bubbled

Indeed. One really interesting aspect of this problem is that by removing the balls, you even out the odds of pulling one color or another as the numbers get big. For example, I ran a simulation 1,000 times where I flipped a fair coin 1,000,000 times and noted abs( H - T ). The average value was 783. But if we put 500,000 black balls and 500,000 white balls in the urn, the average number of balls left after exhausting one color is very close to 2! Coin outcomes remain 50-50. Ball outcomes readjust to b/(b+w) and w/(b+w). This puzzle has revealed so many nice symmetries, I thought I'd share another one. If you have b black balls and w white balls in an urn, the expected number of monochromatic balls you will pick out of the urn before you pick a ball of the other color is the same as the expected number of balls remaining after you have exhausted one color. Pretty interesting.

Indeed. One really interesting aspect of this problem is that by removing the balls, you even out the odds of pulling one color or another as the numbers get big. For example, I ran a simulation 1,000 times where I flipped a fair coin 1,000,000 times and noted abs( H - T ). The average value was 783. But if we put 500,000 black balls and 500,000 white balls in the urn, the average number of balls left after exhausting one color is very close to 2!

Aha!!

I'm not totally convinced by this argument. I obviously believe it. But don't see the logical argument why the number of black balls left scales linearly with the starting number of black balls, given a fixed number of white balls. For instance. It's obvious that if w = 10 and b = 1, then on average, there will be 1/11 black balls left. But let's look at w = 10 and b = 2. I would use this formula: 1/12 * 1/11 * 2 * 2 + 1/12 * 10/11 * 2 * 1 which does equal the expected result of 2/11. The first term is chances a particular black ball is in last position * chances the other black ball is in second-to-last position * 2 ways for that happen * 2 black balls left. The second term is chances a particular black ball is in last position * chances the second black ball is not in second-to-last position * 2 ways for that to happen * 1 black ball remaining. But the statement "If there are originally b black balls, this expectation becomes b/(w+1)" does not convince me of this fact. Maybe I'm missing a simple identity, but I haven't been "aha'ed" just yet.

"aha" still eludes me. I ran a simple Python simulation as well. But I thought I'd share my friend's elegant recursive approach to this problem. The nice thing was he got exact answers, which helps when looking for patterns.

Right answer, marked solved. It's similar to DeGe's result, which is the probability that a particular point is a terminus of the arc. A very simple words-only description also gives the result. In this case it shoul explain why simulating a line segment works also when the ends are joined (circle.) I thought long and hard about a words/logic only answer. So I'd love to see one. I'd find it much more interesting than my prosaic approach.

I ran a simulation of 10,000,000 games to check bonanova's work.

In that case I'd agree with everything you say. Here's my take:

I only see five gifts. This looks like an interesting puzzle, but I's like to know if there's a $1 gift or a $1M gift.

If each player will get the full amount if they happen to choose the same envelope, then Very well done, Bushindo. So, I ran a simulation in Python (I can post my code if anyone's interested). Here are the results that support Bushindo's conclusions: Each game is played 9,131 days (25 years with 4 leap days) and I ran 100,000 games. I wanted to play enough games to have a good chance of having Bob win a game: Bob won 1 game Charles won 99,999 games Bob won a total of $570,809,817,520,192 for an average of $625,133.96 per day! Charles "only" won a total of $433,549,519,222,236 for an average of $474,810.56 per day. But, Charles got his wish, he got to brag that he beat Bob 99,999 times, while they enjoyed their private island sipping margaritas. Here's a follow-up question: Charles is clearly crushing Bob whenever the the envelopes are 1-3, and he's losing all his equity on the very rare case where the envelopes are 3^29-3^30. So, why can't he simply make a small change to his strategy? Keep his exact same strategy, but in the rare case where he sees exactly $3^29, he also switches. Wouldn't this fix his problem? If not, what's wrong with that strategy?

Yes, indeed! But let's assume that Alec has all the money in the world and he's just interested in seeing what strategies are best. I think I know what the two "correct" strategies are given Bob's and Charles's different motivations. I've also run some large simulations, and the results are very interesting. I'll post answers tomorrow, if no one wants to take a crack at it.

Both players will win the full value of the same envelope if they end up with it.

Alec proposes to Bob and Charles that they will play a game over a 25 year period. Each day while Bob and Charles are away from Alec's house, Alec will prepare two envelopes. Alec will flip a fair coin until the coin comes up tails OR until he has flipped his coin 30 times. Based on the number of flips he makes each day, he will prepare the two envelops as follows: If the first flip is tails, he will prepare one envelope with $1 in it and the other with $3 in it. If he flips twice, the two envelopes will contain $3 and $9. Three flips will produce envelopes with $9 and $27 and so on. If he somehow flips 30 times, the larger envelope will contain $2.0589113e+14 and the other envelope will contain $6.8630377e+13. Both Bob and Charles know Alec's procedure and the exact distribution of the possible envelope values. Once Alec has set the two envelopes, he will randomly place one envelope on the left side of a table and the other on the right side. Both Bob and Charles come to the table, Bob is given the right envelope and Charles the left envelope. They are allowed to privately examine the contents of their envelope. They are then given the chance, privately, to switch. They get to keep the contents of the envelope they end up with each day. If one switches and other doesn't, they will end up with the same envelope that particular day. Question #1: Bob wants to maximize his expected value over the course of the entire game. What strategy should he use? Question #2: Charles is less interested in maximizing his EV. He's motivated to end up winning more money than Bob over the course of the 25 years. If he knows Bob's "perfect" strategy, what strategy can Charles use to maximize his chances of ending up ahead of Bob after the 25 years? If Charles maximizes his chances of beating Bob, he will end up ahead of Bob nearly 100% of the time.

I would say no. The switching in and of itself is not what causes the gain in expected value. Assuming that the chances of 500 or 2000 (in your example) are equally likely in the other envelope, then the gain in EV, comes from calculating that value and determining it is higher then 1000, and then switching. Even though you haven't looked into the other envelope, its value can be calculated and switching back would lose the EV gain you made by switching the first time. If you were to know the exact distribution (possible values and frequencies of those values) of all envelopes, then before you look into any envelope, you can assign it an expected value of an unknown random envelope. At that time, switching would not gain any EV, because absent any additional information, the other envelope has the exact same EV. Once you look into the first envelope, by knowing the overall distribution of all envelopes, then adding in the new and valuable information you have gleaned by reveling the contents of the first envelope, you can then make a proper choice as to whether or not to switch.

Assuming that your Uncle has $3000, so he could stuff one envelope with 1000 and the other with 2000: The real issue with all of these puzzles are the definitions. I don't think it's worth anything to switch, unless you have a clearly defined range of possible amounts in the envelopes. The possible range of values must have an upper limit, unless we are to accept a possible infinite envelope (I'm not). Let's say the range of possible values is 1-100 inclusive. And all possible parings are equally likely. Then the actual best strategy is if you open an envelope with 50 or less in it, you switch, otherwise you stand pat. If you're not given a clear range, then a perfect strategy is not possible to construct. For instance, if the range was 1-1,000,000, but you didn't know it, all the value you might gain by switching, will be lost when you draw an envelope higher than 500,000 and switch into an envelope that must be smaller.