blonde poker forum

Was playing a dice game called farkle where you roll 6 dice a little like Yahtzee.

Can someone tell me the odds of rolling 6 dice and getting the same 6 numbers as the previous roll i.e

1 1 1 1 2 2  and then getting 1 1 1 1 2 2 again

I find a figure of 7720/1 but not 100% sure this is correct. Anyone help?

Tricky one.

Sometimes i get this stuff completely wrong, but isn't the odds on rolling the first lot alone 7776/1?? So doing a duplcate roll would be massive.

1/6
1/6 x 1/6
1/36 x 1/6
1/216 x 1/6
1/1296 x 1/6 = 1/7776

yeah thats the figure i got, but wasnt sure.

1/7776 for 6 same numbers

rolling say 111111 then 111111 again is 1/46656

but if say 1,2,2,4,5,5 then its considerably less as each dice has several chances, bar the last dice.....if u know what i mean. but far to many permutations to work out unless u knew the original 6 dice rolled

ahh thanks! that makes sense.. it was 1 1 1 1 2 2 both times..and i said ball park off the top of my head was 50000/1 friend said 7k

Some confusion here......though maybe it is me who is confused.

In ariving at the odds, the first 6 rolls of the die should NOT be used in the calculation - they are simply the starting point. It's a given that if you roll a die 6 times, it will land 6 times......

So, we roll it 6 times, it comes, say, 2-5-3-1-6-5.

Now, we start the second 6 rolls that need to match, & from which the odds are calculated.

I would assume that each roll equates to a 5/1 shot to match the previous. The odds of the sixth roll of the second batch matching the 6th roll of the first batch remain exactly the same, 5/1. The fact that the first 5 rolls of the second set of rolls all matched does not alter those odds of 5/1. It's no different to spinning a coin which comes up "heads" 20 times in a row. The odds on "heads" on the 21st spin are still evens.

lol!

so the odds of getting 1 1 1 1 2 2 again are? :P

its 50:50 - obv

It's 5/1 to roll any number, any time, ever.

So it's 5/1 to roll a "1" on the first roll.

And 5/1 to roll a "1" on the 2nd roll, irrespective of what the first roll was.

Etc.

Or at least, it seems that way to me. Where are the maths experts when you want them?

the odds of rolling the dice independently obv stay the same no matter what, but getting the same roll cumulatively doesn't, esp if you are rolling it in the same order.

The odds of rolling a particular number are 1/6, the odds of rolling a 6 on the last roll of 66666666666666666666666666666666666 are still 1/6 but the odds of rolling all them consecutive sixes do not stay 1/6. No way, no sir, no how. It has to be cumulative.

Please could you clarify if all the dice are rolled at once (working this out would probably need an expert like Tank) or if each dice is rolled separately (easy calc as per TK´s post).

If you wish.

To my mind, each roll is an independent event, absolutely unaffected by any other factor.

There is also the question of the die being in order or not eg is 654321 the same as 123456?

That was the basis of my input.

I have no idea of the odds if each batch of 6 rolls is done "6 at a time". As you say, we'd need someone intelligent, like Tank, or MereNovice, to work that out. We are not worthy.

This may be the funniest thread ever.

Middle of the week man, we are supposed to be taking our foot of the gas. Not getting dished problems like this.

2592/1

(I think  ;) )

its a game of dice where the dice are all thrown at once :P

Yeah then I am pretty sure its 2592/1

Sigh

Nobbed it the first time

3110.4/1

Oh, I seeeee.

My bad, I misread the question, sorry.

this

there are 15 ways to roll 111122 and 46656 total combos of 6 dice

This is an absolute nightmare to work out because your odds on the second set depends on the outcome of the first set.

If for example the first set is 1 2 3 4 5 6 then at least one of your dice is guaranteed to be correct in set 2.

If you roll 1 1 1 1 1 1 in the first set then they could all be wrong.

I don't know if it's even possible to work out tbh but I'm pretty sure it's a fuck load higher than anyone's suggested so far.

Wut?

Its exactly as we said above, 6^6 = 46656 combos, 15 of which are 4 1's and 2 2's

that's what makes it easy to work out, we know set 1 so it's a simple matter of working out the number of permutations

Correct me if I'm wrong, but the math for rolling 1, 1, 1, 1, 2, 2 in any order is

[ 6!/(4!*2!) ] / [ 1/(6^6) ]

So that, twice, as the outcome of the first roll does not influence the outcome of the second roll.

