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Topic: Monty Hall and Let's Make A Deal problem (Read 4710 times)

sr. member
Activity: 322
Merit: 250
March 14, 2011, 08:38:11 AM
#28
Let me explain it in a way you all may understand it.  It's as if Satoshi has a really long straw, all the way from Japan.  Satoshi drinks YOUR milkshake.
Potatoes will be fine.
newbie
Activity: 14
Merit: 0
switch havent you seen 21 Wink?
hero member
Activity: 616
Merit: 500
Firstbits.com/1fg4i :)
Most attempts to solve a block will lead to failure, so would that mean that to increase the odds of you finding the solution before others you should discard the first N attempts before wasting time trying them ?
full member
Activity: 143
Merit: 100
February 20, 2011, 05:44:08 AM
#25
The only time it is good to switch is if you know ahead of time that the game show host will in fact open a door other than your own that does not contain the bitcoins.

On the other hand if the host announces to you that he will do it only after you picked your door, it's better to stay with your door. Why? Because if the host wants to beat you then he will only pick a door if you picked the right door to begin with, in an effort to get you off the right door.


And all of this presumes that Monty Hall knows what is behind the doors.  Or that Monty is even alive.
joe
member
Activity: 64
Merit: 10
February 19, 2011, 06:38:18 AM
#24
The only time it is good to switch is if you know ahead of time that the game show host will in fact open a door other than your own that does not contain the bitcoins.

On the other hand if the host announces to you that he will do it only after you picked your door, it's better to stay with your door. Why? Because if the host wants to beat you then he will only pick a door if you picked the right door to begin with, in an effort to get you off the right door.
full member
Activity: 238
Merit: 100
February 19, 2011, 04:20:11 AM
#23
Although I agree that what you said makes it a better choice to switch.

"You are on the TV show Lets Make a Deal.  There are 1000 Bitcoins that Monty has placed behind one of three doors.  He lets you choose a door.  Then he opens one of the other two doors that he knows does not have the prize behind it.  Now, he asks you whether you want to switch doors or keep the door you first chose.  Statistically, should you stick with your initial choice, or switch?"

Although the first problem in wikipedia as follows:

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"

I do not agree your chances are any better. Just because he knows what's behind the doors does not mean he has any obligation to not open the door with the car.
donator
Activity: 826
Merit: 1060
February 18, 2011, 06:16:58 AM
#22
So you do agree then that Satoshi drinks your milkshake?  

The best analogy I can come up with is this.

Satoshi filled his milkshake, but didn't drink yet because it tastes terrible.

We fill our milkshakes and start to drink. At first they taste terrible, but we keep drinking. Gradually our milkshakes taste better and better.

And by the time Satoshi eventually drinks his milkshake [...] it tastes mighty sweet, thanks to everyone else.
This metaphor was intended to show the symbiotic nature of bitcoin.

We needed Satoshi to invent Bitcoin so that we could benefit from it. But equally, Satoshi needed us to use Bitcoin so that he could benefit from it.

Satoshi could have generated endless bitcoins, but they would be of no value to him unless we were generating and using our own bitcoins. Our actions gave value/sweetness to Satoshi's bitcoins/milkshake.
newbie
Activity: 58
Merit: 0
February 18, 2011, 01:37:36 AM
#21
Yeesh this thread is so cryptic.  And the only message I can decipher from it is very unpleasant.  I hope clarification is forthcoming.

Sorry, no clarification possible.  I had a few beers and was just amusing myself spewing total nonsense.  But, look at the bright side.  When you consider the time and electricity spent on this thread, Bitcoin generation does not seem such a waste of resources.


Oh I get it now, when you made your first post on this forum, I had a 1 in 82 chance of figuring out that you are a pain in the a...you know.

But by giving you the benefit of the doubt for the next 80 posts and reserving my judgement I had a 81 in 82 chance of figuring it out.


We hate you, in the most endearing way.  Grin
hero member
Activity: 482
Merit: 501
February 18, 2011, 12:39:04 AM
#20
Here's an interesting and perhaps non-intuitive consequence of the way generation works. If the difficulty increases, your rate of generation drops. If the increase in difficulty causes some people to drop out of generating, your rate of generation doesn't go up again. When other people drop out, it makes it take longer until the next difficulty adjustment, but you can still expect to generate the same number of coins per day until then, no matter what other people do.

Would it be possible for the clients, and thus the network, to determine that no blocks have been generated in X time, and to decrease the difficulty by Y amount?

it is possible in theory, but there are problems as far as determining "the time", and it would require all (or majority) of clients to be upgraded to the new version.
full member
Activity: 143
Merit: 100
February 18, 2011, 12:22:51 AM
#19
Yeesh this thread is so cryptic.  And the only message I can decipher from it is very unpleasant.  I hope clarification is forthcoming.

