# Binomial Distribution.

1. Jan 28, 2009

### prime-factor

1. The problem statement, all variables and given/known data

A coin can be flipped a maximum of four times

The following conditions exist:

H(first) = $1 H(second) =$2
H(third) = $3 H(fourth) =$4

And first, second, third and fourth, refer to what order one head is obtained.

What is the expected gain
2. Relevant equations

Binomial Distribution.

E(x) = Sum of ((x) . P(x))
3. The attempt at a solution

Drew up a binomial distribution:

Combination are as follows:

HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT

THHH
THHT
THTH
THTT
TTHH
TTHT
TTTH
TTTT

It ends up being (8/16) for each one which is equal to half which is the same for each one. Where have I gone wrong? Please help.

2. Jan 28, 2009

### CompuChip

Nowhere. You said it yourself, it's a binomial distribution. That means - by definition - that each successive throw is independent of all the others, so each time you flip the coin you have 1/2 probability for heads (as you expect from unbiased flipping). Writing out all the $2^4 = 16$ possibilities just showed that explicitly.

3. Jan 28, 2009

### prime-factor

Compuchip:

First, thanks a bunch :). But say this would have been a game of dice with four throws:

e.g.

3(first) = $1 3(second) =$1
3(third) = $3 3(fourth) =$4

The probability of success for getting a 3 is (1/6) and that of a fail is (5/6)

It still ends up the same because it is just going to be: [(5/6)(5/6)(5/6)] . (1/6) each time just in a different order?