Question

Questions on geometric distribution

Alec and Bill take alternate turns at kicking a football at a goal, and their probabilities of scoring a
goal on each kick are p1 and p2 respectively, independently of previous outcomes. The first person to
score allows the other person one more kick. If the other then scores, the game is drawn. If the other
then misses, the first has won the game. Alec begins a game.

Given that p1 = p2 = 1/3, find the expected total number of kicks in the game. [2]

Answer is 4 but I had no idea why. Anyone care to explain? (with calculations possibly)

Answer 1

Other info given is

Show that the probability that
(i) Alec scores first, on his nth kick, is p1(q1q2)^(n−1), where q1 = 1 − p1 and q2 = 1 − p2, [2]
(ii) Alec wins the game is
p1q2/ (1 − q1q2), [3]
(iii) the game is drawn is
p1p2(1 + q1)/(1 − q1q2) . [3]

Answer 2

ok we can do this, lets go slow so i can remember how as we go. if i recall correctly you do not actually sum a geometric series, but rather the derivative of one

Answer 3

FYI, I have no prob showing the all 3 probabilities. But I couldn't for the rest of my life figure out the last question. Any help would be appreciated. :)

Answer 4

oh ok so we do not have show the probabilities from scratch, just compute the expectation.
i should write this on a piece of paper before we start, first thing we do is replace the p by 1/3 and q by 2/3 so we can see what we actually have

Answer 5

i)
\[\frac{1}{3}(\frac{2}{9})^{n-1}\]
ii)\[\frac{2}{5}\]
iii)
\[\frac{1}{3}\] is what i get, is that what you have?

Answer 6

got the same thing.

Answer 7

i am trying to figure out how to use them but i am starting from scratch.

how can the game end in 2 kicks?
goal, not 1/3*2/3
goal goal 1/3*1/3

3 kicks
not goal not (2/3)^2(1/3)
not goal goal 2/3 (1/3)^2

4 kicks
no no goal goal (1/3)^2(2/3)^2
no no goal no 1/3(2/3)^3

does this look right?

Answer 8

ok i see a pattern emerging finally

Answer 9

i think we have so sum two series, one where there is a win and one where there is a draw.

Answer 10

win
2 kicks
\[\frac{1}{3}\frac{2}{3}\] 3 kicks
\[(\frac{2}{3})^2\frac{1}{3}\]
4 kicks
\[(\frac{2}{3})^3\frac{1}{3}\]
n kicks
\[( \frac{2}{3}) ^{n-1}\frac{1}{3}\]

Answer 11

now we have to multiply by the number of kicks and add this up
\[\frac{1}{3}\left(2\frac{2}{3}+3\frac{2}{3}^2+4\frac{2}{3}^3+...\right)\]

Answer 12

inside sum is 8 although that might not be obvious, so the answer so far is
\[\frac{8}{3}\] hmm that means the other one must only by 1/4

Answer 13

i dont see why would the 4 kicks differ from the 2 kicks...

Answer 14

they are essentially the same thing.

Answer 15

now for "draw"
2 kicks,
\[\frac{1}{9}\] 3 kicks
\[\frac{2}{3}\frac{1}{9}\]
4 kicks
\[(\frac{2}{3})^2\frac{1}{9}\]
n kicks
\[(\frac{2}{3})^{n-2}\frac{1}{9}\]

Answer 16

if i read this correctly there are two possibitlites for it ending with 4 kicks, one ending in a win and the other ending in a tie
no no goal no
no no goal goal

Answer 17

first one has probability (2/3)^3(1/3) second on (2/3)^2(1/3)^2

Answer 18

ok maybe i am being stupid and we can add these up

Answer 19

second try
probability it ends in 2 kicks is 1/3
3 kicks 2/3
4 kicks 4/27
5 kicks 8/81

Answer 20

so looks like we are summing up
\[2(\frac{1}{3})(\frac{2}{3})^0+3(\frac{1}{3})(\frac{2}{3})^1+4\frac{1}{3}(\frac{2}{3})^2+...)\] where the
\[=\frac{1}{3}\left(2\times( \frac{2}{3})^0+3(\frac{2}{3})^1+(\frac{2}{3})^2+...\right)\]

Answer 21

and sure enough, the answer is
\[\frac{1}{3}\times 12=4\]

Answer 22

i don't see why shouldn't we calculate the win for Bill... your 'win' is essentially only counting Alex right?

Answer 23

bill can win too right?

Answer 24

yes. and the draw.

