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60 votes
Accepted

Rock, Paper, Scissors and Trump

Looks like 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 ...
Bass's user avatar
  • 77.8k
41 votes

Will a greedy algorithm solve Tatham's Flood?

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 ...
TheGreatEscaper's user avatar
39 votes

Will a greedy algorithm solve Tatham's Flood?

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 ...
dtldarek's user avatar
  • 1,065
30 votes

Shortest game of Monopoly

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 1: Roll 2 (Because ...
William Pennanti's user avatar
29 votes
Accepted

Can you beat random?

Here's how you win in the long run Expected number of losses Thanks to Evargalo for fixing the last calculation.
hexomino's user avatar
  • 137k
20 votes

Can you beat random?

I agree with both answers, but the answer is a little simpler than their explanations. It can be simplified to: Those two rules alone cover every possibility. For completeness:
Stevish's user avatar
  • 1,345
20 votes

Rock-Paper-Scissors but rock scores double

This probably isn't the answer you want but:
fljx's user avatar
  • 16.6k
16 votes
Accepted

Angels & Demons (Open question)

noedne's user avatar
  • 15.4k
16 votes
Accepted

Reverse dots and boxes

I would suggest an alternate (simpler) strategy:
Smock's user avatar
  • 950
16 votes

Rock-Paper-Scissors but rock scores double

James Martin's user avatar
15 votes
Accepted

Which parent should you start playing against?

Mary should play the parent first.
Nuclear Hoagie's user avatar
13 votes
Accepted

Winning strategy in a calendar game

I think that your proposed method is correct. Here's a quick visual representation of the same method (by no means a formal proof):
DqwertyC's user avatar
  • 8,248
13 votes

Angels & Demons (Open question)

I just came back home and was about to draw a diagram illustrating this, but I can't do any better than @noedne already did.
Glorfindel's user avatar
  • 28.1k
13 votes
Accepted

2019 gold coins to share

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 ...
Reinier's user avatar
  • 5,167
13 votes

Which parent should you start playing against?

Mary should play her first game against To prove this, notice that if $n$ were even, For the case presented in the puzzle, with $n$ odd:
Nitrodon's user avatar
  • 289
13 votes
Accepted

Guarantee a win in choosing numbers from extremes of a row of numbers

And the answer is... Why? Now:
Jujustum's user avatar
  • 4,059
12 votes
Accepted

What is the optimal strategy for the Dictionary Game?

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 ...
Jafe's user avatar
  • 78.2k
12 votes
Accepted

A knight chased by three knights

Observation: Implication: Conclusion:
Daniel Mathias's user avatar
12 votes
Accepted

The median game

Consider that Given that, I see play proceeding as follows N.B. I'm assuming here that strategies which are equally optimal for a given player are equally likely to be executed. This seems to be ...
hexomino's user avatar
  • 137k
12 votes
Accepted

Game of the glasses on the windowsill

Here is my evaluation: Conclusion And yet...
Florian F's user avatar
  • 30.6k
12 votes
Accepted

The Ultimate Battle of two players

Solution Define the recovery ratios $R_i$ for player $i$ $$R_1 = \frac{H_1}{A_2} \text{ and }R_2 = \frac{H_2}{A_1}$$ and the "turns-to-die" $T_i$ for player $i$ $$T_1=\Big\lceil\frac{P_1}{...
Benjamin Wang's user avatar
11 votes

Shortest game of Monopoly

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 ...
Psychonaut's user avatar
11 votes
Accepted

Buzzword Bingo Boss Battle!

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 ...
Bass's user avatar
  • 77.8k
10 votes
Accepted

Reverse dots and... cubes? - Snowflake edition

The player who can win is by this strategy:
Deusovi's user avatar
  • 147k
10 votes

Playing blindfolded 2048

Source: I played this game a lot while distracted and this seems to work pretty well. [Sort of unrelated, but 2048 is one of the first C++ programs I wrote. Code is here: link to code.]
DenverCoder1's user avatar
  • 10.5k
10 votes
Accepted

Lap Theory Optimal Betting Game

Their best strategy is Their chances with this strategy are Optimality
Albert.Lang's user avatar
  • 6,195
10 votes
Accepted

How long can you survive at the devil's playground?

You're screwed in constant time, no matter your speed, since the devil has a good strategy. I cannot claim to having found the devil's optimal strategy, but I do claim that there is an upper bound to ...
IvanSanchez's user avatar
10 votes

Can you stop copy Alice?

Edit 2012rcampion observed that we also need to multiply by 2 the distances on the plane Not an answer with a strict proof of the final behavior, but instead a simulation with code that show a ...
10010100102ohno's user avatar
10 votes

Alternative solutions to a Michael Rabin puzzle

In light of the close votes, I am posting an answer, so that people can make a more informed decision about whether the question merits closure. The intended solution is: But why did they do this? I ...
Timothy Chow's user avatar
10 votes

Can you beat random?

Here is a winning strategy: Then at each move there are 3 possibilities: Your expected win is Your probabilities for round n (n>1) are: Counting 1 for a win, 1/2 for a draw and 0 for a loss, ...
Evargalo's user avatar
  • 6,300

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