Math / Probability Help

[jeffh]

15: 0% (since A wins ties with the higher bonus)

Note that the original poster said "all ties indicate failure". In the general case that's useless as it doesn't say who fails, but in the example given I assumed it would be the person using Move Silently who failed, as they were the one trying to actually do something.

I realize this is not the book rule. I was trying to answer the question exactly as it was asked.

This is the reason for the difference between our answers (note that the difference is exactly the chance of a tie).

 

[Thanee]

Note that the original poster said "all ties indicate failure".

Yeah, but that is wrong, therefore I ignored it (but added the explanation how ties are resolved in D&D).

(Editing a note in to make it a bit more clear.)

But you are right, that based on that (ties essentially going to B), it has to work slightly different (i.e. starting one higher at 15 not at 14), and the above written by you or Undead Lincoln is, of course, correct under the given assumption concerning ties (tho, on third glance, Undead Lincoln's formula still seems off by 1 point ).

Bye
Thanee

 

[Pbartender]

I think it's about time we had one of our semi-regular threads about probability, so here goes . . .

I need some assistance from our math experts, please. Is there a formula for establishing the probability of suceeding at an oppossed d20 roll if you know the bonus of each person rolling? For example, if PC A attempts to sneak past NPC B and PC A has a move silently bonus of +10 and NPC B has a listen bonus of +5, I'm assumming that there is a (relatively) simple formula whereby one could determine the likelihood of A's success -- I just don't know enough about math to know and/or devise it. If it affects the formula, assume that all ties indicate failure.

Thanks in advance for your help!

Alright. Here's the answer (modified for a dice game instead of a card game) I gave on my Thermostatistics final exam six years ago...

First, neither player's dice roll is dependant on the result of the other. This means that both players have equal probabilities of rolling any specific number one through twenty on the dice. Therefore the probability for success is based solely on the difference of the bonuses added to the dice rolls.

The mean roll for both players is 10.5. Since the base probability distribution for both players is identical, each player has a base 50% chance of beating the other (ignoring ties, for now).

Now a difference of +20 (1+20=21, better than 20) between the two players guarantees success for the primary roller, and a difference -20 (20-20=0, worse than 1) guarantees failure. So each +1 bonus (or -1 penalty) accounts for 2.5% of the overall probability.

The equation to determine the probability of the winner would look something like:

50% + ([Player A modifiers] - [Player B modifiers]) * 2.5% = Probability of Player A Success.

So using your example above,

50% + (+10 - +5) * 2.5% =
50% + (5) * 2.5% =
50% + 12.5% =
62.5% chance of success.


Now, ties get a little tricky.

Let's look at it another way...

Say both players have a bonus of +0, and one rolls a very average roll of 10 (you can't roll 10.5, so we'll round down, as is proper for D&D etiquette). The other player has 50% chance of rolling higher than 10 (success!), a 45% chance of rolling lower than 10 (failure!), and a 5% chance of rolling exactly 10 (tie!). Whoever is the beneficiary of a tie gains the extra 5% bonus to his chance of success.

What makes ties so tricky isn't so much the calculation, but that whoever gains that 5% bonus can be contingent on any number of parameters.

 

[Thanee]

Pb, that is not correct. I'm not entirely sure where the error lies, but there is one.

I think it might be the independance. The rolls themselves are, of course, independent in the sense, that one roll does not influence the other, but the result depends on the combination of both rolls.

The probability distribution is not linear, it's more akin to a bell curve, where the low differences are weighed higher than the high ones.

Bye
Thanee

 

[Pbartender]

I think it might be the independance. The rolls themselves are, of course, independent, but the result depends on the combination of both rolls.

The probability of success does not depend on the actual results of the rolls, only the distribution of the probabilities... Which are identical and independent for both rollers, since they are using two independent but identical dice. In such an instance (and ignoring any other modifiers to the score), the probability of success for either player is always 50%, with small adjustments for how ou treat a tie.

Trust me, I got full marks on that exam question.

The probability distribution is not linear, it's more akin to a bell curve, where the low differences are weighed higher than the high ones.

No it's not. You're looking at it wrong. You are not rolling 2d20, you are two 1d20's and comparing the results. You don't have one bell curve (2d20 wouldn't be a bell curve anyway, but that's beside the point), you have two linear curves that are identical. The real point is that the shape of the curves don't make a difference, so long as both players are using the same probability curve, and result of one playr's roll doesn't affect the result his opponent's roll.

