hey guys, i cant wrap my head around this math, i was hoping someone can tell me how it can be figured out
if you had a 51% chance to win a gambling game (i guess you would call it investment then), how many times should you play it, so that you have a 10% chance or less of ending up losing more money than you started with?
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i might've worded that wrong. i meant having less money than you started with
the number of successes required to have a net gain is ceiling(trials/2), so we can use the binomial distribution to figure out for a series of increasing numbers of trials, what the probability of achieving the required number of successes or fewer is. however, since it's "or fewer", which includes cases where we potentially have a net loss, we need to look at the opposite probability to ensure that we actually get the value we need.
trials
and jsut to double check - if we redefine this so that "success" means that we lose the game, and we want to keep the "successes" at or below floor(trials/2), then we need
trials
what?! why would his chance of losses increase with more attempts?
it's not, it's decreasing.
at the 62nd attempt, the probability of getting any number of failures but still not have a net loss is slightly above 90%.
thank you. i didn't study binomial probability. should i be able to learn this stuff without any sort of mathematics background beyond grade 12 functions?
also can you calculate this by hand?
what if the chance of winning was 75%? what would the number of attempts be then?
have to know this for investing
>should i be able to learn this stuff without any sort of mathematics background beyond grade 12 functions?
yeah sure, basic stats is pretty easy. if you can handle, say, algebra or pre-calc, you can handle basic stats.
>also can you calculate this by hand?
you could, but i wouldn't want to.
the cumulative density function is:
[math]
\Sigma{}^{\lfloor{}k\rfloor{}}_{i=0} \binom{n}{i} p^{i}(1-p)^{n-i}
[/math]
where k is the maximum number of successes, n is the number of trials, and i is the current number of successes
for, say, 50 trials, you would set k = floor(50/2), and for every separate i from 1 to 50, you'd calculate [math] \binom{n}{i} p^{i}(1-p)^{n-i}[/math] and sum them all
the 8th trial crosses 10%
This sounds like a card counting question in disguise. For a good risk of ruin you want at least 100x your maximum bet, 200x to sit really pretty. There is loads of information on the math behind it and how to do it. There are also simulations you can run on it. I recommend you don't even start because it is a lot of work and blackjack rules around the country are going down the shitter.
woop, i fucked that up.
for 50 trials, you'd set k to floor(50/2)=25 and n to 0.51, and for every i from 1 to 25, you'd calculate that.
actually, im asking this for penny stock investing, which many people say is "gambling", but the odds are better than blackjack if you know the game
if you really want to play around with the stats on this, download and install R. the code I posted in is working R code.
thanks a lot
maybe i'm misinterpreting the problem but that doesn't seem right.
wolframalpha.com
isn't this saying that the probability of doing better than breaking even with 62 trials is only about 0.5?
God damnit you're right. I fucked up and used the wrong function; I thought dbinom() was cumulative density, not point density.
Gambling is haram, infidel
wait what?
how exactly do i use this calc? what do i type into it say if i wanted to know what OP wanted to know but with 75%?
after thinking about this more, your question isn't really well formed
because you've defined your target state as "don't lose more than you win", we're always looking for the midpoint of any number of trials. but since these are all independent trials, that probability at the midpoint is going to be the same every time.
for example: we could model this with the negative binomial distribution, where we're looking for the probability of getting some number of successes in a given number of trials, this time remembering to use the proper function for the cumulative density instead of the point density, but the number of successes changes in lock step with the number of trials.
don't know about wolfram alpha but here it is in mathematica.
basically you're looking for the point that the graph crosses 0.9 on the y-axis. for 0.75 you get above the 90% threshold with only about ~20 trials but for 51% it takes ~4000 trials to guarantee the specified probability of doing better than breaking even.
this calculation could probably be done more easily by using the cdf and the fact that P(x > n) = 1 - P(x [math]\le[/math] n) but oh well.
whoops read the first graph wrong it's far less than 20 trials.