Answer 25

oh sorry, yes i was calculating the win for both
bill wins on odd kicks

Answer 26

for example bill can win on
no goal no

Answer 27

so there are two ways for there to be any number of kicks, win or draw
odd kicks bill can win, even ones alex can win

Answer 28

oh i see

Answer 29

i can write down what i computed if you like, i computed two columns, one for win and one for draw, then added them

Answer 30

but what did you do with the 8/3? i don't see it appearing anywhere in the calculations below

Answer 31

that all was a mistake, i was trying to add the series up separately, before i realized it would be easier to add them together

Answer 32

i was going to add up one series for wins, and one for draws, but it was foolish, i didn't see the pattern when i added them

Answer 33

ok got it.

Answer 34

ok but it doesn't explain how to add
\[\sum n(\frac{2}{3})^{n-2}\] though

Answer 35

if you know how to do it, then we can take it as a given.
it requires a trick
actually two tricks

Answer 36

pretty sophisticated question i would say
best of the day so far

Answer 37

maybe i will reword and and repost it as a challenge

Answer 38

Sry, where did u get the 2(1/3)(2/3)^0+3(1/3)(2/3)+13/3(2/3)^2+...) from ?

Answer 39

As for the sum you would probably need to substitute it with binomial expansion to calculate.

Answer 40

oh ok lets go slow
2 kicks probabilty is
1/3*2/3+1/3*1/3= 1/3
3 kicks
2/3*1/3*2/3+2/3*1/3*1/3 =1/3*2/3
4 kicks
2/3*2/3*1/3*2/3 2/3*2/3*1/3*1/3 = 1/3*4/9
etc

Answer 41

no wait but i added up the equations for n kicks and sum up the thing it doesnt adds to the same amount.

Answer 42

i saw the pattern because the numbers were 1/3, 2/9,4/27,8/81
but in order to make a nice goemetric series i had to factor out a 1/3 to make it work

Answer 43

as for summing
\[\sum_{k=2}^{\infty}n(\frac{2}{3})^{n-2}\] you can use a nice trick

Answer 44

instead of summing that one , find the following sum
\[\sum_{n=1}^{\infty}n(\frac{2}{3})^{n-1}\] which looks exactly like the derivative of
\[\sum_{n =0}^{\infty} (\frac{2}{3})^n\]

Answer 45

problem is i don't get (2/3)^(n-2) by adding up (1/3)(2/3)^(n-1) and (2/3)^(n-2)(1/9)

Answer 46

hold on you lost me there

Answer 47

i just added and saw the pattern
there are two ways to end in n kicks a win which has probability
\[(\frac{2}{3})^{n-1}\frac{1}{3}\] and a loss which has probability
\[(\frac{2}{3})^{n-2}\frac{1}{3}\frac{1}{3}\] when you add you get
\[\frac{1}{3}(\frac{2}{3})^{n-2}(\frac{2}{3}+\frac{1}{3})\]
\[\frac{1}{3}(\frac{2}{3})^{n-2}\]

Answer 48

oh i see no you do not get
\[(\frac{2}{3})^{n-2}\] you get
\[\frac{1}{3}(\frac{2}{3})^{n-2}\]

Answer 49

ok i got it apparently i did\[\sum_{n=1}^{\infty} n(3/4)(2/3)^n \] instead of \[\sum_{n=2}^{\infty} n(3/4)(2/3)^n\]

Answer 50

No wonder i couldn't get the answer. Anyway, Thx for the help.

Answer 51

not sure where the
\[\frac{3}{4}\] comes from

Answer 52

it is the same as yours sat

Answer 53

oh pehaps another trick better than the one i used. expand

Answer 54

i mean explain (please)

Answer 55

\[\frac{1}{3}(\frac{2}{3})^{n-2}=\frac{1}{3}(\frac{2}{3})^{n}\frac{9}{4}=(\frac{2}{3})^{n}\frac{3}{4}\]

Answer 56

oh yes, a much better trick i see

Answer 57

oh no actually i don't see, i was summing
\[\sum nx^{n-1}=\frac{1}{(1-x)^2}\] what for
\[\sum nx^n\] ?

Answer 58

oh i expanded them first then used the same trick 1/(1-x)^2. cept that i don't really recognize the summation form of the binomial expansion.

Answer 59

ok good night then. its 2am in my time. thx for everything.

Answer 60

\[\sum nx^n=x\sum nx^{n-1}\]

Answer 61

pretty tough stuff for only 2 marks

Answer 62

ok but then you are really after
\[\sum nx^{n-1}\] so why not do that at the first step
@Zarkon

Answer 63

you can

Answer 64

If this is in reference to a couple posts above..I was just showing where his 3/4 came from...not that I would do that

Answer 65

ok i just don't know why you would multiply to get
\[nx^n\] and then divide to find what you want. seems like an odd way to do business