cotpi 47 - Extra coin

[cotpi]

Akio has one more coin than Bansi. They throw all of their coins
and count the number of heads. If all the coins are fair, what
is the probability that Akio obtains more heads than Bansi?

--
Originally posted at: http://cotpi.com/p/47/
Correct solutions will be archived at the URL mentioned above.
Solutions to 'Raised to the same power':
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[Ted Schuerzinger]

On Thu, 19 Apr 2012 11:56:34 +0530, cotpi wrote:

> Akio has one more coin than Bansi. They throw all of their coins
> and count the number of heads. If all the coins are fair, what
> is the probability that Akio obtains more heads than Bansi?

Let n = the number of coins Bansi has. p = probability Akio throws more
heads.

If n = 0, p = 1/2
If n = 1, there's a 1/8 chance all three coins will be heads, and a 3/8
chance Bansi throws tails and Bansi throws at least one heads; again, p
= 1/2.
If n = 2:
There's a 1/32 chance of all five heads.
Half the time Bansi throws one heads, half that time, Akio throws two or
three heads.
There's a 1/4 * 7/8 = 7/32 chance that Bansi throws two tails and Akio
throws at least one heads.

Presumably, the probability will always be 1/2. An easier way to state
it would be as follows:

For the first n coins, both Akio and Bansi will throw an expected n/2
heads. There's a 1/2 chance the extra coin will also land heads.

--
Ted S.
fedya at hughes dot net
Now blogging at http://justacineast.blogspot.com

[Mark Brader]

Ted Schuerzinger:

> Presumably, the probability will always be 1/2.

Looks that way.

> An easier way to state it would be as follows:
>
> For the first n coins, both Akio and Bansi will throw an expected n/2
> heads. There's a 1/2 chance the extra coin will also land heads.

That's not a proof; it only accidentally suggests the result of 1/2.
There are more results possible than "Akio gets one more head" and
"Akio and Bansi get an equal number of heads" with probability 1/2 each.
--
Mark Brader | "Earthmen learned how to send ships through space, and
m...@vex.net | so initiated human history, though I suppose there was
Toronto | previous history on Earth." -- Jack Vance, "Emphyrio"

My text in this article is in the public domain.

[Ed Murphy]

Mark Brader wrote:

> Ted Schuerzinger:
>> Presumably, the probability will always be 1/2.
>
> Looks that way.
>
>> An easier way to state it would be as follows:
>>
>> For the first n coins, both Akio and Bansi will throw an expected n/2
>> heads. There's a 1/2 chance the extra coin will also land heads.
>
> That's not a proof; it only accidentally suggests the result of 1/2.
> There are more results possible than "Akio gets one more head" and
> "Akio and Bansi get an equal number of heads" with probability 1/2 each.

Okay, formalizing it:

Label Akio's coins A1, A2, ..., An, A(n+1).
Label Bansi's coins B1, B2, ..., Bn.

For all the coins except A(n+1), there are 2^(2n) ways for them to land,
all equally probable, and these can be divided into three sets:

1) Akio has more heads than Bansi among those coins
2) Akio has the same number of heads as Bansi among those coins
3) Akio has fewer heads than Bansi among those coins

Each way in 1) is the mirror image of a way in 3), so those sets have
the same size.

For each way in 1), Akio has more heads than Bansi overall, regardless
of how coin A(n+1) lands.

For each way in 3), Akio does not have more heads than Bansi overall,
regardless of coin A(n+1) lands. (At most, he might have the same
number of heads as Bansi overall.)

For each way in 2), Akio has a 1/2 probability of having more heads than
Bansi overall, depending on how coin A(n+1) lands.

Thus, the overall probability is
(x * 1) + (x * 0) + ((1-2x) * (1/2)) (for some value of x)
= 2x/2 + (1-2x)/2
= (2x+1-2x)/2
= 1/2

[Alexander Thesoso]

No need for all that thinking and math.
Symmetry.
Probability of more heads = probability of more tails. 1/2

[Mark Brader]

> > Akio has one more coin than Bansi. They throw all of their coins
> > and count the number of heads. If all the coins are fair, what
> > is the probability that Akio obtains more heads than Bansi?

Alexander Thesoso:

> No need for all that thinking and math.
> Symmetry.
> Probability of more heads = probability of more tails. 1/2

You skipped a step, which is:

If Akio has more coins, then he must either get more heads
then Bansi, or more tails. Further, since he has only 1 more
coin, he cannot get more heads *and* more tails. Therefore
he must have more heads *or* more tails, and the probabilities
of those cases must add to 1.

With this addition, Alexander's proof is equivalent to the one
in more symbolic language posted by Rupert Grint.
--
Mark Brader "It flies like a truck."
Toronto "Good. What is a truck?"
m...@vex.net -- BUCKAROO BANZAI

[Harry Vaderchi]

On Thu, 19 Apr 2012 19:58:12 +0100, Mark Brader <m...@vex.net> wrote:


> With this addition, Alexander's proof is equivalent to the one
> in more symbolic language posted by Rupert Grint.

That's the actor; the problem solver is John Grint. (any relation?)

--
[dash dash space newline 4line sig]

Albi CNU

[Mark Brader]

Mark Brader:

> > With this addition, Alexander's proof is equivalent to the one
> > in more symbolic language posted by Rupert Grint.

Harry Vaderchi:

> That's the actor; the problem solver is John Grint. (any relation?)

Okay, let's fix this in terms of symbolic language:

#define Rupert John

There!
--
Mark Brader, Toronto | "I've always wanted to be a mad scientist!
m...@vex.net | Or perhaps just mad!" -- Robert L. Biddle

[Ted Schuerzinger]

On Thu, 19 Apr 2012 17:14:07 -0500, Mark Brader wrote:

> Okay, let's fix this in terms of symbolic language:
>
> #define Rupert John

Don't you mean

s/Rupert/John/g;

or some such? ;-)

[christian.bau]

On Apr 19, 7:26 am, cotpi <puzz...@cotpi.com> wrote:
> Akio has one more coin than Bansi. They throw all of their coins
> and count the number of heads. If all the coins are fair, what
> is the probability that Akio obtains more heads than Bansi?

Quite simple actually.

Akio throws n+1 coins, Bansi throws n coins. We put Akio's last coin
aside, comparing n vs. n coins. Probability that Akio has more heads
is p, probability that Bansi has more heads is the same p, probability
that both have the same number of heads is 1 - 2p. p may be hard to
calculate.

Looking at n+1 vs. n coins, Akio has more heads if he either had more
heads with the first n coins, or if both had the same number of heads
with n vs. n coins and the (n+1)st coin is heads. The probability for
this is p + (1 - 2p) / 2 = p + (0.5 - p) = 0.5.

[Jeffrey Turner]

On 4/19/2012 2:26 AM, cotpi wrote:

> Akio has one more coin than Bansi. They throw all of their coins
> and count the number of heads. If all the coins are fair, what
> is the probability that Akio obtains more heads than Bansi?

Akio has one coin, Bansi none. There's a fifty percent chance that
Akio will toss more heads than Bansi. If there's one answer to the
question, then it must be fifty percent.

--Jeff