PROBABILITY THEORY: Should I put the ball back?

Question

In a distant past where party games are still allowed, a boring mathematician host presented two jars with the same appearance and tells you (another boring but very competitive mathematician): There are two red and one black balls in one jar (named as X) and 101 red and 100 black balls in the other jar (named as Y). He then takes out one ball from one of the jars, showing it to you, and asks you whether to put it back or not. After according to your answer, he takes out again a ball from the same pot, showing it to you. He then now asks you whether the pot he has drawn a ball from in the last time is X or Y.

 

In order to get a larger probability that you guess correctly (since you are highly competitive), should your first answer be to put the ball back or not?

Answer 1

Actually, there should be a choice of whether to put it back or not depending upon whether the first ball is red or black. Let's consider all four choices (put back or don't put back when first choice is red or black); the metric to use for the decision has not been given, so let's say it is the accuracy and assume that it is equally likely to choose either of the 2 jars

if choose with replacement if black

P(jar 1, black, black) = 1/2 * 1/3 * 1/3 = 1/18

P(jar2, black, black) = 1/2 * 100/201 * 100/201 = 5000/40401

5000/40401 > 1/18; select jar 2; misclassification 1/18

P(jar 1, black, red) = 1/2 * 1/3 * 2/3 = 1/9

P(jar2, black, red) = 1/2 * 100/201 * 101/201 = 5050/40401

5050/40401 > 1/9; select jar 2; misclassification 1/9

Total misclassification: 1/18 + 1/9 = 1/6

if choose with replacement if red

P(jar 1, red, black) = 1/2 * 2/3 * 1/3 = 1/9

P(jar2, red, black) = 1/2 * 101/201 * 100/201 = 5050/40401

5050/40401 > 1/9; select jar 2; misclassification 1/9

P(jar 1, red, red) = 1/2 * 2/3 * 2/3 = 2/9

P(jar2, red, red) = 1/2 * 101/201 * 101/201 = 10201/80802

2/9 > 10201/80802; choose jar 1; misclassification 10201/80802

Total misclassification: 1/9 + 10201/80802 = 2131/8978

if choose without replacement if black

P(jar 1, black, black) = 1/2 * 1/3 * 0 = 0

P(jar2, black, black) = 1/2 * 100/201 * 99/200 = 99/804

99/804 > 0; choose jar 2; misclassification 0

P(jar 1, black, red) = 1/2 * 1/3 * 1 = 1/6

P(jar2, black, red) = 1/2 * 100/201 * 101/200 = 101/804

1/6 > 101/804; choose jar 1; misclassification 101/804

Total misclassification: 0 + 101/804 = 101/804

if choose without replacement if red

P(jar 1, red, black) = 1/2 * 2/3 * 1/2 = 1/6

P(jar2, red, black) = 1/2 * 101/201 * 100/200 = 101/804

1/6 > 101/804; choose jar 1; misclassification 101/804

P(jar 1, red, red) = 1/2 * 2/3 * 1/2 = 1/6

P(jar2, red, red) = 1/2 * 101/201 * 100/200 = 101/804

1/6 > 101/804; choose jar 1; misclassification 101/804

Total misclassification: 101/804 + 101/804 = 101/402

If we choose replacement (both with black and red first),

total misclassification = 1/6 + 2131/8978 = 5441/13467 = .4040

If we choose without replacement (both black and red first)

101/804 + 101/402 = 303/804 = .3769

If we choose without replacement with black and with replacement if red

101/804 + 2131/8978 = 19553/53868 = .3630

If we choose with replacement with black and without replacement if red

1/6 + 101/402 = 19553/53868 = 28/67 = .4179

If you have to choose sampling with or without replacement and can't consider the first selection, then the best choice is sampling without replacement, as .3769 < .4040

However, it would be even better if you could select without replacement if the first selection is black and with replacement if the first selection is red, as the probability is .3630, which is less than .3769

Note that, in this case, the second selection matters, both if the first selection is red or black; otherwise, in either the black or the red, the second selection won't matter. You might wonder then, does the first selection matter if we are choosing based on the second selection only; the answer is that it does because the sampling technique changed based upon the first selection.

In the worst case, with replacement with black and without replacement if red, your second selection does not matter, so you might as well not even make the second selection.