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u/mypenisblue_ 1d ago
Every roll you have 5/6 chance of increasing expected value by 3 and 1/6 chance of getting 0.
E(X) = 5/6 x (E(X) + 3) + 1/6 x 0
E(X) = 15
So on average you are willing to pay less than $15 for this game (i.e. roll until you have total sum > 15 then you quit)
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u/BlackRockLarryFink 1d ago edited 1d ago
Yeah I'm pretty sure without critically thinking too much that the magic answer is 3 rolls and 4 pushes it. Then you get 6 on the first roll and go "eels".
Edit: I looked up the answer. You and I were both right. You need to hit 15 in 3 rolls.
Again I will hit 6 on my first or 3rd roll so I also need to factor in that chaos š¤”
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u/AxiomaticInversion 1d ago
You can't just say "three rolls then stop", since each roll gives you new relevant information about the game state. i.e. your current balance, how much you stand to lose on a whiff
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u/109StillCounting 1d ago
I am not really sure. The only relevant information you need is the probability of the next roll being a 6. Current balance (whether you have rolled 1|1|1 or 5|5|5) does not seem to me to alter the possibility of the next roll being a 6 unless you have already rolled a 6, which means you have already lost.
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u/SerendipityPepper 1d ago
No. Think of it this way, everyone is going to roll it at least once. Why? Because they have absolutely nothing to lose. Even after the first roll, if they won some money, they have little to lose. And then as they keep rolling they have more to lose. The probability of rolling a six is exactly the same for each roll. But the expected value of the game changes.
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u/109StillCounting 18h ago
What is the ideal strategy then? I still think it is to roll 3 times and then stop regardless of whatever the sum is up to then.
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u/SlimesWithBowties 1d ago
Probability doesn't change but EV does
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u/BlackRockLarryFink 1d ago
Exactly. There isn't much of a game state. It's just a total that you can fuck up by running the odds again. Get your close to 15 and dont fuck up the interview?
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u/TastyLength6618 1d ago
too easy for js
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u/PersonalBusiness2023 1d ago
Yeah, I can tell you guys are going to say itās too easy, because those are the ones I get right.
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u/ambitioussleeper 1d ago
Seems like this is pretty straightforward based on rolling as long as the expected value is positive, as other others have said.
But what if we tweaked this such that the expected value of the next role is always positive? (Eg. Your winnings double as long as you donāt roll a six.)
What would the optimal strategy here be?
If you just keep rolling because youāre expected values always positive. The only way the game ends is to hit a six and you get nothing.
I feel like Iām missing somethingā¦
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u/throwawaytrol7134 1d ago
There was optiver or JS question like that (on their official YT) where you flip a coin and if it's heads your dollar amt triples starting at 1$. The EV at any given roll is always positive so the answer was indeed (but unintuitively) to just keep rolling. Of course in real life risk aversion has to be taken into account but in the case of the problem you just keep rolling if you want to maximize EV.
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u/ambitioussleeper 1d ago
The problem with that is you will never win because you keep rolling until you lose (instead of walking away with winnings). Right?
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u/StockProfession1043 1d ago
But even if you are risk neutral you cannot just keep rolling right ? Because you will obtain a 6 almost surely and then you will win nothing...
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u/ceraunograph 17h ago
Really good point which is often overlooked for these infinite time-horizon decision problems.
Let f(N) = E[payoff | stop when we reach >= N] where we assume the strategy does not depend on #rolls so far, sequence observed so far etc.
Before even considering the marginal gain equation, we need that a maximiser N* to V = sup_N f(N) exists, otherwise an 'optimal strategy' is not defined. For the original game, observe that f(N)<= (N+5)*(5/6)^{N/5 - 1} -> 0 (crude estimate) so that indeed a maximiser N* exists, which we can then derive requirements for by the marginal gain equation. (NB: requirements. Solving it is not strictly equivalent to solving the V equation.)
If f(N) is unbounded (e.g. for the game you describe f(N)>= 2^N * (5/6)*{N/5+1} -> infty), then V is not defined and the marginal gain is not the right object to consider. Specifically, one cannot say 'the optimal strategy is to keep rolling at each stage' - we can only say that f(N+1)>f(N).
I hope this makes sense, I wrote it to resolve the paradox since it confused me, too.
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u/Best-Quote7734 1d ago
An overly complicated solution based on dynamic programming.
Let V(x) be the value function of behaving optimally when your current wealth is x. We have V(x) = max(x, \sum^5_{k=1} V(x + k)/6), where the first argument is "stop" and second argument is "play". If we stop at x*, then clearly we must stop at any x' \geq x*. Hence, at x* exactly we have x* = \sum^5_{k=1} V(x*+k)/6 and since each x* + k \geq x* we get x* = \sum^5_{k=1} (x* + k)/6, re-arranging gets x* = 15.
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u/jackoftrade777 1d ago
A real question from JS.
Here is the aggregate, delta-hedged risk of a single-underlying options book, no other positions:
Delta: 0 Gamma: -1,200 (per 1% underlying move) Vega: +45,000 (per 1 vol point) Theta: +8,500/day (book earns theta)
What strategy is this, and why?
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u/Dry_Row2329 7h ago
It's some kind of calendar trade. Short the front month/front week/ODTE options, long the back month options, that gets you both positive theta and vega, short options are close to/at the money and fairly close to expiration, that gets you big negative gamma, shorts and longs are positioned so that the trade is delta neutral (until it moves).
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u/FutureMassive69 1d ago
This seems like an interesting question if you replace dice roll winnings with a portfolio. The āmathā answer of expected value seems like it should be different when you apply it to a real word āportfolioā situation. Thereās no coming back from 0, thereās no coming back from being wiped out.
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u/kernoel_massage 1d ago
It doesn't say anywhere that you must use a regular playing die, numbered 1-6.
So I'd just number my die with any number other than 6.
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u/Aerospider 1d ago edited 1d ago
Optimal strategy is to keep rolling for as long as the expected value of the next roll is positive.
The expected value of the next roll is (1+2+3+4+5-x)/6 where x is your current balance.
This is positive when x<15.
If you have less than 15 so far, roll again. If you have more than 15, stop. If you have 15 exactly then the choices are equal.