Math Help Needed: Calculating Probabilities

How can I calculate the odds of succeeding 15 out of 20 tries if there is a 75% chance of success each time?

One way of thinking of it is if I roll a 100 sided die 20 times, what are the odds it will be 75 or less on at least 15 of those rolls.

I’m looking for a general formula that allows me to use different values for each of the above 3 numbers. I can’t find any probability calculation that doesn’t rely on some trick to greatly simplify the problem. Anyone worked this out before?

Hi Kael.

This is all about the Binomial distribution…

http://onlinestatbook.com/2/probability/binomial.html

Hi Kael.

This is all about the Binomial distribution…

http://onlinestatbook.com/2/probability/binomial.html

p(15) = (20!/(15!(20-15)!))(0.75)^15(1-0.75)^(20-15)

Jellyfish is right, I think, but nota bene: His equation gives the odds for exactly 15 success. Not at least 15, but 15, no more no less. If you want a range, you need to use the calculation for each value and add them sums together.

Yes, what Greatatlantic said is totally correct.
So for example, if you wanted to know the probability of succeeding at least 15 tries, you would need this…
p(15)+p(16)+p(17)+p(18)+p(19)+p(20)

You guys step aside. I can handle this. I’ve had training. Kael, to determine the probability, you multiply 75% by 15. Ergo, your odds of success are 1125%.

At least that’s how I would do it.

-Tom

Never tell me the…

You guys are awesome! Thank you so much for everyone who responded (even the jokes!) and special thanks to jellyfish for the answer.

Rather than tally up a bunch of cases, you can approximate the answer with the normal distribution:

OK, I don’t know why I’m struggling with this problem, but I guess I’m out of practice.

Standard fair deck of 52 cards. A player is dealt a (bridge) hand of 13 cards. What are the odds of getting a hand with exactly 6 Spades, 4 Hearts, 2 Diamonds, and 1 Club?

It should be (13 choose 6) * (13 choose 4) * (13 choose 2) * (13 choose 1) / (52 choose 13) = 0.00195919778

How do I write binomial coefficients nicely on a forum?

Edit: Maybe that’s better.


⎛13⎞⎛13⎞⎛13⎞⎛13⎞
⎝ 6⎠⎝ 4⎠⎝ 2⎠⎝ 1⎠
____________
⎛52⎞
⎝13⎠



I approve this message.

Well, I’m still struggling with this conceptually, but a colleague confirmed your equation anymunym.

I guess instead of seeing possibilities, I just keep thinking that there must be some overlap between getting 6 of one suit and 1 of another.

Thank you much, though, for your answer and the work you put into getting the display right.

You’re probably thinking of effect-of-removal. That comes into play when the order matters. Or put another way, when you have information about one or more cards already and want to know how the odds change as-you-go.

You can formulate the question as an in-order thing and manually add up all of the “or the first card could be the other suit, in which case the rest of the equation looks like…” stuff, but the end result is that you’re just writing out longhand what the binomial distribution stuff does. All of those other terms end up canceling out.

That’s a great expanation.

A closed box contains a marble. We know that the marble is white or black, but we don’t know which. A white marble is added to the box, which is shaken. We hear the two marbles clattering around inside the box. You reach into the box (without looking) and pull out a marble. It is white. You can’t tell if it is the original marble, or the one that we added to the box.

What is the probability that the marble still in the box is also white?

66%, but I can’t show my math.

Can you briefly state your reason(s) for that answer, though? Like, a common sense explanation?

Is this similar to the Monty Hall Problem? I watched YouTubes, read the Wikipedia page and even tried deciphering the formula, but still can’t get my brain wrapped around the probabilities involved. My brain always wants to say it should be 50%. There are a few similar scenarios where it does make sense to me, and this looks like another of those scenarios, but my brain would have said 50%.

I’m pretty sure this is going to be an application of Bayes Theorem… though it can be pretty easy to slip it up applying it to any given situation.

This strikes me as somewhat similar to the famous “Monte Hall” problem, which had even mathematicians arguing against the correct conclusion.

So, by drawing a white marble, you either have drawn the second marble, or you’ve drawn the original marble… the odds of either is 50/50. However, given that the marble in your hand is white, that means you’ve either drawn the second marble… or the original marble is white. Really, the question is no longer about what the color of the original marble is, the question is what color is the marble still in the box. If the original marble was white (50/50), the marble is the box is going to white.

That would be my attempt at a common sense explanation. I hope it made some kind of sense.