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The Big Lock-Down Math-Off, Match 24

Welcome to the twenty-fourth match in this year’s Big Math-Off. Take a look at the two interesting bits of maths below, and vote for your favourite.

You can still submit pitches, and anyone can enter: instructions are in the announcement post.

Here are today’s two pitches.

James Tanton – Tie folding and a link to Artin’s conjecture

James Tanton is a prolific maths education thinker and the inventor of Exploding Dots. You can find him on Twitter at @JamesTanton where he regularly posts chin-scratching maths problems, or at jamestanton.com.

Who knew that such simple fun as folding a tie can connect with unsolved problems in number theory, and perhaps even provide some new insights?

Clare Wallace – A princess problem

Clare is a mathematician at Durham University. She tweets at @Clare_L_Wallace.

A few months ago, I was volun-told that it was my turn to present at the postgraduate seminar in Probability and Statistics at my university. Never one to miss an opportunity to gain a talk for my own (much less frequent) Pure Maths seminar series, I agreed on the condition of a joint talk between the two groups. Rather than trying to explain my research to two completely different groups of mathematicians, I set out to find some interesting, unintuitive probability questions I could ask my fellow grad students. The Sleeping Beauty Problem is one of them.

Here’s how it works.

Sleeping Beauty has signed up to be part of our psychology experiment. (No princesses were harmed in the making of this pitch!) She’s going to take a sleeping potion on Sunday night, and go and lie down. Then, the researchers will flip a coin. If it’s heads, they will do the Interview Procedure on both Monday and Tuesday; if it’s tails, they will only do it on Monday.

Interview Procedure:

  • Wake up Sleeping Beauty
  • Conduct an Interview
  • Give her two potions to take: a forgetting potion, and a sleeping potion

The forgetting potion means that Sleeping Beauty will not remember whether or not she’s already been interviewed, but she won’t forget the rules of the experiment.

The question is this: Sleeping Beauty is in an Interview, and she’s asked: “From your perspective, what is the probability that the coin landed tails?”

Here it gets tricky.

The argument for 1/2

Heads and tails were equally likely to start with, and Sleeping Beauty hasn’t learned anything: she always knew she would wake up at some point in the process. She has no new information, so she should stick with what she’d have answered before the experiment:

\[ \mathbb{P}(\text{Tails}) = \frac{1}{2} \]

The argument for 1/3

This one is a bit more involved.

Firstly: If the coin landed heads, it’s equally likely, as far as Sleeping Beauty is concerned, that the interview is taking place on Monday or Tuesday.

So

\[ \mathbb{P}(\text{Monday} | \text{Heads}) = \mathbb{P}(\text{Tuesday} | \text{Heads}) \]

We can use a cheeky Bayes’ Theorem here:

\[ \frac{ \mathbb{P}(\text{Monday and Heads})}{\mathbb{P}(\text{Heads})} = \frac{\mathbb{P}(\text{Tuesday and Heads})}{\mathbb{P}(\text{Heads})} \]

On the other hand, learning that it’s Monday doesn’t tell us anything about the coin toss. So

\[ \mathbb{P}(\text{Tails} | \text{Monday}) = \mathbb{P}(\text{Heads} | \text{Monday}) \]

Our old friend Bayes makes a comeback:

\[ \frac{ \mathbb{P}(\text{Monday and Tails})}{\mathbb{P}(\text{Monday})} = \frac{\mathbb{P}(\text{Monday and Heads})}{\mathbb{P}(\text{Monday})} \]

And now we’ve convinced ourselves that

\[ \mathbb{P}(\text{Monday and Tails}) = \mathbb{P}(\text{Monday and Heads}) = \mathbb{P}(\text{Tuesday and Heads}) \]

Three equally likely options; nothing else can happen. “Tails and Monday” is the same thing as “Tails”, so

\[ \mathbb{P}(\text{Tails}) = \frac{1}{3} \]

This problem caused a lot of trouble amongst the grad students! By the end of a week of lunchtime discussions, we were (almost) all convinced by the 1/3 argument. My friend Kieran Richards took it one step further, and found a way to switch the probabilities. This time, we roll a six-sided die, and wake up Sleeping Beauty twice if we roll 1, 2, or 3, and once if we roll 4, 5, or 6. In the Interview, Sleeping Beauty is told that the roll was even, and asked “From your perspective, what is the probability that the roll was at least a 4?” In this scenario, following the first argument gives you an answer of 1/3, and following the second gives you an answer of 1/2…


So, which bit of maths made you say “Aha!” the loudest? Vote:

Match 24: James Tanton vs Clare Wallace

James with tie-folding
(50%, 23 Votes) 23 Votes
Clare with a right royal problem
(50%, 23 Votes) 23 Votes

Total Voters: 46

This poll is closed.

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The poll closes at 9am BST on Saturday the 30th, when the next match starts.

If you’ve been inspired to share your own bit of maths, look at the announcement post for how to send it in. The Big Lockdown Math-Off will keep running until we run out of pitches or we’re allowed outside again, whichever comes first.

One Response to “The Big Lock-Down Math-Off, Match 24”

  1. Avatar Jeff Jo

    I’m not sure if this is the correct avenue, but there transformed problems that can solve the Sleeping Beauty Problem. (They are really the same transformation, but the presentation is different.)

    1. Use four princesses. Each is given a note card with two words on it. Two of the card say “Heads” and two say “Tails.” One of each of those also says “Monday” and the others say “Tuesday.” The majority of the the same as in the original problem. The same coin is flipped on Sunday Night. On Monday, the interview procedure is performed on any princess whose card says “Monday”, or says the result of the coin flip. This is repeated on Tuesday, with the obvious change that they wake princesses whose card says “Tuesday” instead of “Monday.” The question in the interview is “From your perspective, what is the probability that the coin landed on the face printed on your card?”

    The princess with the card that says “Monday, Heads” is undergoing the original procedure. The others are undergoing an equivalent one. But for each, the question is equivalent to “From your perspective, what is the probability that your note card says a different coin fact than the other two awake princesses?” The answer is clearly 1/3.

    2. One princess this time, but two coins. The interview procedure is performed on Monday if either coin shows Heads. Then on Monday Night, coin #2 is turned over, and the procedure is again perfromed on Tuesday if either coin shows Heads. The question is the same, but refers to coin #1.

    Before it is determined whether to wake her, there are four equally-likely combinations for the current state of the coins: HH, HT, TH, and TT. Since the princess knows she wouldn’t have been wakened if it was TT, that case is eliminated. Their remains only one case where coin #1 is Tails, making the answer 1/3.

    Reply

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