More probability puzzles

[Elixir]

In physical problems one looks to see if a normal distribution is close enough to meet a need.

If your physical condition is a casino operating on razor thin gross margins on huge sums of money, your "need" is to strive for spot on, no matter how close you get.

 

[steve_bank]

In physical problems one looks to see if a normal distribution is close enough to meet a need.

If your physical condition is a casino operating on razor thin gross margins on huge sums of money, your "need" is to strive for spot on, no matter how close you get.

I expect if you ran a few hundred thousand trials on any casino wheel and counted the number of times the ball ends in each slot all slots would not be equal. There would be some variation, but small.

Rerun the experiment on the same wheel and the slot totals would not be the same as the first experiment.

I say that from dealing with electro-mechanical systems. Exact repeatability is difficult if not impossible. Especially when a human operator is involved. An operator spins the wheel and drops a ball, that in itself adds randomness.

There may be a machine to test wheels.

There is probably a statistical standard casino wheel makers have to meet.

 

[steve_bank]

The first step is always bounding the problem, is it solvable in a practical sense.

The number of possible unique sequences for 38 slots is huge,

If you are looking for a pattern that can be exploited at a roulette wheel you would have to run enough trails that each sequence is repeated many times. Even automated with a machine that would take a long time.

An ordinary PC would probably be too slow to crunch all the data. When playing around with statistics and distribution coding testing the code easily runs into execution time problems..


Way back in the 80s before PCs I was crunching data sets by hand with a calculator. At the beginning I was sure I was seeing patterns in the data. Our brains are wired to see patterns, a survival trait.

So, I think finding a pattern at the wheel by observation or by finding a way around probability theory is impossible. The probability on any spin is always 1/38.

Statistically it is no different than shuffling a deck or 38 cards and picking one or picking 1 out of 38 balls from a box.

I saw the term subjectivism applied to gambling and probabilities. Instead of theory the gambler goes by what he observes and draws a conclusion or sees a pattern on which to bet on.

 

[steve_bank]

Somebody had a sign on his office 'In god we trust, all else bring data'.


An easy exermnt that requires a little time.

Buy a roulette wheel, there is a range of prices online.

For 38 slots and a nominal 100 counts per slot, 38*100 = 3800 spins.

Assume 1 minute per spin.

3800 spins = 3800 minutes = 63.3 hours.

Do a little each day, a long term project. Record the number of times the ball ends up in each slot and see how close the counts are.

Repet the experiment and compare results.

 

[bilby]

Somebody had a sign on his office

Does this "somebody" have an account here? A name? Any existence at all, outside your anecdote?

I'm struggling to accept that "somebody" is a trustworthy and reliable source for, well, anything.

 

[steve_bank]

Unless I am missing something in the long run the house always cones out ahead. Statistically over decades of play there may be gamblers who hit a streak of winning.

The colors and layout of the betting area is for show, it masks the underlying probabilities

I can see how someone can go down a rabbit hole looking for combinations of bets that beat the house.

1-Single number bet pays 35 to 1. Also called “straight up.”

Including 0,00 odds are 1/38

2-Double number bet pays 17 to 1. Also called a “split.”

Odds are 2/38

3-Three number bet pays 11 to 1. Also called a “street.”

Odds are 3/38

4-Four number bet pays 8 to 1. Also called a “corner bet.”

Odds are 4/38

5-Five number bet pays 6 to 1. Only one specific bet which includes the following numbers: 0-00-1-2-3.

Odds are 5/38

6-Six number bets pays 5 to 1. Example: 7, 8, 9, 10, 11, 12. Also called a “line.”

Odds are 6/38

7-Twelve numbers or dozens (first, second, third dozen) pays 2 to 1.

Odds are 12/38

8-Column bet (12 numbers in a row) pays 2 to 1.

9-18 numbers (1-18) pays even money.

Odds 18/38, even money but odds less than 50/50

10-8 numbers (19-36) pays even money.

Odds 18/38

11-Red or black pays even money.

12-Odd or even bets pay even money.

Assuming an approximation to a uniform distribution any betting scheme is easily testable.

