Guess the number

I haven’t posted a puzzle in a while.  The following problem has the usual nice characteristics; it works on a cocktail napkin or as a programming problem, via exact solution or simulation, etc.

I am thinking of a randomly selected integer between 1 and m=10 (inclusive).  You are the first of n=3 players who will each, in turn, get a single guess at the selected number.  The player whose guess is closest wins $300, with ties splitting the winnings evenly.

Here is the catch: each player may not guess a number that has already been guessed by a previous player.  As the first player, what should your strategy be?  Which player, if any, has the advantage?  And what happens if we vary m and n?

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Probability of a Scrabble bingo

My wife and I have been playing Scrabble recently.  She is much better at the game than I am, which seems to be the case with most games we play.  But neither of us are experts, so that bingos— playing all 7 tiles from the rack in a single turn, for a 50-point bonus– are rare.  I wondered just how rare they should be… accounting for the fact that I am a novice player?

Let’s focus the problem a bit, and just consider the first turn of the game, when there are no other tiles on the board: what is the probability that 7 randomly drawn Scrabble tiles may be played to form a valid 7-letter word?

There are {100 \choose 7}, or over 16 billion equally likely ways to draw a rack of 7 tiles from the 100 tiles in the North American version of the game.  But since some tiles are duplicated, there are only 3,199,724 distinct possible racks (not necessarily equally likely).  Which of these racks form valid words?

It depends on what we mean by valid.  According to the 2014 Official Tournament and Club Word List (the latest for which an electronic version is accessible), there are 25,257 playable words with 7 letters… but many of those are words that I don’t even know, let alone expect to be able to recognize from a scrambled rack of tiles.  We need a way to reduce this over-long official list of words down to a “novice” list of words– or better yet, rank the entire list from “easiest” to “hardest,” and compute the desired probability as a function of the size of the accepted dictionary.

The Google Books Ngrams data set (English version 20120701) provides a means of doing this.  As we have done before (described here and here), we can map each 7-letter Scrabble word to its frequency of occurrence in the Google Books corpus, the idea being that “easier” words occur more frequently than “harder” words.

The following figure shows the sorted number of occurrences of all 7-letter Scrabble words on a logarithmic scale, with some highlighted examples, ranging from between, the single most frequently occurring 7-letter Scrabble word, to simioid, one of the least frequently occurring words… and this doesn’t even include the 1444 playable words– about 5.7% of the total– that appear nowhere in the entire corpus, such as abaxile and zygoses.

Scrabble 7-letter words ranked by frequency of occurrence in Google Books Ngrams data set.

Scrabble 7-letter words ranked by frequency of occurrence in Google Books Ngrams data set.  The least frequent word shown here that I recognize is “predate.”

Armed with this sorted list of 25,257 words, we can now compute, as a function of n \leq 25257, the probability that a randomly drawn rack of 7 tiles may be played to form one of the n easiest words in the list.  Following is Mathematica code to compute these probabilities.  This would be slightly simpler– and much more efficient– if not for the wrinkle of dealing with blank tiles, which allow multiple different words to be played from the same rack of tiles.

tiles = {" " -> 2, "a" -> 9, "b" -> 2, "c" -> 2, "d" -> 4, "e" -> 12,
   "f" -> 2, "g" -> 3, "h" -> 2, "i" -> 9, "j" -> 1, "k" -> 1, "l" -> 4,
   "m" -> 2, "n" -> 6, "o" -> 8, "p" -> 2, "q" -> 1, "r" -> 6, "s" -> 4,
   "t" -> 6, "u" -> 4, "v" -> 2, "w" -> 2, "x" -> 1, "y" -> 2, "z" -> 1};

{numBlanks, numTiles} = {" " /. tiles, Total[Last /@ tiles]};

racks[w_String] := Map[
  StringJoin@Sort@Characters@StringReplacePart[w, " ", #] &,
  Map[{#, #} &, Subsets[Range[7], numBlanks], {2}]]

draws[r_String] :=
 Times @@ Binomial @@ Transpose[Tally@Characters[r] /. tiles]

all = {};
p = Accumulate@Map[(
       new = Complement[racks[#], all];
       all = Join[all, new];
       Total[draws /@ new]
       ) &,
     words] / Binomial[numTiles, 7];

The results are shown in the following figure, along with another sampling of specific playable words.  For example, if we include the entire official word list, the probability of drawing a playable 7-letter word is 21226189/160075608, or about 0.132601.

