Question

as per estudier
On average how many times do you have to toss a coin to get a run of an odd number of heads followed by a tail?

Answer 1

dang probability

Answer 2

i thought this was simple when i saw it at first, but now i am thinking it is not

Answer 3

To be honest, I don't even understand the question

Answer 4

odd number of head followed by a tail. i take it to mean this.
h t
or
t h t
or
t t h t
h h h t
or
t h h h t
h h t h t

Answer 5

each as odd number of heads followed by a tail, and i think those are the only possibilities for success in the number of tosses for each case

Answer 6

2 tosses: h t
3 tosses: t h t
4 tosses: t t h t or h h h t
5 tosses: t h h h t or h h t h t

Answer 7

then you would multiply and add
\[2\times \frac{1}{4}+3\times \frac{1}{8}+ 4\times \frac{1}{8}+5\times \frac{1}{16}+...\] of course all the work is in the ... part

Answer 8

U have 1/2 squared + fourth + sixth + = 1/3 which is probability to succeed right off the bat so fail means tail after even heads.

Then u have to do that 3 times, it is true, but that is trials not tosses.

Each trial ends with a tail so you can say that the average number of tosses per trial is 1/half or 2.

So 2*3 = 6.

Answer 9

that is what i thought at first too. but while i was out i was thinking about it and i changed my mind

Answer 10

Else x = 1/2 (1+x) +1/4(2+x) + 1/4 *2 -> x=6.

Answer 11

i was thinking you were right. i said 3 and forgot that it was for two tosses and you said six and i thought you were correct, but then i started thinking that we are both wrong. look at what i wrote above for cases of success

Answer 12

A (small) set of trials would seem to confirm that is nearer to 6 than to 3.

Simplest would be to plug it in to a simulator and see..

Answer 13

ok but i am still looking for a reason because look at the possibilities i wrote above.

Answer 14

5 tosses: t h h h t or h h t h t the second one is tail after even heads

Answer 15

yes but it says odd number of head followed by a tail. so
h h t h t works for 5 tosses because you have one head followed by one tail. the last two. in other words i think these are the only two possibilities for success on 5 tosses. and that is why i think this is rather complicated

Answer 16

Then u go on to 5h's and T

Answer 17

i count two ways for 5 tosses. not one way. how about 6 tosses?

Answer 18

That's what I just said, 5 h's and t

Answer 19

oh no

Answer 20

the way i read it, it does not say odd number of heads in a row and and then tails. i just says how long do you have to wait to get a run of and odd number of heads and then a tail

Answer 21

so for example
h h h h h t
is one way
but so is
t t h h h t
6 tosses and the last 4 give you a string of three heads followed by a tail

Answer 22

also
h h t t h t works and so does
t t t t h t

Answer 23

doesn't say what happens on the first bunch of tosses, just that you end up with odd number of heads followed by a tail. so both of those i would count as having a string of odd number of heads followed by a tail

Answer 24

OK, we keep talking about different things, I am talking about probability of success and u are trying to establish the event space....

Answer 25

yes but in order to find the probability of success i need to know what a success is. how many ways to do it on two tosses? we both get one way so probability is 1/4

Answer 26

on three tosses we both get one way
t h t
so probability is 1/8

Answer 27

The 1/3 is based on that u start tossing odd heads followed by a tail....

Answer 28

So that failure is then based on tail after even heads...

Answer 29

in four tosses i count
t t h t
h h h t
so probability is 1/8
if you don't count t t h t then we will get different answers

Answer 30

you see the difference yes?

Answer 31

we have to agree on what counts as a success.

Answer 32

No, I dont think so, u will get same answer by more complicated routye....

Answer 33

hold on. if we do not agree on what a success is then we cannot arrive at the same answer

Answer 34

1/3 is defining success as rolling odd heads followed by tail.

Answer 35

believe me i am not saying i am right and you are wrong, i am just saying that we have to know what a success looks like

Answer 36

If u define success your way, it will not be a third.....