The code can be expanded to add in multiple simultaneous bets and money bet on each, and seeing if you can statistically come out ahead.

Might be doable in a spreadsheet macro.

//Scilab script
clear
clc
ntrials = 100000
count = 0
x = [36 1]
nnums = length(x)
for i= 1:ntrials
    y = grand(1,1,"uin",1,38)
    for j = 1:nnums 
        if(y == x(j))
            count = count + 1
            break
        end
    end
end

mprintf("count  %d  p  %f\n",count,100*count/ntrials)
mprintf("The0retcal p  %f/n",100*nnums/38)

 

[Alcoholic Actuary]

Unless I am missing something in the long run the house always cones out ahead. Statistically over decades of play there may be gamblers who hit a streak of winning.

The colors and layout of the betting area is for show, it masks the underlying probabilities

I can see how someone can go down a rabbit hole looking for combinations of bets that beat the house.

1-Single number bet pays 35 to 1. Also called “straight up.”

Including 0,00 odds are 1/38

2-Double number bet pays 17 to 1. Also called a “split.”

Odds are 2/38

3-Three number bet pays 11 to 1. Also called a “street.”

Odds are 3/38

4-Four number bet pays 8 to 1. Also called a “corner bet.”

Odds are 4/38

5-Five number bet pays 6 to 1. Only one specific bet which includes the following numbers: 0-00-1-2-3.

Odds are 5/38

6-Six number bets pays 5 to 1. Example: 7, 8, 9, 10, 11, 12. Also called a “line.”

Odds are 6/38

7-Twelve numbers or dozens (first, second, third dozen) pays 2 to 1.

Odds are 12/38

8-Column bet (12 numbers in a row) pays 2 to 1.

9-18 numbers (1-18) pays even money.

Odds 18/38, even money but odds less than 50/50

10-8 numbers (19-36) pays even money.

Odds 18/38

11-Red or black pays even money.

12-Odd or even bets pay even money.

Assuming an approximation to a uniform distribution any betting scheme is easily testable.

The code can be expanded to add in multiple simultaneous bets and money bet on each, and seeing if you can statistically come out ahead.

Might be doable in a spreadsheet macro.

//Scilab script
clear
clc
ntrials = 100000
count = 0
x = [36 1]
nnums = length(x)
for i= 1:ntrials
    y = grand(1,1,"uin",1,38)
    for j = 1:nnums
        if(y == x(j))
            count = count + 1
            break
        end
    end
end

mprintf("count  %d  p  %f\n",count,100*count/ntrials)
mprintf("The0retcal p  %f/n",100*nnums/38)

You are begging the question. You assume a uniform distribution, program it as such, then lo and behold the results are uniform. That's a terrible way to 'test an assumption'.

aa

 

[steve_bank]

Unless I am missing something in the long run the house always cones out ahead. Statistically over decades of play there may be gamblers who hit a streak of winning.

The colors and layout of the betting area is for show, it masks the underlying probabilities

I can see how someone can go down a rabbit hole looking for combinations of bets that beat the house.

1-Single number bet pays 35 to 1. Also called “straight up.”

Including 0,00 odds are 1/38

2-Double number bet pays 17 to 1. Also called a “split.”

Odds are 2/38

3-Three number bet pays 11 to 1. Also called a “street.”

Odds are 3/38

4-Four number bet pays 8 to 1. Also called a “corner bet.”

Odds are 4/38

5-Five number bet pays 6 to 1. Only one specific bet which includes the following numbers: 0-00-1-2-3.

Odds are 5/38

6-Six number bets pays 5 to 1. Example: 7, 8, 9, 10, 11, 12. Also called a “line.”

Odds are 6/38

7-Twelve numbers or dozens (first, second, third dozen) pays 2 to 1.

Odds are 12/38

8-Column bet (12 numbers in a row) pays 2 to 1.

9-18 numbers (1-18) pays even money.

Odds 18/38, even money but odds less than 50/50

10-8 numbers (19-36) pays even money.

Odds 18/38

11-Red or black pays even money.

12-Odd or even bets pay even money.

