Re: Interesting (but difficult) question - calculating 'implied' probabilities of a wager
Pavel314
"Anonymous" <no.r...@here.com> wrote in message
news:Ou6dnXpmPox49KTa...@bt.com...
> Here is a hypotheical scenario.
>
> A friend and I decide to visit the local county fair. There is a
> competition to see who can throw a heavy ball the highest. I bet my friend
> that I can throw the heavy metal ball more than X metres high.
>
> He in turn, says "I'll pay you a dollar for every Y centimeters that you
> can throw the ball above X meters - BUT to make it worth my while, you
> have to PAY ME Z dollars for me to take on the bet".
>
> From the above, my friend has calculated (implicitly from the wager he has
> made), the probability of me being able to throw the ball above X metres.
> How may I calculate the probaility, so I can work out the (implied) odds
> of my success?
>
> What methodology/logic/technique may I use to calculate the probability of
> me throwing the ball above X metres (based on the wager given above)?
You're right, this is interesting but difficult. I haven't solved it myself
but I have a few ideas. I cross-posted to sci.stat.math to see if we could
get some help from over there.
You are assuming that you can throw the ball (100 * X) + D centimeters,
where D is the amount by which you exceed 100 * X centimeters. It's not
stated, but let B be the amount of your initial bet that you can do this.
Your friend must also be assuming that your throw will be greater than 100 *
X centimeters or he would have taken your original bet.
Your friend must be assuming that your throw will be less than (100 * X) +
(D / Y) centimeters, where D is the distance by which you exceed X meters
and Y is his $1 incremental bet. Since he's betting Z, he's assuming Z > D /
Y.
I think that what needs to be done here is to map the expectations of each
player at a 95% confidence interval and use Bayesian inference to work out
the joint probability but I don't have time to do that right now, got to run
off to the day job.
Paul
Pavel314
"Pavel314" <Pave...@NOSPAM.comcast.net> wrote in message
news:SoCdnbzFjODOpaHa...@comcast.com...
>
> "Anonymous" <no.r...@here.com> wrote in message
> news:Ou6dnXpmPox49KTa...@bt.com...
>> Here is a hypotheical scenario.
>>
>> A friend and I decide to visit the local county fair. There is a
>> competition to see who can throw a heavy ball the highest. I bet my
>> friend that I can throw the heavy metal ball more than X metres high.
>>
>> He in turn, says "I'll pay you a dollar for every Y centimeters that you
>> can throw the ball above X meters - BUT to make it worth my while, you
>> have to PAY ME Z dollars for me to take on the bet".
>>
>> From the above, my friend has calculated (implicitly from the wager he
>> has made), the probability of me being able to throw the ball above X
>> metres. How may I calculate the probaility, so I can work out the
>> (implied) odds of my success?
>>
>> What methodology/logic/technique may I use to calculate the probability
>> of me throwing the ball above X metres (based on the wager given above)?
I worked on this at lunchtime today. It seems the missing link is the
confidence level you and your friend are placing on their bets. Hopefully,
someone more skilled in statistical reasoning will come to our aid.
Paul
Anonymous
Pavel314 wrote:
Hi Paul, thanks for your feedback. Incidentally, I agree with you - I am
thinking along the same lines - it may not be possible to calculate the
oddds, without knowing the distribution of the height of the throws.
I'll do some more thinking ...
Pavel314
>
>
> Pavel314 wrote:
>
>> "Pavel314" <Pave...@NOSPAM.comcast.net> wrote in message
>> news:SoCdnbzFjODOpaHa...@comcast.com...
>>
>>>"Anonymous" <no.r...@here.com> wrote in message
>>>news:Ou6dnXpmPox49KTa...@bt.com...
>>>
>>>>Here is a hypotheical scenario.
>>>>
>>>>A friend and I decide to visit the local county fair. There is a
>>>>competition to see who can throw a heavy ball the highest. I bet my
>>>>friend that I can throw the heavy metal ball more than X metres high.
>>>>
>>>>He in turn, says "I'll pay you a dollar for every Y centimeters that you
>>>>can throw the ball above X meters - BUT to make it worth my while, you
>>>>have to PAY ME Z dollars for me to take on the bet".
>>>>
>>>>From the above, my friend has calculated (implicitly from the wager he
>>>>has made), the probability of me being able to throw the ball above X
>>>>metres. How may I calculate the probaility, so I can work out the
>>>>(implied) odds of my success?
>>>>
>>>>What methodology/logic/technique may I use to calculate the probability
>>>>of me throwing the ball above X metres (based on the wager given above)?
>
> thinking along the same lines - it may not be possible to calculate the
> oddds, without knowing the distribution of the height of the throws. I'll
> do some more thinking ...
After another lunch hour spent considering the situation, here's what
I've come up with. I still don't have an answer but maybe this will help.
Assume your throw distance to be normally distributed, not an unnatural
assumption.
YOUR BET
You estimate that your average throw, trying to make the longest throw
you can, will be 100 * X + E centimeters, where X meters is the minimum
distance you have to throw to win the bet. You also estimate that the
limit point, X meters, is two standard deviations below your mean throw;
this gives you only a 2.5% chance of losing the bet. So the standard
deviation of your throw is E / 2.
FRIEND'S BET
Assume that our friend has seen you throw in the past and makes the
same estimates that you have on mean and standard deviation. Wanting the
same risk that you put on your bet, he sets the parameters of his bet so
that he doesn't have to pay you unless your throw goes more than two
standard deviations above the estimated mean, 100 * X + 2 * E.
He will pay you one dollar for each Y centimeters you throw past 100 *
X. He will have already collected Z dollars for taking the bet, so to
break even if you throw to two standard deviations beyond the mean we
need:
Z = (2 * E) / Y
Solving for E,
E = (Z * Y) / 2
This puts your mean throw as
( 100 * X ) + ((Z * Y) / 2)
and the standard deviation of your throw as
(Z * Y) / 4
So at least we have the distribution of your throw in terms of your friend's
bet parameters.
I'm sure there's a way to calculate payback by the probability of each Y
segment but I don't have that yet.
Paul
Anonymous
Thats probably too generous (i.e. not likely to happen in the real
world), but for the purposes of illustrating the underlying 'mechanics'
of the problem - ok.
> this gives you only a 2.5% chance of losing the bet. So the standard
> deviation of your throw is E / 2.
Erm, I don't follow, why is the stddev E/2 ?. BTW, you haven't yet
defined E, I was assuming that it was the error term (i.e. a SNV ~ N(0,1))