Answer 37

but i don't think so

Answer 38

But then u will not multiply by 2 at the end either....

Answer 39

right. if i count t t h h h t as a success in 6 rolls and you only count
h h h h h t, then we cannot come up with same answer

Answer 40

We will come up with the same answer, the calculations will be different.

Answer 41

i see 4 ways to get a string of odd number of heads followed by a tail in 6 tosses

h h h h h t
t t h h h t
t t t t h t
h h t t h t

Answer 42

I get 1/3 because tail after even heads is fail in that scenario.....

Answer 43

really? ok then please go slow because i believe you

Answer 44

but only in that calculation, it is just a calculation, an abstraction if u will...

Answer 45

no fine, i just want to see the calculation. i am more or less sure you are right, i just would like to check

Answer 46

SO probabilty of HT is a half squared right?

Answer 47

got that

Answer 48

so maybe we start with
x = 1/4 + ...

Answer 49

and then 3 heads and a tail is to the fourth...

Answer 50

hold the phone

Answer 51

what about
t h t

Answer 52

to me that counts and you are ignoring it it seems

Answer 53

I am only talking about the probability of getting hhht....

Answer 54

but what about t h t? will we come back to it? because the way i read the problem that counts as a success

Answer 55

if you jump right to
h h h t
then i understand exactly what you are going to get
and if then next one is
h h h h h t etc then i agree with you. but i count other ones

Answer 56

I am assuming (possibly wrongly) that I will immediately start tossing heads.

Let me think about it some more, the trials fail after a tail, the rest is picked up in the 2 at the end.

Answer 57

Also, u have yet to pick any holes in the algebraic formulation...

Answer 58

wow i just googled this problem and it is not so straight forward it seems.

Answer 59

although the answer might in fact be 6 (!)

Answer 60

I WAS certain about it, now u have made be have doubts..:-)

Would like to be certain again....

Answer 61

my problem is not with the formula. my problem is with ignoring the possibility of
t h t being a success. if you are going to compute
P(ht) + p(h h h t) + ...
you will certainly be right

Answer 62

My idea is the tail flips are being picked up by the 2, which is also based on three trials ie a 1/3.

Answer 63

If it isn't a third because some other type of calculation, then the number of trials will be different and so will the 2.

Answer 64

Going to watch the box and give my head a rest....)

Answer 65

ok but here is something i nicked from the web. i will read it carefully and see if it makes any sense

Answer 66

How about the easier formula H=(1/2)*(H+1)+(1/4)*2+(1/4)*(H+2), where H is the expected number of flips? I should not be the one to explain it, since I am borrowing the method from luv2fap, but I will, anyway. Let the expected number of tosses be H. Then, if on the first flip we got T, then we “wasted” one flip, and due to the independence of the tosses, we again expect to stop flipping after H turns (the probability of flipping a T is 1/2). Now, if we managed to flip an H, we need to examine the second toss more carefully. In case of a T – we are done and we used exactly 2 flips (the probability, however, to have HT is 1/4 and this is why we multiply 2 by it). In case of an H – we are back to square 1 having wasted 2 flips and expecting to be done after H flips. The probability for HH is 1/4, again.

Answer 67

this certainly gives h = 6 for sure. hope i didn't confuse matters too much, but the reasoning is different i think

Answer 68

How about the easier formula H=(1/2)*(H+1)+(1/4)*2+(1/4)*(H+2)

That is the algebraic formula I gave u before...

Answer 69

yes so it is. sorry for wasting your time. i thought you got this from summing up, but i was mistaken

Answer 70

No, it was interesting discussion, time not wasted...:-)

Answer 71

now i that i looked at the explanation above i see where the formula came from. i wonder why this is the same as just adding the geometric series you get for
h t
h h h t
h h h h h t
etc and then doubling?

good work

Answer 72

It's not just doubling...:-) (Well, it is but it isn't...)