Assuming an approximation to a uniform distribution any betting scheme is easily testable.

The code can be expanded to add in multiple simultaneous bets and money bet on each, and seeing if you can statistically come out ahead.

Might be doable in a spreadsheet macro.

//Scilab script
clear
clc
ntrials = 100000
count = 0
x = [36 1]
nnums = length(x)
for i= 1:ntrials
    y = grand(1,1,"uin",1,38)
    for j = 1:nnums
        if(y == x(j))
            count = count + 1
            break
        end
    end
end

mprintf("count  %d  p  %f\n",count,100*count/ntrials)
mprintf("The0retcal p  %f/n",100*nnums/38)

You are begging the question. You assume a uniform distribution, program it as such, then lo and behold the results are uniform. That's a terrible way to 'test an assumption'.

aa

Ok, then what is the distribution? I knew I could looked it up but that is no fun. It does not cnange anything , any betting scheme can be tested statically to see whether over thousands of spins you will always come out ahead.

Uniform using binomial distribution to caculate expectedvalues assumng Bernouli Trials.

Statically simulating systems I designed with multiple underlying distributions was part of me work. Scilab with its random number generators and statistical functions was my main tool.

When in doubt simulate.

How to Calculate Expected Value in Roulette

The game of roulette is a good example of the application of expected value. We can analyze what the winnings will be if we continually bet on red.

www.thoughtco.com

The Roulette Wheel and its Distributions

You don’t see many run-down, out-of-business casinos, which should serve as a tip-off: almost nobody beats the house in the long run. In this article, I’ll show you - using the binomial distribution and the normal distribution - why that is.

www.flyingcoloursmaths.co.uk

Discrete uniform and binomial distribution functions

Konvexity provides CFA Coaching, Training classes in Gurgaon and Delhi. We provide training to help you build strong foundation in finance.

konvexity.com

Wheels undoubtedly have been formally tested many times.

The problem is getting a large enough number of trials. The same problem with computer pseudo random number generators. As n trials get large the numbers 1-38 app och a true inform distribution, small differences in frequencies..

So, if you wtaced 1000 spins at a casino and bet on what yiu think is a prefed numer on the enxt spin, it is still 1/38.

This is a simple lab yest, I am sure there is data out there on tye net.

https://www1.maths.leeds.ac.uk/~sta6kvm/reprints/Report-AlBaines-1990.pdf

 

[Swammerdami]

Hope this thread gets more activity. Solutions have been posted for NONE of Problems I, II and III. I thought they were mostly easy problems.

Kudos again to Steve for reminding us that the roulette wheel in II involves a 38-valence discrete uniform variate. Apologies if I didn't make that clear in my statement(s) of II. Steve? Can your scilab produce variates on a Cauchy distribution? If it's not built in you can get a Cauchy variate very simply: Set it to the tangent of u, where u is a real uniform variate on the open interval (-3.14159265358979328, 3.14159265358979328)

-----------------------------------

IV Let me mention briefly an oft-ignored approach to correct probability estimation. I will use for example "VERE WROTE THIS" as found in a Cardano cipher in a context likely to relate to the authorship ("Vere" is the main candidate).

I assign it the Roman numeral IV for clarity -- it's quite important, BUT completely unrelated to any of I, II or III.
Encipherments can arise by chance, though a LONG message in a LIKELY key, if sufficiently so will have HUGE probative value. How do we form a guess to the clue's probative value, and how do we then interleave this knowledge with other independent clues?

The method of probability ratios is useful. Here we'll start with the Vere vs NOT-Vere issue, and assume Vere becomes 50x (or 5x) as likely as NOT-Vere per this clue; it is not correlated with the "a priori" probability. We stipulate that the a priori chance of Vere is p (we'll work p = .001, p = .01, p = .1). Here a priori means the reaction to all evidence except the alleged Cardano-cipher clue.

