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# Another random selection puzzlePrev TopicNext Topic

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• Texas
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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 question. The professor starts the experiment with Jar 1. He flips a fair coin. If the coin is Heads, he places a red ball in Jar 1, else if Tails he places a blue ball. If the selected ball is red, he flips the coin again and picks another ball per the stated rule. If the ball is red each time, he continues to sample until he reaches a maximum of 5 balls, then stops and moves on to Jar 2. If any selected ball is blue, he stops sampling any more balls for Jar 1 at that point and moves on to Jar 2. For Jar 2 the setup is the same except that instead of a coin, the professor uses a fair six-sided die. If 1 or 6 is rolled the ball selected is blue. Otherwise, red is selected. Again, this continues until a maximum of 5 red balls is reached or a blue ball is drawn (in which case no more balls are drawn).

Finally, the professor flips a fair coin again, with Heads representing Jar 1 and Tails Jar 2. Within the selected Jar, he picks one ball at random (with the possibility that there may only be one blue ball in the jar).

The professor finally reveals to the student that the selected ball is blue. The jars and starting container are still out of sight of the student. The student is explained the procedure and rules and is asked what the probability is that the selected blue ball came from Jar 1. What is the solution?

• United States
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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) = Ln(4)/Ln(12) ≈ 0.5579

• Texas
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cottoneyedjoe, very nice job! You got the correct answer!

• United States
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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.

• Texas
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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].

• United States
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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.
• The jar contains exactly two different colors.
• The jar contains exactly three different colors.
• The jar contains exactly four different colors.
• The jar contains exactly five different colors.

(Six different colors is not possible)

• 700 light yrs West of Milky Way Galaxy's Center
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September 1, 2019
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Quote: Originally posted by cottoneyedjoe on Aug 26, 2022

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) = Ln(4)/Ln(12) ≈ 0.5579

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

• Texas
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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.

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