More probability puzzles

[Swammerdami]

Inspired by Alcoholic Actuary, I plan on placing some more probability puzzles in this thread. I'll assign them Roman numerals. Please feel free to add other puzzles of your own, labeling them with the next Roman numeral.

I have three "nifty" puzzles to start with, ranging from the quite easy, to one I solve with elementary calculus. (Although derived via calculus, it has an amazingly simple solution, so perhaps there's some elegant intuition or shortcut to find the solution.)


I. Pick a card, any card.

Place a $100 bet on the table; Shuffle an ordinary deck (26 red cards and 26 black cards). and keep it face-down. You will go through the cards one-by-one, keeping the card's color hidden until you announce. For each card announce either "Inspect" or "Red!"; then turn the card over to expose its color.

You must say "Red!" exactly once during this process. (If you "Inspect" 51 times in a row, you are required to select the final card.) If the chosen card is Black you lose your $100. If Red, you're paid $100 for a $200 total.

What is your best strategy? For example, if you say "Inspect, Inspect, Inspect" and the first three cards are Black do you call "Red!" next (the odds are on your side) or do you "Inspect" another hoping that the odds improve even further?

Please hide your solutions in Spoiler tags.

Edited to capitalize Roman, just to please Bilby -- who despite his complaint here doesn't seem to know that proper names are capitalized!

 

[bilby]

I'll assign them roman numerals.

I for one, like Roman numerals.

 

[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?

 

[Alcoholic Actuary]

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?

Spoiler: From a gambling degenerate perspective

And not a mathematician. There sections on the roulette wheel called 'neighborhoods' that consist of 4-6 slots on the wheel next to each other. A dealer (or croupier I guess?) can't hit any particular number or color, or third, or street, or row necessarily. But if you give them a section of the wheel to aim for, they can influence the game based on how fast the wheel is spinning, and when and how fast to throw the ball.

So I guess my strategy would be to select 17 consecutive slots between the zero and double zero and put a $5.88 bet on each of them (100/17) and promise the dealer $5.80 if he can hit that (almost) half of the wheel. (5.88 * 35 = 205.8).

If we assume perfectly random results, the win probability of my strategy is 44.7% compared to betting red/black or even/odd at a 47.4% but I prefer the idea of allowing the dealer to non-randomize the results

aa

 

[Swammerdami]

Interesting, aa. About how many orbits of wheel and ball are there between when the ball is cast and when it settles on a number? (Circa early 1980's there were teams -- Claude Shannon may have been involved -- wearing tiny computers and betting roulette neighborhoods but AFAIK they were operating without dealer connivance.)

In the 1980's I visited Kathmandu's biggest (only?) casino and saw a Wheel of Fortune. (They have these in many Nevada casinos as well.) It's a bit like roulette but the wheel is mounted vertically and the only high payoff is 20:1 which requires that one specific slot (marked "20"). settles at the top of the wheel.

In Las Vegas, betting on the Wheel of Fortune is allowed only BEFORE the wheel is spun, but in Kathmandu betting is permitted until the dealer rings a bell which she does when there's only about ⅔ of a revolution left. I bet the 20:1 when the "20" was about ⅔ a revolution away from the top.

You don't even need to "clock" the wheel to do this! The dealer (who spins the wheel for hours a day) has clocked it, consciously or not, and rings the bell at approximately a fixed amount of time before the wheel comes to a stop. When her hand moves toward the bell you quickly glance at the "20" and if it's in position, make the bet!

I played this way briefly and soon won a 20:1 payoff. I lacked the patience to pursue this after my "proof of concept" victory. (Anyway it didn't seem unlikely that management was quite aware of the issue and changed the procedure whenever they found it being exploited.)

 

[Swammerdami]

All my puzzles here are intended to be pure probability questions. Feel free to make the conversation more interesting and fun by relating real-world aspects, but the intended puzzle and its solution are just mathematical abstractions.

For example, 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 B₁, B₂, B₃, B₄, ... , B₃₃ 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 $10,000 in cash to bet; assume that's all the wealth you have in the world. 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 $10,000 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. :-(

 

[Alcoholic Actuary]

Interesting, aa. About how many orbits of wheel and ball are there between when the ball is cast and when it settles on a number? (Circa early 1980's there were teams -- Claude Shannon may have been involved -- wearing tiny computers and betting roulette neighborhoods but AFAIK they were operating without dealer connivance.)