Sorry, no clarification possible.  I had a few beers and was just amusing myself spewing total nonsense.  But, look at the bright side.  When you consider the time and electricity spent on this thread, Bitcoin generation does not seem such a waste of resources.  And the Milkshake analogy is adequately explained on YouTube . . .
http://www.youtube.com/watch?v=RKQ3LXHKB34
sr. member
Activity: 294
Merit: 252
February 17, 2011, 03:59:33 PM
#18
Here's an interesting and perhaps non-intuitive consequence of the way generation works. If the difficulty increases, your rate of generation drops. If the increase in difficulty causes some people to drop out of generating, your rate of generation doesn't go up again. When other people drop out, it makes it take longer until the next difficulty adjustment, but you can still expect to generate the same number of coins per day until then, no matter what other people do.

Would it be possible for the clients, and thus the network, to determine that no blocks have been generated in X time, and to decrease the difficulty by Y amount?
legendary
Activity: 2506
Merit: 1010
February 17, 2011, 01:12:25 PM
#17
it is amazing the number of intelligent people who do not understand/accept this concept!

I couldn't understand it, until I read:
  http://en.wikipedia.org/wiki/Monty_Hall_problem
donator
Activity: 826
Merit: 1060
February 17, 2011, 09:37:34 AM
#16
Now look at the increase in difficulty of the Bitcoin generation and apply that.

Hmm. This is obviously supposed to be a clue.

Suppose the difficulty goes up by one-half. For example, from 20,000 to 30,000. This will cause your rate of generation to drop by one-third (e.g. from one block every 15 days to one block every 10 days). But I don't see where that leads...

Here's an interesting and perhaps non-intuitive consequence of the way generation works. If the difficulty increases, your rate of generation drops. If the increase in difficulty causes some people to drop out of generating, your rate of generation doesn't go up again. When other people drop out, it makes it take longer until the next difficulty adjustment, but you can still expect to generate the same number of coins per day until then, no matter what other people do.
legendary
Activity: 1291
Merit: 1000
February 17, 2011, 09:16:30 AM
#15
Yeesh this thread is so cryptic.  And the only message I can decipher from it is very unpleasant.  I hope clarification is forthcoming.
donator
Activity: 826
Merit: 1060
February 17, 2011, 07:31:01 AM
#14
So you do agree then that Satoshi drinks your milkshake?  

The best analogy I can come up with is this.

Satoshi filled his milkshake, but didn't drink yet because it tastes terrible.

We fill our milkshakes and start to drink. At first they taste terrible, but we keep drinking. Gradually our milkshakes taste better and better.

And by the time Satoshi eventually drinks his milkshake (using a MtGox dark pool straw costing $1.05) it tastes mighty sweet, thanks to everyone else.

But I find this analogy unsatisfying. Partly because it doesn't reference Monty Hall, but also because there are no flying cars in this scenario.
sr. member
Activity: 429
Merit: 1002
February 17, 2011, 07:22:07 AM
#13
The best explanation for Monty Hall: Make the number of doors 1,000,000. You pick one of the doors and the host opens 999,998 doors that don't have the prize. Your choice: 1/1000,000, the other door: 999,999/1000,000.
full member
Activity: 143
Merit: 100
February 17, 2011, 07:21:00 AM
#12
Therefore, if you stick with "A" you have a one-in-three chance of winning the coins. If you switch, you have a two-in-three chance of winning the coins.

Double like I said.  So you do agree then that Satoshi drinks your milkshake? 
full member
Activity: 183
Merit: 100
February 17, 2011, 07:14:08 AM
#11
Before you explain what you mean, can you provide us with a third metaphor?  Maybe use talking animals for those of us slow learners.
donator
Activity: 826
Merit: 1060
February 17, 2011, 07:10:56 AM
#10
So you agree with the prior two posters then as to the odds?
No, I just agree that you should always switch.

Let's label the box that you choose "A", and label the others "B" and "C".

There are three possibilities:

1. There is a one-in-three chance that the coins are behind "A". In this case, it makes no difference which door Monty opens. You will win by sticking with "A", and lose by switching.

2. There is a one-in-three chance that the coins are behind "B". In this case, Monty must open door "C". You will lose by sticking with "A", and win by switching.

3. There is a one-in-three chance that the coins are behind "C". In this case, Monty must open door "B". You will lose by sticking with "A", and win by switching.

Therefore, if you stick with "A" you have a one-in-three chance of winning the coins. If you switch, you have a two-in-three chance of winning the coins.
full member
Activity: 143
Merit: 100
February 17, 2011, 07:04:24 AM
#9
Let me explain it in a way you all may understand it.  It's as if Satoshi has a really long straw, all the way from Japan.  Satoshi drinks YOUR milkshake.
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