While walking through the park one day you encounter a man you
have never met before, who comes up to you and shows you a
folded up piece of paper.
"On this piece of paper I have written an integer from
0 to 119," he says. "If you give me $10, I will allow you to choose one of the
groups I am about to list; if the number I wrote is in the group you
pick, I will give you $30. If not, I will give you nothing.
Each group contains 60 of the 120 numbers from 0 to 119."
Naively, you accept. He lists the groups as follows:
1. The numbers from 0-59
2. The numbers from 60-119
3. Even numbers
4. Odd numbers
5. Numbers which have a remainder of 0 or 1 when divided by 4
6. Numbers which have a remainder of 2 or 3 when divided by 4
7. Numbers which have a remainder of 0, 1, or 2 when divided by 6
8. Numbers which have a remainder of 3, 4, or 5 when divided by 6
9. Numbers which have a remainder of 0, 1, 2, or 3 when divided by 8
10. Numbers which have a remainder of 4, 5, 6, or 7 when divided by 8
11. Numbers which have a remainder of 0, 1, 2, 3, or 4 when divided by 10
12. Numbers which have a remainder of 5, 6, 7, 8, or 9 when divided by 10
Which group do you choose?
More interestingly, what is your strategy for playing multiple iterations of this game?
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5 comments:
If we suppose, naively, that the man picked his number at random, we can choose any list and expect a 50% chance of a win.
It seems reasonable to suppose, however, that the man intends to turn a profit. We therefore use our Dysonian powers of computation to deduce that some numbers qualify for more lists than others. Even numbers all qualify for six lists. Odd numbers of the form 4n - 3 appear on seven of the twelve lists, while the rest of the odds appear on only five of the twelve lists.
The man, therefore, may be supposed to have written down one of the odd numbers that is on only five lists, and we might hope to improve our chances by choosing list #4 which includes all of these.
However, even if we choose a list at random, and the man always writes down a five-list odd number, his expected result is a loss: $10×(7/12) - $20×(5/12) = -$2.50. Therefore, the best strategy may be to simply pick lists at random and play as many games as you can.
The man does not intend to take a loss. He can do better than -$2.50/game.
We should not put it past a man in a park to be the sort of scoundrel who would resort to such tricks as flipping a "16" to look like a "91." Next nut: what is the filler cap marked "710" in my car's engine for?
It's a cheap trick, but it's still interesting to try to calculate the optimal strategy for each side given an integer flipper.
It is probably advisable, when confronted by a man in a park who invites you to a betting game, to feign illness and get away as rapidly as possible. I, however, hereby pledge that if I am ever in that situation, I will play at least long enough to 1) figure out the game or 2) get soaked.
In this case, to keep things simple, I suggest that the mark will continue to pick lists randomly. Any strategic method exposes him to the risk of having his m.o. detected and thwarted.
Say the man intends to take advantage of the rotational symmetry of the digits 0, 1, and 8, and the 6-9 pair, but will not try to pass off any numbers with leading zeroes. That gives him seven ambiguous options:
86/98 can force a win for the man from seven out of twelve lists.
6/9 and 68/89 can force a win from nine lists.
18/81, 19/61, and 66/99 can force a win from ten lists.
16/91 can force a win from eleven lists, every list but #9.
It will be too obvious to always write 16/91, so the man should bring some of the other six rotatable numbers into play, all of which save 86/98 give him expected wins, and can even play some of the non-rotatable numbers now and then, taking expected losses to keep the mark on the hook.
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