In the 1980's I visited Kathmandu's biggest (only?) casino and saw a Wheel of Fortune. (They have these in many Nevada casinos as well.) It's a bit like roulette but the wheel is mounted vertically and the only high payoff is 20:1 which requires that one specific slot (marked "20"). settles at the top of the wheel.

In Las Vegas, betting on the Wheel of Fortune is allowed only BEFORE the wheel is spun, but in Kathmandu betting is permitted until the dealer rings a bell which she does when there's only about ⅔ of a revolution left. I bet the 20:1 when the "20" was about ⅔ a revolution away from the top.

You don't even need to "clock" the wheel to do this! The dealer (who spins the wheel for hours a day) has clocked it, consciously or not, and rings the bell at approximately a fixed amount of time before the wheel comes to a stop. When her hand moves toward the bell you quickly glance at the "20" and if it's in position, make the bet!

I played this way briefly and soon won a 20:1 payoff. I lacked the patience to pursue this after my "proof of concept" victory. (Anyway it didn't seem unlikely that management was quite aware of the issue and changed the procedure whenever they found it being exploited.)

Haha, I don't recommend this as a get rich quick strategy - If you have to play roulette and have to double your money, that is the way I'd do it. I've played neighborhoods about 5 times and only won money once if you don't count the free drinks. It's more of a way to play the game for a while without losing too much.

I think the ball has to make at least 4 revolutions or the bet is off. Most dealers shoot 20-25 revolutions and call 'no more bets' once they believe the ball can only make 3-4 more revolutions before dropping. Obviously when the ball drops it can bounce off a divider and hit the other side of the wheel and sometimes the pit boss will require the dealer to reverse the ball direction every other spin. Nothing's guaranteed. But as you noted for the big wheel, there is a muscle memory reflex that can subconsciously convert a 'uniform' distribution into a slightly normal, or bi-modal, or other Gamma distribution that significantly curtails the house advantage.

Also interestingly, the story goes that Vegas started putting the results boards up so that people would see (eg) a string of reds and would come over and bet black because 'it's due' and lose some money. However, you might be able to look at the historical data on the boards and see where this dealer's neighborhood tends to be? So you don't necessarily have to collude with the dealer, just play a section of wheel and tip when you win. They'll get the idea.

aa

 

[Swammerdami]

Although IIDB has many smart posters, it appears very few are the sort of "nerd" who enjoys probability puzzles. I'll give some comments and hints on puzzles I, II and III. Come on, Infidels! A wrong answer is better than no answer at all!

Hints.

I.
This is a very easy problem. The solution, albeit very simple, is very elegant; and can be demonstrated with a simple English paragraph that never mentions the word "probability."

II.
Treat it as a pure probability puzzle. Assume no special knowledge or connivance about the roulette wheel. I chose American roulette for the betting because you lose on average the exact same 5.26% (2/38) of your total bet(s) no matter how you bet. There is a popular belief about roulette betting . . . but which is wrong.

Instead of trying to double your bankroll, try to quadruple it. This may yield some insight.

III.
Instead of 33 separate outcomes, suppose there are just two (Yes and No). Solve that and extrapolation to 33 cases will be easy. You may need one dF=0 calculus working, but it's easy high-school calculus.

Suppose that Buffett offers 2:1 odds on Yes, and 1:2 odds on No, and that you determine that betting 10% of your bankroll on Yes is your best move, keeping the remaining 90% of the bankroll unwagered. You spend (10%, 0) on "tickets." You will get the same result by betting 40% on Yes and 60% on No, that is spending your entire bankroll on tickets (40%, 60%). (A 0.30 ticket paying 2:1 will exactly cancel a 0.60 ticket paying 1:2; you get 0.90 net return in either case.)

Simplify the problem posed in III by having just two outcomes.. It's not necessary but solution becomes more elegant when you assume a requirement to invest 100% of your bankroll on tickets.

 

[bilby]

Edited to capitalize Roman, just to please Bilby -- who despite his complaint here

I didn't make any complaints; Just one of my trademark bad puns.

 

[Swammerdami]

Edited to capitalize Roman, just to please Bilby -- who despite his complaint here

I didn't make any complaints; Just one of my trademark bad puns.

Many of your punny funs are Hillaryous. But that one past right over my head.

 

[steve_bank]

1.
You are walking down the street and find a guy with a box on a table. He says there are 4 balls in the box numbers 1 to 4. He will give you 2/1 odds on your guessing the last ball picked from the box.

