Another random selection puzzle

Reply #7

Orange71 — Tue, 30 Aug 2022 22:27:23 GMT

Did you generate an analytical solution for this or perform a Monte Carlo simulation? The tricky part of the question is all the permutations that will result in a number of marbles 12, e.g. a result of 5(Y) + 6(R) + 6(R) is a valid outcome, but it would not be possible to have 6(R) + 6(R) + 5(Y), as the experiment would conclude with the second throw of die.

Reply #6

Stat$talker — Mon, 29 Aug 2022 17:53:14 GMT

Whut'z the Probability of you vs me winning dat Super Lotto Plus , if I converge on Lost Angeless, KHaLiPhOrNyA.? ( People runnin round wit Purple Green hair) .I'm comin out dere next week..!

and I'll be armed wit plenty Bug Spray fa dem Cricketz..

-Stat$talker

Reply #5

cottoneyedjoe — Sat, 27 Aug 2022 17:06:14 GMT

Here's a related puzzle. You have at your disposal a jar, a 6-sided die, and a bunch of white, black, blue, green, yellow, and red marbles. If you roll a 1 you put one white marble in the jar. If you roll a 2 you put two black marbles in the jar. And so on and so on: 3 = three blue, 4 = four green, 5 = five yellow, 6 = six red. You do this till the jar has at least 12 marbles in it.

After you finish, which of the following is more likely than the others?

The jar contains only one color.... [ More ]

Reply #4

Orange71 — Sat, 27 Aug 2022 01:38:10 GMT

I think the series identity should be

[a/(a-1)]ln(a) = (a^n)/(n+1), n=0 to n=

where a in our case represents probability of selecting a blue ball for each flip of the coin or roll of the die. a here must be less than 1.

I agree you did calculate the correct solution for n= for the details of my particular problem statement. It's ln(2)/[ln(2)+ln(3)/2].

Reply #3

cottoneyedjoe — Sat, 27 Aug 2022 00:12:26 GMT

I had forgotten the identity

Ln(1/(1-a)) = (a^n)/n, n=1 to n=

for a 1. It's one of those really elegant formulas. Funny how those transcendentals sneak their way into probability questions.

Reply #2

Orange71 — Fri, 26 Aug 2022 23:56:54 GMT

cottoneyedjoe, very nice job! You got the correct answer

Reply #1

cottoneyedjoe — Fri, 26 Aug 2022 23:30:39 GMT

I got

(661/960)/(661/960 + 646/1215) = 53541/94885 0.564

for the probability of the blue ball coming from Jar 1.

As you increase the maximum balls from 5 to infinity, the probability of drawing a blue ball from Jar 1 converges to Ln(2) [natural logarithm of 2] 0.693, and the probability of drawing a blue ball from Jar 2 converges to Ln(3)/2 0.549. So the conditional probability of a ball coming from Jar 1, given that it's blue, should converge to

Ln(2)/(Ln(2) + Ln(3)/2... [ More ]

Another random selection puzzle

Orange71 — Wed, 24 Aug 2022 18:27:06 GMT

Another lottery-like mathematical challenge...

We have an experiment with a professor and student. The professor has a container that has a very large number of balls, both red and blue in color, with hundreds or more of each. The balls within each color are identical and indistinguishable. In front of him he has two empty jars. Let's call them Jar 1 and Jar 2. The experiment will be done out of sight of the student, who will only see the final result, and will then be presented with a quest... [ More ]