# What is the probability of getting r heads before s tails?

• Eclair_de_XII
In summary: What is the probability of the next two tosses being two heads?##P(H_j\cap H_{j+1})=P(H_j)P(H_{j+1})=p^2##.What is the probability of the next two tosses being two heads?##P(H_j\cap H_{j+1})=P(H_j)P(H_{j+1})=p^2##.
Eclair_de_XII

## Homework Statement

"Toss a coin repeatedly. Denote the event of getting heads or tails on the ##i##-th try by ##H_i## and ##T_i## respectively, where ##P(H_i)=p## and ##P(T_i)=1-p##, for some ##0\leq p \leq 1##. Now denote by ##E## the event of getting ##r## consecutive heads before ##s## consecutive tails. Find ##P(E|H_1)##, ##P(E|T_1)##, and ##P(E)##."

## Homework Equations

Independence: If ##A## and ##B## are events, then ##P(A\cap B)=P(A)P(B)##.

## The Attempt at a Solution

So far, I have denoted ##E## by ##E=(H_{j+1} \cap H_{j+2} \cap ...\cap H_{j+r} \cap T_{k+1} \cap ... \cap T_{k+s})## for some ##j\in ℕ## and for some ##k\in ℕ,\space k>j+r##. The major thing that concerns me at the moment, is the set of ##H_i,T_i## for ##1\leq i \leq j## and the set of ##H_m,T_m## for ##j+r+1\leq m \leq k##. That's really all I could come up with myself.

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with respect to calculating P(E):

hint: it is enough to consider the case where exactly ##n:=(s+r-1)## games are played
What kind of distribution is involved with summing ##n## iid coin tosses? And what portion of this distribution are you interested in?

I think it was the binomial distribution?

Wait a minute...

I've forgotten a very important detail from the opening post: they have to be ##r## consecutive heads and ##s## consecutive tails. So sorry!

Eclair_de_XII said:
Wait a minute...

I've forgotten a very important detail from the opening post: they have to be ##r## consecutive heads and ##s## consecutive tails. So sorry!

What is the probability of the next two tosses being two heads?

StoneTemplePython said:
what portion of the distribution are you interested in and why?

I don't think I understand what you mean by "what portion of the distribution".

verty said:
What is the probability of the next two tosses being two heads?

##P(H_j\cap H_{j+1})=P(H_j)P(H_{j+1})=p^2##.

Eclair_de_XII said:

## Homework Statement

"Toss a coin repeatedly. Denote the event of getting heads or tails on the ##i##-th try by ##H_i## and ##T_i## respectively, where ##P(H_i)=p## and ##P(T_i)=1-p##, for some ##0\leq p \leq 1##. Now denote by ##E## the event of getting ##r## consecutive heads before ##s## consecutive tails. Find ##P(E|H_1)##, ##P(E|T_1)##, and ##P(E)##."

It might be worth checking that we are all agreed on what this question asks. It looks a bit odd to me, but I think it means this:

Event E is: you toss a coin ##r+s## times. The first ##r## tosses are Heads, the next ##s## tosses are Tails.

Calculate ##P(E)##, ##P(E|H_1)## and ##P(E|T_1)##.

However, if the first toss is a tail, then clearly and trivially the first ##r## tosses can't be heads. So, that seems an odd thing to ask. It's a bit like asking: what is the probablity that a team wins all its games, given that it loses its first game?

PS I don't think it's asking: if you toss a coin an indefinite number of times, what is the probability that you eventually get ##r## consecutive heads followed by ##s## consecutive tails.

Eclair_de_XII said:
Wait a minute...

I've forgotten a very important detail from the opening post: they have to be ##r## consecutive heads and ##s## consecutive tails. So sorry!

This is a radically different problem now. It would also explain why your problem is prodding you to calculate conditionals based on a first step analysis of ##P(E|T_1)##, ##P(E|H_1)## in addition to just calculating ##P(E)##

This now seems to be a theory of runs problem. In particular what is the probability of having a run of ##r## heads before ##s## tails if you toss a coin indefinitely (or if you prefer: ##n## tosses, and pass limits). To Be Confirmed: Is this the correct interpretation?

