The Roll of the Dice

[dooya]

Might be a poor choice of words to use the word "dependent" but what I'm trying to point out is that one case he was using a pre-defined number of rolls (3 in this case) of the same die when he was rolling it over and over again for 3 turns.

versus

Throwing 3 dice simultaneously for 1 turn.

Still the same: there is no difference from a statistical perspective. Since dices are thought to be identical (and independent) it makes no difference whether you throw one dice three times or three dices at once.

[Gray_Lensman]

deleted

[DarthMath]

It's possible I comparing Apples to Oranges since I'm trying to apply the 25% example of the event design referred to above. In that case, it is equivalent to a 100 sided die with 1 thru 25 being approx 25% of the results and may not be applicable to a specific exact number coming up on the die roll.

Is that confusing? LOL I think I'll give up before my head explodes with Hobbes.

I do know that the event design sited above worked out as designed thru many turns of repetitive testing.

Indeed, I'm quite sure it's not the same thing. Maybe it's time to give our respective brains some well-deserved rest !!

[Hobbes]

I think I get it now - it just makes it harder when you think of dice.

If you think of flipping two coins - a 1/2 chance and a 1/2 chance does not end up as a 2/2 chance. There is never a 100% 'chance'.

It's 50% on the first coin. If that fails you get one more 50% chance.
So a 75% chance to get a head on two flips.

Brain still hurts.

Cheers, Chris

[arsan]

By the thread title i though this was about boardgames recommendations
Seeing that it involves numbers and calculations... i getting out of here fast!
How i hated mathematics at school!!

[dooya]

It's prpbable I'm comparing Apples to Oranges since I'm trying to apply the 25% example of the event design referred to above. In that case, it is equivalent to a 100 sided die with 1 thru 25 being approx 25% of the results and may not be applicable to a specific exact number coming up on the die roll.

Is that confusing? LOL I think I'll give up before my head explodes with Hobbes.

I do know that the event design sited above worked out as designed thru many turns of repetitive testing.

I think that the event example was the right one, but from the other way around.

Dice example: I know the probability of event X at a single attempt - How is the probability to observe X in three attempts?

Event example: Using 6 attempts, I want to observe X with probability P. - How do I choose the probability of event X in the single attempt?

In both cases it makes no difference whether or not the attempts take place at the same time or sequentially, because the events are independent.

For better understanding an example for statistical dependence: Lets say we have a bag with 6 little balls numbered from 1 to 6. How is the chance to draw the 6 at the first attempt? Of course it is 1/6. Now let us assume we have one ball drawn. Is the chance of drawing the 6 still the same? No it is not: the probability of drawing the 6 in the second attempt is either 0 (because the 6 was drawn at the first attempt) or 1/5 (because the 6 is among the remaining 5 balls in the bag). This is statistical dependence: one event changes the odds of another.

[Hobbes]

I'm going to the pub tomorrow with three dice and £50.

[DarthMath]

I'm going to the pub tomorrow with three dice and £50.

To do what ? Be corrected by Mrs. Hobbes back at home for losing 50£ at a stupid game !!

I've almost forgotten the more important thing I learned studying probabilities : never counts on them in the real life !! Luck is never part of the equations, and yet it's the only truth !!
I think we all have plenty of examples of this simple fact !!

[Ebbingford]

I'm not even going to get involved with thinking about this.
Just reading through the thread has got my old maths teacher mumbling to me in my head about "mutually exclusive events", "factorials", and "6 choose 3".

[caranorn]

Thanks for the headache ;-) ...

And I agree with Dooya. Though possibly people are talking about a number of different things. At first I thought you were talking of throwing 3d6, and asking for the probability of the three dice added together being 6 (a common use in wargamiong, though usually with only two dice). Obviously the discussion is about three dice thrown at once and the probability of one and only one of them showing the 6 face (alt. one die thrown three times, which should be the same as Dooya said)...

Oh that headache :-D ...

[Hobbes]

I suppose what has been troubling me is that I agree with the 42% chance of rolling a six with three dice - but then kept thinking If I did this 100 times I should see about 50 dice showing a six - so 50%.

But I realise both of these are true as you are only concerned with the chance of rolling one six. In some attempts you will get two or three sixes which do not matter for the 42% - but will give you your 50 sixes in the example above. The explanation we found at work is below (I don't really get this though :wacko

In one roll, the probability of rolling a 6 is 1/6.

For two rolls, there is a 1/6 probability of rolling a six on the first roll.
If this occurs, we've satisfied our condition. There is a 5/6 probability
that the first roll is not a 6. In that case, we need to see if the second
roll is a 6. The probability of the second roll being a 6 is 1/6, so our
overall probability is 1/6 + (5/6)*(1/6) = 11/36. Why did I multiply the
second 1/6 by 5/6? Because I only need to consider the 5/6 of the time that
the first roll wasn't a 6. As you can see the probability is slightly less
than 2/6.

For three rolls, there is a 1/6 probability of rolling a six on the first
roll. There is a 5/6 probability that the first roll is not a 6. In that
case, we need to see if the second roll is a 6. The probability of the second
roll being a 6 is 1/6, giving us a probability of 11/36. There is a 25/36
probability that neither of the first two rolls was a 6. In that case, we
need to see if the third roll is a 6. The probability of the third roll being
a 6 is 1/6, giving us a probability of 1/6 + (5/6)*(1/6) + (25/36)*(1/6) =
91/216. Again, this is less than 3/6.

The general formula for rolling at least one 6 in n rolls is 1 - (5/6)^n.

Cheers, Chris

[DarthMath]

So, finally, did you win something at the pub ? ( apart from a good beer, of course !! )

[Hobbes]

So, finally, did you win something at the pub ? ( apart from a good beer, of course !! )

I took your advice and stayed at home with Mrs. Hobbes

Cheers, Chris

[Pocus]

strip-poker?


Ooops sorry Chris!

[Hobbes]

strip-poker?


Ooops sorry Chris!

Are you kidding! Coronation Street was on.