This one is hard.

You play a match, consisting of N coin tosses. You've
been suckered, because the coin is loaded, and your
chance of winning, on each toss, is p < .5
To win the match, you must win a majority of tosses.
You lose on a tie. Even money payoff.
You can choose N. It must be even. What number maximizes
your winning chance, as a function of p? (N = 0 is disallowed)
--
Rich

RichD <r_delaney2001@yahoo.com> wrote:
....
> You play a match, consisting of N coin tosses. [...] your
> chance of winning, on each toss, is p < .5
>
> To win the match, you must win a majority of tosses.
> You lose on a tie. [...]
>
> You can choose N. It must be even. What number maximizes
> your winning chance, as a function of p? (N = 0 is disallowed)

I haven't worked through a sound mathematical approach to this problem,
but computationally it's easy to verify that the solution is to choose
N = 2+2k whenever k/(1+2k) < p <= (1+k)/(3+2k). For any p in (0,1/2)
there is a non-negative value of k satisfying this condition, as p
falls into intervals demarcated by 0/1, 1/3, 2/5, 3/7, 4/9, ..., for
which we choose 2, 4, 6, 8 ...

What makes it easy to verify is that values of the Bernoulli
distribution [Ref 1] are fairly easy to calculate. For a given value
p<0.5, let W_k = probability of a match-win if we choose N=2k, and let
B_k = probability of winning k tosses out of 2k. Then W_k = (1-B_k)/2,
and using k from formula in previous paragraph we can check if W_k is
larger than either of W_(k-1) or W_(k+1). [W_(k-1) not relevant if k=0]

[1] <https://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function>

On 9/14/2023 11:35 AM, RichD wrote:
> This one is hard.
>
> You play a match, consisting of N coin tosses. You've
> been suckered, because the coin is loaded,

For some reason I am thinking of one of Murphy's Law's that states that
a smith and wesson beats four aces.

and your
> chance of winning, on each toss, is p < .5
>
> To win the match, you must win a majority of tosses.
> You lose on a tie. Even money payoff.
>
> You can choose N. It must be even. What number maximizes
> your winning chance, as a function of p? (N = 0 is disallowed)
>

On September 18, Chris M. Thomasson wrote:
>> You play a match, consisting of N coin tosses. You've
>> been suckered, because the coin is loaded,
>
> For some reason I am thinking of one of Murphy's Law's that states that
> a smith and wesson beats four aces.

https://www.youtube.com/watch?v=773E6GPll3A

--
Rich

On September 18, James Waldby wrote:
>> You play a match, consisting of N coin tosses. [...] your
>> chance of winning, on each toss, is p < .5
>> To win the match, you must win a majority of tosses.
>> You lose on a tie. [...]
>> You can choose N. It must be even. What number maximizes
>> your winning chance, as a function of p? (N = 0 is disallowed)
>
> I haven't worked through a sound mathematical approach to this problem,
> but computationally it's easy to verify that the solution is to choose
> N = 2+2k whenever k/(1+2k) < p <= (1+k)/(3+2k). For any p in (0,1/2)
> there is a non-negative value of k satisfying this condition, as p
> falls into intervals demarcated by 0/1, 1/3, 2/5, 3/7, 4/9, ..., for
> which we choose 2, 4, 6, 8 ...
>
> What makes it easy to verify is that values of the Bernoulli
> distribution are fairly easy to calculate. For a given value
> p<0.5, let W_k = probability of a match-win if we choose N=2k, and let
> B_k = probability of winning k tosses out of 2k. Then W_k = (1-B_k)/2,
> and using k from formula in previous paragraph we can check if W_k is
> larger than either of W_(k-1) or W_(k+1). [W_(k-1) not relevant if k=0]
> <https://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function>

Well done. I saw this problem long ago, don't recall the exact solution,
but your reasoning looks right. The reference to the binomial distribution
is a telltale.

Intuitively, for p < .5, you want the shortest possible match: N = 2.
But that's simplistic, and incorrect. Which makes the problem challenging.

Consider a match of N tosses. We seek the optimal N, i.e. any deviation
lowers your winning chance W. We have to consider all possible H/T
outcomes, that's where the binomial distribution enters, via the
coefficients; Probability 101. In most cases, the winner wins by more
than 2 tosses. Then lengthening N by 2 doesn't alter the outcome.

There are only two marginal cases:
i) In case of a tie, N/2, you lose. Then if N --> N + 2, and you win both
tosses, that would switch the result.
ii) In the case you win by 2, if N --> N + 2, the opponent might win both
tosses, thus take the match.

For large N, p < .5 obviously disfavors you. But now, look at the binomial
distribution: the middle coefficient representing a tie, is disfavorable.
For small N, this coefficient becomes significant.

So you seek a balance point N between these extremes, which depends
on p. Lots of tedious algebra -

PS Why must N be even?

--
Rich