Had this in an interview today and I think I boggled it up pretty bad, was at the end of a 1.5 hr interview that consisted of a 45 min options problem before this one, and basically nerves got the best of me.

This was on the phone so a few of these assumptions were unfortunately not articulated very well (by him in my opinion).

The guy also called me 2 hrs later than our call was supposed to be so maybe I'm just more frayed than normal and this is stupid ezpz, but over the phone this really stressed me out; was hoping to hear what you guys thought might be an efficient way to solve this.

Question is as follows:

This is a gambling game.

There is a jar with 100 marbles, 50 red and 50 blue.

One marble is taken from the jar at a time, without replacement. You can see what color that marble is.

At any point you can bet 1$, if it hits red you win you win 9$ plus your 1$ back. If it hits blue you lose and are no longer allowed to wager because you're broke.

((OP Note: The interviewer said 'if you win', and didn't specify that you could not choose red or blue, so I made a derived decision tree based on this and I agonized pretty embarrassingly about this for about 15 minutes before I corrected myself))

However, you can also choose to not bet and not pay your 1$. If you do not bet, and the final marble drawn is red, you win 10$, if it is blue, you win 0$.

How do you play the game? (Assuming your intention is to maximize profit).

I'll give a spoiler with half the answer eventually (I got half before he told me "In an effort to save time, I'm going to end it there", you can tell the interview went pretty miserably for me :/). But I'm 99% sure after the "We're going to have a intra-firm meeting. We will be in touch if there are next steps for you by the end of this week." I was hosed.

He didn't give the other half, but I was curious, for closure's sake.

Thanks guys!

Umm.. Just a guess, but if you win 9 dollars if you bet, but win 10 dollars if you get to the final marble, my guess would be that you have to have odds overcome the difference between the profit of 9 dollars vs 10 dollars, so your odds would have to be a better expected value to win 9 dollars than 50/50 to win 10.

Or am I completely wrong?

I'm not sure I'm interpreting the question correctly, but I think the answer is that it doesn't matter what you do. Your expected earnings is $5. This is because every strategy is equivalent to picking the last marble, which has a 50% chance of being red a priori.

But what if 10 red and 10 blues are taken out, (down to 80 marbles), then 5 blues get taken out in a row? The percentages have now changed, and when you pick you only get paid 9, but if you don't play, you have a chance to win 10.

Like you said, I'm not sure if I'm interpreting this correctly (the post is a bit hard to follow in all honesty)

Nice problem. The best strategy is to never bet. In that case your EV is $5. All other strategies have an EV less than $5.

You might think you can do better if you bet when the percentage of red marbles is greater than 60% since when that condition arises your EV would be greater than $5. However, your EV would be even better by not betting because the last marble has the same probability of being red as the next marble, but you get paid more by not betting. As an extreme example, suppose all of the remaining marbles were red. You still shouldn't bet because you will only make $9 instead of $10 by waiting until the end.

They aren't equivalent. If I bet on the first draw, my EV is $4. Your expected earnings are $5 only if your strategy is to never bet. Any other strategy has an EV less than $5.

Note that if they only paid you $9 when the last marble is red and you never bet, then you could do equally well by betting when all of the remaining marbles are red, but betting any other time would do worse. If they paid you an amount less than $8.98, there there could be a time when it becomes better to bet.

Sigh. Well done Bruce. I thought I had it with my logic. I wonder if they'd think I was an idiot with my answer.

I would be surprised if anyone got that right under the pressure of a phone interview. If they did, they should probably be hired on the spot.

I'm having an issue with the EV though. If the blue to red ratio is above .57, it'll create a 5.13 EV (9 dollars multiplied by .57 probability). That's obviously higher than the 10*.5=5 EV... so why did i miss your logic about not betting when almost all red (say, 9*.9=8.1 EV)?

I dunno.. I'm missing something obvious here.

If your EV is 5.13, you don't compare that to 5, you compare it to the EV that you would get if you wait until the end. The last marble has the same probability of being red as the next marble, as do all the remaining marbles. If you had to place a bet, the last marble would have the same EV of 5.13 as the next marble. But the last one pays better when you don't bet, so it has a higher EV.

It's like in blackjack when the count goes high. On average the count will be the same on the next hand that it is on the last hand of the shoe.

Ok, so if there were 10 marbles left, 9 red 1 blue, its .9. So .9 * $9 = 8.1, much higher than 5, which is .5 * $10, but we shouldn't be comparing 8.1 to 5, it should be 8.1 compared to .9 * $10, which is 9.

Got it. I'm dumb.

If people want some more, here's what my friend sent me from a hedge fund interview he just had. He mentioned his boss made one girl cry in an interview because she couldn't answer some questions (not these, just in general). These aren't too difficult, but kinda fun if you aren't in a pressure situation. It's always easier to answer these sitting at home and you have time to think about it.

"Dude. It was like 5 rounds. They made me compute ytd returns and CAgR in my head on the spot. I had a take home case study that I had to present to like 5 ppl over the course of three hours

there were a couple. one question was, if you have 10% gains for each month the first 6 months of the year and then 10% losses each month for the second half of the year, are you net up or down for the year? What is the YTD return? What if you have that same scenario and someone subscribes in an equal amount of capital after the first 6 months that you started with in the beginning of month 1? Are you up or down for the year? what is the YTD return?

if you start with 20 dollars in capital and 5 years later come out with 70 dollars of capital, what is the CAGR (compound annual growth rate)?"