50X clue weight for Cardano clue. Never mind for now how we finalize these clue weights.
.001 999 --> ÷50 --> 19.98 --> 4.77%
.01 99 --> ÷50 --> 1.98 --> 33.56%
.1 9 --> ÷50 --> .18 --> 84.75%

5X clue weight for Cardano clue
.001 999 --> ÷5 --> 199.8 --> 0.50%
.01 99 --> ÷5 --> 19.8 --> 4.81%
.1 9 --> ÷5 --> 1.8 --> 35.71%

A.Actuary? Do you ever use this sort of arithmetic in your probability calculations?

These examples are just to show how the arithmetic of probability ratios works; elsewhere I argue that if properly constructed this approach will give the best available estimates. In the context of IMDB discussions, such probability analysis is most relevant in the Jesus Historicity threads.

 

[bilby]

Set it to the tangent of u, where u is a real uniform variate on the open interval (-3.14159265358979328, 3.14159265358979328)

Point of information:

Your approximation for π is incorrect in the last digit - it should be 3.14159265358979324

The last digit when cutting the string at this number of significant digits is actually a 3, but it's followed by an 8, so should be rounded up to 4.

 

[steve_bank]

Say the word and I will start a separate thread on gambling.

Swami, all your problems are generally the same underlying form.

Start rolling a die ad record the number of times 1,2,3,4,5,6 occurs.

As the number of trials grow the counts for each number will get close. As the number of trails get lage the probabilities of each number approaches 1/6 but it is never exact for all six numbers. At any point in the count some numers may be a little more than 1/6 and some a little less.

It approaches a true uniform distribution but never gets there. For all practical purposes we consider rolling a die a uniform distribution wen calculating gambling odds.

Same with a roulette wheel.

If you watch a table for a day you might think you see a winning pattern but it is only a small sample of the total number of spins over years.

The house has a small advantage in roulette odds. Syntactically people will win a bet a a table, and some may walk away ahead. But in the long run the house always comes out ahead.

That anyone can and once in a while does win big keeps people coming back to the roulete tables.

If the idds were too low for the gambler nobody would play.

Question, if the numbers on the wheel were in numerical order would it affect the odds? Wahy?

 

[Swammerdami]

Swami, all your problems are generally the same underlying form.

No. NONE of the three problems have anything to do with the topics you're discussing; nor would simulation be relevant to solution. All of my problems posed here involve expected bankroll (or expected logarithm of bankroll) but always with simple analytic expressions: simulation to find means or variances would be a detour in each case.

Solutions are easy in all cases, though III involves an easy application of basic calculus.

Start rolling a die ad record the number of times 1,2,3,4,5,6 occurs.

 

[Bomb#20]

Hope this thread gets more activity. Solutions have been posted for NONE of Problems I, II and III. I thought they were mostly easy problems.

Ok, fine, quit complaining, I'll post a solution to I.

Show

Your best strategy is to say Red. Fifty percent chance of winning. Every other strategy does no better and wastes a lot of time and effort in the process. Saying "Inspect" never changes your odds of winning because you can't do anything useful with the information it gives you about what's left in the deck. The reason counting cards in blackjack gives you an edge over the casino is because you're going to bet more when the odds are in your favor and bet less when the odds are in the casino's favor. If you were required to keep your bet constant, as in this puzzle, then all counting cards would get you is an advance hint about how soon the casino is going to kick your ass.

 

[Swammerdami]

Hope this thread gets more activity. Solutions have been posted for NONE of Problems I, II and III. I thought they were mostly easy problems.

Ok, fine, quit complaining, I'll post a solution to I.

Show

Your best strategy is to say Red. Fifty percent chance of winning. Every other strategy does no better and wastes a lot of time and effort in the process. Saying "Inspect" never changes your odds of winning because you can't do anything useful with the information it gives you about what's left in the deck. The reason counting cards in blackjack gives you an edge over the casino is because you're going to bet more when the odds are in your favor and bet less when the odds are in the casino's favor. If you were required to keep your bet constant, as in this puzzle, then all counting cards would get you is an advance hint about how soon the casino is going to kick your ass.

Thank you. Another way to see this is that when ready to Inspect, the final card (#52) has the same chance as the next card. So you may as well ALWAYS pick the final card. And since initially the #13 card has the same chance as #52, you may as well always pick #13. Or #14. Or #1.


