My Math Forum Probability When Two People Pick Marbles

 Probability and Statistics Basic Probability and Statistics Math Forum

 October 8th, 2017, 03:46 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 608 Thanks: 82 Probability When Two People Pick Marbles Two people have bowls of marbles. Each bowl has 3 green, 3 red, and 3 blue. The two people pick one marble each and note if they picked the same color. This is done without replacement until all the marbles are picked. What is the expected number of times out of 9 that both people picked the same color? Is it as simple is 1/3rd of 9?
 October 8th, 2017, 04:29 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,105 Thanks: 1093 This seems to get pretty involved when you get into it. Via sim I get $P[\text{both pick same color}]=\dfrac 1 4$ that would make the expected value out of 9 $E[\text{both pick same color}]=\dfrac 9 4$ I'll keep looking at it, there might be something simpler.
 October 13th, 2017, 06:29 PM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 608 Thanks: 82 Why is the probability 1/4 and not 1/3? I did simulations of the numbers 1 through 9 appearing once in each simulation. Any of the first three numbers being 1, 2, or 3; any of the fourth through sixth numbers being 4, 5, or; and any of the seventh through ninth numbers being 7, 8, or 9 were counted as being the same color. From 10 simulations of 9 numbers, I got 33 out of 90 to have the same color. The probability of at least 33 out of 90 successes when the probability on each trial is .3333 (I couldn't input the fraction 1/3) is 0.285043975219546 and decreasing the probability of success on each trial to .25 makes the probability of at least 33 successes go down to 0.00926829268929696. That may not be mathematically correct because the nine numbers within each simulation are not independent, but I think we can agree that it would be hard to get at least 33 out of 90 successes if the probability was .25.
 October 13th, 2017, 07:07 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,105 Thanks: 1093 Ok, this time I just did a monte carlo instead of trying to analyze all the possibilities. This time it shows that on average the two will pick matching marbles once per bowlful. Thus the average over 9 trials will be 9 matches. This does match the $\dfrac 1 4$ number I got per pick as there are a total of 4 picks per bowlful. I just messed the expectation by a factor of 4 last post. I can't give you an analysis of this yet. Brute forcing it gets very complicated because of the non-replacement.
October 13th, 2017, 07:51 PM   #5
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,105
Thanks: 1093

I redid the original numeric analysis and obtained the result that the expected value of the number of matched picks per bowl is 1.

I don't know how well you'll be able to interpret this sheet, it's rather heavy in Mathematica speak but I added some comments so maybe you can follow it.

Attached Images
 Clipboard01.jpg (70.4 KB, 20 views)

October 14th, 2017, 03:35 AM   #6
Senior Member

Joined: Oct 2013
From: New York, USA

Posts: 608
Thanks: 82

Quote:
 Originally Posted by romsek This does match the $\dfrac 1 4$ number I got per pick as there are a total of 4 picks per bowlful.
There aren't 4 picks per bowlful. You may be misunderstanding the question. Here's an example with B1P1 meaning Bowl 1 Pick 1, G meaning green, R meaning red, and B meaning blue:

B1P1: R, B2P1: R
B1P2: R, B2P2: G
B1P3: R, B2P3: B
B1P4: G, B2P4: R
B1P5: G, B2P5: G
B1P6: G, B2P6: B
B1P7: B, B2P7: R
B1P8: B, B2P8: G
B1P9: B, B2P9: B

Picks 1, 5, and 9 match. That's 3 out of 9 times, and that's why I guessed the answer to be 3.

October 14th, 2017, 04:24 AM   #7
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,105
Thanks: 1093

Quote:
 Originally Posted by EvanJ There aren't 4 picks per bowlful. You may be misunderstanding the question. Here's an example with B1P1 meaning Bowl 1 Pick 1, G meaning green, R meaning red, and B meaning blue: B1P1: R, B2P1: R B1P2: R, B2P2: G B1P3: R, B2P3: B B1P4: G, B2P4: R B1P5: G, B2P5: G B1P6: G, B2P6: B B1P7: B, B2P7: R B1P8: B, B2P8: G B1P9: B, B2P9: B Picks 1, 5, and 9 match. That's 3 out of 9 times, and that's why I guessed the answer to be 3.
so you're saying the marbles are replaced.

your first post explicitly stated the marbles are not replaced.

October 14th, 2017, 06:27 AM   #8
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,105
Thanks: 1093

Quote:
 Originally Posted by romsek so you're saying the marbles are replaced. your first post explicitly stated the marbles are not replaced.
If the marbles are replaced and each pick is from a bowl of 9 marbles, then
there are clearly 3 picks that match out of 9 total possible picks.

So yes, in the case of replacement there should be

$E[\text{# matches out of 9 picks}]=9 \cdot \dfrac 1 3 = 3$

 October 14th, 2017, 08:10 AM #9 Senior Member   Joined: Sep 2016 From: USA Posts: 444 Thanks: 254 Math Focus: Dynamical systems, analytic function theory, numerics Its easy to see why 3 is unlikely to be the expectation. Consider the first 2 picks. Pick 1 matches with probability 1/3. Now, whether pick 2 matches depends heavily on whether or not pick 1 did. In the case where pick 1 didn't match, they no longer have the same color marbles which makes matching here much less likely than 1/3. In other words, each pick where they don't match, increases the probability of not matching in later picks since their marbles will become increasingly different in color composition. The strategy to understand this is not to go 1 pick at a time but to look at entire sequences of length 9. Fix the first person's sequence and now consider only the second persons sequence. 1 Match: The number of ways the second sequence matches the 1st in exactly 1 position is 9. To put this in a way which helps the analysis more, its actually $\binom{9}{1} = 9$. Since there are 9 positions and we choose 1 to match. 2 Matches: The number of ways the second sequence matches the 1st in exactly 2 positions is $\binom{9}{2}$. Continuing for all 9 match possibilities and noticing that the number of total sequences possible is $\frac{9!}{3!^3}$ where we have modded out by permutations of the 3 colors which should be indistinguishable, you can compute the expectation directly $E = \sum_{k=1}^9 \frac{k \binom{9}{k} 3!^3}{9!} = \frac{48}{35}$ You should carefully check this since these types of problems are easy to make simple mistakes on so this number might be wrong.However, analyzing the full sequences themselves I think is the best way to proceed. Thanks from EvanJ Last edited by SDK; October 14th, 2017 at 08:17 AM.
October 16th, 2017, 06:12 AM   #10
Senior Member

Joined: Oct 2013
From: New York, USA

Posts: 608
Thanks: 82

Quote:
 Originally Posted by romsek so you're saying the marbles are replaced. your first post explicitly stated the marbles are not replaced.
The marbles are not replaced. The two people have separate bowls with the same composition.

 Tags marbles, people, pick, probability

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post EvanJ Probability and Statistics 6 April 18th, 2017 02:18 PM EvanJ Probability and Statistics 2 January 5th, 2016 05:14 AM EvanJ Probability and Statistics 4 December 24th, 2015 06:13 AM EvanJ Advanced Statistics 1 March 19th, 2014 06:55 AM Ataloss Advanced Statistics 3 March 24th, 2012 12:55 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top