111122, 111212, 111221, 112112, 112121, 112211, 122111, 121211, 121121, 121112, 221111, 212111, 211211, 211121, 211112 if anyone's interested

Fair enough.

No we don't.

OP gave set 1 as an example. It could actually be any one of 46656 combos.

yeah but we're all answering the example question

love that your last 2 posts were accepting what hopkin wrote and disagreeing with what I wrote when we'd both made the exact same point

I wasn't accepting what Hopkin said.

I just couldn't be arsed arguing with Hopkin because all he did was repeat a previous post.

My 'fair enough' was just accepting that he's happy to be wrong.

Your point was easier to counter because you put up an argument against what I'd said which gives me something worth responding to.

Oh

Im with you now Mattu

We only answered the example 1 1 1 1 2 2

The odds are dependent on what you are trying to match as they obviously change dependent on the permutations available

Bare with me and its easily sorted, however it is obviously only expressible as a formula

It's easier if you roll the first set and don't actually look at them.

The odds against getting a match is then 1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6

which is 46656 to 1

Pretty sure that's the answer.

100% wrong as you allow nothing for when dice are duplicated

Fair enough.

LOL

So are the odds against rolling 1 1 1 1 1 1 different to rolling 1 2 3 4 5 6?

Are the odds different when you roll the whole bunch compared to 1 at a time?

First question would be yes assuming we are not concerned on the order, second question would be no.

You can roll 123456 and then you can roll 213456 - and they're the same in this scenario as the combination of numbers is what matters not the total number of permutations.

What you said is the same as rolling 123456 followed by 123456 exactly

That's why I said don't look at the first set. It's then 46656 to 1 because that's your total combinations.

As soon as you know what the first set is you have other considerations and it gets difficult.

neil had already cleared up the point you were making and then aciiiid reraised the 111122


so we were all answering the 111122 weren't we?

it'd help if we were all having the same discussion. that's help a lot of threads actually

norrr, it's only that to throw the exact same numbers in the exact same order. we're throwing 6 dice together

1 in 46656 is only true for 111111, 222222, 333333, 444444, 555555 and 666666 as there's only one possible way of throwing them

I'm confused. 

So what are the odds against throwing 111112?

there are 6 ways to throw it so 6 in 46,656

The difference now is that it doesn't matter if your first roll is a 1 or a 2.  And if it is a 1, it doesn't matter if the 2nd roll is a 1 or a 2.   And so on. Your hopes of still rolling 111112 are kept alive.

I'm going to try to answer this myself.

Is it 46656/6 as it could be 111112, 111121, 111211, 112111, 121111 or 211111?

surely the odds of the first roll are irrelevant as you are setting your target (you obv neednt roll)

therefore the question is what are the odds of rolling any specificc 6 no combo

you then just need to clarify whether order is important, (to which i assume the answer is no)


The above howeve ris very different to the question:

what are the odds of rolling the same 6 numbers twice in a row


.............................

ah fk it

Ninja'd

So the answer to the OP is that it's impossible to know the odds until you know what the first set of numbers is?

Yes, but he gives an example set - 111122 to work with.  If he'd given 123456 or 333333 the odds would be different.

nope, there's a definite answer

I think it's 462 in 46,656; 1 in 100.98 so almost exactly 1%

hoping that's right. there are 462 unique sets of numbers that we can make with 6 dice if we ignore the order

Is this a massive level...?

nope the odds are definatly bigger than massive

lol...:) i lost the cunting game anyway (its farkle on facebook)

Is this not wrong also as the chance of the first set being certain numbers is higher than other sets of certain numbers?

462 is correct I think though

I dont think this is impossible

Are we retracting any of this now?
I thought I had explained it a bit more in the second post so a man of your intelligence would get it.
That was the bit I am happy to be wrong about.

You're not allowed to spark a five page discussion and confuse a bunch of simpletons and then fade into the shadows.

I'm really not sure. the 462 and the 46,656 are right, just not sure how we should be combining them or indeed if the 462 is even relevant

I think the 462 is relevant, but you need the odds of rolling each of the 462 numbers and then use this to weight the various probabilities of matching the original numbers?

Because of this:


and this:


and this:


also a bit of this:


as well as combining this:


with this:


I don't think I'll retract anything just yet.

Thanks for giving me the option though.