There's a very simple and pleasant way to do this with renewal theory, but that is extremely far from what you've listed as your relevant equations.

- - - -
What your question is prodding you to do is to condition on the first step/trial. If the first trial is heads, you work through the conditional events to get an equation for ##P(E|H_1)##. Now condition on first step tails -- repeat process and get an equation ##P(E|T_1)##.

You should now have two equations with two unknowns. Solve.

The big idea is if you get heads on your first trial, you either 'win' or you end up in the same place as ##T_1##, but I don't want to give this all away. I'd suggest going back to the drawing board on what sorts of Relevant Equations could be of help here as I think you need to work through the conditioning yourself. Drawing a picture may be of some help.
- - - -

Eclair_de_XII said:
I don't think I understand what you mean by "what portion of the distribution".
I was prodding you to sum over a certain chunk of the binomial distribution. As the problem description has changed, the point I was making is not relevant.

Eclair_de_XII and PeroK
StoneTemplePython said:
This now seems to be a theory of runs problem. In particular what is the probability of having a run of ##r## heads before ##s## tails if you toss a coin indefinitely (or if you prefer: ##n## tosses, and pass limits). To Be Confirmed: Is this the correct interpretation?

Ah! So it's the probability of what happens first: ##r## consecutive heads or ##s## consecutive tails.

That makes sense now.

StoneTemplePython said:
What your question is prodding you to do is to condition on the first step/trial. If the first trial is heads, you work through the conditional events to get an equation for ##P(E|H_1)##. Now condition on first step tails -- repeat process and get an equation ##P(E|T_1)##.

You should now have two equations with two unknowns. Solve.

The big idea is if you get heads on your first trial, you either 'win' or you end up in the same place as ##T_1##, but I don't want to give this all away. I'd suggest going back to the drawing board on what sorts of Relevant Equations could be of help here as I think you need to work through the conditioning yourself. Drawing a picture may be of some help.

Okay, so what I came up with was: ##P(E|H_1)=P(\bigcap_{i=2}^{r} H_i) + P(E|T_1) - P((\bigcap_{i=2}^{r} H_i)\cap (E|T_1))##. Basically, I followed the bolded hint, and used the equation ##P(A\cup B) = P(A)+P(B) - P(A\cap B)##. What I think is going on, is given that the first coin toss returned a head, one of two things can happen for ##E## to be possible:

##A## - you get ##r-1## more heads
##B## - you get a tails and have to start over as if your initial coin flip were a tail

Hence, the probability of ##E## given that ##H_1## has occurred is the probability that either you get ##r-1## more successive heads, or the probability of ##E## given that you get a tails. I'm not sure about how this could apply if someone had to restart the coin flips multiple times... I guess that ##E|H_1## is equivalent to the event that given that any coin flip is a heads, you get ##r-1## more heads, or the event that you mess up on the trial, get a tail, and start over as if the initial coin flip were a tails?

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Okay, now I need to find out what ##E|T_1## is. I have a strong suspicion that ##E|T_1=(\bigcap_{i=2}^{s} T_i)^c\cup## and something with ##E|H_1##, I suppose.

I'm going to guess: ##E|T_1=(\bigcap_{i=2}^{s} T_i)^c\cup E|H_1##.

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Eclair_de_XII said:
Okay, now I need to find out what ##E|T_1## is. I have a strong suspicion that ##E|T_1=(\bigcap_{i=2}^{s} T_i)^c\cup## and something with ##E|H_1##, I suppose.

I'm going to guess: ##E|T_1=(\bigcap_{i=2}^{s} T_i)^c\cup E|H_1##.

If it were me, at this stage I'd drop the set intersection notation and just write out the math formula in terms of multiplying, adding, dividing and exponentiation of ##p## and ##(1-p)##. You then should get two equations that you can solve...

(We could also write these out as states, as in a finite state markov chain which will give numeric answers you could use to check your solution against as well as other insights like exponential decay in probability of not being absorbed, but I don't think they're the right way to get the closed form solution you're after.)

