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Phinnaeus Gage

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Activity: 1918

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Bitcoin: An Idea Worth Spending

Behind one of the following barn doors is a rare Barn Find, with goats behind each of the other three doors:

If you correctly pick the door revealing the Barn Find, you win 200% profit. If a goat is revealed, you lose.

But not to be deterred, you'll have the option to guess again with the remaining three doors if your first choice revealed a goat. You can also opt to pass, whereupon the next player can pick up where you left off. Either way, if the correct door is picked revealing the Barn Find in this round, you win 150% profit.

But not all is lost if you desire to try one final time with the two remaining doors to win 100% profit--true odds at this juncture.

EDIT: Just realized that this negates the premise of the title of this thread, but still something to think about.

dree12

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Activity: 1246

Merit: 1077

Quote from: ?? on ??

More troll math, not so troll:

x = 0.999... (means infinite nines)
10x = 9.999...
10x - x = 9.999... -0.999...
9x = 9
x = 1

PS: I like to use more spaces than I should .

But this is a perfectly valid proof. The infinite series only differ by their first term, and can be subtracted hence.

Phinnaeus Gage

legendary

Activity: 1918

Merit: 1570

Bitcoin: An Idea Worth Spending

Quote from: Luno on January 30, 2013, 01:33:11 PM

If there are three options to choose from to start with, probability is 66% for a win. If there are 4 options and you play: pick switch switch, your probability for a win is 5 out of 8.

So the OP is loosing money, unless there is a catch.

The explanation does not however suggest that this is an ordinary Monty Hall problem:

You choose 2 doors which are removed, and the gamehost promisses that there is a 50% probability for a right choice in the last draw between the remaining 2 doors.

If this is the case, He won't make any money unless the randomization is fake.

After the first choice of the four is eliminated, then it reverts to the Monty Hall Problem.

jwzguy

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Merit: 1002

What is this I don't even

Luno

sr. member

Activity: 504

Merit: 250

If there are three options to choose from to start with, probability is 66% for a win. If there are 4 options and you play: pick switch switch, your probability for a win is 5 out of 8.

So the OP is loosing money, unless there is a catch.

The explanation does not however suggest that this is an ordinary Monty Hall problem:

You choose 2 doors which are removed, and the gamehost promisses that there is a 50% probability for a right choice in the last draw between the remaining 2 doors.

If this is the case, He won't make any money unless the randomization is fake.

Phinnaeus Gage

legendary

Activity: 1918

Merit: 1570

Bitcoin: An Idea Worth Spending

Quote from: ingrownpocket on January 30, 2013, 12:11:33 PM

1. Four options to choose, only one is correct.
2. Choose one, remove it from the list since the probability to choose a wrong option is 3/4.
3. Repeat step 2.
4. Now you have 50% probability ✔ GUARANTEED ✔ to get the right answer.
5. Send 1 BTC to 1profit4U and you'll get the right answer (you must have 50% probability ✔ GUARANTEED ✔)
6. Buy a magnet for improved skillz.
7. Repeat all until rich.

http://en.wikipedia.org/wiki/Monty_Hall_problem

ingrownpocket

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