# Using dice to give a probability of 1 in 50 (0.02) [Cotton Mather related]



## Eoghan (Jun 13, 2012)

The chances of an unvariolated person dying from smallpox if variolated in Boston was 1 in 50 - how do I use dice to represent this?


----------



## VictorBravo (Jun 13, 2012)

It doesn't work out easily, because the cube has six sides, so 1 in 6 probability of a given number. Two tosses would be 1/36, three tosses 1/216, etc.

But if you combine it with coin tosses you can get close:

First, have one pick a number 1-6. Second, have him pick heads or tails. 

The odds of the dice coming up on that particular number, combined with the odds of three coin tosses coming up the heads or tails chosen is 1/6 X 1/2 X 1/2 X 1/2, or 1/48.


----------



## Theoretical (Jun 13, 2012)

Eoghan said:


> The chances of an unvariolated person dying from smallpox if variolated in Boston was 1 in 50 - how do I use dice to represent this?



Board game supply stores (online) sell 10 sided dice. Also there's an app on iPhone called Mach dice that lets you put in a variable number of dice and their sides.


----------



## VictorBravo (Jun 13, 2012)

Theoretical said:


> Board game supply stores (online) sell 10 sided dice. Also there's an app on iPhone called Mach dice that lets you put in a variable number of dice and their sides.



That would work, but you would need to do two tosses: the first you would pick two numbers, and the second only one number:

Odds of one of the two numbers in the first toss = 2/10. Odds of the single chosen number in the second toss = 1/10.

2/10 X 1/10 = 2/100 or one in 50.


----------



## arapahoepark (Jun 13, 2012)

What are you trying to get at exactly?


----------



## SRoper (Jun 13, 2012)

How about randomly drawing one card from a deck and getting the ace of spades? That is 1 in 52.


----------



## Semper Fidelis (Jun 13, 2012)

Hmmm...I'm not sure how this can be done.

I keep thinking of ways to make this work. You could get a 10 sided dice from an old D&D set and roll it twice. 

The probability of rolling a 1 on the first roll and either a 1 or a 2 on the second roll would be .02.

First roll probability is 1/10.
Second roll probability is 1/5.

Overall probability is 1/50.


----------



## toddpedlar (Jun 13, 2012)

Quickest way with dice (normal dice, not something D&D junkies have) to 1/50 (1/54) is to take 3 dice and roll them. The probability of getting 111, 222, 333 or 444 is 1/54. 

Or, to be fun, rolling either 666 or 616 will also occur with a probability of 1/54


----------



## Jack K (Jun 13, 2012)

If you _have_ to do it with a single, six-sided die, the method is to effectively turn your six-sided into a five-sided by declaring before you roll it that one of the sides doesn't count and, if that side cones up, it results in a redo.

So... 

1. Declare that any 6 doesn't count and requires a redo. Chance of rolling a 1 is .20

2. Declare again that any 6 doesn't count and requires a redo. Chance of rolling a 1 is again .20

3. Now let all sides count. Chance of rolling a 1, 2 or 3 is .50

.20 x .20 x .50 = .02 That's exactly 1 in 50!


----------



## Semper Fidelis (Jun 13, 2012)

Jack K said:


> If you _have_ to do it with a single, six-sided die, the method is to effectively turn your six-sided into a five-sided by declaring before you roll it that one of the sides doesn't count and, if that side cones up, it results in a redo.
> 
> So...
> 
> ...



That doesn't solve the problem. The probability of rolling any number is still 1/6. The probability of rolling any number other than 6 is 5/6.


----------



## sastark (Jun 13, 2012)

View attachment 2908


----------



## Semper Fidelis (Jun 13, 2012)

sastark said:


> View attachment 2908


----------



## Rich Koster (Jun 13, 2012)

Take one die and throw it up in the air 49 times and catch it. On the 50th attempt, let it drop to the floor.

OR

Line up 10 dice displaying the 5 to the audience. On a separated place display a sole die displaying the 1. Point to one, than the other to make the point.


----------



## Jack K (Jun 13, 2012)

Semper Fidelis said:


> Jack K said:
> 
> 
> > If you _have_ to do it with a single, six-sided die, the method is to effectively turn your six-sided into a five-sided by declaring before you roll it that one of the sides doesn't count and, if that side cones up, it results in a redo.
> ...



