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After this interview, Jane Street hired this Quant analyst at $220,000–$600,000/year In 21 minutes, you’ll learn what a Tier 1 hedge fund interview is like Here are some of the questions they ask: 1. You have $100, fair coin. Heads - you get 2x your bet back, tails -... show more
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35 Kommentare
retarded, bets it all every time, loses on flip 4, went bust. This is a Kelly Criterion problem. no wonder they resorted to fraud if they are hiring these retards
Dumb. 100 flips is virtually guaranteed to get a loss, meaning betting 100% 100 times is virtually guaranteed to get you to bankrupt.
He would be fired after his answers for question 1. His answer was betting 100% of bankroll and hoping to win 100 coin flips in a row lol
Yea this is terrible response lol what if you lose it all on the first flip? I’d bet $2 and double down every time I lose. At 50/50 odds and 2x the upside vs downside, you’ll come out ahead over course of 100 flips
Just bet a cent each flip and buy booze with the rest. Best possible outcome.
You are guaranteed to go broke if you bet 100% of your bankroll on a 50/50 proposition for 100 trials.
Was the first question “heads you double your bet, tails you lose it all?” It sounds like he heard the question as “heads you win double your bet.” So he’s thinking a $100 bet returns $300 (winning $200, which is double your bet). If that’s the case, he’s right.
So he has 0.00000000000000000000007% chance of winning like 10^50 dollars? What an awful answer
Expected value is zero Nigga think he bill hwang
Kelly criteria nerdsnipe
so the odds of not going bust are astronomically small
Bruh lol. This is embarrassing. 25% bankroll each flip from deriving the kelly formula and solving for the variable to maximize expected log wealth per flip.
I say no coin flips, and put the coin in my pocket. Now, I have $100.25. Boom.
Nope their interviews are like 7 rounds of way more complex problems than this, plus this answer is correct for 1 iteration, not 100. For that it’s prob Kelly fraction, [2(.05)-.05]/2 = 25% of bankroll… maximizing EV while minimizing chances of losing it all.
Interesting satire
They are not accounting for trailing draw down, nor risk of ruin. Bad answer in terms of risk management. To be fair, it is a junior/starting position (or so I hope).
What is my win-condition for #1? Is my goal $10,000 or $1 million or $100 million? All in is correct from a pure EV standpoint, but that comes with excessive volatility.
This is just the Kelly Criterion
Today I learned the Kelly formula. You bet on a coin flip: heads pays 3x your bet, tails you lose it. To max long-term growth, we max: 0.5 ln(1+2f) + 0.5 ln(1-f). (log because it reflects compounding) Set the derivative to zero, solve, you get f = 25%. Bet 25% of bankroll
Not watching the whole vid so maybe he corrects himself later on, but pretty sure it's not optimal to bet all your money (Q2/Q3). Even though betting all your money maximizes your EV for that given single play, if you lose, you're losing the opportunity to play more future positive EV rolls.
Fuck yeah boys I was gonna guess , if it’s 50/50 go all in but I would tier down the risk as I increased wins and get initial back depending how many more times I would bet.
Expected value is 1.0. 2x0.5+0x0.5= 1.0 So this is bet converge on neutral. No memory nor dependencies between flips Bet $1 per flip. No adjustment of strategy since there is no dependency.
Deport them both
I would just bet $1 100 times expecting to lose $50, but winning 50 times at 2 x’s original bet plus getting money back So I bet all 100 lose half and get other half back +100 bucks so in total I walk away with $150 which is basically a 50% return (plus or -5% or so)
Any Polymarket trader could do this problem in 5 minutes 😂
I don’t get it If you bet 100% every time, you either triple or you go to 0. But going to 0 is certain. Someone explain please!
Anyone that says to bet 100% should never be hired, ever. U will go bankrupt eventually,, as there will be “runs” of heads or tails going multiple times in a row,,,,too much of a sample size, 100, to avoid the probabilities of a bad run of luck….. bet 10% to bet,
He answered wrong lmao
@Grok can do all these calculations for (what's the X premium cost?) $80 per year? These fucking jobs are finally cooked. Time is running out. Better milk those employers until they figure it out.
Almost like they are teaching you “yes bet everything all the time”
You bet 0. On embauche un quant pas un joueur.
No , bet 100% only if you chance of winning is greater than 50% in this case .
Wrong. Optimal bet for 100 flips is 25%, where you have a win factor of 2.0 and loss of 1.0. Idiots.
You bet 25% of your pot.
Is this a joke? Betting 100% on each flip guarantees a bust over 100 flips. There is a 1 in 1.26~ nonillion chance of coming out of that strategy with any money at all. Optimal bet has to be something like 2/3 but I honestly don’t know enough math to figure the real value
![A billionaire is 10 million times richer than a student and pays 4 times more for the same infinite jackpot: $30.84 versus $7.79. Jane Street tests this exact instinct. In their published mock interview, a candidate is handed a dice game and asked what it is worth. The answer is never the average roll. It is the value of the option to walk away and roll again. That is the same move Daniel Bernoulli made on a coin in 1738. Here is the coin. A fair coin is flipped until it lands heads. The pot pays $2 if heads comes on flip one, $4 on flip two, $8 on flip three, doubling with no cap. Its expected value is not large. It is infinite: EV = (1/2)(2) + (1/4)(4) + (1/8)(8) + ... = 1 + 1 + 1 + ... = infinity By the rule most people carry, price equals expected value, you should hand over your whole net worth to play once. Nobody does. For 250 years that was called a paradox. It was not. The pricing rule was wrong. Value the outcomes by utility, not dollars. Bernoulli's fix was logarithmic utility, the idea that a dollar matters less the more you already hold: U(W) = ln(W) The fair price c is the amount that leaves your expected utility unchanged: sum (1/2^n) · ln(W - c + 2^n) = ln(W) Solve it for real wealth and the infinite EV collapses to a small, exact number: Wealth $100 -> pay $7.79 Wealth $1,000 -> pay $10.95 Wealth $1,000,000 -> pay $20.87 Wealth $1,000,000,000 -> pay $30.84 Your wealth spans 7 orders of magnitude. The price moves by a factor of 4. None of this is a curiosity about coins. It is why position sizing exists. A payoff tail can be astronomically large and still be worth almost nothing to you, because ruin is priced in logs. The same equation, rediscovered by John Kelly in 1956 as maximizing E[ln(W)], is what tells a trader to bet a fraction of the bankroll instead of the whole thing. Every fund that priced its bets at expected value instead of log wealth eventually met the tail it refused to price. Expected value said infinity. The correct answer was $11. Models over opinions.](https://image.24vids.com/tw-2084986258762158152/amplify_video_thumb/2084985737523380224/img/7blq1hM9WeoIt-xH.jpg)