0.9 * 9 - 0.1 * 1 < 0.9 * 10 - 0.1 * 0

I wrote this out and saw I would never decide to bet.

No reds: 0 * 9 - 1 * 1 < 0 * 10 - 1 * 0

Also, the EV of betting isn't 5.13 in your example where the probability of red is .57. It would be

.57*9 + (1-.57)*(-1) = $4.70

You need the probability of red to be greater than 60% for the EV to become greater than 5. But that doesn't matter because the probability that the last marble is red would be the same.

Don't you get to keep betting if you win? This seems to imply that you do.

I don't think that's what the question meant. It was more like an all or nothing scenario, but what is the better choice as far as probabilities go.

This is the type of question I'd be somewhat scared of in an interview. It's not all that difficult with time, but with someone staring at you or you are on the spot, it tends to be a lot harder. I've had simple questions asked of me and sometimes I freeze up for a second with 5 thoughts in my head at the same time "What does he really mean here? Is this a trick question? What answer will impress him the most? How'd she look naked?" etc. I usually get through it, but I'd be lying if I said I never bombed an interview question.

The confusing part to me was that I didn't recognize that if you wait until the end you EARN $10, whereas if you bet before your expectation is to EARN can only be a maximum of $9 (b/c you have to pay the $1 to play). When I first read the problem it sounded to me like the maximum possible was only $9 earned regardless of the choice.

And this was the other thing I thought was possible when I read the problem. I assumed if you won a bet, you could keep playing until all of the marbles were gone, with the last marble always being a freeroll. If this was the case, there would clearly be a better strategy if the marbles fell in your favor at some point (if they never fall in your favor, then just wait until the end).

The very thing you quoted implies that you only get to bet once. Otherwise, you wouldn't necessarily be broke.

"However, you can also choose to not bet and not pay your 1$. If you do not bet, and the final marble drawn is red, you win 10$, if it is blue, you win 0$."

I have $1. If I bet and lose I now have $0 and can no longer bet because I am broke. If I bet and win I now have $10 and am not broke. Why can't I bet with that money? This is how investing would work in reality.

That's not how the marble futures market works. You can only make 1 investment per 100 draws. This is due to new SEC regulations.

this is a correct way of thinking about it. try and think of it in terms of expected profit (which is a function of expected revenue - expected cost). the part after is slightly incorrect. hint: it involves the one dollar.

this is incorrect. but i agree the question is very suspectly worded. this was how it was read to me verbatim however so i figured it might be fun for you to try and solve it. i will answer any questions i can just as he did answer my questions.

this is not the answer though.

think about extreme (but simple) cases, where for example there were 2 red marbles, 1 blue marble left.

you are correct that when you place a bet you can only win 9 plus your 1 back. aside from that i'm not sure i understand the question.

the answer is there are scenarios in which all blues are taken, or all reds are taken. but these can be accounted for with probability distributions. good work, this is how i envisioned it initially (what happens if 10 and 10 are taken?)

think of it as the sample size shrinks even further and your edge approaches would would be an "ideal" scenario.

in these sorts of logic problems i have noticed that its usually best to think about the extreme cases one way or another as the solutions.

this is the incorrect value, but i like your line of reasoning. expound on it!

this is in essence the solution to half of the problem which i found (along with the EV of bet/no-bet). the other half is calculating the expected profit from either solution (bet or no bet).

he wouldn't tell me what the extension of the solution was but i can tell you that the EV of profit in the scenario of not betting i got (in for example, 2red, 1 blue remaining) was $6.67(no bet) and $5.67 (bet). to do this i just made a probability distribution tree and calculated the possible scenarios:

bet no bet
.66(r)*9, .33(b)*-1 .66(r) [.5red, .5blue] .33(b) [red 1.0]

my theory is basically in line with bruce, in that the nearest you can approach the guaranteed 1$ would only equal it. (for example if all blues were taken).

even in this extreme scenario you would have an expected profit of (if betting) 9$, vs not betting (10$).

in addition, due to the random sampling, i theorize that it would hypothetically never be optimal to bet vs wait because you would expect if you ran this game millions of times for the distribution of the drawn marbles to approach .5r/.5b. in that scenario it is undoubtedly profitable to not bet.

in addition: to those who asked about whether you can bet once. bruce is correct, you can only bet once. i struggled with this (thinking it would be an investment, like soebody had said, where we could maximize a shot once and then continue to bet if that edge remained. however, it is as bruce called it regulated such that you can ONLY BET ONCE. this was again something i had to ask.

there was another qeustion fromt he interview concerning options that was fun that i might post here (right before this question)

thanks guys! let me know if you have any other different answers or see holes in my logic.

It's not incorrect. The EV for the game when you never bet is $5. That's trivial because the last marble has probability 1/2 of being red.

There will be times after some marbles are drawn that your EV for waiting will be greater than $5. It can be as much as $10 when only reds are left. But the EV of waiting will always be greater than the EV of betting because whatever the probability is that the next marble is red, the probability that the last marble is red is the same, and we don't have to bet $1 on the last marble. That's the key point, that the probability for the last marble and the next marble are the same. There are no calculations necessary to reach this conclusion. I've already given the complete argument.


No, the 60% solution is completely irrelevant.


Of course it's $6.67. That's because the probability that the last one is red is 2/3. In the original game, it's 1/2, so the EV is $5. You don't need a decision tree. You're missing the key.