superprismatic

Thanks, Phillip. This is one cool problem. It wasn't clear to me that every permutation on eight things can be formed during this game. So, I wrote a program to check it. Sure enough (modulo programming bugs), every permutation can occur. I'm trying to think about what to do next. In the meantime, I wanted to let you know that someone is working on this pretty hard.

Did you make a mistake on the 4th term?

Actually, I go through the whole deck. No. They are not independent. But that doesn't matter to the expected value. It's the same argument that's used in proving that the expected number of fixed points in a permutation is 1. You could probably find that proof on the web.

I didn't assume that each card has a probability of 1/13 of being in the discard pile. I used a counting argument to prove it. The dependencies you mention have no effect on the expected value.

Yep, you've got it!

Oh, I misunderstood. So here's a corrected one:

Below are 5 matrices. Your task is to orient the matrices so that when you add them up element-wise, you get a 5×5 matrix of all 20s. You may flip them over, rotate them by multiples of 90°, or both. Matrix 1: 1 4 2 4 3 5 8 0 0 10 2 1 2 3 10 9 1 8 0 2 2 1 8 0 1 Matrix 2: 10 4 7 4 5 3 4 10 5 9 0 10 4 9 7 9 3 0 9 0 4 7 0 1 10 Matrix 3: 2 2 7 0 2 9 4 8 0 1 5 0 3 10 10 6 4 2 4 5 1 1 1 0 3 Matrix 4: 5 0 2 0 10 1 8 6 3 8 2 3 2 2 4 5 1 0 9 7 7 0 1 3 1 Matrix 5: 3 3 1 11 5 0 10 2 1 7 3 1 9 0 0 1 4 5 2 1 1 10 8 7 4[/code]

Glad that you appreciated my little joke and like my puzzles. Thanks. Now as to your addition:

Suppose you permute the numbers 1 through 11 amongst themselves on a clock face, leaving the 12 undisturbed. Then pick a starting number. Begin a list with this number. Then, move clockwise around the clockface the number of numbered spots given by the last number on your list. Add this number to your list. Then move clockwise the number of spots given by the last number on the list. Continue in this way until you hit the 12 or a number which is already on your list. Here's an example. Suppose you number the clockface clockwise from the 12 as follows: 12,1,4,2,3,5,6,7,8,9,10,11. And you start at the 2. The next number for the list will be 5, then 10, then 8, etc. Your list will end up 2,5,10,8,3,7,4,6,12. Or, suppose you number the clockface clockwise from the 12 as: 12,1,3,2,4,5,6,7,8,9,10,11. And you start at 1. Your list will then be 1,3,5,10,8,4 and 8 would be next but it is already on the list. Can you permute the numbers between 1 and 11, and pick a starting number, such that your list will contain all 12 numbers? Interesting fact: Less than .0009% of random permutation/start combinations will produce a list containing all 12 numbers.

I'm still thinking of how to attack this analytically.

Yes, you assume correctly.

I meant (a). I said 1M trials, but what I really want is how these compare for a lot of trials -- the long run, if you will.

No. I want a more precise statement of the phrase. If it actually makes any sense, you should be able to rephrase it in several ways. Kindly do so. Otherwise, your OP deserves to be ignored. Insults are not appropriate on this site.

What does "counting pieces that are not even further sub-divided" mean? This is one point that begs for explanation. "even" seems misplaced here. Please elaborate.

Assume you start off with a bankroll of $1,000,000. Consider the following 3 games: Game 1: Flip a biased coin which lands Heads 45% of the time, and Tails the rest of the time. If you play this game, you get a $1 to add to your bankroll if you flip Heads and you lose $1 from your bankroll if you flip Tails. Game 2: There are 2 cases: _______Case 1: If your bankroll modulo 3 is 0, you flip a coin which lands Heads 1% of the time, and Tails 99% of the time. You get $1 to add to your bankroll if you flip Heads, Otherwise you lose $1 from your bankroll. _______Case 2: If your bankroll modulo 3 is 1 or 2, you flip a coin which lands Heads 90% of the time, and Tails 10% of the time. You get $1 to add to your bankroll if you flip Heads, Otherwise you lose $1 from your bankroll. Game 3: Flip and unbiased coin (50% of the time Heads, 50% of the time Tails). If the coin lands Heads, play Game 1; if it land Tails, play game 2. After 1,000,000 plays of each game (always starting with a bankroll of $1,000,000), which of these games should increase your bankroll and which should decrease your bankroll?

You are offered a chance to win $100. All you have to do is play 3 games and win any two consecutive games to get the $100. The hitch is that you must play against two players, A and B, alternately. That is, you must play them in one of two orders, either ABA (A first, B second, A third) or BAB (B first, A second, B third). Now, the probability that you can beat A is known to be PA and the probability that you can beat B is known to be PB. Furthermore, PA is less that PB. Which of the two orders, ABA or BAB, should you choose to give yourself the best chance of winning the $100? Note: In the ABA order, you are playing the one you are most likely to beat (B) only once, while you are playing the one you are least likely to beat (A) twice. In the BAB order, you are playing the one you are most likely to beat (B) twice, while you are playing the one you are least likely to beat (A) only once.

When my daughter was about 4 or 5 years old, I saw her doing odd looking calisthetics in front of a full-length mirror. Then she asked me the question, "Why does the mirror switch left and right, but not up and down?" I had no idea how to answer her in a way she could understand. In fact, I'm not even sure I understand. Does anyone have a clear and concise answer to my daughter's question that a child of that age could understand?

Yep, Captain, I hear what you guys are saying. The OP could have specified this point a bit better. I'm a bit amazed that it made so little difference, 20 vs 22. Cool!

Reflection doesn't seem natural to me for use in saying that two things are the same. I guess I don't like the fact that it makes things which are of different handedness the same. It's all a matter of definition anyway. Nobody can argue what's right and what's wrong here. The OP didn't specify precisely what "different" meant. Thanks for the clarification.