
Coin Puzzle: Predict the Other's Coin - pratikpoddar
http://pratikpoddarcse.blogspot.in/2012/12/predict-others-coin.html
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hmexx
If they both agree to use their own coins result as the guess of the OTHER
person's coin, they should get it right 50% of the time instead of 25%.

Possible tosses:

TT win

HT lose

TH lose

HH win

Counter-intuitive though isn't it.

 _Why does using your own result improve the odds of winning the game!?_

~~~
gabemart
>Why does using your own result improve the odds of winning the game!?

I find the result very confusing as well. I suspect the source of this
counterintuitiveness is that I over simplified the puzzle when I first read
it.

I simplified the puzzle to "A attempts to guess B's coin, and B attempts to
guess A's coin", whereas in fact the true puzzle is "A and B together try to
guess the total set of results".

When A and B guess randomly, they attempt to guess the total set without using
the information they have available to them. When A and B both guess the
results of their own coin, they use the information they have regarding the
set of results (i.e. the results of their own coin) to reduce the problem
space and increase their chances of success. Clearly, if you have heads, you
know the chances of the set of results being heads-heads are much higher than
choosing a random set of results.

In other words, the proposition for A is not "What are the odds of B having
heads given that you have heads?" but rather "What are the odds of the set of
results being heads-heads given that you have heads?"

~~~
_dps
I'm not sure the conditional probability argument explains it. Specifically,
given A=H, the probability of (H,H) is the same as the probability of (H,T) so
that alone shouldn't inform your guess.

I believe the core mechanism at play here is that _coordination_ eliminates
the possibility of one player being right and the other being wrong, because
one can partition the coin outcomes into "We have the same result" and "We
have different results"; by agreeing in advance to only guess "we have the
same result" every time you eliminate the "I'm right, you're wrong" and
"You're right I'm wrong" failure modes (50% of the probability space under
random guessing).

One could accomplish the same by always guessing the _opposite_ of what you
obtain (i.e. always bet on "we got different results" which again has a 50%
probability).

This is related to the "guess your own hat's color" riddle:

"You and a friend are on a game show. The host sets you facing each other at a
table, blindfolded. A hat, known to be either white or black, is placed on
each of your heads. The host removes the blindfolds, and asks each of you to
write down the color of your own hat without communicating with each other. If
_either_ of you guesses correctly, you both win a prize. How do you guarantee
success by pre-communicating a strategy?"

The answer here uses the same partitioning into "either we're the same, or
we're different". One friend agrees to always guess his own hat to be the same
as the friend's, while the other always guesses that his hat is different.

~~~
gabemart
I think that you are correct and I am incorrect.

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raverbashing
Humm let me see if I get this right

\- Assuming everybody is honest

The chance of A getting the guess right is 0.5, same for B, so the joint
probability should be 0.25 (or 1/4)

You're out of 2 dollars (3 - the 1 you get) for something with a chance of 1/4
(in the other 3/4 you're being payed $1) SO

Looks like it's worth playing.

Expected losses: 2 * 1/4 < Expected wins: 1 * 3/4

Edit: see comment below

Edit 2, see the other comment below where the probability of prediction can be
of 1/2, in this case: Expected losses: 2* 1/2, wins: 1/2 so you should not
play

~~~
sksksk
> You're out of 3 dollars for something with a chance of 1/4 (in the other 3/4
> you're being payed $1)

You're out $2 (since you get the $1 from the other players at the beginning,
no matter what)

~~~
raverbashing
Humm makes sense =) Knew I was forgetting something!

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sold
Here is a bit harder version of the puzzle, with 100 players instead of 2.

100 prisoners are given either a blue or a red hat, at random. Each prisoner
is told colors of every hat except their own, and has to make a guess about
his hat with absolutely no communication. They win only if they _all_ guess
correctly. They can agree on a strategy beforehand. What is the optimal
probability of success?

~~~
raldi
Each should guess whatever would make an even number of red hats. That makes
it a 50% chance they're all wrong, and a 50% chance they're all correct --
thus consolidating all their "rightness" into a single block of probability.

~~~
StavrosK
That assumes a uniform distribution. If there were more red hats than blue
hats in general, it would be more advantageous for everyone to always guess
whatever makes an even number of the majority color, as they would win more
than 50% of the time.

~~~
raldi
I'm not following. With 100 hats, an even number of one color means there's an
even number of the other color, too.

~~~
StavrosK
Yeah, sorry, I'm mistaken. If one color is more probable than the other, you
can win more by guessing odd rather than even, but you'd have to know the
distribution. If you got 51 red hats every time, for example, you'd never win
making even hats.

For this to work, however, you'd have to know the exact expected probability
of each hat (and if it was even or odd), which is improbable.

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audreyt
The team of A and B can increase the chance from ¼ to ½ by consistently taking
their output from step 2 and using it as the guess for step 3.

Same thing if they both agree to guess the inverse of step 2.