PeroK
Okay, I got my answer for ##P(E)##, ##P(E|H_1)## ,and ##P(E|T_1)##. It's not perfect, but I hope it'll get me something.

Thank you all for your help, and such.

I'd suggest sharing it... as is often the case in probability, the problem is straightforward once you've figured out the conditioning.

There's really no reason you can't get full credit here.
- - - -
alternative approaches that I know would be to use renewal theory or martingales. Drawing the picture for the finite state markov chain can help with intuition I think, as long as you don't get too lost in the weeds.

Okay, here's what I have:

##P(E|H_1)+ (p^{r-1}-1)P(E|T_1)=p^{r-1}##
##-(1-p)^{s-1}P(E|H_1) + P(E|T_1)=1-(1-p)^{s-1}##

##A = \begin{pmatrix}
1 & (p^{r-1}-1) \\
-(1-p)^{s-1} & 1 \\
\end{pmatrix}##

##det(A)=1+(1-p)^{s-1}(p^{r-1}-1)##

##A^{-1} = \frac{1}{det(A)}\begin{pmatrix}
1 & (1-p^{r-1}) \\
(1-p)^{s-1} & 1 \\
\end{pmatrix}##

##\begin{pmatrix}
P(E|H_1) \\
P(E|T_1) \\
\end{pmatrix} = A^{-1}
\begin{pmatrix}
p^{r-1} \\
1-(1-p)^{s-1} \\
\end{pmatrix}
=\frac{1}{1+(1-p)^{s-1}(p^{r-1}-1)} \begin{pmatrix}
p^{r-1}+(1-p^{r-1})(1-(1-p)^{s-1}) \\
1-(1-p)^{s-1}+p^{r-1}(1-p)^{s-1} \\
\end{pmatrix}##

##P(E)=\frac{1}{1+(1-p)^{s-1}(p^{r-1}-1)}[p^{r}+p(1-p^{r-1})(1-(1-p)^{s-1})+(1-p)(1-(1-p)^{s-1})+p^{r-1}(1-p)^{s}]##

Eclair_de_XII said:
##P(E)=\frac{1}{1+(1-p)^{s-1}(p^{r-1}-1)}[p^{r}+p(1-p^{r-1})(1-(1-p)^{s-1})+(1-p)(1-(1-p)^{s-1})+p^{r-1}(1-p)^{s}]##

Did you try testing this out numerically? I think I've transcribed this right, but for p=0.5 and s=10, r=10, I get probability of 1 which can't be right. I also get probability of 1 for some other cases I tested.

Python:
def the_OP_func_version1(p, s, r):
denominator = 1 + (1-p)**(s-1)*(p**(r-1) - 1)
numerator_0 = p**r
numerator_1 = p*(1-p**(r-1))*(1- (1-p)**(s-1))
numerator_2 = (1-p)*(1 - (1-p)**(s-1))
numerator_3 = p**(r-1)*(1-p)**s
result = (numerator_0 + numerator_1 + numerator_2 + numerator_3)/denominator
return result

- - - -
note you'd probably save yourself some work by just doing elimination and not computing the matrix inverse explicitly, but it's your call.

Also, if you use ##q:= (1-p)## I think that can clean things up quite a bit.

Eclair_de_XII said:
Okay, here's what I have:

##P(E|H_1)+ (p^{r-1}-1)P(E|T_1)=p^{r-1}##
##-(1-p)^{s-1}P(E|H_1) + P(E|T_1)=1-(1-p)^{s-1}##

Can you explain via enumeration, how you got these two equations?

Incidentally your first equation looks right, but the second one looks wrong.

