Why don't we have The Price is Right have an episode where instead of a car, it's 2,000 BTC, and then this will be a relevant conversation 
Why don't you make such a suggestion to the show?
Topic: the three doors problem, increasing the chance of finding the right hash - page 2. (Read 3645 times)
June 15, 2011, 08:11:41 PM
Why don't you make such a suggestion to the show?
June 15, 2011, 07:59:35 PM
Why don't we have The Price is Right have an episode where instead of a car, it's 2,000 BTC, and then this will be a relevant conversation
Why don't NVIDIA and Intel develop a processor that only does 1 hash per second, but it's always the exact right one?
Why don't NVIDIA and Intel develop a processor that only does 1 hash per second, but it's always the exact right one?
June 15, 2011, 06:41:46 PM
Most times I ran it the monty hall implementation ran with the lowest average tries, but when the purely random method won it would usually win by a larger margin than the monty hall problem.
Good, run it a few more hundred thousand times and then run a chi-square analysis.
I think you will find that they are not significantly different in terms of how often they find the hash.
thanks, that's an excellent suggestion. seeing how we're dealing with random numbers and the doors aren't guaranteed to have prizes I don't really doubt that statistically this method is completely unsound. However I didn't know about chi-square tests and I've got other ideas I'd like to map out using this method.
June 15, 2011, 06:22:49 PM
This is awesome!
This is just like the youtube video I saw the other day.
Some dude (solo miner) had noticed that it takes him minimum of about 2 days to solve a block.
So his "solution" to the "wasted" time was to switch to pooled mining for 2 days each time he solves a block, and then switch back to solo mining.
This is just like the youtube video I saw the other day.
Some dude (solo miner) had noticed that it takes him minimum of about 2 days to solve a block.
So his "solution" to the "wasted" time was to switch to pooled mining for 2 days each time he solves a block, and then switch back to solo mining.
June 15, 2011, 06:15:18 PM
In the Monty Hall scenario the host ALWAYS picks the wrong door, and you are using this particular piece of information it to your advantage by switching.
The pool cannot do this (only picks a random door).
Think about it for a moment.
EDIT: By always picking the wrong door he is artificially lengthening the game (the game never ends on round 2), making it a completely different problem/game than the pool.
The pool cannot do this (only picks a random door).
Think about it for a moment.
EDIT: By always picking the wrong door he is artificially lengthening the game (the game never ends on round 2), making it a completely different problem/game than the pool.
June 15, 2011, 05:49:15 PM
Most times I ran it the monty hall implementation ran with the lowest average tries, but when the purely random method won it would usually win by a larger margin than the monty hall problem.
Good, run it a few more hundred thousand times and then run a chi-square analysis.
I think you will find that they are not significantly different in terms of how often they find the hash.
June 15, 2011, 04:59:07 PM
Indeed, this doesn't work. The "host" needs to know which door to open that doesn't have the right thing. The extra odds comes from the information the host gives you. If miners just discard blocks at random, it does nothing to help your odd.
To show you the truth of this, imagine that the host doesn't know anything about where the car is. You pick a door, he opens one, then you switch. 33% of the time you'll have picked the right door to start, 33% of the time, the host will open the right door (ending the game early) and 33% of the time you will switch and find the right door. As you can see, your odds didn't go up by switching doors.
To show you the truth of this, imagine that the host doesn't know anything about where the car is. You pick a door, he opens one, then you switch. 33% of the time you'll have picked the right door to start, 33% of the time, the host will open the right door (ending the game early) and 33% of the time you will switch and find the right door. As you can see, your odds didn't go up by switching doors.
When the host picks the right door at the start you win. No more need to switch.
When the host picks the wrong door, then what exactly is the situational difference between this and the monty hall problem? In both cases you've gotten one wrong door open, and two doors closed. That gives you 2/3 of a chance if you switch.
I wrote a C implementation, source here http://pastebin.com/PN2k6HZP
Basically it makes a hash from a random number between 0-99,999, then the number is found via two methods, each run 99 times.
The first method is an implementation of the monty hall problem.
