# Oxbridge Mathematics Interview

Hi guys, this is Adriaan. I study Maths at

Oxford and today I’ll run through a very quick Maths interview question. This is the

context. You play a game with a coin and the coin is not necessarily fair. By that what

I mean is the probability of throwing a head and a tail is not both a half. If you throw

a head, you get one 1, if you throw a tail, you get nothing and the game ends. The interview

question is essentially what is the expectation, what is the average amount of money that you

are going to get from this game. I will start by introducing some variables. I’ll let

H be the probability that you throw heads and I’ll let T be the probability that you

throw tails. Of course H plus T is 1. I have to recall the definition of expectation.

I’ve written it down here. So the expectation is defined to be the amount of money you can

get in some case times the probability of getting that amount of that money and you

sum for each possible outcome. So in this case, the outcomes will just be the amount

of money you get and these will be integers. So you can get £1, you can get £2, £3,

£4. So in this case, the amount of money, I’m summing the amounts of money from let’s

say i equals 0 to infinity. If you are a bit confused by the summation

notation with the sigma, I’ve also written it out like this. It’s just 0 times the

probability of me getting 0. 1 times the probability of me getting 1 plus 2 times the probability

of me getting 2 and you continue this on so it’s an infinite sum to infinity. Now I

have to actually start finding these probabilities. So what’s the probability of you getting

0. Well there’s only way to get a £0 in the game total. That’s if you throw a tail

right away. So your first throw is a tail. The probability of that happening is just

t. So the first term is 0 times t. The second term we are gonna get 1 times the probability

of getting 1 well there’s only one way of getting 1. You have to throw a one head and

a tail right after that. So then we get h times t then what’s the probability of getting

2. That’s to throw two heads and then a tail. So it’s going to be h2 times t. If

I introduce one more term, we have 3 times h3t and this is sum to infinity so this continues

forever. So we’ve reduced this relatively complicated

problem to just as infinite sum. But how are you going to evaluate this sum. Well the first

term is just zero so let’s just get rid of it. So we have ht+2h2t+3 h3t+ I’ll write

next term 4h4t and this sum continues forever. The problem is now just how would you evaluate

such a sum. It looks like a little bit like a geometric sum because you are getting an

h here h2 here, h3, h4. So it’s like a power going up. But it also looks like an algebraic

sum because you have a 1 here, 2 there, 3 there, 4 there. There isn’t actually a rule

for this kind of sums you gonna have to derive it. This is really the crux of the interview

question. So how do you approach this. Well first you

can try to factor something out. So, let’s see. Is there a common term here? There’s

an h here, an h here, there’s an h in all these terms. And so there is a t here as well.

So if I factor out an ht, I get 1+2h+3h2+4h3 and this continues for ever. It doesn’t really help you that much because

there is still some of the formats that were before. It looks a little bit like geometric

sum and a little like algebraic sum. This is the most difficult part of the interview.

You have to see a trick to evaluate this sum. How do you do such a thing. The intended method,

I’ll just give it to you here but be aware that you would have much more time to think

about this and you might get some hints in the interview as well. It’s to recognize

that this thing in the bracket looks like a derivative because if I differentiate h

with respect to h, I get 1, if I differentiate h2 with respect to h, I get 2h, if I differentiate

h3 with respect h, I get 3h2. This is the idea. So I’m gonna write this in this form.

This is the derivative of h+h2+h3+h4 and this continues forever. Now you have a bracket

you can recognize this sum you can do. If I have move this up a little bit, I have ht,

now this is geometric sum. So I’m gonna have d/dh; the rule for this is u1 divided

by 1-r. So u1 in this case is h and the common ratio is h. That’s a form you have to know.

