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Two Tricky Probability Problems

By Moloy De posted Sat May 13, 2023 05:12 AM

  
Problem 1: 
 
What is the probability, P, a family has 2 girls, given they have 1 girl?
 
I first saw this problem when trying to learn about Bayes’ Theorem. “Easy!” I thought. “Given that they already have one girl… P(2 girls | 1 girl) = P(girl) = 50%.” I was shocked when I saw the answer actually is 1/3.
 
“Where did I go wrong?!” I spluttered. 5 minutes pass. The math was so simple, so elegant, I couldn’t possibly have made a mistake. Until it hit me.
 
ALREADY. “Given that they already have one girl.”
 
I had unwittingly made an assumption about the order of the data. I assumed they had one girl first, thus believing the only possible kid combinations were GG and GB, meaning P(2 girls | 1 girl)=50% is correct! Except, nowhere does it say the “given girl” had to be first, meaning the correct possibilities are BG, GB, GG and P(2 girls | 1 girl) = 1/3.
 
Lesson: Don’t implicitly assume the order of the data
 
 
Problem 2: 
 
You are given $1 dollar and a magic coin. If you flip heads, you double the money. If it lands tails, you halve it. What is the expected amount of money after 1 flip?

This question comes from All of Statistics, a fantastic read by Dr. Wasserman.
 
When first approaching this problem, I figured that the probability the coin lands heads and tails is 50% each so half the time it comes up heads, the other half tails. Thus, the money would double to $2 and halve back to $1 or halve to 50 cents and then double to $1. Thus, I concluded 𝔼[money after 1 flip] = $1.
 
I was wrong.
 
The correct answer is a weighted average of the outcomes: 𝔼[d after 1 flip] = double ⋅  P(heads) + halve ⋅ P(tails) = $2 ⋅ P(heads) + $0.5 ⋅ P(tails) = $1.25
 
Hold up. How could the money be $1.25? The money can only double or half so it must be a multiple of $1 e.g., $0.5, $1, $2, $4. Intuitively, my logic seems sound so where did I go wrong?
 
It turns out, I was thinking about the “expected value” incorrectly in two ways:
 
The expected value actually means, “If I were to repeat this experiment infinite times, what is the average of all of those trials?”  My original mistake was to keep flipping the same coin — i.e., double to $2 then halve to $1. Instead, each individual flip should be independent of the others.

Counterintuitively, the mean can be a value that is impossible for the distribution. For example, the money can never be $1.25 — it will only double to $2 or halve to $0.5, yet the “average/mean/expected value” is a weighted sum of the outcomes. Another example: in a Bernoulli distribution the result can only be 0 or 1, yet the average/expected value of the distribution is 𝔼[X] = 0.5
I still wanted to prove it to myself, so I simulated the experiment for 100,000 coin flips, getting 𝔼[X] = $1.2497 ≈ $1.25
 
The expected value/mean/average of an experiment means “repeat this experiment many times — what’s the average across all trials?” Also, note that each individual trial should be independent of the others.

The mean/average/expected value can be a value that’s not even in the distribution itself!

Thank you for reading! 


QUESTION I: Start a game paying $1 in a pot. If Head appears the host doubles the money in the pot. If Tail appears the host takes away all the money in the pot and you need to start afresh putting $1 in the pot. What would be your strategy for the game to maximize your return?
QUESTION II: You are at zero. If head appears you go to one and if Tail appears you go to minus one. You continue your movement while tossing afresh once you complete a move. Would you ever be able to come back to zero?

REFERENCE: Two Tricky Probability Problems Blog

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Olakunle Olaniyi
16 days ago

In this game, the outcome of each round is purely random, so there's no strategy that can guarantee maximizing returns over the long term. However, we can discuss a strategy that aims to minimize losses and potentially capitalize on winning streaks.

One approach to consider is the concept of "Kelly Criterion." The Kelly Criterion is a formula used in gambling and investing to determine the optimal size of a series of bets or investments to maximize wealth over time. In our simplified game, we can adapt this concept by betting a fraction of our current wealth on each round.

Let's denote:


  •  as the probability of getting a Head (assuming it's 0.5 in this fair coin toss game).

  •  as the probability of getting a Tail (

    ).

  •  as the reward (2 times the initial bet if a Head appears, 0 if a Tail appears).

The Kelly Criterion formula is given by:



where:



 is the fraction of wealth to bet on each round.

In our case,

and

. Plugging these values into the formula:



This means that the optimal fraction of wealth to bet on each round according to the Kelly Criterion is 0.25, or 25%.

So, one strategy to maximize returns over time would be to consistently bet 25% of your current wealth on each round of the game. This strategy aims to balance between maximizing potential gains while minimizing potential losses, assuming the game is fair and outcomes are independent.

However, it's important to note that even with this strategy, there's no guarantee of making a profit in the long run due to the inherent randomness of the game. It's possible to experience both winning streaks and losing streaks, and the outcome of each round remains uncertain.