It _does_ solve the problem. Maybe I didn't explain it well enough.

If you don't count your roll and instead try again anytime you get a 6, then your probability of ending up with a 1 is 1/5. If 6's don't count and always result in a redo, it's as if the die had no 6's. Then each of the other five sides has an equal 1/5 probability of being your result.

Admit it, folks. I solved the puzzle with a single 6-sided die. I want a prize.


----------



## CharlieJ (Jun 13, 2012)

If you're just looking for a good illustration, forget the die. Picking a random card out of the deck is close enough and will communicate the point very clearly. Anything complicated won't carry the proper punch.


----------



## VictorBravo (Jun 13, 2012)

CharlieJ said:


> If you're just looking for a good illustration, forget the die. Picking a random card out of the deck is close enough and will communicate the point very clearly. Anything complicated won't carry the proper punch.



I agree, simplest and most elegant way to demonstrate it, but Scott up in post #6 already came up with that one.


----------



## Eoghan (Jun 14, 2012)

Semper Fidelis said:


> Hmmm...I'm not sure how this can be done.
> 
> I keep thinking of ways to make this work. You could get a 10 sided dice from an old D&D set and roll it twice.
> 
> ...



... and you think as a member of the puritanboard of good standing I have a Dungeon and Dragons set !!!!!!!!!!!!


----------



## Eoghan (Jun 14, 2012)

*My gut says not*



Jack K said:


> Semper Fidelis said:
> 
> 
> > Jack K said:
> ...






Many years ago the show "the ten thousand dollar question" (?) finished with the contestant being offered three doors to chose from. A, B or C. Once he had chosen one the host opened a door and showed there was nothing behind it. Should he change his original choice?

The answer is always yes!

In the original setup he had a 1 in 3 chance of being right.

In the new setup he has a 1 in 2 chance IF HE CHOOSES THE OTHER DOOR!


----------



## Eoghan (Jun 14, 2012)

I left my book on probability at school hence the resorting to PB.

However

If you roll a dice and look for evens, evens, evens and a six = 0.5 x 0.5 x 0.5 x 1/6 = 0.02 near enough (with a few decimal places 0.020833333333333333)

The idea is to develop some sort of way of showing your chances of dying from variolation as opposed to being unprotected (1/6)


----------



## Eoghan (Jun 14, 2012)

sastark said:


> View attachment 2908



I wouldn't want to play golf with you Seth...


...come to think of it I wouldn't want to play golf full stop!


----------



## SRoper (Jun 14, 2012)

Jack K said:


> Semper Fidelis said:
> 
> 
> > Jack K said:
> ...



I understood your solution. It works.

Think about the simpler case of flipping a coin. If you say you are going to ignore tails and keep flipping until you come up heads, what is the probability that you will have heads at the end?


----------



## SRoper (Jun 14, 2012)

Eoghan said:


> Jack K said:
> 
> 
> > Semper Fidelis said:
> ...



This is called the Monty Hall problem. It's a great puzzle, but it doesn't apply to Jack's solution, though.


----------



## Jack K (Jun 14, 2012)

Eoghan said:


> I left my book on probability at school hence the resorting to PB.
> 
> However
> 
> ...



That is indeed very close to .02, and quite simple.


----------



## arapahoepark (Jun 14, 2012)

I believe if you have a TX calculator you could do this on there under probability.


----------



## Constantlyreforming (Jun 14, 2012)

actually, since probability is connected to the decree of God, isn't all probability limited to 50-50......either yes or no......it will happen or it won't?


----------



## arapahoepark (Jun 14, 2012)

Constantlyreforming said:


> actually, since probability is connected to the decree of God, isn't all probability limited to 50-50......either yes or no......it will happen or it won't?



So rolling a dice, either you roll it or you don't?


----------



## Semper Fidelis (Jun 14, 2012)

Jack K said:


> Semper Fidelis said:
> 
> 
> > Jack K said:
> ...



Jack,

It does not solve the problem. The dice itself does not change to a five sided dice simply because you've chosen to ignore a 6. The probability of it landing on any side of the dice is still 1/6. It doesn't matter whether you ignore 6 as your dice roll. Dice don't land on 1's with any more frequency just because you reject the 6.