StoneTemplePython said:
second one looks wrong

This is what I did:

##P(E|T_1)=P[(\bigcap_{i=2}^{s} T_i)^c\cup (E|H_1)]=P(\bigcap_{i=2}^{s} T_i)^c) + P(E|H_1) - P(\bigcap_{i=2}^{s} T_i)^c)P(E|H_1)=[1-P(\bigcap_{i=2}^{s} T_i)] + P(E|H_1)-[1-P(\bigcap_{i=2}^{s} T_i)]P(E|H_1)=[1-(1-p)^{s-1}]+P(E|H_1)-[1-(1-p)^{s-1}]P(E|H_1)=[1-(1-p)^{s-1}]+(1-p)^{s-1}P(E|H_1)##

Of course, this is based off the assumption that ##E|T_1=(\bigcap_{i=2}^{s} T_i)^c \cup (E|H_1)##, so it's likely this assumption is incorrect.

Eclair_de_XII said:
Of course, this is based off the assumption that ##E|T_1=(\bigcap_{i=2}^{s} T_i)^c \cup (E|H_1)##, so it's likely this assumption is incorrect.

Why would you need an assumption that is likely incorrect?

As I've suggested before, all you need to do is enumerate at most ##s## tosses, where the first one is tails. I'll define ##q:= 1-p##. I find it convenient to interpret in terms of getting a reward of 1 when the run of r heads occurs before the run of s tails and a reward of 0 otherwise. (I'd then calculate the expected reward, but the expectations operator looks awkwardly like the event E so I'm leaving that out.)

In the case that player one loses, this is given by ##q^s##, and there is no reward for losing. But we could also have ##\{qp, q^2p, q^3p, ..., q^{s-1}p\}##, each of which puts us in a state that is identical to having had a heads on the first toss.

We should forget about commutativity for a minute and literally this should be read as ##q^k p= \text{tails on first k tosses then one heads}##. Each one of these corresponds to an event and each one is mutually exclusive and hence additive. If you want you can label each event as number of tails (k) until first heads.

So the expected reward when the first toss is tails is ##\sum_{k=1}^{s-1} p q^k P(E\big \vert H_1)##... why is this true? Note: I wrote "when the first toss is" here...

if you prefer we can condition on the first toss being tails and hence in this conditional world you'd have

the expected reward conditioned the first toss as tails is ##\sum_{k=1}^{s-1} p q^{k-1} P(E\big \vert H_1)##

it's just a small book-keeping difference.
- - - -
another way to view the above is in terms of the geometric distribution for tosses until first heads, so for completeness you could think of it as if you always toss a coin until seeing heads, given that the first toss was tails, it's just there's no reward if your toss number happens to be ##\geq s##

##= \sum_{k=1}^{\infty} p q^{k-1}\cdot \text{Expected Reward at toss k} =\big( \sum_{k=1}^{s-1} p q^{k-1} \cdot P(E\big \vert H_1)\big) + \big(\sum_{k=s}^{\infty} p q^{k-1}\cdot 0\big)##

where you have an expected reward of ##P(E\big \vert H_1)\big)## for that first slice of the summation, and you get zero reward for all the 'later arrivals'.

- - - -

Of course, there's some other work to be done like simplifying this sum into a finite geometric series, and combining with the other equation and solving.

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Eclair_de_XII said:
Okay, here's what I have:

##P(E|H_1)+ (p^{r-1}-1)P(E|T_1)=p^{r-1}##
##-(1-p)^{s-1}P(E|H_1) + P(E|T_1)=1-(1-p)^{s-1}##

##A = \begin{pmatrix}
1 & (p^{r-1}-1) \\
-(1-p)^{s-1} & 1 \\
\end{pmatrix}##

##det(A)=1+(1-p)^{s-1}(p^{r-1}-1)##

##A^{-1} = \frac{1}{det(A)}\begin{pmatrix}
1 & (1-p^{r-1}) \\
(1-p)^{s-1} & 1 \\
\end{pmatrix}##

##\begin{pmatrix}
P(E|H_1) \\
P(E|T_1) \\
\end{pmatrix} = A^{-1}
\begin{pmatrix}
p^{r-1} \\
1-(1-p)^{s-1} \\
\end{pmatrix}
=\frac{1}{1+(1-p)^{s-1}(p^{r-1}-1)} \begin{pmatrix}
p^{r-1}+(1-p^{r-1})(1-(1-p)^{s-1}) \\
1-(1-p)^{s-1}+p^{r-1}(1-p)^{s-1} \\
\end{pmatrix}##

##P(E)=\frac{1}{1+(1-p)^{s-1}(p^{r-1}-1)}[p^{r}+p(1-p^{r-1})(1-(1-p)^{s-1})+(1-p)(1-(1-p)^{s-1})+p^{r-1}(1-p)^{s}]##

I am not sure if your answer is correct. I get a slightly different-looking answer, obtained in a different way.