Probability that 7 randomly drawn tiles form a word, vs. dictionary size.

Probability that 7 randomly drawn tiles form a word, vs. dictionary size.

A coarse inspection of the list suggests that I confidently recognize only about 8 or 9 thousand– roughly a third– of the available words, meaning that my probability of playing all 7 of my tiles on the first turn is only about 0.07.  In other words, instead of a first-turn bingo every 7.5 games or so on average, I should expect to have to wait nearly twice as long.  We’ll see if I’m even that good.

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Risk of (gambler’s) ruin

Suppose that you start with an initial bankroll of m dollars, and repeatedly make a wager that pays $1 with probability p, and loses $1 with probability 1-p.  What is the risk of ruin, i.e., the probability that you will eventually go broke?

This is the so-called gambler’s ruin problem.  It is a relatively common exercise to show that if p \leq 1/2, the probability of ruin is 1, and if p > 1/2, then the probability is

(\frac{1-p}{p})^m

But what if the wager is not just win-or-lose a dollar, but is instead specified by an arbitrary probability distribution of outcomes?  For example, suppose that at each iteration, we may win any of (-2,-1,0,+1,+2) units, with respective probabilities (1/15,2/15,3/15,4/15,5/15).  The purpose of this post is to capture my notes on some seemingly less well-known results in this more general case.

(The application to my current study of blackjack betting is clear: we have shown that, at least for a shoe game, even if we play perfectly, we are still going to lose if we don’t vary our bet.  We can increase our win rate by betting more in favorable situations… but a natural constraint is to limit our risk of ruin, or probability of going broke.)

Schlesinger (see Reference (2) below) gives the following formula for risk of ruin, due to “George C., published on p. 8 of ‘How to Win $1 Million Playing Casino Blackjack'”:

(\frac{1 - \frac{\mu}{\sigma}}{1 + \frac{\mu}{\sigma}})^\frac{m}{\sigma}

where \mu and \sigma are the mean and standard deviation, respectively, of the outcome of each round (or hourly winnings).  It is worth emphasizing, since it was unclear to me from the text, that this formula is an approximation, albeit a pretty good one.  The derivation is not given, but the approach is simple to describe: normalize the units of both bankroll and outcome of rounds to have unit variance (i.e., divide everything by \sigma), then use the standard two-outcome ruin probability formula above with win probability p chosen to reflect the appropriate expected value of the round, i.e., p - (1-p) = \mu / \sigma.

The unstated assumption is that 0 < \mu < \sigma (note that ruin is guaranteed if \mu < 0, or if \mu = 0 and \sigma > 0), and that accuracy of the approximation depends on \mu \ll \sigma \ll m, which is fortunately generally the case in blackjack.

There is an exact formula for risk of ruin, at least as long as outcomes of each round are bounded and integral.  In Reference (1) below, Katriel describes a formula involving the roots inside the complex unit disk of the equation

\sum p_k z^k = 1

where p_k is the probability of winning k units in each round.  Execution time and numeric stability make effective implementation tricky.

Finally, just to have some data to go along with the equations, following is an example of applying these ideas to analysis of optimal betting in blackjack.  Considering the same rules and setup as in the most recent posts (6 decks, S17, DOA, DAS, SPL1, no surrender, 75% penetration), let’s evaluate all possible betting ramps with a 1-16 spread through a maximum (floored) true count of 10, for each of five different betting and playing strategies, ranging from simplest to most complex:

  1. Fixed basic “full-shoe” total-dependent zero-memory strategy (TDZ), using Hi-Lo true count for betting only.
  2. Hi-Lo with the Illustrious 18 indices.
  3. Hi-Lo with full indices.
  4. Hi-Opt II with full indices and ace side count.
  5. “Optimal” betting and playing strategy, where playing strategy is CDZ- optimized for each pre-deal depleted shoe, and betting strategy is ramped according to the corresponding exact pre-deal expected return.

Then assuming common “standard” values of $10,000 initial bankroll, a $10 minimum bet, and 100 hands per hour, the following figure shows the achievable win rate ($ per hour) and corresponding risk of ruin for each possible strategy and betting ramp:

Win rate vs. risk of ruin for various betting and playing strategies.

Win rate vs. risk of ruin for various betting and playing strategies.

