Prime

Chances are, most of the $100 bills would end up in the hands of very excited people below and not on the ground. By the way, have you chosen a specific building for that important experiment in probability? When?

Not only I forgot about half dollar coin in my previous post, I also made few arithmetic errors in calculating probabilities. Although, the answer who has a better probability is the same.

“Simple” is a relative term. To me it would be a simple programming code, if it was feasible to traverse all variations. Alas, I am not going to live long enough to see my laptop running through 419 possible arrangements of coins. (My laptop is old and slow.) Nonetheless, I’ll go beyond the OP question and give the exact probability. Oops, I forgot about 50-cent coin. Back to the drawing board.

Maybe I am misinterpreting the OP, but...

... Thanks. I am not clear on the calculation you've requested. In the case of (10,9,7) the first two cards add up to 19. We don't need an aid of a computer to calculate the average payoff of the 4th draw in case of (10,9,7). It is (364-10-9-7)/49 ~ 6.9, meaning we must stay on (10,9,7) and even more so -- on (10,9,8). However, that's a moot point, since the recursive function comes back to the case of (10,9) with the value of the draw of ~8.5653, whereupon it "decides" it is better to stay on (10,9) and discards the value of the draw. Later on the function comes back to the case of (10) with the value of the draw of ~9.5431 and again discards it in favor of 10 in hand. Conversely, in the case of the first card of (9), the function returns the average payoff for the draw with the best strategy of ~9.5818, meaning that drawing is better. Actual program that I ran spills out into Excel spreadsheet optimal payoffs for all first two card combinations, all first card draws, and the final result. Also showing best staying card in each case.

Was this result calculated? The first two cards didn't seem to have that much effect in simulations. The code calculates the exact probability for the average payoff with the best strategy (unless I messed up.) Both simulation and code must take into account the cards leaving the deck. Also, must discard invalid (in terms of strategy) variations. E.g., a draw like (13, 13, 8). The criteria for the staying card on the third turn can be established analytically (see the spoiler.)

Cards leaving the deck must be accounted for. Whether the difference between the exact and an approximate strategy is significant depends on how you view extra few $$ gained after thousands of games played. (See my next post.)

To calculate the payoff for the best dynamic strategy, we need a computer program, which traverses all variations choosing optimal path. In this case such program is relatively simple, while variations aren’t that many. Thus the payoff for the best strategy is found. To produce a formulation of the strategy complicates matters a little bit. However, in this case with the maximum of only 4 card draws, the best strategy can be expressed in simple terms as well.

You can not beat me at this game, because after finding the best strategy, I will only play for the winning side. That said....

I came to know that shuffle by the name “perfect shuffle.” Of course, you should start dropping the cards from the top half, so that the top card does not end up in the same position after the shuffle.

The house must be a regular rectangle, per OP. For those who's never seen Baba Yaga's hut, copy and paste this: домик бабы-яги картинки (or you could go with "Baba Yaga's hut") into Google's image search. With such house construction, the dog's freedom is limited by the length of the leash until the time Baba Yaga gets hungry...

In that case, it could be anything at all. After all, the man is blind, and there is nothing in the OP to suggest he can see two moves ahead. However, By what rules? I haven't played in any official tournament in awhile. Have the rules changed recently? From what I remember, if you touch your opponent's piece first, strict adherence to "touch move" would require capturing that piece if possible.

In that case, it could be anything at all. After all, the man is blind, and there is nothing in the OP to suggest he can see two moves ahead. However,

You are using a different notation. Other people here use algebraic notation, where columns are numbered "a" through "h" from left to right (on the White's side); and rows are numbered 1 through 8 from White's side to Black's. Your notation is called "descriptive". When you say qb8!!, it could be confused with algebraic Qb8. In descriptive notation correct designation of the move you suggest would be: Q-QB8?? (Queen to Queen's Bishop 8).

However, there are many people out there who do not need a chessborad with pieces to play chess, and those people don't have to be actually blind.

"Q.E.D" means "which had to be demonstrated" and typically refers to the concluding statement(s) of a proof. "WLOG" points to a relation between statements/formulas. Like, if it is good for Goose than without loss of generality it must be good for Gender. Or, let's say, in an induction proof: if the rule holds for a randomly chosen number WLOG it must hold for any other number. Q.E.D.

We had a recent discussion of Lagrange Interpolating Polynomial here: http://brainden.com/forum/index.php/topic/15537-closed-form-expression-on-steroids/?p=327698

I don't see a stipulation that each box truly corresponds to one of the labels. (A box could contain 2 creams and 1 chocolate, or nuts and bolts.) With that stipulation, the problem would be exactly as http://brainden.com/forum/index.php/topic/484-truth-in-packaging/?hl=%2Btruth+%2Bpackaging as well several other similar topics. Here Paralogic has given the answer.