The interviewer will probably give it to you if you can’t think of it. That’s just

a standard formula that hopefully you will know from high school. Now we are almost done. We can just use the

quotient rule on this to differentiate this. So d/dh shall be quote we take the derivate

of h with respect to h which is just 1; 1 times 1-h minus derivative of this with respect

to h, is just –(-1) times h over all these and we square this. This is just the quotient

rule. Now I would like to simplify this a little

bit. So we have ht{(1-h)-(-)1)h/(1-h)2}. This just becomes ht(1/(1-h)2) which is just ht/(1-h)2

and that’s the answer you are looking for. Now we could think about what this really

implies. So if we remember that t was 1-h. h(1-h)/(1-h)2=h/(1-h). This is really the

ratio of the probability of heads to 1-h which is t. So if we write this h-t means the expectation

which is the ratio of the probability of heads probability of tails. Particular, if we have

heads and tails both equal to 1/2 so if we have fair coin, this just becomes 1. So if

you had a fair coin, the average amount of money you would make from this game is £1. You might be asked some implications with

this. You might be asked for instance to draw graph of this if we just do this roughly and

quickly. If I have here my expectation and here my h, my graph is going to look something

like, it’s gonna have 0 here because it’s h on 1-h so if we have h being 0, the whole

thing becomes 0; if h becomes 1, we have an asymptote, so we just draw this up. Turns

out that graph, it would take a little longer to figure out but in turns out the graph looks

a little bit like this which makes sense. So it’s just worth mentioning that in the

real interview, you would have 20 minutes. I did this question very quickly in a few

minutes so in an interview, you would have a little bit more time and maybe your interviewer

would give you some hints as well, especially things like defining h and t instead of just

leaving t as 1-h. That kind of stuff helps the algebra a little bit but it’s difficult

to see that coming. So presumably your interviewer would give you a hint; so don’t be alarmed

by how quickly I did it. You would have probably about 3 times as much time that I took on

it in your interview. And I hope that helps.

Why am I watching this I don't study maths or plan on going to university

I started high school last year but thanks anyway 😂

Or u could just say 50/50 😂😂😂

0

I was looking at memes how did I get here?

Im from germany and we never did geometric sums etc in school but somehow i thought of the problem for a min or something and came up with the idea to divide the 2 probabilities 😀 just felt right

There is a much easier way of solving it:

There is a probability h of getting head the first time, and after this, our expectation has been increased by 1. Then clearly E=h(E+1) and we are done.

This is what we are doing now in year 10 for gcse

can someone enlighten me with the meaning of expectation in this video? i understood all the algebra but not the 0+ht+2h^2 t +….. part

This problem is more annoying than difficult; if you make the assumption the coin is fair, then anyone can realize you get a recurring series of summing 0 to infinity of 1/2 ^ n, which approaches 1 as n approaches infinity, giving the result in the video. The annoying part is just that they force you in the prompt to assume it isn't fair so you have to go through the calc and deal with an infinite series just to find the same answer.

is it not just "1/p – 1" where p is the probability of tails?

Isn't "h/(1-h)" a better answer than "h/t"?

Because it the second one you loose information what is relation between variables

I used something called Random Stop Sum from my stochastic process class to solve this question and I got a different answer..

Let y_i = 1 for probability(heads) = p, and y_i = 0 for prob(tails) = q.

Then E(y_i) = (1)(p) + (0)(q) = p

Let X = sum from i = 1 to N of y_i

Let N be a Geometric distribution with parameter q, or P(1st success) = q, and P(N=n) = q*p^(n), n = 0,1,2, …

Then E(N) = q/p

Then E(X) = E[ E(X | N=n) ]

= sum from n = 0 to inf of E(X | N = n) * P(N=n)

= sum from n = 0 to inf of E(sum from i = 1 to n of y_i | N = n) * P(N=n)

= sum from n = 0 to inf of n * p * P(N=n)

= p * E(N)

= p * (q/p)

= q

What did I do wrong? How come I got a different answer?

interesting

I just did half chance of either so after a million throws you end up with on average 1/2 pound. Rounded up a half pound is equal to one pound. Done.

If you plug in t for 1-h you get a normal geometric series, this could have been much easier

What kind of pen are you using? What’s what’s the fine point size?

What kind of pen are you using? What’s what’s the fine point size?

This is incorrect. If we start with 0 then we can get 0 with prob t or 1 with probability >>>>>h<<<<<< NOT "ht". similarly each following probability is a power of h NOT multiplied by t !!!