There are plenty of interesting things to note and investigate here.  The idea is that we can pick a maximum acceptable risk of ruin– such as the red line in the figure, indicating the standard Kelly-derived value of 1/e^2, or about 13.5%– and find the betting ramp that maximizes win rate without exceeding that risk of ruin.  Those best win rates for this particular setup are:

  1. Not achievable for TDZ (see below).
  2. $20.16/hr for Hi-Lo I18.
  3. $21.18/hr for Hi-Lo full.
  4. $26.51/hr for Hi-Opt II.
  5. $33.09/hr for optimal play.

Fixed basic TDZ strategy, shown in purple, just isn’t good enough; that is, there is no betting ramp with a risk of ruin smaller than about 15%.  And some betting ramps, even with the 1-16 spread constraint, still yield a negative overall expected return, resulting in the “tail” at P(ruin)=1.  (But that’s using Hi-Lo true count as the “input” to the ramp; it is possible that “perfect” betting using exact pre-deal expected return could do better.)

References:

  1. Katriel, Guy, Gambler’s ruin probability – a general formula, arXiv:1209.4203v4 [math.PR], 2 July 2013
  2. Schlesinger, Don, Blackjack Attack: Playing the Pros’ Way, 3rd ed. Las Vegas: RGE Publishing, Ltd., 2005
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A harder birthday problem

It is a well-known non-intuitive result that in a group of n=23 people– conveniently the size of a classroom of students– the probability is at least 1/2 that k=2 or more of them share a birthday.  This is a nice problem for several reasons:

  1. Its solution involves looking at the problem in a non-obvious way, in this case by considering the complementary event that all birthdays are distinct.
  2. Once the approach in (1) is understood, computing the answer is relatively easy: the probability is 1-(365)_{n}/365^n, where (x)_{n} is the falling factorial.
  3. The answer is surprising.  When I ask students, “How many people are needed for the probability of a shared birthday to exceed 1/2?”, guesses as high as 180 are common.
  4. Picking on the not-quite-realistic assumption of all d=365 birthdays being equally likely actually helps; that is, with a non-uniform distribution of birthdays, the probability of coincidence is higher.

But what about larger k?  For example, suppose that while surveying your students’ birthdays in preparation for this problem, you find that three of them share a birthday?  What is the probability of this happening?

The Wikipedia page on the birthday problem doesn’t address this generalization at all.  There are several blog and forum posts addressing the k=3 case specifically, by grouping the prohibited cases according to the number of pairs of people with shared birthdays.  This is generalized to arbitrary k \geq 2 on the Wolfram MathWorld page; the resulting recurrence relation is pretty complex, but it’s a nice exercise to prove that it works.

Probability that at least (2,3,4,5) people share a birthday, vs. group size.

Probability that at least (2,3,4,5) people share a birthday, vs. group size.

The motivation for this post is to describe what I think is a relatively simpler solution, for arbitrary k, including Python source code to perform the calculation.  Let’s fix the number of equally likely possible birthdays d=365, and the desired number k of people sharing a birthday, and define the function

G(x)=(1+x+\frac{x^2}{2} \ldots + \frac{x^{k-1}}{(k-1)!})^d

Then G(x) is the exponential generating function for the number of “prohibited” assignments of birthdays to n people where no more than k-1 share a birthday.  That is, the number of such prohibited assignments is n! times the coefficient of x^n in G(x).

(When working through why this works, it’s interesting how often it can be helpful to transform the problem into a different context.  For example, in this case, we are also counting the number of length-n strings over an alphabet of d characters, where no character appears more than k-1 times.)

The rest of the calculation follows in the usual manner: divide by the total number of possible assignments d^n to get the complementary probability, then subtract from 1.  The following Python code performs this calculation, either exactly– using the fractions module, which can take a while– or in double precision, which is much faster.

import numpy as np
from numpy.polynomial import polynomial as P
import fractions
import operator
 
def p(k, n, d=365, exact=False):
    f = fractions.Fraction if exact else operator.truediv
    q = 0 if n > d * (k - 1) else P.polypow(np.array(
        [f(1, np.math.factorial(j)) for j in range(k)], dtype=object),
        d)[n]
    for j in range(1, n + 1):
        q = q * f(j, d)
    return 1 - q
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Strong induction

What is the best way to explain induction to a student?  That is, given a true-or-false statement P(n) involving a natural number n \geq 0, we would like to prove that the statement is true for all such n \geq 0.  How does one prove such a claim, when it seems to require checking an infinite number of cases?  My motivation for this thinking-out-loud post involves a particular form of induction whose presentation in some textbooks may cause students some confusion (or at least, it did cause confusion in at least one student).

