If A randomizes his coins, a, and his totals guess, x = a,a+1,a+2,a+3, then he wins 1/4 of the time. But I think B can then do better. If B sees A has guessed x=6 then B knows A's number is 3. B then knows what to guess and wins 3/4 of the time (whenever B's number is not 3). If B sees x=5 then B knows A's number is not 0 or 1. If B sees x= 0 then B knows A's number is 0. If B sees x=1 then B knows A's number is not 2 or 3. A's guess gives information to B for all guesses other than 3.


PairTheBoard

Ooops. I overlooked something. I'll recheck and post.

OK, I was right the first time. For a second I thought I had overlooked something, when I had actually factored it in.

First I'll make a note to myself that there are 4 combinations of coin choices for each player and the 16 total combinations are :

0+0 = 0 (1 comb for 0)
0+1 = 1+ 0 = 1 (2comb for 1)
0 + 2 = 2+ 0 = 1 + 1 (3comb for 2)
0 + 3 = 3 + 0 = 1+ 2 = 2 + 1 (4comb for 3)
1 + 3 = 3 + 1 = 2 + 2 (3comb for 4)
2 + 3 = 3 + 2 (2comb for 5)
3 + 3 (1comb for 6)




If both players randomize equally between 0,1,2 and 3, playerA is justified in guessing 3 with 1/4 certainty. Can player B randomize in such a way as to make 3 less possible if A is randomized?

NO. The chance is always 1/4 if player 1 randomizes perfectly.

Then.. Player B has to pick between 2 and 4 with 3/16 chance.

Can either player improve on this?

I'd say no. If player A chooses 2 or 4 he not only sacrifices 4/16 for 3/16, he also allows player B to choose 3 for his gain.




There is absolutely no gain by picking any particular number of 0 1 2 or 3, as long as you randomize. If you take a look you'll notice that in the calculations for the sum total of 3, the numbers 0 1 2 and 3 appear each 2 times, so there is no difference.

Edit: I had missed one thing - if B has picked 0 he has to pick 2, if he has picked 3 he has to pick 4.

FINAL ANSWER : Optimal strategy is to randomize the initial pick perfectly for both players and for A to choose always 3 and for B to always choose 2 or 4 (and if he has chosen 0 or 3 he has no such choice). There is no mixed strategy involved, since initial picks don't influence the sum total.



P.S. That's one ****ty game.


P.P.S. I can't believe how slow you guys are. Only EvilSteve kind of solved it, but erroneously enough for me to claim being the first. If I find any of you have picked on my latest thread you'll never hear the end of it!

Actually, now that I'm rereading what EvilSteve said, he is quite wrong.

This, "knowing your hand" thing tilted me hard, because I disproved its significance immediately and then thought I hadn't for quite some time. Man, am I stupid.

I should add one more thing.

This is an optimal solution. This means that we assume no knowledge of opponent's frequencies.

If we have sufficient knowledge of opponent's tendencies we can pick the number immediately. A certain mathematical formula can be specified which would judge whether you go for optimal or for a guess. I'm not going to do that (I'm not even sure I could; I'm way too rusty) If there's a 26-25-25-24 distribution, you still go optimal. If there's 50-50-0-0 distribution and you have the zeroes giving you your optimal number, then you pick one of the 50s.

I think B can do better than 3/16 against A's strategy. B can still randomize his coins. But if he sees his coin number is 0 he guesses 2, with 1/4 chance A's coin number is 2. If B sees his coin number is 3 he guesses 4 with 1/4 chance A's coin number is 1. If B sees his coin number is 1 or 2 he can guess either 2 or 4 with 1/4 chance A's coin number is what he needs to be right.


PairTheBoard

Ah. Wait a minute.

Yes, I was wrong.

A must choose 3, because he doesn't give up any information doing so. B can choose any of 3 options and they all have same value.

I feel like an idiot now.

So, there's plenty of room for leveling. A must call 3 as to not give up information, but B can play RPS and try to guess in any of his 4 spots.

Yup, I'm an idiot. It's been confirmed and now official.

At least I'm content that everyone else felt confused by it. I guess I got mislead by the overall numbers, to miss something so obvious.

What is wrong about my answer (1st on the page) and the others similar to it?

I fail to see how B improves on his 1/4 chance of winning?

Well first Rhaegar thought I "kind of solved it", then decided I was "quite wrong", but apparently never noticed that I was using a different interpretation of the rules and therefore wasn't addressing the same problem at all.

But yeah, I'm pretty sure acehole60 gave the correct strategy in the first response, based on the consensus rule set where there's no winner unless somebody guesses the total exactly.

Apologies for the ambiguity in the OP. Banzai knows more about the game than me, so I hope Post #15 cleared this up.

Post #2 in the thread showed that A can guarantee doing as well as B. It remains to show that B can do as well as A. This is very unsurprising, but one explicit strategy for B to complete the proof of the "solution" of the game is:

Choose 0,1,2,3 coins uniformly at random. (This means A can't win more than 1/4). Then
(a) if A doesn't choose 3, then guess 3 without having to bother to look how at many coins you chose
(b) if A does choose 3, then look at how many coins you chose and if you chose 0 guess 0; if you chose 1 guess 2; if you chose 2 guess 4; if you chose 3 guess 6. This gives B precisely a 1/4 chance of winning regardless of A's strategy - even if A has not chosen uniformly at random.

Thus the solution is the game is equally fair to both players.

OP stated the game is played over several turns.
If Player A detects player B is not at random choosing his 3 numbers: I guess player B becomes vulnerable for exploitation.
So I guess Player B has to choose both coin and numbers at random:

Coin = 0: 0,1,2
Coin = 1: 1,2,4
Coin = 2: 2,4,5
Coin = 3: 4,5,6

Player B should play:
1/12 -> 0
2/12 -> 1
3/12 -> 2
3/12 -> 4
2/12 -> 5
1/12 -> 6

Probably there are more optimal strategies for Player B (cfr Pokerfarian).

Rhaeger was saying B could only do 3/16 so my reply to him was to show B could do 1/4.


I didn't understand what you were saying when I first read your first post. Having now gotten up to speed a little better I can see your first post made perfect sense and was right on.


PairTheBoard

Yes, I was comically (tragically) wrong.