Probabilities - question posed by Kessler.

[hutmaster]

What is the correct answer?

[sonatine]

66.6%

[tyde]

LOL JEWS

[1marley1]

66.6%

2 boxes left. 1 is 100% silver one is 100% gold.

Why not 50%?

[sonatine]

66.6%

2 boxes left. 1 is 100% silver one is 100% gold.

Why not 50%?

we know its not the silver/silver box so we discard the 2 silvers from the aggregate pool of remaining coins, leaving us;

2 gold + 1 silver.

im not sure any other data matters but its possible that theres some tweak where the fact that we didnt get a silver means the odds are somehow higher that the remaining box is gold/gold.. because probability is weird like that... but ultimately it seems less stochastic than that at face value.

note im not a math guy and kessler is so im probably wrong and its not that easy.

[Sandwich]

Not a math guy, but the Q presupposes you pick a G coin first, so…
Half the time you choose the GG box. You have 100% possibility of picking G next.
Half the time you choose the GS box. You have 0% possibility of picking G next.
100% x .5 + 0% x .5 = 50%, amirite?

BUT are chances twice as high you picked the GG box since if you picked the GS box, half the time you’d pick S and it wouldn’t count toward the question. Maybe 66.6 is correct after all?

[sonatine]

for anyone who wants the answer:

 

https://www.toppr.com/ask/en-us/ques...ii-contains-1/

its 66.6%

[dwai]

I find it probable you're a fuckin nerd

[YUUP]

You can only pick gold gold from one box so 33.33?

[nunbeater]

this is the old Monty Hall problem with extra steps

[aayjay]

[Dan Druff]

You basically just need to figure out which box it was, and not what you're going to draw going forward. Because once you figure out which box it was, you know with 100% certainty what you're drawing next.

If you drew a gold coin, that means you picked one of three which existed. Since two are in one box, and one is in another, the probability is 2/3 that the gold coin came from the gold-gold box, and 1/3 that it came from the gold-silver box. The probability is obviously zero that it came from the silver-silver box.

This means that the probability of getting another gold is 2/3 (66.6%), since that's the probability of having chosen the gold-gold box.

Another way of looking at it is simply saying there are 3 gold and 3 silver, and you're removing 1 gold (the one you picked) and 2 silver (the ones in the silver-silver box). This leaves 2 gold and 1 silver, so again it's 2/3.

Q.E.D.
S.F.O.

[hutmaster]

You basically just need to figure out which box it was, and not what you're going to draw going forward. Because once you figure out which box it was, you know with 100% certainty what you're drawing next.

If you drew a gold coin, that means you picked one of three which existed. Since two are in one box, and one is in another, the probability is 2/3 that the gold coin came from the gold-gold box, and 1/3 that it came from the gold-silver box. The probability is obviously zero that it came from the silver-silver box.

This means that the probability of getting another gold is 2/3 (66.6%), since that's the probability of having chosen the gold-gold box.

Another way of looking at it is simply saying there are 3 gold and 3 silver, and you're removing 1 gold (the one you picked) and 2 silver (the ones in the silver-silver box). This leaves 2 gold and 1 silver, so again it's 2/3.

Q.E.D.
S.F.O.

Yes this is the correct answer. Surprisingly when Kessler posted it with almost 500 replies to his poll, only 32% answered correctly.

[Ryback_feed_me_more]

Answer has to be 2/3. Question is what are the odds of getting a second gold coin out of the same bag you pulled from the first time. The answer is 2 outta 3. Either way its the bag with the second gold coin meaning 100% on that bag or its the one with the silver coin remaining alone so out of 3 coins remaining possible its 2 golds and a silver total. the silver silver bag is automatically removed as a possibility.

[gut]

Was Kessler meaning the question to be just math, or IRL?

I doubt anyone would lay 3/2 in this scenario if they were drawing, being that its a coin flip (pun kind of intended) at that point.

[1marley1]

Was Kessler meaning the question to be just math, or IRL?

I doubt anyone would lay 3/2 in this scenario if they were drawing, being that its a coin flip (pun kind of intended) at that point.

Tell me you don’t understand probability without telling me you don’t understand probability…

[gut]

Was Kessler meaning the question to be just math, or IRL?

I doubt anyone would lay 3/2 in this scenario if they were drawing, being that its a coin flip (pun kind of intended) at that point.

Tell me you don’t understand probability without telling me you don’t understand probability…

So you would lay 3/2 everytime?