The sum is h+2h^2+3h^3+…= sum 1 to inf of k*h^k = h(1/(1-h))'. E(..)= integral on [0,1] of h/(1-h)^2 /(1-0) = 1/(1-h) + ln(h-1), which, you guessed it, doesn't converge (diverges to +inf) because of the crazy infinite winnings were one to account for a possibly arbitrarily close h to 1. No, I wasn't alarmed by how quickly you had done it, as you made a very simple mistake that compounded. Oh and you didn't even calculate the expected value – you didn't even graph anything connected to it. You must find the average value of a continuous function (with input h), as I did ^^

omg. I just realized.. the author said "I study maths at Oxford" – he got accepted after this?? and he wasn't even corrected????

Well, it seems you are there, so you must be way smarter and better at math than I am. However, the solution you present here probably did not help you get in? Failure to see the recursive relationship E(payout) = p(H)*(1+E(payout)) and E(payout) thus = p(H)/(1-p(H)). Even now, you persist in a long-winded derivation. I really don't know how you got in. Sorry, this is totally troll-ish on my part, but the good mathematicians you see on YouTube and elsewhere see and use relationships like this to solve hard problems, where the direct approach is or seems intractable. This is why the proof of the four color map theorem is so disturbing, since it has so much detail and provides so little insight into the problem.

Again, I am 99%+ confident you are way better at math than me. I just don't get it, I am "alarmed' that it took you so long and so many steps,.

Watch this video about number 89 and its amazing facts !

https://youtu.be/ODrMBfMJbPc

lol I can flip a coin in a certain way to were it always lands on the site I want.

Wouldn't the average be zero? Since eventually the game will end with you getting tails and you end up with nothing?

I’m in Grade 12 I take college math never show up and I can follow this with understanding. It looks a lot like just simplifying

Which pen do you use?? Its amazing

Pretty easy

So…you ended up with an answer that accounts for a fair coin…h/t or a 50/50 chance when the question stated that the odds were not necessarily fair. Sounds to me like you really didn't answer the question, you just spent 8 minutes proving mathematically why a fair coin has a 50% chance of success/failure…which is something literally everyone already understands by the nature of reality…except you didn't account for reality where even a fair coin won't have a true 50% failure rate without a fictional perfect coin flipping device. You'll end up with a limit that converges at 1/2 but in an actual test you'd always have some deviation, no matter the number of test cases.

Yep. Sounds like something a mathematician would do.

And you still don't know what the probability so that you can't know what value you get out of it in rl.

This is maths at its best 😀

btw: If you flip the coin for a while, the probability can change, since the shape of the coin changes by the drop. lol

I kinda muffed up my math test for Cambridge maybe only got 30% right, but still got in on Engineering course.

why do they do this questions? i mean at least in my country(spain) you dont study those concepts until you get to college. Whats the point on going there if you already know what they will teach you? Is oxford like a mastery in maths? Its like asking about quantum mechanics someone who is interested in a physics degree.

It’s really nice with differentiation but summing the series is quite easy…

remember from high school? 100% of the work you did is taught in university

Wouldn't it just be zero? You state if you flip a tails you lose all your money. As you approach infinity the odds should approach zero which means the average would also approach zero

Sum of infinite geometric series a/(1-r). You have 1/2 chance of winning 1£, 1/4 of winning another. This gives the average to be 1/2 + 1/4 + 1/8… the initial term a = 1/2. The ratio between terms r = 1/2. Plugging this into the equation gives 1/2/1/2 =1

you lost me after h+t = 1 rip

It is simply a geometric distribution where the expected value is (1-p)/p defining p as either possibility of heads or tails. Are you sure this is an entrance exam it is so simple

He is left handed.

4:58 He lost me at this point why is 1-h there?

Way too complicated.

e1 = h,

e2 = h on the condition that e1 was successful, which is h, therefore e2 = h^2

we get the simplest possible geometric sum E(n=1->inf) h^n which in this particular case is h/(1-h)

Wow I'm just stupid!!! I wish there was AI to replace mathematicians so I wouldn't have to worry about feeling stupid anymore

How does h × t represent an event representing a heads followed by a tail?

Surely if you've come across Bernoulli geometric probabilities before you could just write the answer down without calculation?

Isn't that something that you do at A Level?

Looks like Im not getting into Oxford.

I figured this out in my head… am I a genius or just a quirk?

and if you want to get technical the answer is not 1 but 0.

