Math Question Help

[Acidmouse]

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?

[millidonk]

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

[Acidmouse]

yeah thats the figure i got, but wasnt sure.

[geordieneil]

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

[Acidmouse]

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

[tikay]

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.

[TightEnd]

lol!

[Acidmouse]

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

[outragous76]

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

its 50:50 - obv

[tikay]

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

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?

[millidonk]

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.

[kukushkin88]

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

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).

[tikay]

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.

If you wish.

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

[doubleup]

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

[tikay]

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

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).

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.