Often the first exposure to induction is the so-called “weak” form: to show that P(n) is true for all n \geq 0, it suffices to prove the following two claims:

  • Base step: P(0) is true.
  • Induction step: For all n \geq 0, if P(n) is true, then P(n+1) is true.

That is,

(P(0) \land \forall n \geq 0 P(n) \Rightarrow P(n+1)) \Rightarrow \forall n \geq 0 P(n)

So far, so good.  Although this is technically all that we need, that this is referred to as “weak” induction suggests that there is a stronger form.  There is… sort of.  In his Applied Combinatorics (see Reference 1 below), Tucker skips weak induction altogether, presenting induction in the following “strong” form:

  • Base step: P(0) is true.
  • Induction step: If P(0), P(1), P(2), \ldots, P(n-1) are true, then P(n) is true.

The idea is that the induction hypothesis is stronger, which can make the resulting proof simpler; in the process of proving P(n) in the induction step, we have at our disposal not only the assumed truth of P(n-1), but also all of the other previous statements.  (I say “sort of,” and wrap “weak” and “strong” in quotes, because the two forms are actually equivalent.  Strong induction isn’t really any stronger in the sense of more “proving power”; statements provable using strong induction are also provable by weak induction, and vice versa.)

Notice that the induction step above, quoted directly from Tucker, lacks some rigor in that there is no explicit universal quantifier for n.  That is, for what values of n do we need to demonstrate the induction step?  This is important, because it turns out that if we are more careful, we can economize a bit and demonstrate that P(n) is true for all n \geq 0 by proving the single more compact claim:

  • Strong induction “single-step”: For all n \geq 0, if P(k) is true for all k<n, then P(n) is also true.

That is,

(\forall n \geq 0 (\forall k<n P(k)) \Rightarrow P(n)) \Rightarrow \forall n \geq 0 P(n)

Velleman (Reference 2) takes this presentation approach, which certainly has elegance going for it.  But he then goes on to “note that no base case is necessary in a proof by strong induction [my emphasis].”  This is where I think things can get confusing.  It’s certainly true that, once we have demonstrated the implication in the single-step version above, then we know that P(0) is true.  As Velleman explains, “plugging in 0 for n, we can conclude that (\forall k<0 P(k)) \Rightarrow P(0).  But because there are no natural numbers smaller than 0, the statement \forall k<0 P(k) is vacuously true.  Therefore, by modus ponens, P(0) is true.  (This explains why the base case doesn’t have to be checked separately in a proof by strong induction [my emphasis]; the base case P(0) actually follows from the modified form of the induction step used in strong induction.)”

There is nothing strictly incorrect here.  What may be misleading, though, is the repeated assurance that no special attention is required for treatment of the base case.  When writing the actual proof of the implication in the single-step version of strong induction, the argument may “look different” for n=0 than it does for n>0.  I think Wikipedia actually does the most admirable job of explaining this wrinkle: “Sometimes the same argument applies for n=0 and n>0, making the proof simpler and more elegant.  In this method it is, however, vital to ensure that the proof of P(n) does not implicitly assume that n>0, e.g. by saying “choose an arbitrary k<n” or assuming that a set of n elements has an element [my emphasis added].”

From a teaching perspective, what are good examples of induction proofs that highlight these issues?  This is a good question.  When discussing strong induction, the most common example seems to be the existence part of the fundamental theorem of arithmetic: every integer greater than 1 is the product of one or more primes.  Here we really can make just one argument, so to speak, without treating the first prime number 2 as a special base case (interestingly, Wikipedia does address it separately).

In a course on graph theory, I think another nice example is the proof that any round-robin tournament contains a Hamiltonian path.  Again, the single-step version of a strong induction argument works here as well (although we have to be a little more careful about what happens when some constructed subsets of vertices end up being empty).

Finally, another common example where strong induction is useful, but special treatment of base case(s) is required, involves postage stamps: show that it is possible to use 4- and 5-cent stamps to form every amount of postage of 12 cents or more.

References:

  1. Tucker, Alan, Applied Combinatorics, 6th ed. New York: Wiley and Sons, 2012
  2. Velleman, Daniel J., How to Prove It: A Structured Approach. New York: Cambridge University Press, 2006
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How many dice?

The figure below shows an “unfolded” view of a typical 6-sided die (d6), used everywhere from board games to casinos:

An "unfolded" view of a standard d6.

An “unfolded” view of a standard d6.

