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MAT199: Math Alive Voting, Power, and Sharing Ian Griffiths Mathematical Institute, University of Oxford, Department of Mathematics, Princeton University

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A dinner party To celebrate the start of the Math Alive lectures we decide to go out to dinner. Because there are a lot of us, Winberies can only offer a set menu. They give us a choice of chicken or salmon.

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A dinner party To celebrate the start of the Math Alive lectures we decide to go out to dinner. Because there are a lot of us, Winberies can only offer a set menu. They give us a choice of chicken or salmon. A quick poll shows that more of us prefer salmon than chicken. So we go with salmon.

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A dinner party To celebrate the start of the Math Alive lectures we decide to go out to dinner. Because there are a lot of us, Winberies can only offer a set menu. They give us a choice of chicken or salmon. A quick poll shows that more of us prefer salmon than chicken. So we go with salmon. When we call the restaurant they inform us that the salmon is off and it has been replaced with beef. We do another quick poll and find that more of us prefer the chicken than the beef.

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A dinner party To celebrate the start of the Math Alive lectures we decide to go out to dinner. Because there are a lot of us, Winberies can only offer a set menu. They give us a choice of chicken or salmon. A quick poll shows that more of us prefer salmon than chicken. So we go with salmon. When we call the restaurant they inform us that the salmon is off and it has been replaced with beef. We do another quick poll and find that more of us prefer the chicken than the beef. We call back to confirm we would like chicken and Winberies inform us that the chicken is now off but the salmon is back on. The choice is now salmon or beef. Do we need to do another poll to make sure the majority get the one they prefer?

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Nicolas de Caritat – Marquis de Condorcet 1743 – 1794

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Full English breakfast 1743 – 1794

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Summary of lecture 1 Paradoxes can arise when ranking preferences (our Winberies dinner experience). A clear winner is called a Condorcet winner. If there isnt a clear winner there are various ways of selecting the winner: Plurality – the choice with the most first places wins. Plurality with run-off – the two people with the most and second-most first places are pitted against one another in another election. Sequential run-off (the Hare system) – eliminate the bottom candidate at each round. Borda count – assign a score for top choice, second choice, and so on, and add up the total scores. The winner is the one with the highest score. Voting and social choice

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Holiday destinations Cornwall

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Holiday destinations Scotland

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Holiday destinations Jersey (the real one)

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Kenneth Arrow Born in 1921

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Table showing fraction of non-Condorcet winners

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Summary of lecture 2 If there isnt a clear (Condorcet) winner then the way in which you determine the winner often leads to a different result. You cannot produce a universal fair voting scheme (proof by Kenneth Arrow). …but provided there arent too many choices, it is likely that you get a Condorcet winner anyway so there is no issue. You can share a cake between two people in an envy free way by letting one person cut and the other choose. Sharing a cake between three people in a fair way is straightforward. Sharing a cake three ways in an envy free way is harder…

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Summary of lecture 3 A cake may be shared three ways in an envy-free way. Generalization for four ways is harder (Brams and Taylor, 1992). There is no clear way of generalizing this for five or more people. We may divide up several items two ways in an envy-free way (e.g., a divorce settlement). We can divide up several items more than two ways in a fair but not envy-free way (e.g., dividing up an estate).

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The Math Alive Barbecue

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To be given a burger at the Math Alive Barbecue you must be able to call out the colour of the hat you are wearing. You must call out one at a time (but in no required order). You all stand in a queue facing forward. You are not allowed to look behind you. You may discuss a strategy before the barbecue. What is the strategy that guarantees at least all but one members calling the correct colour? The Math Alive Barbecue

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Strategy Lab member 1 counts the number of red hats he sees and calls out red if he sees an even number and blue if he sees an odd number. 1 2 3 4 5 6 7 8 The Math Alive Barbecue

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Strategy Lab member 1 counts the number of red hats he sees and calls out red if he sees an even number and blue if he sees an odd number. 1 2 3 4 5 6 7 8 Red The Math Alive Barbecue

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Strategy Lab member 1 counts the number of red hats he sees and calls out red if he sees an even number and blue if he sees an odd number. Member 2 counts the number of red hats he sees. He sees an odd number (3) so knows that his hat must be red and so calls out red and guesses correctly. 1 2 3 4 5 6 7 8 Red The Math Alive Barbecue

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Strategy Lab member 1 counts the number of red hats he sees and calls out red if he sees an even number and blue if he sees an odd number. Member 2 counts the number of red hats he sees. He sees an odd number (3) so knows that his hat must be red and so calls out red and guesses correctly. 1 2 3 4 5 6 7 8 Red The Math Alive Barbecue

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Strategy Lab member 1 counts the number of red hats he sees and calls out red if he sees an even number and blue if he sees an odd number. Member 2 counts the number of red hats he sees. He sees an odd number (3) so knows that his hat must be red and so calls out red and guesses correctly. Member 3 counts and sees an odd number too (3) so knows his hat is blue. This proceeds to the end of the line. 1 2 3 4 5 6 7 8 Red Blue The Math Alive Barbecue

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Bending light Light will always take the fastest route between two places.

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