Two players independently choose 0, 1, 2 or 3 coins. The object is to guess the total number of coins chosen.

Player A says his guess out loud. Then Player B says his guess, which must be different. Then the total is verified.

Players take it in turns to be Player A in subsequent rounds.

I'll give it a shot:

Player A obv has the worst of the deal. But if he randomizes his pick and guess "3", he levels the playing field and both have 1/4 to win. Seems like this is the best player A can do, and with this strategy, it doesnt matter what player B does (EDIT: he shouldnt do something stupid, tho, he should just pick a number and guess something that is above/at his number and not 3).

I came to the same conclusion, neither player has the advantage this way (EV player A = EV player B).
Player B should also pick his number_B at random and I guess choose at random 1 of these 3 numbers (number_B, number_B+1, number_B+2, number_B+3 -> 3 excluded). Otherwise his play is exploitable in the long run.

Result:
0.25: player A wins
0.25: player B wins
0.5: no winner

It's called Spoof and is a drinking game here in the UK:

http://www.google.co.uk/#hl=en&outpu...w=1024&bih=471

I'm really not sure, I can't see any strategy being beneficial, other than a pattern recognition and meta-game considerations. Perhaps a pro rock-paper-scissors guy can elaborate on that somewhat.

on random choices guessing 3 has a 4/16 chance and any other pick has less then that so player A should always say 3. Player A has a first move advantage over player B since Player B has to say a different score that obv. has a smaller chance to win.

ok, I just saw I overlooked that you know your hand... :P

Like others said, its A=1/4 B=1/4.

Lets show this given strategy is optimal. Let's let a be A's secret number, x be the guess, and similarily b,y is player B's secret number and guess.

First, notice that after A makes a normal* guess, (by normal I mean that given x,a , that there exists a number from {0,1,2,3}, where if b equals that number, then A wins) B knows one number that A's secret number isn't, namely that a != x-b. Then, atleast one of the other three choices must be chosen 1/3 of the time. So B guesses that choice + b. So B wins atleast 1/3 of the time if it is his turn.

Second, suppose it is A's turn and that B chooses his number randomly. Then at most he can win 1/4th of the time since his guess is equivalent to guessing b. So A's winrate was bounded above by 1/4.

Finally, lets resolve the *starred issue. If A doesn't make a normal guess, then his winrate is 0, and B's winrate is bounded below 0, so A doesn't improve his equity to make bad guesses.

OOPS I DIDNT PROVE IT WAS OPTIMAL

Why isn't a random guess a 1 in 7 chance of being correct?

Since you know your number, you know certain other numbers are impossible. So if you picked a 2, guessing 0 is silly.

^ Thanks, in my defence it is early in the morning here.

Mm I think you have to massage it but a similar argument to mine I think will show its optimal.

I need clarification on the rules. Do you have to guess the exact number to win and there's no winner if neither player guessed correctly, or do you win by guessing closer to the actual number than your opponent (and a tie if you're both equally close)? For the following I'll assume the win condition is being the closest.

So if A picks randomly from {0,1,2,3} and guesses 3 for the total, player B could make his choice a coinflip between 0 and 3. Then B would guess 2 for the total if his choice was 0, and 4 for the total if his choice was 3. With this strategy B wins 3/4 of the time. And actually against A's strategy, if A never adjusts, there's no need for B to randomize. He could pick either 0 or 3 every time.

Against the randomized version of B's strategy described above, A could adjust by making his choice a coinflip between 0 and 3, while still guessing 3 for the total. So now both A and B win half the time.

Is there any way for B to improve against A's new strategy? Since we're looking at exploiting a fixed strategy, we only need to look at pure strategies for B. If B chooses 0, A's guess of 3 will be exactly correct half the time so that's no improvement for B. If B chooses 1, the total is a coinflip between 1 and 4. The best B can do with his guess is to win half the time so no improvement there. If B chooses 2, the total is a coinflip between 2 and 5. Again the best B can do with his guess is win half the time. And if B chooses 3, A will be exactly correct half the time. Conclusion: B has no way to improve.

So we have a Nash equilibrium. A and B each do a coinflip between 0 and 3 for their choices, A guesses 3 for the total, and B guesses either 2 or 4 depending on whether his choice was 0 or 3. No advantage for either player.

I didn't address all of A's possible strategies though. A doesn't have to guess 3, but if he guesses lower than 3 (against B's strategy of choosing by coinflip either 0 or 3), then B is guaranteed a win if B chose 3, by guessing 3 (and may also win some of the time when B chose 0). So B wins at least 50% when A guesses lower than 3. A similar argument holds when A guesses higher than 3 (B wins by guessing 3 whenever B chose 0). It was already shown that A can win at least half the time by always guessing 3, so A does not benefit by guessing anything else.

Yes, you must guess correctly to win. If all players guess incorrectly the game continues.