[ROLLIN.IO] Rollin 2016 New Year // Contests & Giveaways!! / CLOSED.

[ndnh]

On-thread Trivia!




aaand the question for today is:

7. At the Rollin game show, you are given three wooden boxes:
one containing two Bitcoins, another box containing two Litecoins, and one containing a Bitcoin and a Litecoin. (assume all are physical ones)

You pick a wooden box at random and draws out a Bitcoin. What is the probability that the second coin is also a Bitcoin? (2 mBTC) How did you arrive at the solution? (4 mBTC) What is this classic problem called? (1 mBTC)

Post your answer(s) with your Rollin account. Rollin account should be at least level 2 OR the bitcointalk account full member and above. First to answer correctly, to the point wins the prize.

Remember, no edits.

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[LoyceV]

7. At the Rollin game show, you are given three wooden boxes:
one containing two Bitcoins, another box containing two Litecoins, and one containing a Bitcoin and a Litecoin. (assume all are physical ones)

You pick a wooden box at random and draws out a Bitcoin. What is the probability that the second coin is also a Bitcoin? (2 mBTC) How did you arrive at the solution? (4 mBTC) What is this classic problem called? (1 mBTC)

I'm going to go with my math-gut-feeling

The probability is 66.667%

You know for sure you don't have the box with 2 litecoins.
That leaves two options:
1. you have the box with 2 bitcoins
2. you have the box with 1 bitcoin and 1 litecoin
Before getting a coin out of these 2 boxes, there is a 75% chance to get a bitcoin, and 25% chance to get a litecoin. You got a bitcoin already, which makes it more likely you have the bitcoin-box than the bitcoin/litecoin-box.

I have no idea what this classic bitcoin and litecoin problem is called
Username Loyce

[ndnh]

You might be correct (or wrong). But I won't say it till I get 2 more attempts.

[Akai_bank]

The probability is 2/3....in fact the thing that you pick a BTC coin don't change probability of the choose of the box.
If for example we have:
Box 1: BTC 1 -btc2
Box 2: ltc 1 -ltc 2
Box 3: BTC 3 - ltc 3

The extracted BTC could be only one of BTC 1, btc 2, BTC 3, so possible cases for the BTC are 3/3.
If I pick one of 3, the P is now 2/3.

[LG_]

I think its 50/50 change ?
Cause if you picked one bitcoin there is only 2 boxes that contains one for more btc's

[fumblingperch]

username: fumblingperch

You pick a wooden box at random and draws out a Bitcoin. What is the probability that the second coin is also a Bitcoin?
You get 50%

How did you arrive at the solution?
because in moment you have pick 1 Bitcoin already, only can be box 1 or 3

What is this classic problem called?
don't know

[sweetswoebegone]

Probability is 1/3

because the chance for the first coin a bitcoin and a seconds a bitcoin again, it's only the box 1 between 3 options

[whome007]

This is awesome! Thanks for the contests and prizes!!

my user: whome007

keyword:  XMAS

Thanks!!

[ndnh]

I'm going to go with my math-gut-feeling

The probability is 66.667%

Yes, the correct answer is 2/3

You win 2 mBTC

You know for sure you don't have the box with 2 litecoins.
That leaves two options:
1. you have the box with 2 bitcoins
2. you have the box with 1 bitcoin and 1 litecoin
Before getting a coin out of these 2 boxes, there is a 75% chance to get a bitcoin, and 25% chance to get a litecoin. You got a bitcoin already, which makes it more likely you have the bitcoin-box than the bitcoin/litecoin-box.

You have said that there is 3/4 chance of picking a bitcoin and 1/4 of picking a litecoin once you remove the other box and that it is more likely to have the bitcoin-box than the bitcoin/litecoin box, but how likely?

You win 1 mBTC.

I have no idea what this classic bitcoin and litecoin problem is called

My version. hehe

[ndnh]

The probability is 2/3....

Well, arguing that 66.667% is not equal to 2/3, you win 1 mBTC

in fact the thing that you pick a BTC coin don't change probability of the choose of the box.
If for example we have:
Box 1: BTC 1 -btc2
Box 2: ltc 1 -ltc 2
Box 3: BTC 3 - ltc 3

The extracted BTC could be only one of BTC 1, btc 2, BTC 3, so possible cases for the BTC are 3/3.
If I pick one of 3, the P is now 2/3.

That is correct. But the last part is a bit vague, but your approach (the beginning) is correct.

You win 2 mBTC

[ndnh]

The other solutions are not correct.

This on-thread trivia is closed.

This is an example of Bertrand's box paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox


Bertrand's box paradox is a classic paradox of elementary probability theory. It was first posed by Joseph Bertrand in his Calcul des probabilités, published in 1889.

There are three boxes:

a box containing two gold coins,
a box containing two silver coins,
a box containing one gold coin and a silver coin.
After choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, it may seem that the probability that the remaining coin is gold is 1⁄2; in fact, the probability is actually 2⁄3.


You can read the explanation for the solution here : https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox#Box_version


Simply stated, the Bitcoin you picked could be of either Box 1 or Box 3. There is thus 2/3 chance that Box 1 is the box picked and 1/3 that Box 3 is the one picked. Therefore, there is 2/3 chance that the other one is also a Bitcoin.

[Akai_bank]

Thanks for contest and prizes, was very funny!

Username: Akai
Code: XMAS

[llllllll]

Code: A1

User ID: llllllll

Suggestion on improvement

1. make a timer countdown on the faucet claim, so it is easier for people to know when to click on the claim button

2. setup a happy hour of maybe 10k satoshi per claim on faucet on certain timing for 1 hour per day

Rollin.io is the best dice site with good community

All the best for rollin.io in year 2016, hoping to see more and more new players join in the community and start rolling the dice with us on rollin.io

[ndnh]

On-thread Trivia!




aaaand the question for today is:

8.   nvm



Post your answer(s) with your Rollin account. Rollin account should be at least level 2 OR the bitcointalk account full member and above.
First to answer correctly, to the point wins the prize.

Remember, no edits.

[Wendigo]

Hmm no one both have a 50/50 to win cause they are 2 left?

[Shadew]

both of them ( max and sax) increased their chance of winning from 1/3 to 1/2
because fax already won as rex said to max which is left to a head to head chance
 
username : Shadew

[Himea]

Max and fax increase their chance to win to 1/2

Username : himea

[llllllll]

Fax and Max chances increase from 1/3 to 1/2

Sax already won the game as Max begs Rex to tell him the identity of the person who has lost the draw. "If Fax won a prize, give me Sax's name. If Sax won a prize, give me Fax's name.

so actually that left only Fax and Max

Sax is happy about it when Max tell him the result is because he has already won the lottery

so the lottery is now between Fax and Max only

their chances went up from 1/3 to 1/2 because there is only 2 person left.

User ID: llllllll

[Jeremycoin]

Max's and Sax's chance of winning whose went up.
From 1/3 to 1 (100%) because they must be win the prize, cos the winner is 2 person. Which means that they will surely got the prize
BTW, it's actually from 2/3 not 1/3. Cos there's 2 winner not just 1

username : Jeremyroll

[ndnh]

Ah, sorry.. the question is

Whose chance of losing went up?..

fixed it.

I'll simplify the names and sorry none of the answers is the expected answer.


Looks like the question is not correct in the first place lol.

Anyway, the expected answer is not 1/2 : 1/2. Well, the question contradicts itself more than once, I will replace it with another question.

Edit: I couldn't find a winner.
It is pretty complicated and some answers ignore the part of dice entirely or that Max has only 50% of the chance that Sax has.