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Author Topic: game theory question  (Read 2949 times)
jgcblack
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« on: September 24, 2014, 08:27:42 PM »

We have a unique lowest bidder auction.
http://en.wikipedia.org/wiki/Unique_bid_auction

This means that your bid can be anything from £0.01 upwards, you pay £1 for your bid and you do not have to pay your bid to win.

After everyone has bid the unique and lowest bid is the winner.  In this game we have 400 players with an expected play rate of 80% Out of those players around 50% will play twice with around 5% playing 5 times.

So we should have 400 × 0.8 = 320.
320×0.5 ×2 = 320 +160 +(0.05×5) = 496 total bids.

The players are of mixed abilities but effectively we are assuming only us and maybe one other understand the game.  The rest will take random bids at anything from 0.01 up to 9.43 (I saw that bid.... wff!?)

As such I'm looking for how to work out how many bids we need to make to "block up the numbers" and allow our next bid to win.

I think that 50 should be plenty bearing in mind that my guess is over 90% of bids will either be completely useless or under 0.20p. (So we have from 0.01-0.50)


Does anyone know how to work out this problem?
« Last Edit: September 24, 2014, 08:37:12 PM by jgcblack » Logged

Eso Kral
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« Reply #1 on: September 24, 2014, 08:34:56 PM »

The new John Black aka KingPush is my nap to sort this out for you bud as I'm more of a feel type punter. Smiley
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« Reply #2 on: September 24, 2014, 09:19:29 PM »

I thnk it is impossible without knowing other's budgets.  I think your strategy looks fine.

I'd always stop at a number ending in 1 or 6.  The thinking is people say I'll throw 50 entries at that, rather than I'd throw 51 at it.  I'd probably just start at 11p and finish at 61. Don't know the right answer, but it feels good.

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Most of the bets placed so far seem more like hopeful punts rather than value spots
jgcblack
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« Reply #3 on: September 24, 2014, 09:23:56 PM »

I thnk it is impossible without knowing other's budgets.  I think your strategy looks fine.

I'd always stop at a number ending in 1 or 6.  The thinking is people say I'll throw 50 entries at that, rather than I'd throw 51 at it.  I'd probably just start at 11p and finish at 61. Don't know the right answer, but it feels good.



I think blocking out 0.01-0.10 is boring but key.

I estimate that I should be north of 97% with 50 bids in a 496 bidder game.
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Tal
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« Reply #4 on: September 24, 2014, 09:26:10 PM »

The players are of mixed abilities but effectively we are assuming only us and maybe one other understand the game.


Copied from every staking thread ever.
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jgcblack
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« Reply #5 on: September 24, 2014, 10:14:05 PM »

The players are of mixed abilities but effectively we are assuming only us and maybe one other understand the game.


Copied from every staking thread ever.

Stop owning and be useful please Mr genius
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jgcblack
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« Reply #6 on: September 24, 2014, 10:17:12 PM »

A direct comparison to the "birthday problem" is as follows...

496÷365 = 1.35 × 50 = 67.5


However we don't have bids as evenly distributed as birthdays.

Ours will be skewed to the 0.30 and under range.... and the random £2.56 type bids.
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« Reply #7 on: September 24, 2014, 10:29:36 PM »

A direct comparison to the "birthday problem" is as follows...

496÷365 = 1.35 × 50 = 67.5


However we don't have bids as evenly distributed as birthdays.

Ours will be skewed to the 0.30 and under range.... and the random £2.56 type bids.

So you have a good idea how to solve the problem, but start a thread anyway?  OK

Little known fact, the answer to the birthday problem is 3 in my family.
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kinboshi
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« Reply #8 on: September 24, 2014, 10:29:44 PM »

How do you know there is only one other who "gets it", and are you discounting him going for the same strategy as you?
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jgcblack
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« Reply #9 on: September 24, 2014, 10:33:40 PM »

How do you know there is only one other who "gets it", and are you discounting him going for the same strategy as you?

Guess, based on knowing the majority of the players personally, and having been close enough to see the players interacting with the game.  Only one person mentioned strategy out loud,  acting on it is another thing.  He didn't.

« Last Edit: September 24, 2014, 10:36:34 PM by jgcblack » Logged

titaniumbean
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« Reply #10 on: September 24, 2014, 10:39:24 PM »

How do you know there is only one other who "gets it", and are you discounting him going for the same strategy as you?

Guess, based on knowing the majority of the players personally, and having been close enough to see the players interacting with the game.  Only one person mentioned strategy out loud,  acting on it is another thing.  He didn't.



yeh It seems arrogant but people just don't get this stuff.

watch this - http://oyc.yale.edu/economics/econ-159


episode 1 focuses on a very basic 'game' concept and is put to a room full of people who got accepted into yale. the bulk of the room just don't get it, some of them really really don't get it even after the strategy theory starts being discussed. 


you have to then work out how exploitably they will use the info that only you and they understand the game. as per people who play GTO in poker vs people you don't need to.
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kinboshi
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« Reply #11 on: September 24, 2014, 10:56:37 PM »

How do you know there is only one other who "gets it", and are you discounting him going for the same strategy as you?

Guess, based on knowing the majority of the players personally, and having been close enough to see the players interacting with the game.  Only one person mentioned strategy out loud,  acting on it is another thing.  He didn't.



You know all 400 people and know their knowledge and understanding of game theory?
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jgcblack
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« Reply #12 on: September 24, 2014, 11:11:41 PM »

How do you know there is only one other who "gets it", and are you discounting him going for the same strategy as you?

Guess, based on knowing the majority of the players personally, and having been close enough to see the players interacting with the game.  Only one person mentioned strategy out loud,  acting on it is another thing.  He didn't.



You know all 400 people and know their knowledge and understanding of game theory?

I know a lot of them... And I heard and saw their actions when bidding. (Stall next door)
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MintTrav
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« Reply #13 on: September 24, 2014, 11:39:42 PM »

After everyone has bid the unique and lowest bid is the winner.  In this game we have 400 players with an expected play rate of 80% Out of those players around 50% will play twice with around 5% playing 5 times.

So we should have 400 × 0.8 = 320.
320×0.5 ×2 = 320 +160 +(0.05×5) = 496 total bids.

Wouldn't it be 544 bids (if you treat yourself as additional to the 320 players)? I don't suppose that changes anything.
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jgcblack
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« Reply #14 on: September 25, 2014, 03:43:30 AM »

After everyone has bid the unique and lowest bid is the winner.  In this game we have 400 players with an expected play rate of 80% Out of those players around 50% will play twice with around 5% playing 5 times.

So we should have 400 × 0.8 = 320.
320×0.5 ×2 = 320 +160 +(0.05×5) = 496 total bids.

Wouldn't it be 544 bids (if you treat yourself as additional to the 320 players)? I don't suppose that changes anything.

True but we're working out how many bids we need to make  so haven't placed any yet
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