Monty Hall and Let's Make A Deal problem

[ronaldmaustin]

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?

Does anyone see the correlation between this scenario and Bitcoins?

[Nefario]

You should switch, your first choice is a one in 3 choice, when he removes one of the doors it turns into a one in 2 choice, greater odds so you should switch.

Many people would say there is no differencebut the odds are better ifyou switch.

[Cusipzzz]

Nefario is correct, but it is amazing the number of intelligent people who do not understand/accept this concept!

[ribuck]

Then he opens one of the other two doors that he knows does not have the prize behind it.

Monty cannot open the door you selected, so his action cannot reveal any further information about what's behind that door. But his action does narrow down the options for the other doors, so you should always switch.

Does anyone see the correlation between this scenario and Bitcoins?

Um, no. Do enlighten us please.

[ronaldmaustin]

Then he opens one of the other two doors that he knows does not have the prize behind it.

Monty cannot open the door you selected, so his action cannot reveal any further information about what's behind that door. But his action does narrow down the options for the other doors, so you should always switch.

Does anyone see the correlation between this scenario and Bitcoins?

Um, no. Do enlighten us please.

So you agree with the prior two posters then as to the odds?

[eMansipater]

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?

Does anyone see the correlation between this scenario and Bitcoins?

Is it that, like Monty, many people who just gave away 1000 Bitcoins will be feeling pretty foolish soon?  Just joking you people that applies to, those early trades were an important part of building the bitcoin economy, as are trades today.

[mpkomara]

blockexplorer could probably help you track down which door the bitcoins were behind.

[ronaldmaustin]

No, it is simply that the odds do not increase from 1 in 3 to 1 in 2.  They increase from 1 in 3 to 2 in 3.  You double your chances by switching, instead of the mere 50 percent increase that everyone supposes.  Now look at the increase in difficulty of the Bitcoin generation and apply that.

[ronaldmaustin]

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.

[ribuck]

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.

[mpkomara]

Before you explain what you mean, can you provide us with a third metaphor?  Maybe use talking animals for those of us slow learners.

[ronaldmaustin]

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? 

[sirius]

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.

[ribuck]

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.

[Bimmerhead]

Yeesh this thread is so cryptic.  And the only message I can decipher from it is very unpleasant.  I hope clarification is forthcoming.

[ribuck]

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.

[Stephen Gornick]

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

[BitterTea]

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?

[ronaldmaustin]

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

[nanotube]

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.