Dice roll probability

[alluvial]

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

See my previous post and extrapolate.
If you could explain your reasoning, then we can see whose reasoning is flawed.

[doodab]

See my previous post and extrapolate.
If you could explain your reasoning, then we can see whose reasoning is flawed.

You fail to take into account that rolling 1 is a success regardless of what you roll on the reroll.

You have a 2/6 chance of a success on the first roll, plus a 1/6 chance of a second roll with 2/6 chance of success, plus a 1/36 chance of a third roll with 2/6 chance of success, and so on. Write that out then factor out the 2/6 common term, you get

2/6 x ( 1 + 1/6 + 1/36 + ...) = 2/6 x 6/5 = 2/5

[Ticktock]

Enjoy your game MUN.

You've certainly kept one lot of geeks occupied for an evening.

[alluvial]

You fail to take into account that rolling 1 is a success regardless of what you roll on the reroll.

You have a 2/6 chance of a success on the first roll, plus a 1/6 chance of a second roll with 2/6 chance of success, plus a 1/36 chance of a third roll with 2/6 chance of success, and so on. Write that out then factor out the 2/6 common term, you get

2/6 x ( 1 + 1/6 + 1/36 + ...) = 2/6 x 6/5 = 2/5

You neglect to consider that rolling a 2 will prevent any further rolls. You have a 2/6 chance of getting a result with the first roll but you have to consider that a 1 will cause a further roll and then you have to consider the possible outcomes of the subsequent rolls.

What I gave was the probability of getting one success only which can be achieved in two ways, rolling a 2 with the first roll followed by a roll that gives not 2 and not 1.

If you roll a 1 on the first roll, then you roll again and then the probability of only getting one success is then:
1/6 + (1/6 * 4/6) which is the probability of getting the 1 plus the probability of not getting a 1 or 2.

To get two successes you have to first roll a one followed by a two (as the rolls then stop) or a 1 then a 1 then not 1 or 2, and the probability then:

Probability of first 1 is 1/6
Probability of second 1 is dependant on the first roll and is then 1/6 * 1/6
Probability of 2 on the second roll is 1/6 * 1/6

But rolling the second one causes a third roll and to only have the two successes, then this must not be a 1 or 2. The probability of this is 1/6 * 1/6 * 4/6

So the probability of only two successes is then the prob of the sequences: 1-2 and 1-1-(not 1 and not 2)

P(1st seq.) = 1/6 + (1/6 * 1/6)
P(2nd seq.) = 1/6 + (1/6 * 1/6) +((1/6+(1/6 * 1/6)) * 4/6)

So prob of two successes is: the sum of these.

To get three success you will need the following sequences:

1-1-1-(not 1 and not 2) or 1-1-2 and I can't be arsed to work that out.

[doodab]

You neglect to consider that rolling a 2 will prevent any further rolls. You have a 2/6 chance of getting a result with the first roll but you have to consider that a 1 will cause a further roll and then you have to consider the possible outcomes of the subsequent rolls
.

This is all taken into account. Only rolling a 1 results in a further roll. That is where the series 1 + 1/6 + 1/36 + ... comes from.

What I gave was the probability of getting one success only which can be achieved in two ways, rolling a 2 with the first roll followed by a roll that gives not 2 and not 1.

So why keep telling me my formula for the average number of successes per die is wrong then? Especially when you keep saying you can't be arsed to work out the answer properly.

I'll try and sort the answer to the odds problem out in the morning, I'm a bit tired now, been a long day.

[EternalOptimist]

For all practical purposes, its dumb to consider an infinite number of throws. It's possible, but gamers dont think that way, and they certainly dont play that way.

