Paradoxical games and a minimal model for a Brownian motor

American Journal of Physics

Volumn 73, Issue 2, 2005, Pages 178-183

  Astumian, R Dean ^a  

^a University of Maine   (United States)

Author keywords

[No Author keywords available]

Indexed keywords

EID: 33747108377     PISSN: 00029505     EISSN: None     Source Type: Journal    
DOI: 10.1119/1.1801191     Document Type: Review

References (14)

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10. This example is a realization of the simplest version of Simpson's paradox with dice. Take one 10-sided die, and one 20-sided die (available at many novelty and gaming stores). Your number on the 10-sided die is 1 (your opponent has 2-10), and on the 20-sided die your numbers are 1-16, and your opponents are 17-20. You each take one of the two dice and roll it as many times as it has sides. The winner is the one who rolls his/her numbers on their die more frequently. If you take the 10-sided die, you expect your number to come up once out of the ten rolls (10%), while one of your opponents numbers will on average appear on his/her die 4 out of the 20 rolls (20%). Clearly, you will generally lose this game. On the other hand, if you take the 20-sided die, your number will appear 16 times out of the 20 rolls on average (80%), but your opponents number on the 10-sided die will appear 9 out of the 10 times (90%). Once again, you lose. However, if you both roll the 10-sided die ten times, and the 20-sided die 20 times, you expect your number to appear a total of 17 times out of the total of 30 rolls (57%), and your opponents number will appear a total of 13 times in the 30 rolls (43%). You win! For more on Simpson's paradox, see the web site 〈http://plato.stanford.edu/entries/paradox-simpson/ 〉.
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14. O. Percus and J. Percus, "Can two wrongs make a right? Coin-tossing games and Parrondo's paradox," Math. Intell. 24, 68-72 (2002).