A coin is tossed repeatedly. If it comes up heads Pascal gets a point but if it comes up tails Fermat wins a point. The first person to win three points is the winner and receives the prize of £12.
Unfortunately the game had to end abruptly after three tosses of the coin. Pascal had two points and Fermat had one point. They decided to share the £12 in a ratio that matched the probability of them winning the game if it had continued.
How should they divide the £12?
You may be surprised at the correct answer as it is not £8 and £4!
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The £12 should be divided in the ratio of the probabilities of the two players winning which can best be shown in a simple tree diagram.
Alternatively consider the following equally likely scenarios for the outcome of the next two tosses:
Three of the four possibilities would make Pascal the winner so the prize should be shared in the ratio 3:1 meaning that Pascal should receive £9 and Fermat should receive £3.
Eric Levy, United States
Thursday, November 17, 2016
"Thank you for the podcasts! I really enjoy the puzzles. This relates to the 5/29/15 podcast re: coin flipping game that was stopped before completion. The flips when stopped were two Heads and one Tail. You indicate that the options on next 2 tosses are HH, TH, HT, and TT. Since the game stops when one person reaches 3 points, wouldn't HH and HT be the same, as the second flip isn't needed? This seems to match your ultimate answer of 75% and 25%, though ... with the person with Tails only winning with TT, which is 25% chance. I get the same answer but intermediate steps differ. Am I looking at this incorrectly? Thanks!"
Thursday, November 17, 2016
"Dear Eric, Thanks so much for your observations and you are completely correct. The only reason I chose to list the next two outcomes was to produce ‘equally likely’ outcomes making the arithmetic very slightly easier. I’m glad you enjoy the puzzles."
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