Who can go the lowest?

[BMAD]

A game token costs $10 to play.  The pay out is $100.  You can purchase multiple entries if you desire.  For each entry you purchase, you must pick the lowest positive number that no one else picks.  If there are ten people, including yourself, seeking to purchase tokens, what is your strategy?

[BobbyGo]

Spoiler

Assuming that any positive number is valid, I'd buy one token and pick:

0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

as my number.

 

[Buddyboy3000]

Using the same method that BobbyGo took, my number would be a...

Spoiler

...1/googolplexianth. A googolplexianth is a one followed by a googolplexian of zeros. This number can also be written out as 10^10^10^10^10^10. So, the smallest positive number I can pick is (1/googolplexianth).

 

[phil1882]

i'd probably pick

1/tree(3)

 

[harey]

Bad news guys, I win:

Spoiler

https://en.wikipedia.org/wiki/Signed_zero

But a mathematician will not agree.

[Pickett]

I guess I took this question to be something different. Rather than just "what's the smallest positive value you can think of?" for which I would go with something like: 

1^{-(3↑↑↑↑...↑↑3)} 
or in other words: 1 over Graham's number (or simply "G")...but why stop there? why not go with
1/G↑↑↑↑↑↑↑...↑↑↑↑↑↑↑↑G

That interpretation of the OP isn't as interesting or fun because it just gets ridiculous... The way I took the original question was pick the smallest positive INTEGER that no one else picks...what is your strategy when up against 9 others doing the same (all allowed to pick however many tokens as they want)?

So, if that's the case, it becomes a much more challenging problem. Obviously it doesn't make sense to purchase more than 9 tokens, as with 10 the BEST you could hope for is breaking even...I would probably purchase the following numbers (and my rationale next to them):

• 1 - I'm assuming I will lose this $10, but maybe everyone else will think someone else picks this and therefore no one else does!
• 2 - Might as well try this one as well, but again, most likely will lose this $10 as well...
• 7...ish - Some relatively low number that hopefully no one else picks...I would assume I'd lose this.
• 10 - Let's assume everyone buys their 9 tokens and picks 1-9...10 would be the first number that no one would pick...
• 16...ish - Again, some pretty low number, that's pretty much just a shot in the dark and you hope no one else picks it. 
• 42 - If by some crazy chance everyone buys 9 tokens and all are duplicated, this is the lowest possible number that wouldn't be duplicated...plus isn't it really the answer to everything?

At this point, I've already bet $60 in hopes of winning $100...and really, the odds are still not in my favor...so to me, it's not worth the risk and so my ultimate strategy would be to not play and "break even" :c)

[BMAD]

yes, i meant integer

[Quantum.Mechanic]

Since the integer requirement, there haven't been any replies, so I'll go with an obvious, and not-likely-to-be-optimal one.

Spoiler

Powers: Buy a single ticket, and randomly choose X >= 1, where P(X) = 2^-x. So P(1) = 1/2, P(2) = 1/4, etc. 

 

[CaptainEd]

I’m the only one who doesn’t know how the game works. Here are a couple of questions that may help me understand.

do we all play simultaneously? Or are purchases made sequentially (so that each player is able to meet the requirement of avoiding numbers others have chosen)?

Once everyone has played, how is it determined who has won? If more than person chooses a number, is that number removed, and the lowest number chosen by only one person is the determiner?

[BMAD]

On 12/24/2017 at 12:16 AM, CaptainEd said:

I’m the only one who doesn’t know how the game works. Here are a couple of questions that may help me understand.

do we all play simultaneously? Or are purchases made sequentially (so that each player is able to meet the requirement of avoiding numbers others have chosen)?

Once everyone has played, how is it determined who has won? If more than person chooses a number, is that number removed, and the lowest number chosen by only one person is the determiner?

Each person is betting $10 that they can pick the lowest positive integer that is not picked by anyone else.  Each time an individual chooses a number they bet $10.  Once everyone is satisfied that they picked enough numbers (as they can pick more than one) they show their choices.  The individual with the lowest number that was not picked by anyone else, wins $100.  You are competing against nine other logical and equally wealthy people, what would be your strategy to win the prize?

Edited January 2, 2018 by BMAD

[CaptainEd]

Thanks, BMAD, I understand now

[Wilson]

[spoiler="collusion"]bribe the other nine,less than $10 each, not to pick a number[/spoiler]

[plasmid]

Not fully calculated because the math gets extremely messy and complicated pretty fast, but a description of how to go about it

Spoiler

The best strategy in a game like this will obviously depend on the other players’ strategy. There’s no great way of determining how best to play against them if they could just arbitrarily do anything. So I’ll try to describe what I would consider an optimal strategy for this game if there are 10 players that each buy one ticket. Intuitively, if everyone were to buy more than one ticket then they would collectively be paying >$100 for a $100 payout and there will be no good strategies in that situation. I’ll also describe the strategy in terms of probability of picking each number, so a strategy could be defined as a vector of (P(1) (probability of buying a ticket numbered 1), P(2) (probability of buying a ticket numbered 2), P(3) ...)

 

If there are 10 people (including yourself) purchasing tickets, what value of P(1) leaves everyone with a 10% chance of winning if they pick 1? In order for you to win if you pick 1, none of the other nine players can pick 1, and that will happen with probability (1-P(1))⁹. For a 10% chance of winning if you pick 1, that means

0.1 = (1-P(1))⁹

0.774264 = 1-P(1)

P(1) = 0.225736

If everyone else buys a "1" ticket with probability less than that, then you would have a winning strategy if you buy a 1 ticket. If everyone else buys a "1" ticket with probability greater than that, then they'll have a <10% chance of winning and you're probably better off buying a different number that might give you a >10% chance of winning.

 

With that value for P(1), what value of P(2) would give you a 10% chance of winning if you pick 2? In order to win by picking 2, two things need to happen: you need none of the other nine players to win by being the sole player to pick 1, and you need no one else to pick 2.

The odds that no one else will pick 1 and that no one else will pick 2 are

(1-P(1))⁹ x (1-P(2))⁹

The odds that N players will pick 1 (where N>1) and no one else will pick 2 are

P(1)^N x (1-P(1))^9-N x ₉C_N x (1-P(2))^9-N

So the odds that you’ll win if you pick 2 are

Sum{N=0, 2, 3, … 9} P(1)^N x (1-P(1))^9-N x ₉C_N x (1-P(2))^9-N

That’s getting to be more than I want to power through with calculations, but is in principle solveable if you do all of the grizzly math in that equation to find a value for P(2) that gives you a 10% chance of winning. And it demonstrates the principle of how to calculate the remaining P(x) terms.

Although a better answer might be that if there are 10 people playing a game with at least $10 entry fee and $100 payout with a non-zero chance that no one will win and the house will keep everything if everyone ends up picking a number in common with at least one other player, then your best move is to not play.