One strategy would be to
The opponent will naturally soon realise what's happening. But it won't help. Here are the possible results from Alicia's point of view:
So whatever the opponent chooses, the expected value will favour the strategy given above. By the law of large numbers, the average amount gained from a single throw
The reason this ...
It's definitely not optimal! Here's a counterexample:
Start from the white area.
Greedy algorithm would make you do all the colours on the top right (5 moves: yellow, green, dark green, blue, dark blue), then the red (1) and then all the other colours (6 moves), making a total of 12 moves.
There is an obviously more efficient method: orange, yellow, green, ...
That problem is NP-hard, so an efficient strategy to calculate the optimal moves would be a major breakthrough in computer science. Of course, there might be a greedy strategy, but not an efficient one, e.g. that works in exponential time.
To prove that it really is NP-hard, we will reduce vertex cover to your problem.
Let $G$ be the input graph. We will ...
If you are trying to lose the game as quickly as possible then you can do it in 1 turn and 2 rolls of the dice. This is using the UK version as I'm not sure about the US one.
Roll 2 (Because of the doubles):
The chance of this is:
All but Mr. Bun will attend the tea party, with Duke Froggington II as the guest of honor. The total number of animals who will be attending the party is $(6*12+2)-1=73$.
Looking at some specific cases:
One animal left: he will definitely nominate himself.
Two animals left: each animal knows that if he nominates himself, there will only be one left, and ...
Yes, X can win.
To simplify things I'm going to take advantage of your rules that O cannot win, so that X doesn't need to worry about O getting 1000 in a row.
Just consider a 1 dimensional game, chose an origin, and label the locations with integers in order.
Define the "bin $K$" as the set of spaces $x$ such that $1000\ K \le x < 1000 (K+1)$.
I am not myself a perfect logician, but here the behaviour I observed at my own tea parties, and the explanation my own favourite nounours gave as years passed.
Party of one
This issue was quickly solved before I got my second plush. Nounours was the single invitee, and behaved as well as a bear does at a party.
Party of two
When Dr. Tortue joined us ...
Let's assume every player knows the exact same list of words. Player 1 can ensure a loss for player 3 on the first turn by selecting the second word from the list. This allows player 2 to pick the first word, leaving no possible words for player 3. (Equivalently, player 1 can pick the second-to-last word, allowing player 2 to pick the last one.)
Say the ...
I can ensure that I receive at least
To get this amount of coins, I could apply the following strategy:
My friend can also ensure that I do not get more coins than this in the following way:
So it appears that I should have chosen a fairer game to distribute the coins...
No, the game given in the OP is not the shortest possible Monopoly game (using the American edition). As with the UK edition solution given by William Pennanti, a two-player game can be lost in a single turn with two rolls of the dice:
Player 1, turn 1, roll 1:
Player 1, turn 1, roll 2:
The probability of this game:
So, we are playing misère Tic-tac-toe with the option of taking more than one turn in a row. Seems quite interesting, so let's take a stab at it.
The flow of the narrative here is a bit messy, because the answer was not all written in one go. Since a full rewrite would be too time consuming, I've added some chapter breaks to give at least some structure to ...
Let us suppose that each player can bid any integer equal to or above the current minimum bid. Then:
Barry O'Neill develops a simple formula for this strategy in a 1986 article in the Journal of Conflict Resolution.
If we play by the exact rules listed here--first player must bid \$5 and each bid must increase by an increment of one dollar--I do ...
In the unrealistic case of there being two players, both knowing the same $n$ words then it really is a type of Nim game. If $n$ is odd, then the first player can always win, and if $n$ is even then the second player can always win.