We can model the process as a discrete-time stationary Markov chain with states H1, H2, ..., Hr and T1, T2, ..., Ts. The process at the completion of some toss is in state Hj if the toss has just resulted in the jth 'head' in a run of 'heads' (so in state H1 we have just gotten our first 'head' in a potential run of 'heads'). The states Tk have the same type of meaning, but for 'tails'. The two states Hr and Ts are "absorbing", meaning that the game ends when one of them is first reached.

The transitions are simple: when in state H1, we go to state H2 if the toss is a 'head' and to state T1 if a 'tale'. For ##1 \leq i \leq r-1## we go from state Hi to H(i+1) if 'heads' and to state T1 if 'tails'. Similarly, we go from state Tk to T(k+1) if 'tails' and to H1 if 'heads'.

So, now all we need to do is compute the first-passage probabilities for H1->Hr and T1->Hr. The process starts in state H1 with probability p and in state T1 with probability q = 1-p.

If we let ##x_i## be the probability of eventually reaching Hr, starting from state Hi, and ##y_j## be the probability of reaching Hr, starting from state Tj, then, we have:
$$\begin{array}{l}x_i = p x_{i+1} + q y_1, \; i=1,2, \ldots, r-1\\ x_r = 1\\ y_j = q y_{j+1} + p x_1, \; j=1,2, \ldots, s-1 \\ y_s=0 \end{array}$$
These equations lend themselves to a nice solution for ##x_{r-1}##, then ##x_{r-2}## , ##\ldots, ## then ##x_1## in terms of ##p, q## and ##y_1##. So, we have a fairly simple equation for ##x_1## in terms of ##y_1##. Similarly, we obtain a reasonably simple solution for ##y_1## in terms of ##x_1##. Now we have two equations that can be solved to find ##x_1## and ##y_1##. The answer is ##\text{Answer} = p x_1 + q y_1.##

Everyone, thanks for your help and perspectives on how to solve this problem, but I really need to find an expression for ##P(E|T_1)## in addition to ##P(E)##.

Eclair_de_XII said:
Everyone, thanks for your help and perspectives on how to solve this problem, but I really need to find an expression for ##P(E|T_1)##...

I basically gave you this in post #19... I actually felt like I gave away too much there. What questions do you have?

- - - -
edit:

Eclair_de_XII said:

also, for avoidance of doubt here:

I signed off on your ##P(E|H_1)##... If you somehow solve for ##P(E)## using whatever method, then you already know ##P(E|T_1)##. I.e. via Total Probability you have ##P(E) = p \cdot P(E|H_1) + (1-p) \cdot P(E|T_1)##which I'd actually write using indicator random variables and call Total Expectation:
##P(E) = \mathbb{E}\big[\mathbb I_E\big] = \mathbb{E}\Big[\mathbb{E}\big[\mathbb I_E\big \vert \mathbb I_{\text{Toss 1 is Heads }}\big]\Big] = p \cdot \mathbb{E}\big[\mathbb I_E\big \vert \mathbb I_{\text{Toss 1 is Heads} = 1}\big] + (1-p) \cdot \mathbb{E}\big[\mathbb I_E \big \vert \mathbb I_{\text{Toss 1 is Heads} = 0} \big]##
##= p \cdot P(E|H_1) + (1-p) \cdot P(E|T_1)##

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StoneTemplePython said:
I basically gave you this in post #19... I actually felt like I gave away too much there. What questions do you have?

You seemed to be giving the expected values of ##E|T_1## but I can't seem to draw a correlation between that and ##P(E|T_1)##.

Eclair_de_XII said:
You seemed to be giving the expected values of ##E|T_1## but I can't seem to draw a correlation between that and ##P(E|T_1)##.

