probability
Assume that we have three boxes:
Box1 contains a red ball and a blue ball
Box2 contains a green ball and a yellow ball
Box3 contains a black ball, a white ball and a gray ball
The probability of choosing Box1 is 0.5, Box2 is 0.3 and box3 is 0.2.
If we draw five Balls randomly with replacement. Considering the drawings with the highest probabilities, how many ways can have sum of probabilities higher than a specific threshold, e.g. >90%?
Here’s another approach. [Added: Also, see more on this approach here]
Equivalently, we can do the following.
Reach into the first box and write “S” on one ball (the one to swap). Now choose a ball from the first box. If it is not the “S” ball, that’s your ball. The probability of this happening is $20\over21$, and the ball you choose will be white with probability $10\over21$. If you do choose the “S” ball (you do this with probability $1\over21$), discard it by throwing it into the second box and then choose a ball at random from the second box (now containing an extra ball), and that’s your ball. If you had to do this, the probability of a white result is ${12+{10\over21}\over26}$, because if we add a randomly-chosen ball from box 1 to box 2, the number of white balls in box 2 effectively increases from $12$ to $12+{10\over21}$ and the number of balls in box 2 increases to $26$. Therefore the total probability you want is
$$p = {20\over21}\cdot{10\over21}+{1\over21}\cdot{12+{10\over21}\over26}={2731\over5733}\approx 0.4763649.$$
P.S. I don’t see an easy way to adapt this approach for two swaps.
F = First Box, S = Second Box, W = White Balls, B = Black Balls
$$P(F|W)=\frac {P(F\cap W)}{P(W)}=\frac {P(F)\times P(W|F)}{P(F)\times P(W|F)+P(S)\times P(W|S)}=\frac {\frac 12 \times \frac 23}{\frac 12 \times \frac 23 + \frac 12 \times \frac 13}=\frac 23$$
Best Answer
In the table below draws are grouped into Types by how many balls from each box they have and are ordered by descending probability.
CP - Color Permutations, e.g. Type 20 has draws with $2^13^4$ different ball color orders from each box.
DP - Draw Probability, e.g. each draw in Type 2 occurs with $p=\frac{0.5^40.3^1}{32}$
OC - Order Combinations, e.g. in Type 3 there are $\binom{5}{2}$ combinations of box orders, affecting the total number of draws that fall in this Type
T - Total number of draws of this Type, CP times OC
S - Sum of probabilities
The answer is: to exceed 90%, out of 16807 possible, we need 8653 most likely draws. All 7744 draws of Types 1 through 13 and any $\lceil\frac{0.9-0.87728}{0.000025}\rceil=909$ draws of Type 14, which feature 1 ball from Box 1 and 2 each from the others.