----------



## Hilasmos (Jun 14, 2012)

Semper Fidelis said:


> The probability of it landing on any side of the dice is still 1/6. It doesn't matter whether you ignore 6 as your dice roll.



If you put a red, yellow, and black ball in a bag, what is the probability that you will end up with a red ball vs. a black ball if you continue to draw balls until you get a red or black ball?


----------



## Semper Fidelis (Jun 14, 2012)

Hilasmos said:


> Semper Fidelis said:
> 
> 
> > The probability of it landing on any side of the dice is still 1/6. It doesn't matter whether you ignore 6 as your dice roll.
> ...



I don't know if you're asking this as a new problem or if you think the example is relevant to what I stated.

That is a different problem. Every dice roll is an individual event and the probability of any given roll is _independent_ of a previous roll. No matter how you slice the problem, if you ask "What is the probability that a dice will roll on a particular side" then the answer is _always_ 1/6 if you have a six-sided dice.


----------



## Hilasmos (Jun 14, 2012)

Semper Fidelis said:


> I don't know if you're asking this as a new problem or if you think the example is relevant to what I stated.
> 
> That is a different problem. Every dice roll is an individual event and the probability of any given roll is independent of a previous roll. No matter how you slice the problem, if you ask "What is the probability that a dice will roll on a particular side" then the answer is always 1/6 if you have a six-sided dice.



I thought it was illustrative of Jack's resolution, so I don't think it is different. The probability that you will end up with a red ball vs. a black ball is 50%, despite the yellow ball. The probability that you will draw a red ball on any given draw is 1/3, or land on side of a die is 1/6, which is understood, but that isn't the question being asked. The question being asked is, what is the % chance I will land on a 1 before a 2,3,4, or 5 if I continue to roll the dice until I land on a 1, 2, 3, 4, or 5. The answer is 20%.


----------



## Jack K (Jun 14, 2012)

Semper Fidelis said:


> Jack,
> It does not solve the problem. The dice itself does not change to a five sided dice simply because you've chosen to ignore a 6. The probability of it landing on any side of the dice is still 1/6. It doesn't matter whether you ignore 6 as your dice roll. You've only created a conditional probability. Given that I reject 6, the probability of 1-5 is:
> P(x)*P(not 6) = 1/6*5/6 = 5/36



Ah, but the question is NOT "what's the probability of getting a 1 on the next roll?" The question is an altogether different one: "If I consistently keep rejecting any 6 I may roll, what's the probability of rolling a 1 before I get a 2, 3, 4 or 5?" The answer to that is 1/5.

And Will's example does illustrate the same principle. Thanks, Will!


----------



## SRoper (Jun 14, 2012)

Rich, I think you must be misunderstanding Jack's solution. Here is the algorithm:

Begin
Roll die
If die = 6, goto Begin
End

So what values can die have at the end of the algorithm? Only 1, 2, 3, 4, 5. What is the probability distribution of these five possibilities? It is uniform, so the probability of ending with die = 1 is 1/5.

Here's another way to look at it. With this algorithm, what is the probability of getting a 1 within one iteration (one roll)? It is 1/6 as you point out. What is the probability of getting a 1 within two iterations? Well there are two ways to do this; roll a 1 on your first roll or roll a 6 on your first roll and a 1 on your second roll. So the probability is (1/6) + (1/6)*(1/6) or (1/6) + (1/6)^2 or 7/36. Three iterations? (1/6) + (1/6)^2 + (1/6)^3 = 43/216. Already after three iterations we are at 0.199. If you work out the series, you will find it converges to 1/5.


----------



## Semper Fidelis (Jun 14, 2012)

Jack K said:


> Semper Fidelis said:
> 
> 
> > Jack,
> ...



No, as I pointed out, Will's example is unrelated. As I have already noted every dice rolls probability of rolling a particular number is completely unrelated to any dice roll that came before.


----------



## Hilasmos (Jun 14, 2012)

Giving a right answer to the wrong question doesn't make sense.


----------



## Jack K (Jun 14, 2012)

SRoper said:


> Rich, I think you must be misunderstanding Jack's solution. Here is the algorithm:
> 
> Begin
> Roll die
> ...