They will consider a number of 'rounds' or 'throws' before the probability of a further 'round' becomes remote.
This number of 'rounds' will be the same, or similar, no matter how many dice are involved

if you consider a remote cut off of one in a hundred.

the gamer takes six dice = two hits (1,2)
second round, one dice = 33% chance of a hit 16% chance of another round
third round, one dice = 5% chance of a hit 3% chance of another round
last round, one dice = 1 % chance of a hit, no chance of another round. too remote

six dice = 2.4 hits



the gamer takes twelve dice = four hits (1,1,2,2)
second round, two dice = 66% chance of a hit 33% chance of another round
third round, one.five dice = 7.5% chance of a hit 5% chance of another round
last round, one dice = 1 % chance of a hit, no chance of another round. too remote

twelve dice = 4.7 hits

so the number of 'rounds' is the practical limiting factor. number of hits on six sided dice is about

say, 2.5 for every six cubes rolled.


if you want to get a more forensic answer, you would have to build a recursive model and average the answer over a number of runs

[NotAllThere]

With one dice. The probability of scoring certain number of points is:
0 => 3-6 => 4/6
1 => [2,1+3-6)] ( rolling a 2, or 1 and 3-6) => 1/6 + 1/6x4/6 = 5/18
2 => 1,[2,1+3-6)] ( rolling a 1 and then a (2 or 1 and 3-6) => 1/6 x (5/18).
3 => 1,1,[2,1+3-6)] = 1/(6^2) x 5/18
...
n => 1/(6^n) x 5/18

If you have more than one dice, your probabilities get very complicated due to combinations. I'll think about that later.

[No2politics]

You fail to take into account that rolling 1 is a success regardless of what you roll on the reroll.

You have a 2/6 chance of a success on the first roll, plus a 1/6 chance of a second roll with 2/6 chance of success, plus a 1/36 chance of a third roll with 2/6 chance of success, and so on. Write that out then factor out the 2/6 common term, you get

2/6 x ( 1 + 1/6 + 1/36 + ...) = 2/6 x 6/5 = 2/5

I agree with doodab.

I had mistakenly assumed that if you roll a 2 you get another go, but you only get another go when you roll a 1.

[No2politics]

You neglect to consider that rolling a 2 will prevent any further rolls. You have a 2/6 chance of getting a result with the first roll but you have to consider that a 1 will cause a further roll and then you have to consider the possible outcomes of the subsequent rolls.

What I gave was the probability of getting one success only which can be achieved in two ways, rolling a 2 with the first roll followed by a roll that gives not 2 and not 1.

If you roll a 1 on the first roll, then you roll again and then the probability of only getting one success is then:
1/6 + (1/6 * 4/6) which is the probability of getting the 1 plus the probability of not getting a 1 or 2.

To get two successes you have to first roll a one followed by a two (as the rolls then stop) or a 1 then a 1 then not 1 or 2, and the probability then:

Probability of first 1 is 1/6
Probability of second 1 is dependant on the first roll and is then 1/6 * 1/6
Probability of 2 on the second roll is 1/6 * 1/6

But rolling the second one causes a third roll and to only have the two successes, then this must not be a 1 or 2. The probability of this is 1/6 * 1/6 * 4/6

So the probability of only two successes is then the prob of the sequences: 1-2 and 1-1-(not 1 and not 2)

P(1st seq.) = 1/6 + (1/6 * 1/6)
P(2nd seq.) = 1/6 + (1/6 * 1/6) +((1/6+(1/6 * 1/6)) * 4/6)

So prob of two successes is: the sum of these.

To get three success you will need the following sequences:

1-1-1-(not 1 and not 2) or 1-1-2 and I can't be arsed to work that out.

The question asks you to work out the expected number of successes. What you are doing is working out the probability of 1 success, or 2 successes. Do you plan on doing this literally for ever and adding up the answer!?

[No2politics]

With one dice. The probability of scoring certain number of points is:
0 => 3-6 => 4/6
1 => [2,1+3-6)] ( rolling a 2, or 1 and 3-6) => 1/6 + 1/6x4/6 = 5/18
2 => 1,[2,1+3-6)] ( rolling a 1 and then a (2 or 1 and 3-6) => 1/6 x (5/18).
3 => 1,1,[2,1+3-6)] = 1/(6^2) x 5/18
...
n => 1/(6^n) x 5/18

If you have more than one dice, your probabilities get very complicated due to combinations. I'll think about that later.

The approach of writing down the probabilities for each n is not the way to go. This is because there are infinitely many possible values for n, as you could keep rolling a 1 for ever.

You need to write it as a geometric series like what doodab did.

Having more than one die doesn't complicate things. If you can work out the expected number of successes if you have one die (which is 0.4), the if you have two die you simply double it.