Math Question Help

[StuartHopkin]

This may be the funniest thread ever.

[millidonk]

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

[StuartHopkin]

2592/1

(I think   )

[Acidmouse]

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

[StuartHopkin]

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

Yeah then I am pretty sure its 2592/1

[StuartHopkin]

Sigh

Nobbed it the first time

3110.4/1

[tikay]

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

Oh, I seeeee.

My bad, I misread the question, sorry.

[gatso]

Sigh

Nobbed it the first time

3110.4/1

this

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

[EvilPie]

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.

[StuartHopkin]

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

[gatso]

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

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

[zerofive]

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.

[gatso]

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

[EvilPie]

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

Fair enough.

[EvilPie]

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

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

No we don't.

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