Now THAT is deep math! Nice to have it confirmed another way.


----------



## Semper Fidelis (Jun 14, 2012)

SRoper said:


> Here's another way to look at it. With this algorithm, what is the probability of getting a 1 within one iteration (one roll)? It is 1/6 as you point out. What is the probability of getting a 1 within two iterations? Well there are two ways to do this; roll a 1 on your first roll or roll a 6 on your first roll and a 1 on your second roll. So the probability is (1/6) + (1/6)*(1/6) or (1/6) + (1/6)^2 or 7/36. Three iterations? (1/6) + (1/6)^2 + (1/6)^3 = 43/216. Already after three iterations we are at 0.199. If you work out the series, you will find it converges to 1/5.



Scott,

This is another solution altogether. This does not relate to Jack's claim. What you have asked is what is the sum of probabilities that a 1 will be rolled for every dice roll. It has nothing to do with whether one excludes a side of the dice. 


Sent from my iPad using Tapatalk


----------



## Semper Fidelis (Jun 14, 2012)

By the way Scott, I understand what Jack has said but as I have said the probability that a dice will land on 1 is not dependent on whether you decided to exclude 6. If I ask what the probability is that each roll will land on 1 then the fact that I won't accept 6 as an answer has no bearing on the fact that there is still only a 1/6 probability. 


Sent from my iPad using Tapatalk


----------



## AThornquist (Jun 14, 2012)

Brother Rich, 

I'm glad that after several Engineering degrees and years of statistics usage you have found a system that works for you. However, not all of us believe in that sort of math, and though we're glad you found what's true for you, it just isn't true for everybody. We all have to find our own equations in life without imposing them on others. Besides, the equations you're using have been a part of many great evils in this world, so it's not as if they are without their own problems.

Whatever our equations are, our solutions will always be equal. Group hug, anybody?


----------



## Hilasmos (Jun 14, 2012)

Or better yet, as they say, maybe we should all put our money where our mouths are? 

Does Rich want to offer "fair odds" at 5/1 for every time I roll a 1 on a 6 sided die, but if I roll a 6 the bet doesn't count?


----------



## Semper Fidelis (Jun 14, 2012)

Hilasmos said:


> Or better yet, as they say, maybe we should all put our money where our mouths are?
> 
> Does Rich want to offer "fair odds" at 5/1 for every time I roll a 1 on a 6 sided die, but if I roll a 6 the bet doesn't count?



Of course. I'll let you pay for 72 rolls of the dice at $72.

Roll that dice 72 times and give yourself $5 every time you roll a 1.

Guess what, it doesn't matter that you've decided to exclude the 6 because you're only going the roll a 1 twelve times on average.

Keep the $60 and send me $12 you have left on the table.


----------



## Semper Fidelis (Jun 14, 2012)

OK, Jack, I thought about a way to state what you were trying to get at accurately.

Given that a dice roll is not a 6, the percentage of rolls that are a 1 is 20% or 1/5. In the example I just cited, 60 out of 72 rolls are 1-5 and 12 are 1.

In other words, you don't need to exclude the 6 as not "counting", you can just say that, of the rolls that are not 6, what is the percentage of rolls that are a 1?

In this case it is not based on the probability of a roll because you're not going to get the roll to be anything different than 1/6 no matter how you slice it. You would need to simply note that, of the rolls you accept, the ratio of 1's to all other rolls is 1/5. As you stated the problem initially you said:


> 1. Declare that any 6 doesn't count and requires a redo. *Chance of rolling a 1 is .20*
> 
> 2. Declare again that any 6 doesn't count and requires a redo. *Chance of rolling a 1 is again .20*


As I noted, declaring that a 6 doesn't count doesn't change the *chance of rolling a 1*. It is still 1/6.

There is no way to state the problem that changes the chance of rolling a 1.

As I noted, you can appeal to normal distribution of rolls from 1-5 but that's different than stating the probability of a single event which is always independent of previous events.


----------



## SRoper (Jun 14, 2012)

Rich, it seems like you are claiming that Jack said that the probability of getting a 1 on an individual roll is 1/5 if you ignore 6. I don't see him saying that. Rather, he was saying the chance that you end up with a 1 at the end of his algorithm is 1/5. Jack seemed to indicate that my solution is an accurate restatement of his solution.