Each of the six faces is uniquely labeled with the integers 1 through 6.  But that’s not all; note that the values on opposite faces always sum to 7.  This is a standard arrangement applied more generally to other types of dice as well.  My set of Platonic solid dice– with n sides for n \in {4,6,8,12,20}— all have this same property.  Let us call such an n-sided die standard if all opposite faces sum to n+1.

Problem 1: How many different standard 6-sided dice are there?  That is, in how many ways can we label the faces with distinct integers 1 through 6, with opposite faces summing to 7?

Problem 2: What if we relax the standard (constant opposite sum) property?  That is, in how many ways can we label the faces with distinct integers 1 through 6, with no other restrictions?

Problem 3: Same as Problem 2, but for the other Platonic solids (d4, d8, d12, d20)?

I think these are great examples of problems for students that straddle whatever line there may be between mathematics and computer science.  Problems 1 and 2 can be solved as-is “by hand.”  (And they are interesting in part because the answers are perhaps surprisingly small numbers.)  But the usual mathematical machinery involved just counts the number of dice in each case; it’s an interesting extension as a programming problem to actually enumerate (i.e., list) them, display visual representations of them, etc.

Problem 3 is more challenging; the machinery is the same, but the larger numbers involved require some amount of automation in the housekeeping.

Now for what I think is a really hard problem motivating this post (Edit: although after a response from a reader in another forum with a very elegant solution, perhaps this is not as difficult as I thought it might be!):

Problem 4: Same as Problem 1, but for the other Platonic solids– that is, how many different standard (d4, d8, d12, d20) are there, with constant opposite sums?

 

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Distribution and variance in blackjack (Part 2)

Introduction

This is a follow-up to the previous post describing the recently-developed algorithm to efficiently compute not just the expected value of a round of blackjack, but the entire probability distribution– and thus the variance– allowing the analysis of betting strategies.  This time, I want to look at some actual data.  But first, a digression motivated by some interesting questions about this analysis:

Why combinatorics?

I spend much of my time in my day job on modeling and simulation, where the usual objective is to estimate the expected value of some random variable, which is a function of the pseudo-randomly generated outcome of the simulation.  For example, what is the probability that the system of interest performs as desired (e.g., detects the target, tracks the target, destroys the target, etc.)?  In that case, we want the expected value of a {0,1} indicator random variable.  The usual approach is to run a simulation of that system many times (sometimes for horrifyingly small values of “many”), recording the number of successes and failures; the fraction of runs that were successful is a point estimate of the probability of success.

Usually, we do things this way because it’s both easier and faster than attempting to compute the exact desired expected value.  It’s easier because the simulation code is relatively simple to write and reason about, since it corresponds closely with our natural understanding of the process being simulated.  And it’s faster because exact computation involves integration over the probability distributions of all of the underlying sources of randomness in the simulation, which usually interact with each other in a prohibitively complex way.  We can afford to wait on the many simulation runs to achieve our desired estimation accuracy, because the exact computation, even if we could write the code to perform it, might take an astronomically longer time to execute.

But sometimes that integration complexity is manageable, and combinatorics is often the tool for managing that complexity.  This is true in the case of blackjack: there are simulations designed to estimate various metrics of the game, and there are so-called “combinatorial analyzers” (CAs) designed to compute some of these same metrics exactly.

My point in this rant is that these two approaches– simulation and CA– are not mutually exclusive, and in some cases it can be useful to combine them.  The underlying objective is the same for both: we want a sufficiently accurate estimate of metrics of interest, where “sufficiently accurate” depends on the particular metrics we are talking about, and on the particular questions we are trying to answer.  I will try to demonstrate what I mean by this in the following sections.

The setup

Using the blackjack rules mentioned in the previous post (6 decks, S17, DOA, DAS, SPL1, no surrender), let’s simulate play through 100,000 shoes, each dealt to 75% penetration, heads-up against the dealer.  Actually, let’s do this twice, once at each of the two endpoints of reasonable and feasible strategy complexity:

For the first simulation, the player uses fixed “basic” total-dependent zero-memory (TDZ) strategy, where:

  • Zero-memory means that strategy decisions are determined solely by the player’s current hand and dealer up card;
  • Total-dependent means that strategy decisions are dependent only on the player’s current hand total (hard or soft count), not on the composition of that total; and
  • Fixed means that the TDZ strategy is computed once, up-front, for a full 6-deck shoe, but then that same strategy is applied for every round throughout each shoe as it is depleted.