And now for II and III.

 

[steve_bank]

Swami, all your problems are generally the same underlying form.

No. NONE of the three problems have anything to do with the topics you're discussing; nor would simulation be relevant to solution. All of my problems posed here involve expected bankroll (or expected logarithm of bankroll) but always with simple analytic expressions: simulation to find means or variances would be a detour in each case.

Solutions are easy in all cases, though III involves an easy application of basic calculus.

Start rolling a die ad record the number of times 1,2,3,4,5,6 occurs.

Correct, they are all easy if you know theory and have experience. [removed]

I posed a simple calculus problem on tangent lines to a curve which you did not solve. I know tangent lines because I used calculus on the job periodically and know the type of problem. Calculus was required to be able to read texts.

That you did not solve the problem just says you do not have experience with calculus, that is all. It says nothing about your capabilities.

If you work through the gambling optimization problem you posted don't post it as a puzzle which most people will not be able to solve.

Post your step by step solution as a tutorial for somebody to learn from.

When you said take the derivative and solve for 0 I knew exactly what you were talking about. Basic calculus that appears in engineering texts, solving for a maximum. I am also familiar with Maximum Likelihood Estimators from probability theory, same principle.

A proability problem,

We all use the atrimetic mean as an estimator. If I am meaurng a voltge I might take 3 measremnts and avgerge then. We seem to do intnctively.

What is the basis for usng the arthmetic mean as the best estimator for a normal distribution? Solutions are on the net.

A related problem in maximization in basic electric circuits is the Maxim Power Transfer Theorem which requires calculus and 1st 2nd derivative tests to prove. Finding te solution

Maximum power transfer theorem - Wikipedia

en.wikipedia.org

 

[Bomb#20]

II. I need a fix!

You have $100 but the man in the coon-skin cap in a pig pen wants $200 for your fix of Foximoxijox. You need that fix; you need it NOW. You happen to be in a casino near a roulette wheel. Parlaying the $100 into $200 on that wheel is your only option. $199 would be of no use to you. No, you can't go looking for a Twenty-One game or a Craps table. It's Roulette or nothing.

The casino management is happy to cater to your needs. They'll spin and settle very quickly for you, if you intend some long-series martingale or such. They'll let you bet any amount, even fractions of a penny if you want. But they will not extend you credit. You can only bet the cash you have; when that's gone it's all over. How should you play to maximize your chance of getting the $200?

It's American roulette: there are 38 numbers on the wheel, each equally as likely as any other. You can bet a single number and get paid 35-to-1, bet any 9 numbers and get paid 3-to-1, or bet 18 numbers (even, odd, red or black) and get paid 1-to-1. (No en prise like the hoity-toits in Eueope; you're in Vegas.) How should you bet? Does it matter?

Yes, it matters. How should you bet? Define "should". If you mean, what's the optimal strategy, I have no idea how to go about proving a strategy is optimal when the space of possible strategies is so huge and so complicated. But if you mean people can post the best strategies they've found so far and challenge one another to improve on them, I can get behind that. The obvious strategy, just bet the whole wad on black and have done with it, gives you a 47.37% chance of getting your Foximoxijox. This obviously can't be improved on since the house edge is infamously 5.26%. But oddly, that turns out not to be the case -- there are infinitely many goofy strategies that do just a little better. Here's the top performer I've found...

Spoiler

Calculate the bet you need to get your cash up to $200 if you win at 35-to-1 odds. Win and leave, lose and repeat. You can do this 24 times in a row before you don't have enough money left to make the next bet. If you lose 24 times you'll have $3.38 left from your original $100. Since the casino is so accommodating, split that into 11 bets of 30.73 cents apiece and bet them all on different numbers. If all 11 bets lose, it's all over and you don't get your fix. If one of them wins, now you have $11.06 and you're back in the bet-enough-to-get-to $200 game. Rinse and repeat. Your overall chance of getting your fix is 48.07%.

