Binomial Distribution: Expected Gain for Flipping a Coin Four Times

[prime-factor]

Homework Statement

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

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

What is the expected gain

Homework Equations

Binomial Distribution.

E(x) = Sum of ((x) . P(x))

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.

 

[CompuChip]

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.

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.

 

[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?