I find it preferable to think in terms of indicators sometimes, but they are ultimately Bernouli random variables so you have

##\mathbb{E}\big[\mathbb I_E\big \vert \mathbb I_{\text{Toss 1 = Tails}}\big] = P(E|T_1)##
- - --
edit:
note that with a little contemplation you see that the reward function I setup in post 19 has either a payoff of 0 or 1... it too is a bernouli, so that tells you something...

So... ##P(E|T_1)=p(1-p)^{s-1}P(E|H_1)##...?

Eclair_de_XII said:
So... ##P(E|T_1)=p(1-p)^{s-1}P(E|H_1)##...?

That is awfully close... but slightly off. Maybe a cancellation error when you telescoped the finite geometric series?

Erm... ##P(E|T_1)=P(E|H_1)(p+1-(1-p)^{s-1})##...?

Eclair_de_XII said:
Erm... ##P(E|T_1)=P(E|H_1)(p+1-(1-p)^{s-1})##...?
so referencing post 19 the key thing to hone in on is

##= \sum_{k=1}^{\infty} p q^{k-1}\cdot \text{Expected Reward at toss k} =\big( \sum_{k=1}^{s-1} p q^{k-1} \cdot P(E\big \vert H_1)\big) + \big(\sum_{k=s}^{\infty} p q^{k-1}\cdot 0\big)##
##= P(E\big \vert H_1) \cdot \big( \sum_{k=1}^{s-1} p q^{k-1}\big) ##

what steps are you taking to simplify that finite geometric series?

What I'm doing is:##\sum_{k=1}^{s-1} pq^{k-1}=p(1+\sum_{k=1}^{s-1} q^{k})=p(1+\frac{1-q^{s-1}}{p})=p+1-q^{s-1}##.

Eclair_de_XII said:
What I'm doing is:##\sum_{k=1}^{s-1} pq^{k-1}=p(1+\sum_{k=1}^{s-1} q^{k})=p(1+\frac{1-q^{s-1}}{p})=p+1-q^{s-1}##.

the issue is that

##\sum_{k=1}^{s-1} pq^{k-1} = p\big( 1+ q + q^2 + ... q^{s-2}\big) \neq p\big( 1+ q + q^2 + ... q^{s-1}\big) =p(1+\sum_{k=1}^{s-1} q^{k})##

So is it ##\frac{p}{q}(\frac{1-q^{s-1}}{p})=\frac{1-q^{s-1}}{q}##, then...?

Eclair_de_XII said:
So is it ##\frac{p}{q}(\frac{1-q^{s-1}}{p})=\frac{1-q^{s-1}}{q}##, then...?

I'm not sure how you're getting to this? You should be able to telescope this quite easily. I wrote out the derivation in a couple of lines but I decided not to post it yet. You need to show some work here.

This is how I'm getting it: ##\sum_{k=1}^{s-1} pq^{k-1}=\frac{p}{q}\sum_{k=1}^{s-1} q^{k}=\frac{p}{q}(\frac{1-q^{s-1}}{1-q})=\frac{p}{q}(\frac{1-q^{s-1}}{p})##.

Eclair_de_XII said:
This is how I'm getting it: ##\sum_{k=1}^{s-1} pq^{k-1}=\frac{p}{q}\sum_{k=1}^{s-1} q^{k}=\frac{p}{q}(\frac{1-q^{s-1}}{1-q})=\frac{p}{q}(\frac{1-q^{s-1}}{p})##.

Two fundamental issues:
1.) You still have not shown your work for the telescoping process
2.) ##\sum_{k=1}^{s-1} q^{k}\neq (\frac{1-q^{s-1}}{1-q})##
this obviously cannot be true. Selecting ##q = 0## doesn't work because ##0 \neq 1##. Now select some small ##\delta \gt 0## and set ##q := \delta##. The left hand side is approximately 0 while the right hand side is approximately 1. This is not an equality.

I don't know what you mean by telescoping process. Do you mean that the finite sum is a telescoping series?

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