And just because I can't leave the little bit of hand-waving alone at the end of my previous post, here is solution to the series:


----------



## Jack K (Jun 14, 2012)

Semper Fidelis said:


> As I noted, declaring that a 6 doesn't count doesn't change the chance of rolling a 1. It is still 1/6.



Of course the chance of rolling a 1 on any single roll is 1/6. But it isn't a single roll event if a 6 comes up. I think the part you're missing from what I said is that if you get a 6 it not only doesn't count, you also get a redo. You get an extra roll. That extra roll gives you another chance to end up with a 1 in the end. You have to add in that probability as well.

This is the way Scott is approaching it in his proof. That's worded differently from how I first said it, but it uses the same rolls of the dice and yields the same result.


----------



## Semper Fidelis (Jun 14, 2012)

Well Scott, he did say that the chance of rolling a 1 is .2.

I don't see how your algorithm relates to Jack's original premise whether you or Jack believes it does. How does your geometric series relate to the idea that, if I roll a 6, then the dice roll doesn't count?

I don't need to even start with excluding any sides with your geometric series.

I simply state that the sum of all probabilities that a dice roll lands on 1 successively is .02 if I roll the dice an infinite number of times.


----------



## Jack K (Jun 15, 2012)

Semper Fidelis said:


> Of course. I'll let you pay for 72 rolls of the dice at $72.
> Roll that dice 72 times and give yourself $5 every time you roll a 1.
> Guess what, it doesn't matter that you've decided to exclude the 6 because you're only going the roll a 1 twelve times on average.
> Keep the $60 and send me $12 you have left on the table.



Again, the missing element here (and the reason you won't make any money) is that he'll get more than 72 rolls. You can't know going in how many total rolls he'll have, because every time a 6 comes up he earns another roll. That's what I mean by a "redo" if you roll a 6. That's the key point in my solution. You can't leave that out.


----------



## Semper Fidelis (Jun 15, 2012)

Jack K said:


> Semper Fidelis said:
> 
> 
> > As I noted, declaring that a 6 doesn't count doesn't change the chance of rolling a 1. It is still 1/6.
> ...



Jack,

I will simply state again that your statement (the part I bolded) is sloppy. The chances of rolling a 1 is 1/6. Again, as I just pointed out to Scott, his geometic series has nothing to do with excluding a 6. That series is true whether or not I exclude a number from the roll. Let's say, for instance I decide to say that all rolls that are not 1, don't count. The odds of rolling a 1 on each roll is still 1/6. Since I keep rolling a 1, every roll counts because, theoretically, I'll never roll anything other than a 1 so every roll counts.

The odds for the sum of the probabilities for getting 1 on every roll still adds up to .02 so his algorithm is not related to your particular theorem.

It doesn't take a geometric series to demonstrate the idea. A single sample notes that 5/6 rolls are not 6 and 1 out of those 5 is 1. Nevertheless, the probability that and individual roll is 1 is 1/6. As I keep pointing out, the individual roll probability for a 1 is unaffected, it is the relationship of the distribution of the probability of a 1 related to the probability of all rolls not a 6 that is key.


----------



## Jack K (Jun 15, 2012)

Semper Fidelis said:


> Well Scott, he did say that the chance of rolling a 1 is .2.



Ah. So what you want me to say, to communicate to you better, is that the chance of *ending* on a 1 is .2

If you roll a 6, the event does not end. If the event never ends on a 6 (6 always gives you another roll), then you can only end on a 1, 2, 3, 4 or 5. And each has an equal chance. So the chance of ending on a 1 is .2.

Does it makes sense stated that way?


----------



## Semper Fidelis (Jun 15, 2012)

Jack K said:


> Semper Fidelis said:
> 
> 
> > Of course. I'll let you pay for 72 rolls of the dice at $72.
> ...



