MrRubix's Riddle/Problem Thread: Rebirth

[emerald000]

Re: MrRubix's Riddle/Problem Thread: Rebirth

Yeah, the first answer is 1/573.

But for the second one, do we suppose the rubbing alcohol is 100% ethanol, or something like 90% ethanol-10% water? Assuming it is 100% pure, the final concentration is given as C=1/(1+x) where there is 1 unit of concentration at the start and there is x units removed from the tube.

And for my riddle, you may PM me or AIM me the solution if you don't want to make sure it is OK before posting it.

[ieatyourlvllol]

Re: MrRubix's Riddle/Problem Thread: Rebirth

The answer to the second riddle is one (simplified from a series of bivariable expressions/equations).

I'm pretty sure the Konami riddle is based on dependent probability. I might go ahead and verify the answer later.

[emerald000]

Re: MrRubix's Riddle/Problem Thread: Rebirth

Clearer answer for #2:

If we put the water in the alcohol, we get, at the end, a concentration of 1/(1+x) of water in the water tube and 1/(1+x) of alcohol in the alcohol tube.

EDIT: Did the coin flip mathematically and it gave 49/621 or around 7.89%.

[ieatyourlvllol]

Re: MrRubix's Riddle/Problem Thread: Rebirth

My calculation (using pure alcohol) was as follows:

{(v-x+[(x*[x/(v+x)])/v])/[(v*[v/(v+x)])/v]} = 1(:1 ratio)

That value would probably change (and the system would gain another variable) if the rubbing alcohol had a partial concentration.

[MrRubix]

Re: MrRubix's Riddle/Problem Thread: Rebirth

The answer to the first one is not merely 1/573. It is a conditional probability. You know you flipped a certain number of heads. You don't, however, know where they came from.

As for the alcohol problem, if you're using a bunch of algebra, you're working too hard. You are right though that the final concentrations are identical.

When amount X is poured, the final concentration of alcohol in the alcohol tub is V/(V+X). After mixing and pouring back, the concentration of alcohol in the alcohol jug does not change again because no water is added. However, when the diluted alcohol is poured back into the water tub, the concentration of water in the water tub changes from 100% to V/(V+X). So, again, the final concentrations are the same.

At the end of the process, both tubs contain the same volume of fluid as they did at the start. The only way for the concentration of alcohol to have changed from 100% is if some alcohol was displaced by water. Vice-versa for water. Volume is conserved (both total volume and volume in each tub), so all that has happened is that identical quantities of water and alcohol have traded places (and these identical quantities are slightly less than X). So, by symmetry, the concentrations of alcohol in the alcohol tub and water in the water tub must be identical.

[ieatyourlvllol]

Re: MrRubix's Riddle/Problem Thread: Rebirth

Yeah, algebra really wasn't necessary. I just like doing things mathematically. 8)

The simplest method would be to visualize extremes. You could just as well say that the amount 'X' that you pour from Tub 1 is equal to either 0 or the volume of liquid. In both cases, it's easy to see that the concentrations will end up the same. As a bonus, any intermediary value of 'X' would have to abide by the same laws of proportional equivalence.

I wanna take a crack at the Konami riddle, but I'm too busy right now. =(

[emerald000]

Re: MrRubix's Riddle/Problem Thread: Rebirth

EDIT: Did the coin flip mathematically and it gave 49/621 or around 7.89%.

Is it the right answer?

And why is there a volume variable in your equations? Concentrations are going to be the same no matter what is the initial volume. If you have twice the water, you'll have twice the alcohol, thus simplifying themselves...

[MrRubix]

Re: MrRubix's Riddle/Problem Thread: Rebirth

No, 7.89% is not correct

"The simplest method would be to visualize extremes. You could just as well say that the amount 'X' that you pour from Tub 1 is equal to either 0 or the volume of liquid. In both cases, it's easy to see that the concentrations will end up the same. As a bonus, any intermediary value of 'X' would have to abide by the same laws of proportional equivalence."

That's also a very good way to look at it -- nice job

emerald: It's there for ease of understanding. You can solve the riddle without any math -- just understanding the underlying concept is sufficient to solve it.

[dooty_7]

Re: MrRubix's Riddle/Problem Thread: Rebirth

here is a stab in the dark for the coin flip one...although I will end up sounding retarded :P

The probability of flipping seven heads in a row on a standard coin is 1 / 2^7 (or just 0.5^7) and for the two headed coin the probability is obviously 1.

So the probability that the coin you picked is the 2 headed coin is
(572/573) * (1/2^7) + (1/573)*(1) = 0.009544 which is greater then just 1/573

EDIT: On second thought, I think this solution just gives the probability of getting 7 heads in a row lol

[MrRubix]

Re: MrRubix's Riddle/Problem Thread: Rebirth

If you're going for the Konami riddle, you need to approach it conditionally. There is a very well-known statistical/probability tool for solving these kinds of problems. You KNOW you already got 7 heads in a row. Therefore you aren't solving for "the probability of getting 7 heads" in the strictest sense. You're being asked "you ALREADY got 7 heads. Did these heads COME FROM a fair coin or the two-headed coin?"


Dooty: Yes, that would be right had I asked "What is the probability of getting 7 heads in a row?"
P(getting 7 heads) =
Splitting this event into mutually exclusive events:
P(getting 7 heads fair) + P(getting 7 heads with the double headed coin) =
P(choosing fair coin)*P(getting 7 heads with fair coin) + P(choosing double headed coin)*P(getting 7 heads with double headed coin) =
(572/573)*(1/2)^7 + (1/573)*1

[emerald000]

Re: MrRubix's Riddle/Problem Thread: Rebirth

God I am stupid. Put 7^2 instead of 2^7 in my calculation...

Real answer is 32/175 or around 18.29%.

[MrRubix]

Re: MrRubix's Riddle/Problem Thread: Rebirth

Correct, emerald

What's your solution setup?


Here is mine (highlight to read):

You can solve this with a very straightforward application of Bayes' Rule:

P(Two headed quarter | 7 heads in a row) =
P(Two headed quarter and 7 heads in a row)/P(7 heads in a row) =
P(7 heads in a row | Two headed quarter)P(Two headed quarter)/P(7 heads in a row) =
P(7 heads in a row | Two headed quarter)P(Two headed quarter)/[P(7 heads in a row | Two headed quarter)P(Two headed quarter) + P(7 heads in a row | NOT getting the two headed quarter)P(NOT getting the two headed quarter)] =
(1 * (1/573))/(1 * 1/573 + (1/2)^7 * 572/573) = 0.182857143

[emerald000]

Re: MrRubix's Riddle/Problem Thread: Rebirth

I did exactly the same thing. Good ol' Bayes theorem.

And does anyone is going to attempt mine?

[ieatyourlvllol]

Re: MrRubix's Riddle/Problem Thread: Rebirth

Mine was as follows (highlight):

1 - (572*[1/(2^7)])/{(572*[1/(2^7)])+[1*(1/1)]}

[MrRubix]

Re: MrRubix's Riddle/Problem Thread: Rebirth

I am gonna have to get harder riddles for this thread. Prepare to start drowning in tears, gentlemen.