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045Simplified poker with three cards, A, K and Q: each player antes 1, you are dealt one card and I am dealt another. You may bet 1 or check; if you bet, I call or fold, and if you check the higher card wins the antes. How often should you bluff with the Q, and how often should I call with the K, in equilibrium?Games and strategic reasoningHardOld Mission CapitalNew York · 2022
Try it first
How often should I call a bet when I hold the K?
Show the worked solution
Bluff with the Q one time in three, and call with the K one time in three. You always bet the A and always check the K; I always call with the A and fold the Q. A Q bluff risks 1 more to win the 2 antes, so I must defend two thirds of hands facing it: the A gives half, the K calling one time in three gives the rest. The game is worth 1/18 per hand to you.
Which hands are easy, and where is the real decision?
Clear away the obvious hands first. With the A you always bet, because you win whether I call or fold. With the K, betting only gets called by the A and folds out the Q, which you beat anyway, so you check. As the caller, I always call with the A and always fold the Q. The whole game comes down to two mixed choices: how often you bluff with the Q, and how often I call with the K.
In equilibrium the bettor always bets the A, always checks the K and bluffs the Q one time in three, while the caller always calls the A, folds the Q and calls with the K one time in three; together the A and the K calls defend two thirds of hands facing a bluff.
How do the indifference conditions fix both frequencies?
A teacher who spot-checks homework faces the same logic: check every paper and time is wasted, never check and everyone copies, check at the right rate and copying stops paying. In equilibrium each player mixes at the rate that makes the other indifferent between their two options. Your Q loses 1 by checking. Bluffing loses 2 against my A, and against my K loses 2 if I call and wins 1 if I fold. The two are equal only when I call with the K one time in three. My K loses 1 by folding; calling loses 2 against your A and wins 2 against a bluff, which is worth -1 only when you bluff one time in three.
The relationship
$\underbrace{\tfrac12(-2) + \tfrac12\big(c(-2) + (1-c)(1)\big)}_{\text{Q bluffs}} = -1 \Rightarrow c = \tfrac13 \qquad \underbrace{\frac{-2 + 2b}{1 + b}}_{\text{K calls}} = -1 \Rightarrow b = \tfrac13$
c how often the caller calls with the K
b how often the bettor bluffs with the Q
-1 the payoff of the alternative: checking the Q or folding the K, losing the ante
What it says in wordsEach frequency is set so the opponent's two choices are worth the same.
Check against the pot-odds rule. A bluff risks 1 extra to win the 2 antes, so the caller must defend 2/(2 + 1) = 2/3 of the hands facing it; the A already covers half, and the K calling one time in three covers the other sixth. That two thirds is the number people misremember as the K's calling rate. Put the strategies together and the bettor, who acts first with more information about their own hand, earns 1/18 of a chip per hand. The limitation: with a bigger bet or more cards the ratios change, but the method, indifference on both sides, carries over.
Where candidates lose it
The common loss is setting the K's calling rate to two thirds. Two thirds is the total defence the caller needs against a bluff, and the A already provides half of it, so the K calls only one time in three.
The second is never bluffing because the Q cannot win a showdown. A player who never bluffs lets the caller fold every K to a bet, and the A's bets stop earning. The bluff is what gets the A paid.
What the interviewer asks next
What is the value of the game to each player?
The bet size doubles to 2. How do the bluffing and calling frequencies change?
Now the caller may also bet after a check. What changes?
Asked at Old Mission Capital, Quantitative Research, New York, 2022 (Wall Street Oasis): Asking to find the game theory optimal strategy in a simplified poker game
081A six-chamber revolver holds two bullets in adjacent chambers. The cylinder is spun, the interviewer pulls the trigger on himself and it clicks on an empty chamber. It is now your turn. Do you want him to spin the cylinder again first, or pull straight away?Games and strategic reasoningCoreSchonfeldCentral · 2022
Try it first
Which gives you the better chance of surviving?
Show the worked solution
Do not spin: pulling straight away survives with probability 3/4, against 2/3 with a spin. The click tells you the hammer sat on one of the four empty chambers. Because the two bullets are adjacent, the four empties form a run, and only the last empty in that run is followed by a bullet. A spin throws that information away and resets you to 4 empties out of 6.
What does the click actually tell you?
Think of a row of six houses where two neighbours keep dogs. You knocked at a random house and no dog barked. If you now try the next house along, you are only in trouble if you had knocked on the one house sitting just before the dogs. The click narrows the hammer's position to the four empty chambers, and the question becomes how many of those four have a bullet immediately after them. With the bullets side by side, the empties run 3, 4, 5, 6 in firing order, and only chamber 6 hands over to a bullet.
With two adjacent bullets, only one of the four empty chambers is followed by a bullet, so pulling straight away after a click survives 75% of the time, while a fresh spin survives only 66.7%, four empties out of six.
The relationship
$P(\text{survive}\mid\text{click, no spin}) = \frac{\#\{\text{empties followed by an empty}\}}{\#\{\text{empties}\}} = \frac34 \;>\; P(\text{survive}\mid\text{spin}) = \frac46 = \frac23$
empties followed by an empty chambers 3, 4 and 5 in firing order
4/6 the survival chance of a fresh, random chamber
What it says in wordsWithout a spin you are conditioning on the click, which helps; a spin forgets it.
Does the answer depend on the bullets being adjacent?
Completely, and that is the follow-up most interviewers ask. If the two bullets are not next to each other, the empties split into two runs, two empties now sit in front of a bullet, and pulling straight away survives only 1/2, worse than the 2/3 of a spin. The same count works for bullets one apart or directly opposite: in both cases two of the four empties are followed by a bullet. So the rule is not spin or do not spin; it is count the empties that border a bullet and compare with a fresh spin.