 

[Swammerdami]

II. I need a fix!

You have $100 but the man in the coon-skin cap in a pig pen wants $200 for your fix of Foximoxijox. You need that fix; you need it NOW. You happen to be in a casino near a roulette wheel. Parlaying the $100 into $200 on that wheel is your only option. $199 would be of no use to you. No, you can't go looking for a Twenty-One game or a Craps table. It's Roulette or nothing.

The casino management is happy to cater to your needs. They'll spin and settle very quickly for you, if you intend some long-series martingale or such. They'll let you bet any amount, even fractions of a penny if you want. But they will not extend you credit. You can only bet the cash you have; when that's gone it's all over. How should you play to maximize your chance of getting the $200?

It's American roulette: there are 38 numbers on the wheel, each equally as likely as any other. You can bet a single number and get paid 35-to-1, bet any 9 numbers and get paid 3-to-1, or bet 18 numbers (even, odd, red or black) and get paid 1-to-1. (No en prise like the hoity-toits in Eueope; you're in Vegas.) How should you bet? Does it matter?

Yes, it matters. How should you bet? Define "should". If you mean, what's the optimal strategy, I have no idea how to go about proving a strategy is optimal when the space of possible strategies is so huge and so complicated. But if you mean people can post the best strategies they've found so far and challenge one another to improve on them, I can get behind that. The obvious strategy, just bet the whole wad on black and have done with it, gives you a 47.37% chance of getting your Foximoxijox. This obviously can't be improved on since the house edge is infamously 5.26%. But oddly, that turns out not to be the case -- there are infinitely many goofy strategies that do just a little better. Here's the top performer I've found...

Yes! American Roulette is chosen for the example since the house edge (vigorish, commission) is well-known to be constant; many assume that how you bet doesn't matter. But as you point it, it DOES. And -- subject to an obvious nit -- the strategy you post is provably optimal.

A simpler way to see the point is to try to quadruple your money. If you take a 3-1 shot, your chance is 9/38. If you take a 1-1 shot and let it ride once if you win, your chance is (18/38)*(18/38). The first number is clearly larger. (Taking the 3-1 bet, while better than 1-1, is not optimal. Optimality is in the solution Bomb#20 spoiler-tagged.

One way to get the point intuitively is to note that total house vigorish is proportional to your total action. Quadrupling your $100 on a 3-1 shot makes your total action $100. But with 1-1 and let it ride, your average total action is $100 + ($200*(18/38)) = $194.74, almost twice as much action, so you're average loss to the house is almost twice as much. (Your total loss will always be exactly $100 or minus $300 -- it's the probability weights that change.)

I've added a minor improvement in red to Bomb's spoilered solution.

Spoiler

Calculate the bet you need to get your cash up to $200 if you win at 35-to-1 odds. Win and leave, lose and repeat. You can do this 24 times in a row before you don't have enough money left to make the next bet. If you lose 24 times you'll have $3.38 left from your original $100. Since the casino is so accommodating, split that into 11 bets of 30.73 cents apiece and bet them all on different numbers. If all 11 bets lose, it's all over and you don't get your fix. Bet the entire $3.38 on a single number; if you win you have $121.68 If one of them wins, now you have $11.06 cents and you're back in the bet-enough-to-get-to $200 game. Rinse and repeat. Your overall chance of getting your fix is 48.07%.

If fractional bets are allowed, your overall chance is 48.089%

 

[Bomb#20]

I've added a minor improvement in red to Bomb's spoilered solution.

Huh. I tried that and it seemed not as good -- that's what I get for not running enough iterations.

 

[Swammerdami]

Kudos to Mr. Bomb for solving I and II.

I
I is easy but perhaps counter-intuitive. It's tempting to think "It doesn't hurt to Inspect the first card. If it's Black my odds immediately improve! I'll be less happy if it's Red but things may still work out. It's already slightly better than even-money that the deck will revert to toss-up after the very next card."

You're right that "It doesn't hurt ..." But it doesn't help either.

II
The general principle for this problem is that when the house "vigorish" is constant, it is ALWAYS best to make the highest-payout bet. Many people with good intuition about gambling topics get this wrong. As so often, looking at an EXTREME case may give insight.