The game is fair, you get to reach in and pick the balls. You bet $1 a game for 100 games. How much do you win or loose?

2.
A deck of cards is randomly shuffled. There are 3 players and a dealer. The dealer deals fairly from the top of the deck.

The dealer deals a card for himself and in turn for the players in order dealer-1-2-3 until each person has 5 cards.

What is the probability of the 3rd card dealt to the 2nd player is an ace of spades?

 

[steve_bank]

1.
You are walking down the street and find a guy with a box on a table. He says there are 4 balls in the box numbers 1 to 4. He will give you 2/1 odds on your guessing the last ball picked from the box.

The game is fair, you get to reach in and pick the balls. You bet $1 a game for 100 games. How much do you win or loose?

2.
A deck of cards is randomly shuffled. There are 3 players and a dealer. The dealer deals fairly from the top of the deck.

The dealer deals a card for himself and in turn for the players in order dealer-1-2-3 until each person has 5 cards.

What is the probability of the 3rd card dealt to the 2nd player is an ace of spades?

Spoiler

Both these problems and most of the probability puzzles posted take the same general form, random sampling.

In #1 intuitively there are only 4 choices, so the last pick can only be ball 1,2,3 or 4.

Analytically there are 4! possible sequences, 24 sequences. The probability of 1,2,3 0r 4 as the last pick is 6/24 = 1/4.You can test it by writing 1,2,3 and 4 on ping pong balls or poker chips, putting then in a box, shaking, and reaching in pick 4 balls or chips 100 times.

4. 3. 2. 1.
4. 3. 1. 2.
4. 2. 3. 1.
4. 2. 1. 3.
4. 1. 3. 2.
4. 1. 2. 3.
3. 4. 2. 1.
3. 4. 1. 2.
3. 2. 4. 1.
3. 2. 1. 4.
3. 1. 4. 2.
3. 1. 2. 4.
2. 4. 3. 1.
2. 4. 1. 3.
2. 3. 4. 1.
2. 3. 1. 4.
2. 1. 4. 3.
2. 1. 3. 4.
1. 4. 3. 2.
1. 4. 2. 3.
1. 3. 4. 2.
1. 3. 2. 4.
1. 2. 4. 3.
1. 2. 3. 4.

In #2 it is not sampling from a box full of cards. The deck is randomized. The sequence of the deck from top to bottom is analogous to random sampling. Randomize the deck or sample from a loose deck of cards in a box.

The problem scenario is misdirection. The 3rd card dealt from the deck to player 2 is the 11th card in the deck. Given a deck randomized by a shuffling machine what is the probability of card 11 being an ace of spades?

If you have time to spend another experiment. Do a good shuffle and record card 11. Repeat. The problem is there are 52 choices so you have to do a lot of trials fora statistical significance.

An alternate experiment would be dump the cards in a box, shake, and pick 11 cards.

 

[steve_bank]

It looks like the two problems may be related to the Binomial Distribution.

Binomial distribution - Wikipedia

en.wikipedia.org

 

[bilby]

The problem scenario is misdirection. The 3rd card dealt from the deck to player 2 is the 11th card in the deck. Given a deck randomized by a shuffling machine what is the probability of card 11 being an ace of spades?

1 in 52. The position you draw the card from is completely irrelevant as the deck is in random order. There are 52 cards, one is the ace of spades, the chance of the card at any given position in the deck, immediately after the shuffle, being that Ace is 1 in 52.

 

[Swammerdami]

1 in 52. The position you draw the card from is completely irrelevant as the deck is in random order. There are 52 cards, one is the ace of spades, the chance of the card at any given position in the deck, immediately after the shuffle, being that Ace is 1 in 52.

I am humiliated that you can solve this problem yet not solve my Problem #I.
Would it have helped if I'd specified that my ace was also spayed?

 

[steve_bank]

Where the ball stops on a roulette wheel is an example of a chaotic system.

Small changes in the initial conditions result in wide variations in results over time. The initial conditions can never be exactly duplicated such that the results are repeatable and predicable.

The wheel is spun putting energy into the system. The initial angular velocity will never be exactly the same.

The time between the spinning of the wheel and when the ball is released varies.

The ball's initial velocity, starting position, and momentum varies.