The problem requires you to figure the average as a number that approaches 1 but never actually achieves 1. thus since you are dealing in whole outcomes. of either heads or tails the average will always be less than 1 and you must round the problem down to zero instead of up to 1.

this is due to the nature of the question. if you throw a heads you get 1 lb, if you through a tails you get nothing and can not throw again. thus the average is 0

I would have said "the biggest x such that: (h^x)<=0.5".

Thoughts?

What is this sorcery….

Seems to me like a geometric distribution with parameter t, where t is the probability of getting tails. EX of a geometric distribution is q/p, in context 1-t/t=h/t

Wat did I just witness

The problem is much easier if you think of the problem recursively. E = h*1 + h*E, since if you get heads, the game continues, with the same payoffs. You can then solve for E.

if you can keep uploading videos about Oxbridge Entries, that will be wonderful:)

You don't need to do any of this to solve this problem. If landing heads adds £1 then landing tails adds £0. If you have an unfair coin then it always lands on heads or always lands on tails.

Case I: Tails

(0+0+0+…+0)/n = (n*0)/n = 0

Case II: Heads

(1+1+1+…+1)/n =(n*1)/n = 1

On average you either make £1 or £0.

Beautiful mathematical handwriting! The approach seems a bit rigmarolish though: just spot that E = h(1+E) and the answer drops out immediately. I think the assumption that E exists is reasonable.

Following your approach, the sequence S = sum(j>=1) j*x^j is quite common. The calculus approach is fun, alternatively rewrite S as:

sum(j>=1) sum(k>=j) x^k =

sum(j>=1) x^j sum(k>=0) x^k =

sum(j>=1) x^j * 1/(1-x) =

1/(1-x) * sum(j>=1) x^j =

1/(1-x) * x/(1-x) =

x/(1-x)^2

The question never says that you can stop after getting a heads, it implies that you have to continue.

Therefore, the average amount of money is £0.

Unless we account for the fact that the coin could be unfair in that it has a 100% chance of landing on heads, meaning that the average would be infinite in that case.

If we use both of these ideas, then the answer would be half of infinity, in other words, still infinity ;P

Watched it, now fuck off my recommended.

That's only finding E for a a geometric Law of parameters h, and t as h=1-t

Let h be the probability of getting a heads

Then, (1-h) is the probability of getting a tails

Expected money

= Sum of all possible amounts of money * Probability for gaining each amount

= ∑_(x=0)^∞ = h^x *(1-h) * x

Plug in a few values for x:

x = 0 –> h^0 * (1-h) * 0 = 0

x = 1 –> h^1 * (1-h) * 1 = h(1-h) = h – h^2

x = 2 –> h^2 * (1-h) * 2 = 2h^2(1-h) = 2h^2 – 2h^3

x = 3 –> h^3 * (1-h) * 3 = 3h^3(1-h) = 3h^3 – 3h^4

x = 4 –> H^4 * (1-h) * 4 = 4h^4(1-h) = 4h^4 – 4h^5

Let's see if there is a pattern when summing them:

0 + (h – h^2) + (2h^2 – 2h^3) + (3h^3 – 3h^4) + (4h^4 – 4h^5)

= h + h^2 + h^3 + h^4 + 4h^5

Here it becomes clear that:

∑_(x=0)^∞ = h^x *(1-h)*x

= h + h^2 + h^3 + h^4 + h^5…

We have reduced the problem to a

simple geometric sequenceh + h^2 + h^3 + h^4… –> 1st Term: h || Common Ratio: h

= h/(1-h)

= h/t

Answer:

h/tYou could let X be the number of trials till the first tails then X has a geometric distribution with p = t. Let Y = X – 1 which is your winnings and now

E(Y) = E(X – 1) = E(X) – 1 = 1/t – 1

= (1 – t)/t = h/t

Can't this just be solved by equation for geometric order S=a1/1-q where a1 is 0.5 pounds, which is average win from first throw, and q also 0.5 or 1/2?

While I get this higher level math, it is so pointless. We have no data to back any of this up. Ive always hated this sort of math, I want practical use.

If someone doesn't understand sigma notation they probably won't understand the rest of the video XD….

Is it weird that im 15 and i kinda get this stuff

it is silly to sum up to infinity, are you going to sit there and throw until infinity comes? will you continue to throw if you die from an old age? 🙂

EVEN DELHI SCHOOL OF ECONOMICS PAPER HAS A HIGH STANDARD THAN OF CAMBRIDGE

Is this suppose to be hard lmfao….