Yes, but I didn't offer him that. I offered him $72 for 72 rolls and wanted to see if he'd take me up on it. In fact, this is all theoretical but assuming a perfect distribution, the bet will only be made 1 out of 5 times so, out of 72 rolls the bet would be made 60 times and it would be even money. Again, however, I point out that what you stated in your original statement was sloppy. The individual roll probability is 1/6 so the distribution works out that 1/6 of all rolls is a 1, 1/6 of all rolls is 1/6. If I exclude all 6's then I get:

P = P(1)/1-P(6) = 1/5= (1/6)/(1-1/6) = (1/6)/(5/6) = 1/5


----------



## VictorBravo (Jun 15, 2012)

I'm still boggled by the idea of a 10-sided "dice." It has bothered me all day trying to figure out such a polyhedron. Then I looked up pictures on the internet and see examples that show polar symmetry, but only on one axis. For some reason that bothers me, although I guess they do produce random results the way a coin toss does.

Anyway, thanks Scott for the pretty series. I did something similar in my notepad earlier today and it looks Vic-typically ugly. What sort of equation editor do you use?


----------



## Semper Fidelis (Jun 15, 2012)

Jack K said:


> Semper Fidelis said:
> 
> 
> > Well Scott, he did say that the chance of rolling a 1 is .2.
> ...



No, as I've noted, the issue is distribution. You've taken out a set of rolls but you haven't changed the distribution of rolls that land on a 1. The chance of ending on a 1 is 1/6 with any roll. The distribution of all rolls that are 1 out of all rolls that are either 1,2,3,4, and 5 is .2.


----------



## Semper Fidelis (Jun 15, 2012)

VictorBravo said:


> I'm still boggled by the idea of a 10-sided "dice." It has bothered me all day trying to figure out such a polyhedron. Then I looked up pictures on the internet and see examples that show polar symmetry, but only on one axis. For some reason that bothers me, although I guess they do produce random results the way a coin toss does.
> 
> Anyway, thanks Scott for the pretty series. I did something similar in my notepad earlier today and it looks Vic-typically ugly. What sort of equation editor do you use?



Lot of 10-sided dice,d10,D&D RPG WoD Dungeons Darkness | eBay


----------



## VictorBravo (Jun 15, 2012)

Semper Fidelis said:


> Lot of 10-sided dice,d10,D&D RPG WoD Dungeons Darkness | eBay



I think those are of the debil. Two hexahedrons glued together. Just not natural. . . .


----------



## Jack K (Jun 15, 2012)

Semper Fidelis said:


> The chance of ending on a 1 is 1/6 with any roll.



Not quite. The chance of *rolling* a 1 is 1/6 with any roll. But the chance of *ending* is not, because a 6 won't end it. There are only 5 possibilities that end things. So the chance of ending with any one of those is .2. Hence, the chance of _ending_ on a 1 is .2.


----------



## Eoghan (Jun 15, 2012)

*Rolling dice - skill or chance*

If you roll the dice and choose the axis of rotation do you not push the odds to 1/4. So if I have the 1 and 6 on the ends, and imagine there is a hole drilled through from one side to the other if this is the axis of rotation as the dice rolls across the table it tilts the odds towards 1/4 for a "3".

This effectively makes dice rolling more of a "skill"?

If I could give a consistent throw, could I control the momentum imparted to give say three rotations and then stop on a "3"?

This was the reason I believe that some board games provide a cup to shake the dice before rolling. I know that generating random numbers is a huge problem for scientists and that there is some sort of natural phenomenon that is the closest they have gotten. (help needed here) I do know that one of the reasons the Puritans objected to games of chance was that they saw it as a denial of providence/skill?


----------



## Hilasmos (Jun 15, 2012)

Semper Fidelis said:


> Of course. I'll let you pay for 72 rolls of the dice at $72.
> Roll that dice 72 times and give yourself $5 every time you roll a 1.
> Guess what, it doesn't matter that you've decided to exclude the 6 because you're only going the roll a 1 twelve times on average.
> Keep the $60 and send me $12 you have left on the table.



First, in gambling terms, 5/1 (5 to 1) pays 6, that is why I called it fair-odds. So after 72 rolls I will be at 72 dollars return, assuming even distribution. 

Second, I said when I roll a 6, the bet doesn't "count." Again, in gambling terms this is called a push, and when it happens in blackjack or craps, the money stays on the table. So, if you want to play the game as I described it (which is accurately describing Jack's position, as he affirmed), you will be paing me 5/1 odds on a 4/1 bet. I wouldn't advise moving to vegas to be a game designer.