In other words, the basic TDZ player represents the “minimum effort” in terms of playing strategy complexity.  Then, at the other extreme, let’s do what is, as far as I know, the best that can be achieved by a player today, assuming that he (illegally) brings a laptop to the table: play through 100,000 shoes again, but this time, using “optimal” composition-dependent zero-memory (CDZ-) strategy, where:

  • Composition-dependent means that strategy is allowed to vary with the composition of the player’s current hand; and
  • “Optimal” means that the CDZ- strategy is re-computed prior to every round throughout each shoe.

(The qualification of “optimal” is due to the minus sign in the CDZ- notation, which reflects the conjecture, now proven, that this strategy does not always yield the maximum possible overall expected value (EV) among all composition-dependent zero-memory strategies.  The subtlety has to do with pair-splitting, and is worth a post in itself.  So even without fully optimal “post-split” playing strategy, we are trading some EV for feasible computation time… but it’s worth noting that, at least for these rule variations, that average cost in EV is approximately 0.0002% of unit wager.)

The cut-card effect

Now for the “combining simulation and CA”: for each simulated round of play, instead of just playing out the hand (or hands, if splitting a pair) and recording the net win/loss outcome, let’s also compute and record the exact pre-round probability distribution of the overall outcome.  The result is roughly 4.3 million data points for each of the two playing strategies, corresponding to roughly 43 rounds per 100,000 shoes.

The following figure shows one interesting view of the resulting data: for each playing strategy, what is the distribution of expected return as a function of the number of rounds dealt into the shoe?  The gray curves indicate 5% quantiles, ranging from minimum to maximum EV, and the red and blue curves indicate the mean for basic TDZ and optimal CDZ- strategy, respectively.

Expected value of a unit wager vs. round, for fixed basic TDZ strategy (left) and CDZ- strategy optimized for the current depleted shoe (right).

Distribution of expected value of a unit wager vs. round, for fixed basic TDZ strategy (left) and CDZ- strategy optimized for the current depleted shoe (right).  Five percent quantiles, from minimum to maximum, are shown in gray, with the mean in red/blue.

There are several interesting things happening here.  Let’s zoom in on just the red and blue curves indicating the mean EV per round:

Estimated expected value of a unit wager vs. round, for fixed basic TDZ strategy (red) and optimal CDZ- strategy (blue).

Estimated expected value of a unit wager vs. round, for fixed basic TDZ strategy (red) and optimal CDZ- strategy (blue).

First, the red curve is nearly constant through most of the initial rounds of the shoe.  Actually, for every round that we are guaranteed to reach (i.e., not running out of cards before reaching the “cut-card” at 75% penetration), the true EV can be shown to be exactly constant.

Second, that extreme dip near the end of the shoe is illustration of the cut-card effect: if we manage to play an abnormally large number of rounds in a shoe, it’s because there were relatively few cards dealt per round, which means those rounds were rich in tens, which means that the remainder of the shoe is poor in tens, yielding a significantly lower EV.

Keep in mind that we can see this effect with less than 10 million total simulated rounds, with per-round EV estimates obtained from samples of at most 100,000 depleted shoe subsets!  This isn’t so surprising when we consider that each of those sample subsets indicates not just the net outcome of the round, but the exact expected outcome of the round, whose variance is much smaller… most of the time.  For the red TDZ and blue CDZ- strategies shown in the figure, there are actually three curves each, showing not just the estimated mean but also the (plus/minus) estimated standard deviation.  For the later rounds where the cut-card effect means that we have fewer than 100,000 sampled rounds, that standard deviation becomes large enough that we can see the uncertainty more clearly.

Variance

So far, none of this is really new; we have been able to efficiently compute basic and optimal strategy expected return for some time.  The interesting new data is the corresponding pre-deal variance (derived from the exact probability distribution).  The following figure shows an initial look at this, as a scatter plot of variance vs. EV for each round, with an overlaid smoothed histogram to more clearly show the density:

Variance vs. expected value of a unit wager, for fixed basic TDZ strategy (left) and optimal CDZ- strategy (right).

Variance vs. expected value of a unit wager, for fixed basic TDZ strategy (left) and optimal CDZ- strategy (right).

It seems interesting that the correlation sense is roughly reversed between basic and optimal strategy; that is, for the basic strategy player, higher EV generally means lower variance, while for the optimal strategy player with the laptop, the reverse is true.  I can’t say I anticipated this behavior, and upon first and second thoughts I still don’t have an intuitive explanation for what’s going on.

 

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