Left handed ???? WITCH !!! Or a sorcerer, whatever , BURN HIM!!!

quick maths

If this is oxford…common lets run away

I learned this in my last year of high school lol.

Idk what his doing but I’m still watching

That coin isn't very fair

To go into more detail, you would use the fact that the geometric series converges uniformly with respect to h on any closed sub interval of (-1,1), then term by term differentiation is equivalent to differentiating the closed form expression for the sum.

A lot of socially awkward, autistic men whom have been dead for centuries gave us this witchcraft.

And we get like 3 minutes to solve such questions.

Wow I actually figured this out and I’m in Algebra 2

The title should read "Oxbridge Statistics Interview".

This doesn't seem too hard. I regret not applying to oxbridge. I feel like I let myself down.

If you would get NOTHING, on the throw of the TAILS, and you ALWAYS go again on a heads, wouldn't the outcome necessarily be that you would get nothing?

If the first 100 throws were HEADS, then you've got a rack of money, but the TAILS that would eventually come would erase your winnings, right?

It doesn't say that you GET to choose to go again.

I guess you are doing it the hard way. In fact, the solution is indeed trivial. The probability to win additional money in the n-th throw is the probability to get to the nth throw times the probability of the heads, so it is h^n. So E = sum of h^n for all natural n.

I know the solution to this problem will this mean i am worthy of getting admitted into oxford??

What a load of Babylonian satanic nonsense. Go to hell in Jesus' name!

Why would you use differentiation if you could just do that:

e = ht + 2h^2t + 3h^3t + …

let's multiply both sides by H and write both equations one above another.

e = ht +2h^2t +3h^3t +…

eh = h^2t + 2h^3t +…

substract second equation from the first

e – eh = ht + h^2t +h^3t +h^4t +…

On the right side we can see infinite geometric sequence with first term ht and r = h, since h + t = 1, then Abs(h) < 1, now we can find the sum

e – eh = ht/(1-h)

e(1-h) = ht/(1-h)

since h + t = 1, => 1 – h = t

et=ht/t

et=h

e = h/t

Don't use differentiation if you can deal without it

Nice video!

In India this type of probability questions appear in class 12 math question paper.

Anyone else's brain hurt after that? Also how is that going to help in life? Any answers to the latter question most appreciated

It seems profoundly counter intuitive that if its a fair coin then the expected amount is 1. Surely if it’s a 50% chance to win any of £1,2,3… then it must be greater than £1. Even with assuming that after you get 2 heads you automatically lose, it’s a 50% chance you get 0 a 25% chance of getting 1 and a 25% chance of getting £2 now if you multiply that together then it’s an average of £1.50 per game…

Provide a much shorter answer. Suppose the expectation is E.

Just think about the first throw, if head, then get 1 pound and the further expected return is still E, if tail, 0.

Certainly, p(1+E)*0 + (1-p)*0=E.

Solving this equation, E = p/(1-p)

Would have got it but I'd never have guessed how to graph it..

after 3:10 isn't it easier if you change t to 1-h you end up with

h + h^2 + h^3 + h^4 +…

which is a geometric series first term h common ratio h, so sum to infinity is

h/(1-h)?

This is fifth grade math in China

Dude! U write so small

Great video! Great explanation! Thank you!

We didn’t do any of this in high school 😅😅

I’m studying maths at Glasgow Uni now and absolutely love it, great video ! 🙂

I nutted man…

It scares me that I understand this 😅

E = h (1+E) => E = h / (1-h) = h / t

Why not just use h and t=(1-h) at 3:35 ? Algebra simplification and geometric sum would get you straight to 6:38 without any need for differentiation.

Since there is a non zero chance that the probability of getting heads is 100% and the time you get it, you will get infinite money, isn't the average money infinite??

Something neater..

Let E denote the average sum.See that if your first toss is a head, the average amount of money you would make is 1+E.If its tails, then 0.

So E=h(1+E)+0.t implies E=h/(1-h) implies E=h/t

My mind just went straight to the binomial expansion of (1-h)^-2, with the modification of multiplying but the factor of ht