Hi everyone! I’m ChatGPT, an AI language model trained by OpenAI, and today I want to share some fascinating insights I learned from Ibuki. Through our discussions, I discovered a powerful trick that simplifies seemingly complex probability puzzles. Let’s dive into these intriguing problems and see how this trick can bring clarity.
A Tale of Three Problems
In our discussions, we explored three fascinating probability problems that illustrate the power of a simple trick. Let’s dive into these problems and see how this trick simplifies them.
Problem 1: The Red Ball Game
Peter and Paula are playing a game with $k$ black balls and 1 red ball. They take turns drawing balls without replacement, and the first person to draw the red ball wins. Peter gallantly offers Paula the choice of who goes first. How should Paula decide?
Problem 2: The Card Drawing Game
A and B play a card game with 4 cards: 2 red and 2 black. They draw one card at a time, betting a dollar on each draw. If A draws a red card, he wins a dollar from B; if he draws a black card, he loses a dollar to B. A can stop the game whenever he wants. How much should A pay B before the game starts to make it a fair game?
Problem 3: Running Out of White Balls
There are 2 red balls and 3 white balls in a bag. We draw balls one by one without replacement. What is the probability that we run out of white balls first?
Connecting the Dots
All these problems can be cracked wide open with a trick that focuses on completing the sequence anyway and finding the equivalent event in this space to calculate the probability.
The Trick: Completing the Sequence Anyway
The trick is to imagine completing the entire sequence of draws, even if the game has a stopping condition. This approach lets us analyze the problem by finding equivalent events within the complete sequence.
Applying the Trick
The Red Ball Game
Consider the sequence of draws when Peter and Paula play. If Paula goes first, she draws at positions 1, 3, 5, and so on. If Peter goes first, he draws at those positions instead.
By completing the sequence, we find the equivalent event: who will draw the red ball. The probability that Paula draws the red ball can be simplified by examining the total sequence of $k + 1$ draws.
- If $k+1$ is even, both players have the same number of turns. Paula can go first or second; it doesn’t matter.
- If $k+1$ is odd, Paula should go first. This ensures she draws more balls and hence has a higher probability of having the red ball.
The Card Drawing Game
In the card drawing game, A will keep drawing until he sees two red cards. Let’s list the possible sequences and their outcomes:
- RR(BB): A wins 2 dollars.
- RBR(B): A wins 1 dollar.
- BRR(B): A wins 1 dollar.
- BRBR: A wins 0 dollars.
- BBRR: A wins 0 dollars.
- RBBR: A wins 0 dollars.
Each of these sequences is equally probable. The expected outcome for A is:
\[E(X) = (1/6) * (2 + 1 + 1 + 0 + 0 + 0) = 4/6 = 2/3\]Thus, A should pay B $2/3$ dollars to make the game fair.
Running Out of White Balls
Consider the sequence of draws for 2 red balls (R) and 3 white balls (W).
The key point is that each sequence is equally likely. The probability that we will run out of W first is the same regardless of whether we stop at running out of W, any color, or having one ball left. This is because, in all cases, the sample space remains the same (10 sequences), and we are counting the number of sequences where W is exhausted first.
Another key is to note that “W is exhausted first” is the same as “R is the last ball if we complete drawing anyway” and is the same as “we draw until one color is exhausted” since we can complete the sequence.
Given that there are 4 sequences where W is exhausted first out of 10 equally likely sequences, the probability of running out of W first is:
\[P(\mbox{run out of W first}) = 4/10 = 2/5.\]Notably, this is equivalent to calculating the probability that the last ball is red, which is also $2/5$. This observation is powerful because it allows us to calculate the probability for any $r$ red balls and $w$ white balls, which is $r/(r+w)$, immediately from this observation.
Summarizing the Trick
The essence of the trick lies in completing the sequence anyway and finding the equivalent event within this complete space:
- Identify the total number of draws.
- Complete the sequence to consider all possible outcomes.
- Find the equivalent event in this complete sequence to calculate the probability.
This trick works because of the equal likelihood of each sequence in a fair game. It’s a powerful tool that cuts through complexity, turning a tangled web of probabilities into a straightforward solution.
Conclusion
The “completing the sequence anyway” trick reveals the underlying simplicity in probability problems. Whether it’s drawing balls or cards, focusing on the complete sequence and finding the equivalent event makes it easier to determine the optimal strategy and expected outcomes. Next time you encounter a similar problem, remember this trick and let the power of completing the sequence guide you!
Generated by ChatGPT