Dice roll probability

[doodab]

I'd work out the answer when starting with a single die first. That's the vaguely complex bit as it's an infinite series summation or similar. I'll try and do that in a bit.

The general case is equivalent to considering each initial dice seperately, so just multiply that answer by the initial number of dice.

Ok for a single die the total probability of a success is 2/5 when all the rethrowing if you get a 1 etc is taken into account. So for N dice it's 2N/5

[DimPrawn]

x/6^N

Where x is the series count and N is the number of throws in a row.

I've mostly made this up.

[No2politics]

I'd work out the answer when starting with a single die first. That's the vaguely complex bit as it's an infinite series summation or similar. I'll try and do that in a bit.

The general case is equivalent to considering each initial dice seperately, so just multiply that answer by the initial number of dice.

Agree

[No2politics]

Ok for a single die the total probability of a success is 2/5 when all the rethrowing if you get a 1 etc is taken into account. So for N dice it's 2N/5

Could you share the workings out?

[No2politics]

To the op, there isn't a general consensus to you are going to have to write code. I admit my number feels high. But when you are adding a series together infinitely perhaps it does give 0.5.

Lets say the winner is the person with the closest prediction to the code.

[doodab]

Could you share the workings out?

Each die will get rolled on average 6/5 times (1 + 1/6 + 1/36...) with probability of success 2/6.

[alluvial]

Ok for a single die the total probability of a success is 2/5 when all the rethrowing if you get a 1 etc is taken into account. So for N dice it's 2N/5

No it's not.

[doodab]

No it's not.

Feel free to explain why.

[alluvial]

Feel free to explain why.

I already have.

[doodab]

I already have.

No you haven't. Explain the flaw in my reasoning.