Thus, following this logic, out of four flips, you can expect to get one head (and three tails).

This theoretic problem can be solved using the fact that the probability of the coin displaying heads is 1/4. The probability of the coin displaying heads being 1/4 means that it is probable the coin will display heads once out of every four outcomes. Since the problem only asks for head to be displayed once, the coin can be expected to be flipped 4 times for the head to be displayed once.

It is given that the probability of getting an head is (1/4).It means that if the coin is flipped 4 times, one outcome MIGHT be head and the rest will be tails.He needs to get an head as the outcome. Thus he will have to flip the coin at least once.Then again if the previous is not an head............ like this 4 times he will have to flip. Out of these if none comes out to be an head, then the probability ratio will be hampered (It will be hampered if you take in consideration only the 4 flips; although it will not become wrong as the last 2 flips of a total 8 flips will give the same ratio); thus one HAS TO BE a head out of all these flips CONSIDERING 4 FLIPS.Thus he will have to flip the coin ATMOST 4 times to get an head (although he might luckily get an head before that).The expected answer is 4.

(There are of course possibilities of not getting a single head in 4 flips and so one will need to extend ones observation to 8 flips and so on..).

probability of head is 1/4. if we toss a coin 4 time the probability of a head is 1/4+1/4+1/4+1/4=1

The probability is $1/4$ . In 100 , it is 25 %. so theoretically , to get 25 head I have to flip the coin 100 times. so to get 1 head I have to flip the coin 4 times.

the probability of getting a head is 1/4

let the no .of coin flips he would need to get 1 head is x

so 1/4*x= 1

by solving we will get x=4

hence no of coin flips required is 4.

Let T be the time for the first success in a Bernoulli trials process B(n,p). The expected value is: E(T ) = 1 • p + 2qp + 3q^2 + …= p(1 + 2q + 3q^2 +…)=p(1/(1-q)), where q=1-p. Differentiating this formula, we get E(T ) =p(1/(1-q)^2)=p/p^2=1/p. So, for p=1/4 the expected value is 4.

For every flip the chance for flipping a heads is 1/4, add this number four times for four flips and you get a 100% probability.