# Tag Info

42

It seems to me that there's a simpler solution than the one accepted above. Step 1: Step 2: Step 3: The point here is that

31

One possibility: This works because: EDIT: Inspired by @trolley813's answer here is a way to recycle the rejected entropy:

28

OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that. The simplest somewhat-plausible way to handle #1 ...

25

I see, it took me too long to fininsh my drawing, but let me present it as additional material to sousben's answer:

19

There is a way that doesn't use the dice (3+ coin-flips): Flip the coin three times and record 1 or 0 for heads or tails. If all three numbers are zero start again. Otherwise read the 1s and 0s as a binary number. Its value will be from 1 to 7 inclusive with each number having an equal probability of occurring. Here's a summary of Mike's answer (2+ dice-...

18

Yes, you can do it like this: Why does this work?

17

The row of coins can be fully removed if it has the following property: Proof: Solving strategy:

16

This gambling problem is the famous St. Petersburg paradox. It is a paradox because The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the ...

14

Generalization. There are $K$ coins on the table and one player can pick as many as M coins at once.

13

Using winning/losing position analysis you can tell that the correct move is to take: You can work this out iteratively. If you have 1 coin and it is your turn you will obviously lose. Thus 1 is a losing condition. We can then create a set of winning conditions: You can then use that to work out more losing conditions. You can then extend this logic to ...

11

I would pay

10

Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Then This makes a total of:

9

For my answer I'm considering the sides of the coin to either have a star or be blank for ease of writing. I also will write out the number for peoples name, and use the character for guesses to try and keep things legible. (Ex. Mr Three guessed 4). Also when typing out sequences, + means star and - means blank. (Ex ---- means no stars / all blanks) I agree ...

8

There are several nuances to this question. First of all, it asks how much you are willing to pay, not what price is fair. Second, you have to understand, that even if a game is fair, that does not mean that It is reasonable to play it. For example, if someone offers me a one in a million chance to win a million dollars for 1, I will take it. It seems ...

8

I would not pay anything. I would not play. I would encourage you to not play. Are you doing okay? I'm willing to help you out of you need help. I would offer you a hug. You are my best friend, and you live right next to me. Any outcome of this game that would be monetarily meaningful to either of us would also most likely be highly damaging to our ...

7

The first toss is a star (*) or not (-). Mr Two immediately loses. The second flip can be either, so we have **, *-. -*, or --. Also on that flip, The third and fourth flip and To sum:

7

X should pick up Because Examples:

7

One Dollar One dollar Which should look something like the following image:

7

This works because

6

Roll the die twice, to generate a random number between 0 and 35. Specifically, subtract one from each roll, and have the first roll be the "tens" digit of a number in base 6, and the second roll be the ones digit. If the result nonzero, return the remainder of the result (mod $7$). If the result is zero, then repeat until it is nonzero. This gives one of ...

6

It is possible, but only if you test 5 coins the first time. The difficulty is that, if all the tests up to a certain point have answered No, and there are $k$ coins remaining, then there are $k\choose 2$ combinations remaining. This limits how evenly we can divide the remaining combinations with each test. Specifically, for 6 coins, there are $15 < 2^... 6 First puzzle in 4 moves: Second puzzle in 5 moves: The way I solved these is I do not know if these are optimal. I would not be surprised if shorter solutions are possible. 6 6$16, Consider what else we know: You are my best friend The currency you're using is US dollars You own a home (namely the one next to mine) What do these things imply? Because I'm your best friend, I won't want you to have to move away by you having to sell your house in order to pay me. We're in the USA The median price of a home in the USA is around \$...

6

5

5

Solution where all fake coins weigh the same and all regular coins weigh the same (and all fake are lighter). Label the fakes ABCDEFG, regulars abcdefg. Test A vs. a. See that A < a, and therefore A is fake and a is real. Test aBC vs. Abc. See that aBC < Abc. Since A > a, we have 1 unit of difference going to the left, so it must take at least ...

5

This is a slightly lateral-thinking answer, and is definitely against the spirit of the question, if not its letter. But there's a lot to recommend this approach, enough that I felt it was worth providing. Now, I need to begin by admitting that this approach is only convenient if you happen to live in Westeros, from Game of Thrones. (And if you do, my ...

5

Here's an answer that may take more steps to complete but is a lot more intuitive to think about: Roll the die and flip the coin simultaneously. If the coin lands on heads, read the die value. If the coin lands on tails and the die lands on 1, get the value 7. Else, repeat the process.

5

I think that Therefore, Of course,

Only top voted, non community-wiki answers of a minimum length are eligible