Suppose you have $38 and your two choices are (1) to bet it all on a single number, say 17; or (2) to bet $1 on each of the 38 numbers. What's the difference?

In either case, your expected gross return is exactly $36; your expected loss is $2. But in case (1) you might walk away with $1368 (winning $1330 net profit). In case (2) you will always have a net LOSS of exactly $2 no matter what.

Gambling must balance RISK and REWARD. With "strategy (2)" there's no chance for any reward. This segues into the background for problem III.

III
I'm afraid some might have been turned off, since the problem asks you to maximize the (weighted) GEOMETRIC mean of your resultant bankroll, while most gambling problems assume you're maximizing the "expectation" (the weighted ARITHMETIC mean). But in fact the  Kelly criterion is used to balance risk and reward and indeed leads to EXACTLY the maximization posed in Problem III. The need for such a criterion is well-known to every professional gambler (or investment portfolio manager) even if the phrase "weighted geometric mean" is just gobble-dy-gook to them.

I like III. It's an "important" problem with an elegant solution. That solution seems like it should be a named theorem though AFAIK Google knows nothing of it!

I'll post a solution to III in a day or two if no one beats me to it. The following comment may help a little, so I've enclosed it in spoiler tags.

A very simple application of calculus is helpful for III. The solution I will show uses a Lagrange multiplier. Recently I posted about this frequently-used technique in the Calculus thread. I was NOT thinking of problem III when I made that post (unless perhaps sub-consciously) but the mention reminded me of the elegant solution to problem III.

But you can solve III without Lagrange multiplier or any other "fancy" technique; especially if you start with the Yes/No (N=2) case rather than the N=33 (Superbowl) case originally posed.

 

[Swammerdami]

Puzzle III is phrased in terms of the Super Bowl, but no knowledge of American football is relevant. The player's knowledge of football is encapsulated in his probability estimates (p₁, p₂, p₃, p₄, ...) Instead of betting on the Super Bowl, the bet could be on how many felony convictions some specific politician has before the end of 2024, or whatever.

III. Betting on the Super Bowl

There are 32 teams that might win the Super Bowl, so we will say that the future is partitioned into 33 possibilities: E₁, E₂, E₃, E₄, ... , E₃₃. I've added a 33rd case, "None of the above" to accommodate weirdnesses and ensure we have a perfect partition. Your task will be to determine the bets x₁, x₂, x₃, x₄, ... , x₃₃ you should place to maximize your risk-adjusted reward.

Warren Buffett once offered $1 billion to anyone who guessed the NCAA men's basketball tournament perfectly, so we'll impose on him again to fashion this betting problem. Both he and you each come up with 33 probability guesses p₁, p₂, p₃, p₄, ... , p₃₃ where p_k is the guessed probability that team #k will win the Super Bowl. (To avoid confusion Buffett's estimates will be denoted q₁, q₂, ... rather than p₁, p₂...) Buffett reveals his final estimates BEFORE you finalize yours, so you're free to use his expertise to modify your own guesses. Buffett's 33 probabilities must sum to exactly 1. The same constraint applies to yours.

Buffett offers to accept wagers without vigorish or commission. If he judges Dallas to be 25% to win, he offers 3:1 odds. If you spend $10 for a Dallas "ticket" the ticket will be worth $40 if Dallas wins. More generally, Buffett offers (1-q) : (q) for any event whose likelihood he judges to be q.

If your probability estimates are exactly the same as Buffett's you may as well buy no tickets at all. Instead you'll be trying to exploit the flaws in Buffett's estimates. Assume that you have total confidence in your own probability estimates.

You have 1 unit of money to bet; assume that's all the wealth you have in the world. (Perhaps 1 unit = $10,000.) After you place your bets, and the NFL season and Super Bowl play out, all but one of your tickets will be worthless. You'll have A dollars where A is the value of your winning ticket (if any) plus the change left over if you didn't spend the entire 1 unit on tickets. Your goal is to maximize the Expected Value of the Logarithm of A. (This is the same as maximizing the "weighted geometric mean" of the various outcomes.) Note that this is different than many betting puzzles where the goal is to maximize the expected value of A itself (or the weighted arithmetic mean). However this is NOT a perverse objective. This criterion for reward-vs-risk is well-known, and is often called the "Kelly Criterion." It is associated with John L. Kelly Jr. and, centuries earlier, with Daniel Bernoulli.

