Probability question...

I think it's easiest to think about it like this.

Probability of AT LEAST one 4 is the same as 1 - Probability of NO 4 on any dice. So for one die, it is 1 - .75, or .25, two dice, it is 1 - (.75 * .75), or .4375. For three dice, it is 1 - (.75 * .75 * .75) or .578125.

[edited to remove incorrect formula, see Oracle's message below for the right formula]

Mark

ensor is that really so?

Where did you find the info you posted? It has been proven maytime that the outcome of each die roll has no influence on any previous roll or future roll. So I am not sure what angle yo uare coming from with this statement.

I guess I am still confused about this whole area.... But it has been a long time since I used my mathematics skills (BS in math, but the BS in computer science won out and I now program at a place that needs little if any mathematics).

The outcome of any one die roll has no effect on the next, that's true. However we're not looking at the outcome of a single roll.

The chances of rolling a 4 on at least 1 of X dice is not the same as the chances of rolling a 4 on exactly 1 of 5 dice, nor the same as rolling a 4 on any particular die. In statistics there are things called Unions and Intersections which is what we're talking about here.

I think my previous post might illustrate it better since I listed the different outcomes of rolling 2d4. Does that help any?

I mean, obviously if you roll 3,000 dice your chances of at least 1 of them being a 4 are better than if you only roll 1 die, right?

For the record, 1-(3000*((4-1)/4)) = -2249, so I suspect that equation is wrong... I believe ensor meant to say 1-((s-1)/s)^n). By the way, 1-[((4-1)/4)^3000] = 1-1.5268...E-375 which is basically 1. So if you roll 3000 d4's it's pretty safe to say you're guaranteed at least one 4.

Zzzzz,

The info I posted came out of my head, I don't really have a reference, but here's a cool page about Dice from Mathworld, http://mathworld.wolfram.com/Dice.html.

If you look at it as if you want to know the probability that THIS die roll will be a 4, then you're absolutely right, it will always be .25. But if you want to know if at least one of the 7 die rolls was a 4, you multiply the independent probabilities together. As in sedjtroll's post with listing out all the possibilities.

I think we're both right, we're just answering different questions.

I think Seth is right...

It should be:

1-((1/S)^R)

Where S=The number of sides and R = the number of rolls.

The more times you roll, the smaller (1/S)^R gets, which, when substracted from the certain probability of 1, yields a higher percentage with each roll. Way to go Seth and Ensor.

ZZZ, I think you were confusing the probability of rolling a certain # two times in a row. In that respect, you are correct. I.E. With regards to a roullete wheel, you have the same chance of spinning 17, even if you had just spun 17 on the previous spin.

-Darke

Yeah I think this might be my problem. But again all of my math skills are rusty... and to think I spent hours and hours doing proofs on this type of stuff, advance calc, abstract geometry..... I hate when something I learned goes bye-bye! What a waste!

Ah well thanks to those of you who helped me realized my loss of another college subject! ;)

Whenever I see probability questions like this, my first reaction is to just build 1000 test cases in c++ and then derive the formula from there. Which gets me in all sorts of trouble... :)

For instance, for 2dS: 1/S + ((1 - 1/S) * 1/S).

And for 3dS: 1/S + ((1 - 1/S) * 1/S) + (1 - ((1 - 1/S) * 1/S) * 1/S).

Which is a lot easier to work out in for loops...

... and now I have a wonderful dice rolling probability sim with absolutely no use.... I suppose I'll just tuck it away next to my Let's Make A Deal hack.

I think we all got it wrong..

It should be:

1-((S-1)/S)^R)

Again where S = the number of sides and R is the number of Rolls.

Where as before I said:

and Oracle said:

HTH,
Darke

(Too many cooks in the kitchen)

Consider the (S-1)/S portion of that equation:

(S - 1) / S =
S/S - 1/S =
1 - 1/S

Now sub this back into your equation in place of of (S - 1) / S and we get:

1 - (1 - 1/S) ^ R

replace the R with an N because it doesn't matter what letter we us to represent the number of dice rolled:

1 - (1 - 1/S) ^ N

QED

Heheheh another one in my boat!

Now hiring crewman for a voyage to Mathville! Openings still available, everyone is welcome because we want a merry bunch of lads and lassies!!!!

Here's a better way of looking at the 4d4 example.

Roll the dice one at a time. As soon as you roll the first 4, you can stop because regardless of what the rest of the dice show, you still have at least one 4.

After 1 die, you have a 25% chance of having rolled a 4 so far.

That means there's a 75% chance you'll need to roll a second die, and if you do, there's a 25% chance it will show a 4, so there's a 75% * 25% = 18.75% chance of getting a 4 on the second die Add this to the 25% above, and the chances of getting a 4 on the first 2 dice are 43.75%

That leaves a 100% - 43.75 = 56.25% chance you'll need to roll a third die, with a 25% chance. 56.25% * 25% is 14.0625%. Add this to the previous 43.75, and the chances of getting a 4 on the first 3 dice is 57.8125%

Repeat it once more for the 4th die: 100% - 57.8125% = 42.1875%. 42.1875% * 25% = 10.5469%. Add this to the 57.8125% above and get 68.36%.

There is your 68%.

The equation 1 - (1 - 1/S)^N is a closed-form expression for the finite geometric series I've just started.

Jason

Man, I thought I was good at math, but this whole thread made me feel dumb.

There should be a disclosure at the beginning of any such thread: "May cause migraine from thinking to hard".

OOOOUUCH!

Roll the dice one at a time. As soon as you roll the first 4, you can stop because regardless of what the rest of the dice show, you still have at least one 4.

After 1 die, you have a 25% chance of having rolled a 4 so far.

That means there's a 75% chance you'll need to roll a second die, and if you do, there's a 25% chance it will show a 4, so there's a 75% * 25% = 18.75% chance of getting a 4 on the second die Add this to the 25% above, and the chances of getting a 4 on the first 2 dice are 43.75%

That leaves a 100% - 43.75 = 56.25% chance you'll need to roll a third die, with a 25% chance. 56.25% * 25% is 14.0625%. Add this to the previous 43.75, and the chances of getting a 4 on the first 3 dice is 57.8125%

Repeat it once more for the 4th die: 100% - 57.8125% = 42.1875%. 42.1875% * 25% = 10.5469%. Add this to the 57.8125% above and get 68.36%.

There is your 68%.

The equation 1 - (1 - 1/S)^N is a closed-form expression for the finite geometric series I've just started.
Ah, indeed, I remember now having understood it that way in the past.

On the "any given 4," I really meant "average 4." Which is sort-of more true, but still not really.

Thanks!

Again...

This is the main reason why I head down the wrong road when it comes to this topic. Because there is no guarantee that the 5 rolls of a d4 will produce a sequence that contains a 4. Mind you this is an incorrect assumption, thus my confusion. Since over an infinite number of rolls a 4 should occur within the sequence (of infinite rolls) an equal number of times (as compared to all other outcomes) and since you are increasing the number of rolls from 1 to 5, you are increasing the likelyhood of producing a sequence that contains a 4 as one of the outcomes.

WOW, who would have guessed such a large thread on something that is actually very trival!