Today's problem is hard.

In a casino, you play a match, a string of games,
against the house. It's a stochastic process, your
chance of winning a game is p, a given parameter.

The rules are simple:
The match is N games, each game is one point.
p < .5
The player may select the match length N.
N must be even.
To win, you must by at least 2 points; a tie is a loss.

Why play, as an underdog? It's a casino promotion,
you get a prize if you win, pay nothing if you lose,
a free ride!

Compute the match length which maximizes your winning
chance, as a function of p.

An exact solution requires some tedious algebra, so
don't bother with that, unless you're obsessive. Just
sketch the key ideas, and a path to the solution.

--
Rich

RichD <r_delaney2001@yahoo.com> wrote:
> Today's problem is hard.
>
> In a casino, you play a match, a string of games,
> against the house. It's a stochastic process, your
> chance of winning a game is p, a given parameter.
>
> The rules are simple:
> The match is N games, each game is one point.
> p < .5
> The player may select the match length N.
> N must be even.
> To win, you must by at least 2 points; a tie is a loss.
>
> Why play, as an underdog? It's a casino promotion,
> you get a prize if you win, pay nothing if you lose,
> a free ride!
>
> Compute the match length which maximizes your winning
> chance, as a function of p.
>
> An exact solution requires some tedious algebra, so
> don't bother with that, unless you're obsessive. Just
> sketch the key ideas, and a path to the solution.

As a key idea, choose N = 2m if (m-1)/(N-1) < p < (m+1)/(N+1).
(If p equals one of those fractions, choose N or the appropriate
neighbor.) Note, in the 0/1 < p < 1/3 case, N is 2. Now, as
for a path to the solution ... :)

On Sun, 10 Oct 2021 16:47:29 -0700 (PDT), RichD
<r_delaney2001@yahoo.com> wrote:

>Today's problem is hard.
>
>In a casino, you play a match, a string of games,
>against the house. It's a stochastic process, your
>chance of winning a game is p, a given parameter.
>
>The rules are simple:
>The match is N games, each game is one point.
>p < .5
>The player may select the match length N.
>N must be even.
>To win, you must by at least 2 points; a tie is a loss.
>
>Why play, as an underdog? It's a casino promotion,
>you get a prize if you win, pay nothing if you lose,
>a free ride!
>
>Compute the match length which maximizes your winning
>chance, as a function of p.
>
>An exact solution requires some tedious algebra, so
>don't bother with that, unless you're obsessive. Just
>sketch the key ideas, and a path to the solution.

Note that the probability of success is equal to the sum of
probability of winning exactly N/2+1 and the probability of winning
exactly N/2+2 and ... and the probability of winning exactly N.

Using the notation C(n,k) to indicate the number of ways to choose k
objects out of n, the probability of winning exactly j games is
P(j) = C(N,j) * p^j * (1-p)^(N-j)
and
P(success) = sum(j=N/2+1 to N) [P(j)]

This is trivial to calculate for any given N and p.

Testing over p from .01 to .49 and N from 2 to 100 produces the
following:
range of p optimum N
.01 - .33 2
.34 - .39 4
.40 4 or 6
.41 - .42 6
.43 - .44 8
.45 10
.46 12
.47 16
.48 24 or 26
.49 50
which matches James' solution once you correct the second term to
m/(N+1) and use <= instead of < both times.

Even when p = .49, you win the prize only 38.8% of the time. OK for a
free promotion I guess.

--
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On October 11, James Waldby wrote:
>> In a casino, you play a match, a string of games,
>> against the house. It's a stochastic process, your
>> chance of winning a game is p, a given parameter.
>> The match is N games, each game is one point.
>> p < .5
>> The player may select the match length N.
>> N must be even.
>> To win, you must win by at least 2 points; a tie is a loss.
>> Compute the match length which maximizes your winning
>> chance, as a function of p.
>> An exact solution requires some tedious algebra, so
>> don't bother, unless you're obsessive. Just
>> sketch the key ideas, and a path to the solution.
>
> As a key idea, choose N = 2m if (m-1)/(N-1) < p < (m+1)/(N+1)
> Note, in the 0/1 < p < 1/3 case, N is 2. Now, as
> for a path to the solution ... :)

Somebody cheated and found a shortcut.

https://www.nytimes.com/1980/04/30/archives/miss-ruiz-loses-her-title.html

Show your work -

--
Rich

On October 12, Barry Schwarz wrote:
>> In a casino, you play a match, a string of games,
>> against the house. It's a stochastic process, your
>> chance of winning a game is p, a given parameter.
>> The match is N games, each game is one point.
>> p < .5
>> The player may select the match length N.
>> N must be even.
>> To win, you must win by at least 2 points; a tie is a loss.
>> Compute the match length which maximizes your winning
>> chance, as a function of p.
>> An exact solution requires some tedious algebra, so
>> don't bother, unless you're obsessive. Just
> >sketch the key ideas, and a path to the solution.
>
> Note that the probability of success is equal to the sum of
> probability of winning exactly N/2+1 and the probability of winning
> exactly N/2+2 and ... and the probability of winning exactly N.
> Using the notation C(n,k) to indicate the number of ways to choose k
> objects out of n,

[ snip mind numbing computations ]

um, the problem asks for N, as a function of p.

If a problem requires brute force, it won't get posted.
There are things called insights and concepts.
Sorry amigo, you're a computer programmer, not a mathematician.

However, you're on the right track. One must consider the combinatorics;
but in the abstract, not any particular numerics.

That is, for N games, there are 2^N possible win/loss sequences.
Then consider the set of wins for the player, and how to represent that sum,
using the combination formula and series.

Then note that N is even. If N odd, the problem fails, the obvious
N = 2 is correct. But for N even, that isn't necessarily so, which
makes it interesting. What's the difference? Stare at the combinatorics
for each case, something should jump out.

Finally, think about the definition of optimal, in this context;
if N = Q is optimal, what can you say about other values of N?

That's enough clues -

--
Rich

On Wed, 13 Oct 2021 19:54:00 -0700 (PDT), RichD
<r_delaney2001@yahoo.com> wrote:

>On October 12, Barry Schwarz wrote:
> >> In a casino, you play a match, a string of games,
>>> against the house. It's a stochastic process, your
>>> chance of winning a game is p, a given parameter.
>>> The match is N games, each game is one point.
>>> p < .5
>>> The player may select the match length N.
>>> N must be even.
>>> To win, you must win by at least 2 points; a tie is a loss.
>>> Compute the match length which maximizes your winning
>>> chance, as a function of p.
>>> An exact solution requires some tedious algebra, so
>>> don't bother, unless you're obsessive. Just
>> >sketch the key ideas, and a path to the solution.
>>
>> Note that the probability of success is equal to the sum of
>> probability of winning exactly N/2+1 and the probability of winning
>> exactly N/2+2 and ... and the probability of winning exactly N.
>> Using the notation C(n,k) to indicate the number of ways to choose k
>> objects out of n,
>
>[ snip mind numbing computations ]
>
>um, the problem asks for N, as a function of p.

Numbing is in the eye of the beholder. The computations served to
validate James solution.

>If a problem requires brute force, it won't get posted.
>There are things called insights and concepts.

Yes there are. I freely admit I have no idea how to attack the
problem at that level. That is why the best I could contribute is a
partial verification of someone else's contribution. Would you hold
the same opinion if I had found an exception?

>Sorry amigo, you're a computer programmer, not a mathematician.

You seem to think the two are mutually incompatible.

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