In Vegas known I believe 'card counters' are banned from casinos. At Black Jack they have the probabilities memorized and by seeing shown cards can get better odds at choosing to stand pat or take another card. I think casinos use 2 decks at Black Jack tables to make it harder.

There are chronic gamblers at craps who think they see patterns in the rolls of the dice dice.

Vegas casinos publish the actual demonstrated odds for all the games. When I looked at it before the odds are consistent from month to month with a little variation.

 

[steve_bank]

These problems are all variations on a theme,


Put 10 red and 10 black chips in a box and shake Pull one out and the pliability of black or red is 5/10. Put it back in the box and shake, and pick another. The odds are still 5/10.

That is called sampling with replacement.

Pull the first one out and the pliability of black or red is 5/10. Don't put it back in the box and shake, and pick another. If the first pick was black, the odds for black on the second pick are 4/9, and the odds for red are 5/9. 4/9 + 5/9 = 9/9 = 1 as it should be. At any point the sum of probabilities must equal 1.

This is called sampling without replacement.

Shuffle a deck and put them in a line face down. Turn the first card over. The odds for black or red are 26/52. Pick a second card anywhere in the line. If the first card is black the probability of a black card is 25/51, and the odds of red are 26/51. Sampling without replacement.

No matter how you frame the problem or slice and dice it the probabilities are always
(n red left)/(n total red and black left) and (n black left)/(n total red and black left)

Way back in the 80s when I was reading statistics texts I did the simple experiments to convince myself it worked.

So when I dealt with real problems I always imagined it as sampling balls from a box. Easy to visualize.

Same with parts of QM when I took a night class in modern physics.

 

[Alcoholic Actuary]

Where the ball stops on a roulette wheel is an example of a chaotic system.

Small changes in the initial conditions result in wide variations in results over time. The initial conditions can never be exactly duplicated such that the results are repeatable and predicable.

The wheel is spun putting energy into the system. The initial angular velocity will never be exactly the same.

The time between the spinning of the wheel and when the ball is released varies.

The ball's initial velocity, starting position, and momentum varies.

All true, but none of those is proof that roulette delivers results according to a 'Uniform' distribution - just that they follow some random distribution. The payout pattern (heavily in favor of the house though it may be) is based on uniformly distributed outcomes.

aa

 

[Alcoholic Actuary]

These problems are all variations on a theme,


Put 10 red and 10 black chips in a box and shake Pull one out and the pliability of black or red is 5/10. Put it back in the box and shake, and pick another. The odds are still 5/10.

That is called sampling with replacement.

Pull the first one out and the pliability of black or red is 5/10. Don't put it back in the box and shake, and pick another. If the first pick was black, the odds for black on the second pick are 4/9, and the odds for red are 5/9. 4/9 + 5/9 = 9/9 = 1 as it should be. At any point the sum of probabilities must equal 1.

This is called sampling without replacement.

Shuffle a deck and put them in a line face down. Turn the first card over. The odds for black or red are 26/52. Pick a second card anywhere in the line. If the first card is black the probability of a black card is 25/51, and the odds of red are 26/51. Sampling without replacement.

No matter how you frame the problem or slice and dice it the probabilities are always
(n red left)/(n total red and black left) and (n black left)/(n total red and black left)

Way back in the 80s when I was reading statistics texts I did the simple experiments to convince myself it worked.

So when I dealt with real problems I always imagined it as sampling balls from a box. Easy to visualize.

Same with parts of QM when I took a night class in modern physics.

The basis of the gameshow Deal or No Deal.

aa

 

[steve_bank]

The definition of a random variable is no correlation between events.

Put 10 balls in a box, shake, pick one, and put it back. Pick again. The first pick does not affect the odds of the second picks.

Uncorrelated random events.

I read through Knuth's Semi Numeral Algorithms. There is actually controversy ver what random is. He said you define what it is, and anything that fits the definition is than random.

A uniform distribution is a theoretical definition. Anything that closely approximates a uniform distribution by test is considered uniformly distributed.

I am sure roulette has been subject to thorough formal testing.

On a given wheel there could certainly be patterns, but I'd think picking them out would require a lot of data.

All casinos care about is that it is inform enough that it is not easily hacked and the odds are consistent enough so the house always comes out ahead.

Same with casino grade dice. The depth of dimples are adjusted so the missing mass is the same on each side for balance. Are the odds for a single die exactly 1/6? Probably not.

By the way aa, if I may ask did you take the actuarial exams?