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An unusual quant qn from McKinsey

Odeh

Can someone please offer an explanation and solution to this problem?

The question has been adapted for this forum. It originally comes from a McKinsey case.

Below is a description of expected probability of success, by stage, in a biotech firm's R&D pipeline.

The R&D of a drug has several phases (1-4). As a drug goes through each stage, there is a probability of it suceeding:-

Phase 1 ---(70%)---> Phase 2 ---(40%)---> Phase 3 ---(50%)---> Phase 4 ---(90%)---> successful marketing & sales

This firm believes that the likelihood of success of its primary drug candidate can be improved by investing an additional $150 million in a larger Phase 2 trial. The hope is that this investment would raise the success rate in Phase 2, meaning that more candidate drugs successfully make it to Phase 3 and beyond. By how much would the Phase 2 success rate need to increase in order for this investment to break even?

Assume that if the drug is successfully marketed and sold, it would be worth $1.2 billion (that is, the present value of all future profits from selling the drug is $1.2 billion).

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Odeh replied on 09/18/2017

This is the solution offered by McKinsey. (The link for the whole case is http://www.mckinsey.com/careers/interviewing/globapharm)

Investment would need to increase the probability of success in Phase II from 40 to 80 percent (that is, increase of 40 percentage points). There are multiple ways to approach this calculation. One method is shown here:

  • If a candidate drug passes Phase II, then it has a 50% x 90% = 45% chance of being successfully marketed and sold. Since a successful candidate drug is worth $1.2 billion, a candidate drug that passes Phase II is worth 45% x $1.2 billion = $540 million.
  • To break even (that is, to make the $150 million investment worthwhile), the value of the candidate drug that passes Phase II would need to increase to $540 million + $150 million = $690 million. This means, the probability of combined success in Phase I and II would need to increase by (150/540) = 28 percentage points.
  • So the current probability of Phase I and II, that is, 70% x 40% = 28% would have to increase by 28 percentage points, to 56%. In order to come up to 56%, Phase II probability would have to increase from 40% to 80% (70% x 80% = 56%).
  • This seems like a very big challenge, as an increase by 40 percentage points means that the current probability of 40% needs to double.
Francesco replied on 09/20/2017
Ex BCG | MBB Specialist | #1 Expert for meetings done (1000+) and recommendation rate (100%)

Hi Odeh,

As mentioned, there could be multiple ways to face this problem. A simple way would be to compare the outcome of two scenarios, one with the investment and one without and put them equal:

Step 1: Calculate the expected outcome without investment

In this case your expected outcome would be 1200M times 12.6% (where 12.6%=70%*40%*50%*90%); thus

V1=1200*0.126=151.2

Step 2: Calculate the expected outcome with investment

In this case your expected outcome would be 1200M times x (where x=70%*y*50%*90%, in which y is the new outcome of Phase 2) minus the investment required of 150M; thus

V2=1200*x-150

Step 3: Put the results equal

Now putting the two outcomes equal you get

V1=V2 <->

<->151.2=1200*x-150<->

<->301.2=1200*x<->

<->x=25.1%

Now solving for y:

70%*y*50%*90%=25.1%<->

<->y=79,68%

that is, almost twice the current amount.

Hope this helps,

Francesco

Florian replied on 09/18/2017

Hi,

the case seems like a relatively straightforward case of expected returns.

You know that the outlay of R&D cost $150m is real money. Therefore that money has the same value, as the payoff you get from drugs that make it through all phases (because that is also real without any discount for expectation)

You need to go backwards from the expected payoffs, and estimate the value of all drugs in trial in the various phases (of which you know some are not going to make it.)

In Market: 1.2b
In Phase 4: 1.2b / 90% = 1.33b
In Phase 3: 1.33b/50% = 2.66b
In Phase 2: 2.66b/40% = 6.66b
In Phase 1: 6.66b/70%= 9.5b

Now, the value of the drugs in the first phase does not change. However, we need to break even, meaning we need to make back the $150 investment in our 'in Market' phase. Therefore, the new value for In Market is: 1.35b.

Applying the same backwards logic results in:

In Market: 1.35b
In Phase 4: 1.35b/90%= 1.5b
In Phase 3: 1.5b/50%= 3b
In Phase 2: 3b/40% = 7.5b
In Phase 1: 7.5b/70% = 10.7b

As I mentioned above, the value of 10.7b is irelevant, as this would imply that we give more drugs into the development funnel. However, the question is to broaden the funnel, i.e. how do we get with the potential market of 9.5b (if all drugs tested became a success through all phases) to 1.35b (what we need to recoup investment). We are given the fact that the improvement happens in Phase 2, so we solve for the formula:

9.5b*x = 7.5b, x=0.79

Therefore, the answer shold be that the success rate in phase 2 would need to increase from 40% to 79%, an increase of almost 100%, to recoup the $150m investment.

On a sidenote, maybe worth mentioning that the drugs that make it through all phases have an NPV of 1.2b. That means it can take quite a long time to recoup the money (at 5% interest rate, it is about 20 years worth of returns, of around 60m each year. It depends whether the company has the same risk appetite for return on their investment cost, i.e. is the NPV value of the last stage a valid measure of the return. I assume it is in this case scenario, but in real life it may differ).

Cheers,

Simon replied on 10/02/2017

The sucess rate of Phase II need to be 0.800 to break even the extra investment. Here is my thinking process using the concept of expected value (EV).

The concept of EV can be used to calculate the net expected earning of a particular gambling game, given the investment (INV), profit, and the success rate (SR):

  • EV= -INV + SR x Profit

,where SR=SR1 x SR2 x SR3 x SR4 accounts for the four phases in this case. Straightfowrdly, one would have no change in the expected value (dEV=0) if,

  • dINV=dSR x Profit

,where "d" denotes the change "delta". Here we know the profit is 1.200B and the change of investment is dINV=$0.150B, which gives dSR=0.125. With the unchanged sucess rate in phase I, III, IV, the change of sucess rate in phase II dSR2 need to be:

  • dSR2= dSR / (SR1 x SR3 x SR4) = 0.125 / (0.700 x 0.500 x 0.900) = 0.400

to offset the extra investment. Therefore, the sucess rate of Phase II need to be 0.800=SR2+dSR2=0.400+0.400.

(edited)

Anonymous A replied on 09/18/2017

The logic behind is a decision tree where on first stage you have probablity to move further of 70% an get 0 (or lose investment) of 30%. Then again 40% to move and 60% to have 0 -> 50% to move 50% to have 0 -> 90% to win $1200 and 10% to have 0. THen if you put $150 as investmetn you start to have -150 instead of 0 on each end. By multiplying % on win/lose of the branch and sum up results you go backward from 1200 to the beginning: 1200*0,9+(-150)*0,1=1065 -> 1065*0,5+(-150)*0,5=457,5 -> 457,5*0,4+(-150)*0,06 and so on...

Hope it's clear enough..

(edited)

Anonymous A replied on 09/18/2017

Looks like there should be some information on initial R&D invetments for the product. Because if you assume all initial investments as 0, it's alreade profitable to try.

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