Please ask questions if any of this is unclear. :-(

The description supposes N=33 possible outcomes, but N can be any finite plural integer; and a specific example below will use N=3.

So we are given the bookie's probability estimates ∑q_i = 1 and our own estimates ∑p_i = 1. If w is the winning outcome, a winning ticket for which we pay x_w will be worth (x_w / q_w).

We want to derive our optimal bets x_i with 0 ≤ ∑x_i ≤ 1. Specifically we want to maximize the weighted geometric mean

Maximize ∏ ((x_i / q_i + R)^p_i)​

where R = 1 - ∑x_i is the residual bankroll. Since logarithm is monotonic, this is equivalent to maximizing the expected logarithm of bankroll:

Maximize ∑ (p_i (log (x_i / q_i) + R))​

Note that R may become troublesome. It would be nifty if R=0.

Remark 1.
If we set our bets as x₁ = q₁, x₂ = q₂, x₃ = q₃, etc. the net effect will be the same as buying no tickets at all. All but one of the tickets will become worthless, but the winning ticket will be worth (x_w / q_w), which will be 1, since we've used x_i = q_i for our bets.

Similarly, if we set x₁ = R⋅q₁, x₂ = R⋅q₂, x₃ = R⋅q₃, etc., the value of these tickets will sum to R both before and after the event.

Example:
Suppose our probability estimates are

p₁ = ⅙​

p₂ = ⅓​

p₃ = ½​

And the bookie's estimates are

q₁ = ½​

q₂ = ⅓​

q₃ = ⅙​

As we will see below, our optimal bets are

x₁ = 0​

x₂ = 2/9​

x₃ = 4/9​

Applying Remark 1 we see that the bet set

x₁ = R/2​

x₂ = 2/9 + R/3​

x₃ = 4/9 + R/6​

is also optimal (for any 0 ≤ R ≤ 3/9)

Thus there are an infinity of solutions. In this example, any 0 ≤ x₁ ≤ 1/6 will be optimal, depending on the other x_i. This will also interfere with solution.

The remedy for both problems is simply to enforce ∑x_i = 1; i.e. to insist that R = 0. Remark 1 assures us that this is possible. Now we have a unique solution and a simple solving method using a Lagrange multiplier:

∂ (∑ (p_i log (x_i / q_i)) + λ ⋅ (∑x_i - 1)) / ∂x₁ = 0​

∂ (∑ (p_i log (x_i / q_i)) + λ ⋅ (∑x_i - 1)) / ∂x₂ = 0​

∂ (∑ (p_i log (x_i / q_i)) + λ ⋅ (∑x_i - 1)) / ∂x₃ = 0​

...​

∂ (∑ (p_i log (x_i / q_i)) + λ ⋅ (∑x_i - 1)) / ∂x₃₃ = 0​

∑x_i = 1​

Terms without x_j are irrelevant to ∂ / ∂x_j so these equations simplify dramatically

∂ (p_j log (x_j / q_j) + λ ⋅ x_j) / ∂x_j = 0 ; for j = 1,2,3,...,33​

∑x_i = 1​

Taking the indicated derivative, ∂ (p_j log (x_j / q_j) + λ ⋅ x_j) / ∂x_j = 0 becomes

p_j (q_j / x_j) / q_j + λ = 0 ; for j = 1,2,3,...,33​

The q_j terms cancel! We are left with 33 equations of the form

p_j / x_j = - λ​

Since ∑x_i = 1 and ∑p_i = 1 it is easily seen that λ = -1 and every x_j is equal to the corresponding probability estimate p_j.

Does this surprise? The bookie's odds q_j are irrelevant to the optimal bets (when R=0 is enforced).
We can simply ignore the bookie's odds and buy tickets in exact proportion to our own probability estimates.