The second method is purely random. Neither ever tests the same number twice.
example output
Quote
results for hashes randomly found with doors
found behind door 1: 46
found behind door 2: 53
average number of tries: 48972.77
results for hashes randomly
average number of tries: 52223.12
found behind door 1: 46
found behind door 2: 53
average number of tries: 48972.77
results for hashes randomly
average number of tries: 52223.12
Most times I ran it the monty hall implementation ran with the lowest average tries, but when the purely random method won it would usually win by a larger margin than the monty hall problem.
June 15, 2011, 03:15:56 PM
Who's going to show you there is a goat behind a door?
You sir have hit the nail on the head.In a weird coincidence, I just explained the monty hall problem to my daughter today. *Checks for hidden microphones*
There is a specific term for this kind of phenomenon, but it escapes me at the moment. I believe the term is Dutch or some other similar language. It basically states that upon learning new information or discussing something long forgotten the likelihood of hearing about it fairly shortly afterwards is increased.
June 15, 2011, 02:41:48 PM
Who's going to show you there is a goat behind a door?
You sir have hit the nail on the head.In a weird coincidence, I just explained the monty hall problem to my daughter today. *Checks for hidden microphones*
June 15, 2011, 02:38:08 PM
Indeed, this doesn't work. The "host" needs to know which door to open that doesn't have the right thing. The extra odds comes from the information the host gives you. If miners just discard blocks at random, it does nothing to help your odd.
To show you the truth of this, imagine that the host doesn't know anything about where the car is. You pick a door, he opens one, then you switch. 33% of the time you'll have picked the right door to start, 33% of the time, the host will open the right door (ending the game early) and 33% of the time you will switch and find the right door. As you can see, your odds didn't go up by switching doors.
To show you the truth of this, imagine that the host doesn't know anything about where the car is. You pick a door, he opens one, then you switch. 33% of the time you'll have picked the right door to start, 33% of the time, the host will open the right door (ending the game early) and 33% of the time you will switch and find the right door. As you can see, your odds didn't go up by switching doors.
June 15, 2011, 02:27:22 PM
In the Monty Hall problem, the host has knowledge that the player does not. In bitcoins, the pool does not have this knowledge.
Monty Hall does not open one of the 2 remaining doors at random, he shows you one which does not contain a car because he knows which one has the car in it.
If you do what you suggest you would just increase variance by a wide margin which is what pools are looking to reduce.
Monty Hall does not open one of the 2 remaining doors at random, he shows you one which does not contain a car because he knows which one has the car in it.
If you do what you suggest you would just increase variance by a wide margin which is what pools are looking to reduce.
June 15, 2011, 02:12:16 PM
You misinterpret the statistics behind purely random hashing.
June 15, 2011, 01:58:10 PM
Who's going to show you there is a goat behind a door?
June 15, 2011, 01:34:02 PM
Many of you may be familiar with the three doors problem, aka the Monty Hall problem. (info on wikipedia).
Basically it states that you have three doors presented to you, one which holds a car behind it. You pick one giving you 1/3 of a chance of having the right answer, but then you're presented what's behind one of the doors you didn't pick, and it's not the car. If you switch your pick, your chances of being right increases to 2/3, instead of your original chance of 1/3.
Here's the results of a simulation
http://upload.wikimedia.org/wikipedia/commons/thumb/0/0c/Monty_problem_monte_carlo.svg/583px-Monty_problem_monte_carlo.svg.png
So what if we give miners the monty hall problem? We could increase the chances of finding the right hash by a nice margin.
So in a pool you could effectively have it so 90% of miners ditch 1/3 of the hashes that need to be tested, and 10% of the miners work on the ditched ones.
Basically it states that you have three doors presented to you, one which holds a car behind it. You pick one giving you 1/3 of a chance of having the right answer, but then you're presented what's behind one of the doors you didn't pick, and it's not the car. If you switch your pick, your chances of being right increases to 2/3, instead of your original chance of 1/3.
Here's the results of a simulation
http://upload.wikimedia.org/wikipedia/commons/thumb/0/0c/Monty_problem_monte_carlo.svg/583px-Monty_problem_monte_carlo.svg.png
So what if we give miners the monty hall problem? We could increase the chances of finding the right hash by a nice margin.
So in a pool you could effectively have it so 90% of miners ditch 1/3 of the hashes that need to be tested, and 10% of the miners work on the ditched ones.
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