Interview questions

Jimmy Snyder

Indeed, I was wrong about #10 the russian roulette and about #2, the die game.

MarcoD

I am not sure about the die game, it depends on how you read the question. If you get a return each round, I think the answer is seven, if you are allowed to 'modify' the outcome by throwing the die again, 3.5 should be correct.

BobG

It doesn't say you have to bet the same amount on each game.

BobG

Good point. (Except it isn't necessarily just two flips, since 0 heads isn't counted as a flip.)

MarcoD

I know. Assuming that they won't sneek a question in there which has no simple solution. I tried the following: winning means at least 3 out of 5, which means you need a minimum of $25 to double up three times. I tried some variants, doesn't work.

Jimmy Snyder

This is correct.

This is not correct.

This is correct.

The probability that the game will end after a single toss is .5 and the expected value in that case is 2, i.e. the average of 1, 2, and 3. The probability that the game will end after two tosses is .25 and the expected value is 7, i.e. the average of 4, 5, and 6 for the first toss and of 1, 2, and 3 for the second. The expected value for the game is:

[tex]\frac{2}{2} + \frac{7}{4} + \frac{12}{8} + \frac{17}{16} + ...[/tex]
[tex]= \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + ...[/tex]
[tex]+ \frac{5}{4} + \frac{10}{8} + \frac{15}{16} + ...[/tex]
[tex]= 2 + \frac{5}{4} + \frac{5}{8} + \frac{5}{16} + ...[/tex]
[tex] + \frac{5}{8} + \frac{10}{16} + \frac{15}{32} + ...[/tex]
[tex]= 2 + \frac{5}{2} + \frac{5}{4} + ... = 7[/tex]

AlephZero

I read it as "you want to make a profit of either 200 or 0", i.e. find a hedging strategy so you end up with either 300 or 100.

MarcoD

I am not sure you just gave an argument which is an algebraic rewording/reshuffling of my original argument. (I.e., you can add the expected value/outcomes of the turns or the outcomes of the games.)

MarcoD

Actually, I am sure. You just used a different analysis to end up with a different series to end up with the same outcome.

(It becomes a lot easier if you write down the 'probability tree.' There are more manners of defining a series on it.)

If you write down a tree, you can see you're calculating the limit of:

[tex](\frac{1}{6} + ... + \frac{6}{6}) + \frac{3}{6}((\frac{1}{6} + ... + \frac{6}{6})+ \frac{3}{6}((\frac{1}{6} + ... + \frac{6}{6})+ ... ))[/tex]

which is the fixed point of

[tex]\phi = (\frac{1}{6} + ... + \frac{6}{6}) + \frac{3}{6}(\phi)[/tex]

is

[tex]\phi = \frac{7}{2} + \frac{1}{2}\phi[/tex]

is

[tex]\frac{1}{2}\phi = \frac{7}{2}[/tex]

is


[tex]\phi = 7[/tex]

MarcoD

Actually, for completeness, let's see where you're reasoning -half a chance on 1,2,3, half a chance on 4,5, 6 and repeat- ends up:

[tex](\frac{1}{2}(\frac{1}{3} + ... + \frac{3}{3}) + \frac{1}{2}((\frac{4}{3} + ... + \frac{6}{3})+ (\frac{1}{2}(\frac{1}{3} + ... + \frac{3}{3}) + \frac{1}{2}((\frac{4}{3} + ... + \frac{6}{3})+ ... ))[/tex]

(It's the same series, if you rewrite a bit, but then again.)

is the solution to

[tex]\phi = \frac{1}{2}(\frac{1}{3} + ... + \frac{3}{3}) + \frac{1}{2}((\frac{4}{3} + ... + \frac{6}{3})+ \phi)[/tex]

is

[tex]\phi = 7[/tex]

Beaten to death by now, I guess.

(Actually, the thing is equivalent to 50% chance on a 2, and 50% on a 5 after which you repeat the process.

[tex]\phi = \frac{2}{2} + \frac{5}{2} + \frac{1}{2}\phi[/tex]
[tex]= \frac{2}{2} + \frac{7}{4} + \frac{10}{4} + \frac{1}{4}\phi[/tex]
[tex]= \frac{2}{2} + \frac{7}{4} + \frac{12}{8} + \frac{15}{8} + \frac{1}{8}\phi[/tex]
[tex]= \frac{2}{2} + \frac{7}{4} + \frac{12}{8} + \frac{17}{16} + \frac{20}{16} + \frac{1}{16}\phi[/tex]

Which is the reverse of your argument.

Now completely and utterly beaten to death.)

(A hundred and forty one manners to write the constant 7. Sheesh.)

MarcoD

Impossible, I assume five double or nothing bets can be placed, which means you cannot hold on to a 100 in case you lose all of five bets.

diogenesNY

Numbers 11 and 12 are curveballs.

#11 is _absolutely impossible_ to meaningfully answer with the information provided. There are several undefined variables, and even if one assumes that each and every other particular detail of the options in question are identical, one _might_ infer that the purchase price on the option with the higher strike price was _possibly_ lower, but even then.... well.... no, not really. No way to meaningfully address this question. Not at all.

#12 Another bit of misdirection. Options aren't priced that way. While some analytically guidance is taken from such things as the underlying security's volatility, timeliness, previous trading range, historical volume and changes thereof, and a variety of other factors..... The actual price is arrived at by a negotiation between the underwriter of the options, usually also the seller, and the prospective purchaser. This may happen rapidly.... not necessarily in person even, but it is ultimately a negotiated sale.

If the option is a market traded option, once it is trading, the market (if sufficiently liquid) determines the price.... this is usually some combination of any 'in the money' value of the option (if the option is in fact in the money) plus some 'opportunity vigorish'... however the thing to keep in mind:

This is a negotiated wager between parties presumably unrelated to the company.