Probability of Die Landing on Value After x Days

[εllipse]

I'm sure this is the kind of question this forum gets a lot, but I looked at a few of the recent probability questions and they were all homework questions dealing with numerical values and such, so forgive the bland question.

If I throw a die once a day, what formula can I use to judge the probability of the die landing on a predetermined value (say a 4) after a certain number of days? It's been a while since I had any coursework on probability, so all I've got right now is a little intuition. Of course, each day there would be a 1/6 chance of the die hitting the 4, but after 6 days it seems like there should be a fairly good probability of the die having hit 4, and with even more days the probability should increase but never reach certainty. Surely there's a formula for this, could anyone point me in the right direction?

 

[juvenal]

Probability of success = 1/6.
Probability of failure = 5/6.

Probability of at least one success in n independent trials = 1 - probability of zero successes
= 1 - (5/6)^n. This number approaches 1 as n approaches infinity.

 

[εllipse]

Thank you very much; that's exactly what I wanted.

 

[Vuduman78]

Does anyone know if there is a single formula for if the probability changes in a set manner.

I figured out the above formula while tinkering around on my own. My goal is to figure out the probability of a given even if the chances start out at 10% then increase by 5% each time, what the given chance of the event occurring at any given iteration. To begin with I started with a flat 10% chance, and figured out the above formula. Then found this while trying to check it. I used that knowledge to calculate the given chance by hand of any particular iteration, and they are as follows for iterations 1-19. (I did round most of these)

.1 .235 .388 .541 .6787 .791155 .8747 .932 .966 .9845 ..99379 .997821 .99935 .99984 .999967 .99999512 .99999951 (1 - (2.44 x 10^-8)) then 1.

I believe I calculated all these right. I got to each of them in the same manner as doing the above equation by hand over a lot of iterations. I knew that for instance the 10% repeated probability had to approach 1 asymptotically for instance. and hand calculations showed that to be true. Once I realized I was just multiplying .9 times it self for each iteration the formula was easy to devise. The one for the growing probability is not so easy. I provided the numbers so you can check any theory you come up with. Or if someone knows a proven formula that would be awesome too. I just cannot figure out any kind of elegant formula to express the change, like i could with the flat 10%. This may be stupid easy for someone on here so I figured I'd post it. I am a philosophy major, because I am too far into change to math now. But I love mathematics which is why I am tinkering around with this. Thanks for any help you can give.

 

[scalpmaster]

Law of large numbers discounts everything. Even if dice shows only a certain no. ∞ number of times, you can't say it's loaded because there's always ∞^∞ more times.:tongue:
http://en.wikipedia.org/wiki/Inverse_gambler's_fallacy
http://en.wikipedia.org/wiki/Gambler's_fallacy

 

[Vuduman78]

I do not understand how this is pertinent. For what I am trying to figure out, the probability increases conceivably until it hits a certitude of 100% chance. I already calculated the probabilities, I just feel like there has to be a more elegant way of doing it than I did. That is what I am trying to figure out.