# Need help with dice probability

1. Jan 3, 2004

### stojakapimp

I'll try to make this short, but this thread deals with the dice game bunco.

Briefly, in the game, there are many different rounds, and in each round, you are trying to roll a certain number with 3 dice. So if we are going for 2's, we roll 3 dice and get points for how many 2's show up.

If you end up rolling all 3 dice of the number that you are going for in that round, then it is called a Bunco. So in our example, we are going for 2's, and if we roll three 2's, then we get a bunco. If we roll three 4's, though, it is not a bunco.

So my question is what the probability is that you will roll 9 bunco's throughout 5 games (you go for each dice number 5 times, so 30 rounds altogether).

I know that the probability of rolling 3 of a kind is 1 out of 6^3, so one of 216. But then you have to do that 9 times...so multiply that by 9. But then you have to take into account that it has to be the correct number you are rolling for, and that's where I get confused.

Anybody know the answer?

2. Jan 4, 2004

### WannabeG133

I'm also trying to figure this out with Stojakapimp and one thing I think you might find helpful is that it took us about 2 hours 10 minutes to complete 30 rounds. So if you feel pretty confident about a number for average rolls in a game then multiply that by 2 or so seconds a turn and so on to come up with 2 hours 10 minutes, or 130 minutes. I think another thing that might be important is that we rolled for each number 5 times, which might be another multiplier added to the mix. If anyone could figure this out for us, you'd be considered 1337 for life. Or a genius if you're not a gaming nerd. Thanks again.

3. Jan 4, 2004

### Sonty

you mean to multiply that 9 times as the event has to occur 9 times.
The fact that you'll get the right number is included in the 1/216. Let's say that for a certain round you need the number n, then you have the probability 1/6 to find it on each dice, so 1/216 in total, as you said.
So my guess is $$P=(\frac{30}{6^3})^9$$. I'll check with the book.

Last edited: Jan 4, 2004
4. Jan 4, 2004

### WannabeG133

I've got a question for you. Why is it 30 over 6? If that's the 30 representing the number of rounds played then wouldn't it actually only be 5, which is the number of times you rolled for each number? This is where I got stuck the other night when we were trying to figure this out.

5. Jan 4, 2004

### Sonty

How many times do you roll the dice? The number of times you roll the dice should be up there.

6. Jan 4, 2004

### WannabeG133

How many times you roll the dice is one of the variables, which is where I think x should go. Because honestly, how are we supposed to know how many times you roll without knowing the odds of getting a bunko, multiplying it by 9, then diving that number by 30. That should give you the total number of rolls per round. Problem is that we can't get the numbers we need to input this, so this is more like a system of equations which is a huge chunk of Algebra 1. We need to decide on 2 variables, then solve for them and maybe be able to figure out other variables. My friend is travelling back down to UCSD today and won't get there until tonight, but he and I looked at your equation at church today and it's a gigantic number. I got too tired of freehanding it so we took out his calculator and estimated it'd be a couple million over a couple ten trillion.

7. Jan 26, 2004

### turin

I'm no math person, but I have taken statistics. I seem to remember something about correlation in lecture (knocks the cobwebs off the brain). I think that the particular number for which you are going is inconsequential. My reasoning is one of lack of correlation from one roll to the next. Each time you roll, you are going for a particular number. Assuming fair dice, you will have a 1/6 chance per di, regardless of the number for which you are going. The correlation issue enters in this respect: no matter what you rolled, or for what number, previously, you will still have a 1/6 chance per di for the next roll, again regardless of the number you want.

In short, the desired number (and even the fact that you are alternating desired number) should be invisible to the probability because the rolls are uncorrelated.

I think the only important issues are:
number of times you roll the dice.
number of successful rolls for which you want a probability.
probability of a successful roll.
the fact that the rolls are uncorrelated.

I'm assuming that you roll the dice 30 times. So, you have 30 "slots" in which you can put a success or a failure. I think you have to consider how many different ways can you can arrange your successes and failures in these 30 slots. Then, are you asking about at least 9 successive rolls, or exactly 9?

For exactly 9, I think you raise the prob of success to the 9th power, and then multiply that by the number of different ways you can arrange the successes/failures in the 30 different "slots," recognizing that the successes and failures are identical (another issue that I forgot about).

I'm still scratching my butt to figure out how to handle the at least 9 case. For sure, the probability of at least 9 would be greater.

Last edited: Jan 26, 2004
8. Jan 26, 2004

### chroot

Staff Emeritus
The number being sought is irrevelant. All that matters is that, for any particular number sought, the probability of a bunco is 1/216.

For n trials, each with probability t of success, the probability of exactly x successes is given by:

$$\left(\begin{array}{cc}n\\x\end{array}\right) t^x(1-t)^{n-x}$$

where t is 1/216 and n is 30.

The probablity of at least 9 successes in 30 trials is the sum of this from 9 to 30:

$$P = \sum_9^{30} \left(\begin{array}{cc}n\\x\end{array}\right) t^x(1-t)^{n-x}$$

The chance of at least 9 buncos in 30 rounds is very small, only about 1 in 78 quadrillion.

The chance of at least 1 bunco in 30 rounds is more reasonable, about 13%. The chance of at least 2 buncos is only 0.8%.

- Warren

9. Jan 27, 2004

### turin

Good stuff, chroot. I was wondering, can the origin of these three factors here be meaningfully identified? It is pretty obvious from where the tx comes, but it is not obvious (to me) from where the other two come.

10. Jan 27, 2004

### NateTG

turin:
Let's say we have some result that occurs with probability $$p$$. Then $$q=1-p$$ is the probability that it does not occur.

Now, let's say that we're looking for the probability of the event occuring $$m$$ times in $$n$$ trials, so, what we can do, is look at all possible $$n$$ event sequences that contain $$m$$ events, and then take the sum of the probabilities of each of the sequences occuring individually.

Now, the probability for any one of the sequences occuring will be:
$$q^{n-m}p^{m}$$
since there will be $$m$$ events that result in $$p$$ and $$m-n$$ events that result in q.

And there will be $$\left(\begin{array}{cc}n\\m\end{array}\right)$$
sequences, since each sequence 'chooses' $$m$$ of $$n$$ events to result in $$p$$.

Since the probailities are exclusive, we can add them together. That means that the total probability is:
$$\left(\begin{array}{cc}n\\m\end{array}\right)q^{n-m}p^{m}$$

Which is chroot's formula.

11. Jan 28, 2004

### turin

Thanks alot Nate. That makes sense to me, at least on a cursory level.

12. Feb 2, 2004

### WannabeG133

Turin, it seems you don't understand the game, but Chroot answered it quite nicely and proves that the night we tried to do this freehand I was closest when I was getting in the trillions before finishing all the equations up.

The 30 rounds means that you roll the dice x number of times until the head table gets 21 points or a bunko each round. It's not just 30 times rolled, otherwise the odds would be insanely unimaginable. The guy would have had to sell his soul to the devil to have 1/3 of his rolls be buncos. So that wasn't how the game works, but thanks for trying. You're still more knowledgeable about this kind of stuff than me as I am only in Plane/Solid Geometry.

13. Jul 2, 2008

### sudhir.kochha

5. In 120 throws of a single die , the following distribution of faces was obtained
Faces 1 2 3 4 5 6 Total
f 30 25 18 10 22 15 120

Do these results constitute an example of the “ equal probability”