To illustrate in an evenly distributed 6 roll sequence

1 pays 5/1: returns 6

2,3,4,5 returns 0

6 is a push: returns 1

Roll Return Invested 
1 6 1
2 0 1
3 0 1
4 0 1
5 0 1
6 1 1

6 7 6

Total invested: 6
Total returned: 7
ROI: + 14.2%


----------



## Semper Fidelis (Jun 15, 2012)

Hilasmos said:


> I wouldn't advise moving to vegas to be a game designer.


I would advise spending more time filling your brain with things other than how Vegas operates.


----------



## Hilasmos (Jun 15, 2012)

Semper Fidelis said:


> I would advise spending more time filling your brain with things other than how Vegas operates.



Well, we all live in God's world and follow the same mathematical laws, vegas was merely illustrative of where you were going awry; but unfortunately, I have spent the better part of my younger life in that degenerate environment.


----------



## SRoper (Jun 15, 2012)

VictorBravo said:


> I'm still boggled by the idea of a 10-sided "dice." It has bothered me all day trying to figure out such a polyhedron. Then I looked up pictures on the internet and see examples that show polar symmetry, but only on one axis. For some reason that bothers me, although I guess they do produce random results the way a coin toss does.
> 
> Anyway, thanks Scott for the pretty series. I did something similar in my notepad earlier today and it looks Vic-typically ugly. What sort of equation editor do you use?



I think it is because a ten-sided die is not a platonic solid, although I guess that doesn't explain why we readily accept a coin toss.

I used the equation feature in Word 2007. You'll notice that the second statement is underlined for grammar correction. That's because I ended up taking a screenshot of everything since PB only allows you to post up to two pictures. Didn't notice it was underlined in the screenshot until it was too late for me to care.

Rich, I will have to reply this evening when I can give your response more consideration.


----------



## SRoper (Jun 15, 2012)

Semper Fidelis said:


> Well Scott, he did say that the chance of rolling a 1 is .2.
> 
> I don't see how your algorithm relates to Jack's original premise whether you or Jack believes it does. How does your geometric series relate to the idea that, if I roll a 6, then the dice roll doesn't count?
> 
> ...



Rich, go back to the algorithm I stated. The series is showing the work for why you will end up with a 1 at the end of the algorithm with probability 1/5.

Here's how it works. What is the probability of ending with a one within one iteration? It is 1/6 since there is a one in six chance of rolling a 1 with a six-sided die. This is the first term in the geometric series. Now, if we roll a 1,2,3,4, or 5 at any time we are done. We stop rolling the die. We got an acceptable result. Perhaps this is the part that you are missing.

Now let's say we roll a 6 on our first roll. In this case, we are not done. We need to go on to a second iteration. What is the probability that we got to a second iteration? It is 1/6 since there is a one in six chance of rolling a 6. Now what is the probability of rolling a 1 on the second iteration? It is 1/6. What is the probability of rolling a 1 on exactly the second iteration of the algorithm? It is the probability of rolling a 6 on the first iteration times the probability of rolling a 1 on the second iteration or (1/6)(1/6) or (1/6)^2. This is the second term of the series. And on it goes. The entire series will give the probability of ending the algorithm on a 1.

I'm going to have to end here since we are packing up for a camping trip. No internet, but I will have pen and paper to work out some series Vic-style!


----------



## Eoghan (Jun 18, 2012)

*throwing four dice - different from throwing one dice four times?*

If you are looking for an even number (1/2) three times and then a six (1/6) that gives a probability of close to 0.2. My question is what happens if you roll four dice at the same time. The probability of getting a six goes down to 4/6 (?) or does it?


----------



## Semper Fidelis (Jun 18, 2012)

Eoghan said:


> If you are looking for an even number (1/2) three times and then a six (1/6) that gives a probability of close to 0.2. My question is what happens if you roll four dice at the same time. The probability of getting a six goes down to 4/6 (?) or does it?



No, the probability is 0.517746914. If it was the sum of probabilities then if you had 6 dice you would have a 100% chance of rolling a 6 and if you had 7 you would have more than a 100% chance.

The probability of rolling a 6 in n rolls is 1 - (5/6)^n. The probability of not rolling a 6 in for each roll is 5/6 so the probability of constantly not rolling a 6 is (5/6)^n.


----------