What is the general lesson for a trading interview?
A random reset destroys information, and information has value only if the structure of the problem lets you use it. Here the structure is clustering: the bullets sit together, so a safe chamber is likely followed by another safe one. Markets have the same feature in volatility: a calm day tends to be followed by a calm day, so conditioning on what just happened beats assuming each day is a fresh draw. Say that link in one line after the arithmetic.
Where candidates lose it
The common error is to treat both options as a fresh draw and say it makes no difference, or to say a spin is safer because it resets the odds. Both ignore the click, which is the one piece of information you were given.
The second loss is getting 3/4 without seeing that it hinges on adjacency. Say that with the bullets apart the answer flips to spin, and give the count, two bordering empties out of four.
What the interviewer asks next
The two bullets are placed in random chambers, not necessarily adjacent. Spin or not?
Three adjacent bullets and a click. Spin or not?
After two clicks in a row without spins, what is your survival chance on the third pull?
Asked at Schonfeld, Quantitative Research, Central, 2022 (Wall Street Oasis): Coding, requires to know DP and divde and conquer., Russian Roulette
093Three dice: red has faces 2, 6 and 7; green has 1, 5 and 12; blue has 3, 4 and 8, each face appearing twice. You and I each pick a die and roll once, and the higher number wins. Which die do you want, and does it matter who picks first?Games and strategic reasoningCoreBelvedere TradingChicago · 2022
Try it first
Which die is best against the other two?
Show the worked solution
No die is best: red beats green, green beats blue and blue beats red, each with probability 5/9. So who picks first matters a great deal. Let me choose, then take the die that beats mine and win 5/9 of the time. If you are forced to pick first, every choice loses 5/9 of the time against an opponent who knows the cycle.
How do you work out who beats whom?
Each matchup has only nine equally likely pairs of faces, so write the 3 by 3 grid and count. Red against green: red's 2 beats only the 1, while its 6 and 7 each beat the 1 and the 5, for 1 + 2 + 2 = 5 wins out of 9. Do the same for the other two pairs and every matchup comes out 5 to 4: red over green, green over blue, blue over red. It is rock, paper, scissors built out of dice, and in rock, paper, scissors nobody asks which hand shape is best.
Counting the nine face pairs in each matchup shows red beats green, green beats blue and blue beats red, each in 5 of 9 cases, so the three dice form a cycle and the second player can always pick a die that wins 5/9 of the time.
The relationship
$P(R>G) = \tfrac{1+2+2}{9},\quad P(G>B) = \tfrac{0+2+3}{9},\quad P(B>R) = \tfrac{1+1+3}{9}, \text{ each } = \tfrac59$
P(R > G) the chance red's roll beats green's
1 + 2 + 2 the wins for red's faces 2, 6 and 7 in turn
What it says in wordsCount, face by face, how many of the opponent's three faces each face beats, and divide by nine.
Why does green lose to red when green has the higher average?
The averages are red 5, green 6 and blue 5. Winning is about how often, not by how much. Green's 12 wins every time it shows, but it shows only a third of the time, and green's other two faces, 1 and 5, lose to both of red's high faces. A higher mean and a higher chance of winning are different things, and the gap between them is the whole puzzle. Change the rules so the winner collects the difference between the two numbers, and green's expected margin against red is 6 - 5 = +1: now you want green against red, and blue against red is a dead heat at 0.
Where does a trader meet the same thing?
Head-to-head comparisons need not line up into a ranking. Strategy A can beat strategy B on more days than not, B can beat C, and C can beat A, whenever one of them earns its money in rare large wins, as green does. Before choosing between strategies, decide whether you care about how often you win or how much you make, because under the first a cycle like this one means there may be no best choice at all. The limitation of the puzzle is that it is one roll; over many rolls with the total score counted, the mean matters more and green's 12 starts to pay.
Where candidates lose it
The common slip is choosing green because its average, 6, is highest. The game pays for winning, not for margin, and green loses to red five times in nine.
The second loss is answering the first question and missing the second. Because the dice form a cycle, the real answer is strategic: insist that your opponent picks first. Saying that unprompted is what the interviewer is listening for.
What the interviewer asks next
If each player rolls their die twice and adds the results, does the cycle still hold?
Design three dice whose faces sum to the same total and still form a cycle.
With three players each taking one die, can any die be favoured against both others?
Asked at Belvedere Trading, Prop Trading, Chicago, 2022 (Wall Street Oasis): You have 3 dice: red has 2, 6, 7; green has 1, 5, 12; blue has 3, 4, 8.

c	how often the caller calls with the K
b	how often the bettor bluffs with the Q
-1	the payoff of the alternative: checking the Q or folding the K, losing the ante

empties followed by an empty	chambers 3, 4 and 5 in firing order
4/6	the survival chance of a fresh, random chamber

P(R > G)	the chance red's roll beats green's
1 + 2 + 2	the wins for red's faces 2, 6 and 7 in turn

Try it first

Which hands are easy, and where is the real decision?

How do the indifference conditions fix both frequencies?

Where candidates lose it

What the interviewer asks next

Try it first

What does the click actually tell you?

Does the answer depend on the bullets being adjacent?

What is the general lesson for a trading interview?

Where candidates lose it

What the interviewer asks next

Try it first

How do you work out who beats whom?

Why does green lose to red when green has the higher average?

Where does a trader meet the same thing?

Where candidates lose it

What the interviewer asks next