~~~
datdatruth
This does not work because the text says "Players A and B build a team, they
have one fair coin each" NOT one fair coin total.

~~~
theonewolf
_but_ they can communicate a strategy ahead of time in step 1.

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datdatruth
Spoiler. The below contains an exact response

\-------------------------- The fact that there is no communication ensures
that whether A is right and whether B is right are completely indpendent.
Conceptually, predicting a coin flip is like making a statement "1" or "0"
which will then be XOR'd with a 1 or 0 from a securely, randomly, uniformly
generated OTP of which no copies exist. In other words, the plaintext is
immediately lost forever and you just have the ciphertext.

As a result of this, we must truly consider that A being right is a 50/50
proposition. It is also indpendent of B being right.

Thus we have the following four cases:

A right, B right - Result value * Percent chance = EV

0, 0 = +1 * 0.25 = +0.25

0, 1 = +1 * 0.25 = +0.25

1, 0 = +1 * 0.25 = +0.25

1, 1 = -2 * 0.25 = -0.5

\------------------------------- sum of above: +0.25. Therefore, as long as
each team member is independently predicting their own coin (which means that
their prediciton will be xor'd by a random bit) C should play this game long-
term.

(The values of +1 is because each round starts with the team giving C +1. If
at the end of the roudn C must return 3 then this is +1 -3 = -2 for the
round.)

Now here is another interesting question. What if under the same conditions A
and B both try to predict C's coin toss, of which there is only one? Should C
now play? Here is the answer is: "No", because A can predict heads, B can
predict heads, and then it looks like this: A right, B right - Result value *
Percent chance = EV

0, 0 = +1 * 50% = 0.5

0, 1 = +1 * 0% = 0 } not possible

1, 0 = +1 * 0% = 0 }

1, 1 = -2 * 50% = -1

\-------------------

-0.5

In this case, C should not play. This is because in this case the events are
not truly independent, there is a way to break the 25% 25% 25% 25% into 50%
and 50% - namely by picking the same prediction together.

\--------------------------

~~~
joshka
The problem with this analysis is that with the "choose the same as your flip
strategy", the case where A is right and B is wrong do not occur. Likewise
when A is wrong and B is right does not occur. The results of each coin flip
are independent, however the strategy produces a non-independent result.

    
    
      A, B, GuessA, GuessB, Winner, AmountC, Chance, EV
      H, H, H,      H,      AB,     -3,      0.25,   -0.75
      H, T, H,      T,      C,      +1,      0.25,   +0.25
      T, H, T,      H,      C ,     +1,      0.25,   +0.25
      T, T, T,      T,      AB,     -3,      0.25,   -0.75
    

Sum(EV)=-1 per round So C should not play

~~~
gabemart
C loses 2, not 3, when he loses (he gets one from AB at the start of each
round). C's EV should therefore be -0.5

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noloqy
To answer the actual question: whether C should play this game depends on C's
estimated probability of A and B coming up with a winning strategy. In the
real world, there would be many people with whom I would play this game.

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iotasquared
Both A and B guess the same as the results of their own toss. Their answers
are correct if the tosses are HH and TT, 50% chance.

So no, C should not play this game.

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gmcrews
Dysfunctional teams never can agree. By that I mean, if A decides to use the
strategy "guess what I flip"; B decides on the strategy "guess opposite what I
flip". And vice versa. These teams can never win! So for dysfunctional teams,
C can play the game no matter the payout.

~~~
mchannon
Actually I've done the hash on this approach and this still results in a
losing game for C.

Instead of HH and TT being winners with this approach, HT and TH become
winners with this approach. Any combination of opposite or agreement will
result in the same net odds.

The way for C to win is for A or B to treat it randomly.

A very interesting experiment would be to test how long it would take a random
human subject A with no contact with a consistent B (but a memory for past
results) to figure out how to win (in cases where they were told winning was
possible, impossible, or neither).

~~~
gmcrews
Yes, but my point was that one team member uses HH & TT, while the other team
member uses HT & TH. This is always a loss for the team regardless of how the
flips turn out.

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cousin_it
Maybe you could do even better than the obvious "guess the same as your
coinflip" strategy, by using quantum tricks.

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jgj
The question should really be rephrased to "Will C make money if he/she plays
this game?" Maybe C doesn't like money, has too much of it but really only
likes giving it out after coordinated acts of random chance. Then C would love
this game regardless of A and B's strategy. Maybe C spends all of his/her free
time at RPS tournaments, throwing singles at the winners like a rap video.

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elijahmurray
Anyone find a solution? I have an idea but want to check.

~~~
datdatruth
Yes. Break it down into the possible cases, assign probability for the case,
and value for the case. Multiply each probability by its value and sum these
up.

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TheAmazingIdiot
Hmm. If A and B are unwitting investors, and C is a quant with inside
information, OF COURSE C should take this "bet".

It's not like C would ever everrrr....Lie....

And when G(overnment) comes around to fix this evil fraud, they fine for 10%
of the profits. Everybody but A and B make out like a bandit. Sound like '09,
doesn't it?

(tongue in cheek, only because there were multiple vald analysees of this
question elsewhere in the comments.)