 

[Thanee]

That's why I said "more akin to a bell curve".

The difference of probability between a +17 and a +18 difference of bonus is far lower than between a +1 and a +2 difference of bonus.

(EDIT: What I meant is, if you have D = A's Bonus - B's Bonus ~ the difference between the two bonuses, then the chance for A to win the opposed roll against B raises more between D=1 and D=2 than between D=17 and D=18. It would be the same, if you rolled only one d20 against a fixed DC, but you roll two d20 and compare the results, so the result depends on two dice rolls not just one, and therefore the probability of success is not linearily distributed.)

Anyways, I'm 100% sure, that your approach must be wrong.
There must be something different to your exam question then, if that one was right.

Bye
Thanee

 

[Pbartender]

The difference of probability between a +17 and a +18 difference of bonus is far lower than between a +1 and a +2 difference of bonus.

No. If you are rolling 1d20, it the difference between a +17 and +18 is the same as the difference between +1 and +2.

Plot out the curves of 1d10, 2d10 and 3d10 (set to "roll total == 0", then plot the probability curve) using the nifty dice roller Ferret found, and you'll see exactly what I'm talking about...

1d10 is a flat line... all numbers have the same chance of popping up.
2d10 looks line a pyramid... linearly increasing probabilities until you reach the mean, and then linearly decreasing probabilities.
3d10 and finally you have a bell curve.

Ok... So think about it this way...
One player rolls any set of dice. If he rolls that set of dice many, many times, there's an average score that emerges. That average score IS the 50% mark. If another player rolls the exact same set of dice many, many times, he will end up with the same pattern and the same average score. Every time the second player rolls those dice, he has a 50% chance of beating the average score of the first player. The converse is also true of the first player with respect to the second.

I don't know how else to explain it, but surely works that way.

 

[Voodoo]

Anyways, I'm 100% sure, that it's wrong.
There must be something different to your exam question then, if that one was right.

Sorry Thanee, but PB is right, well so ses my Maths and Statistics degree.

 

[Pbartender]

Sorry Thanee, but PB is right, well so ses my Maths and Statistics degree.

Good to know my Physics and Chemistry degree didn't steer me wrong.
I'm finishing up a 12 hour midnight shift, and for a second I though I was going loopy.

 

[Thanee]

You look only at 40 cases (each with a 2.5% probability), while there are 400 cases (20 * 20) each with a 0.25% probability (all such combinations of the two dice are equally probable).

Look at this matrix (A - A wins / B - B wins / T - Tie).
It contains every possible combination of the two d20 rolls.
Each entry has the same probability (1/400 or 0.25%).

B\A  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
 1   A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
 2   A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
 3   A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
 4   A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
 5   A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
 6   T  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
 7   B  T  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
 8   B  B  T  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
 9   B  B  B  T  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
10   B  B  B  B  T  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A
11   B  B  B  B  B  T  A  A  A  A  A  A  A  A  A  A  A  A  A  A
12   B  B  B  B  B  B  T  A  A  A  A  A  A  A  A  A  A  A  A  A
13   B  B  B  B  B  B  B  T  A  A  A  A  A  A  A  A  A  A  A  A
14   B  B  B  B  B  B  B  B  T  A  A  A  A  A  A  A  A  A  A  A
15   B  B  B  B  B  B  B  B  B  T  A  A  A  A  A  A  A  A  A  A
16   B  B  B  B  B  B  B  B  B  B  T  A  A  A  A  A  A  A  A  A
17   B  B  B  B  B  B  B  B  B  B  B  T  A  A  A  A  A  A  A  A
18   B  B  B  B  B  B  B  B  B  B  B  B  T  A  A  A  A  A  A  A
19   B  B  B  B  B  B  B  B  B  B  B  B  B  T  A  A  A  A  A  A
20   B  B  B  B  B  B  B  B  B  B  B  B  B  B  T  A  A  A  A  A

There are 280 A's, 105 B's and 15 T's. Each have a 0.25% chance of occurance.

There's a 70% chance that A wins, a 26.25% chance that B wins and a 3.75% chance of a tie.

If the ties go to B (as the original poster said), A has a 70% chance to win.
If the ties go to A (as D&D does it), A has a 73.75% chance to win.

Bye
Thanee