Probability question 1 die vs many

My memories from school seems to have disappeared (hey it was years ago) so I’d like some help on a basic probability problem.

Let’s say my opponent gets to roll one die (in the particular gaming problem I am trying to calculate the odds for it’s a D20 but you can assume a normal D6 if you wish) and I get to roll several dice (4 in this particular instance).

How do I calculate the odds of at least one of my rolls being higher than his one roll (or the probability that his one roll is better than all 4 of mine)?

What if he gets to roll one dice plus a certain amount, eq a D20+5 and I roll normal D20s?

The easiest way, conceptually, to approach this is to realize that “the odds of at least one of my rolls being higher than his one roll” is the exact opposite of “the odds of none of my rolls being higher than his one roll.” Therefore, the odds of the first are exactly one minus the odds of the second.

If you don’t know beforehand what your opponent’s roll is, then you have to figure out what the odds are for one of your rolls to beat his. You don’t say what happens in the case of a tie roll, but I’m going to assume, since a tie isn’t “higher,” that that’s not good enough. In this case, there are 400 possible combinations (20[sup]2[/sup]) of his D20 and your D20, of which 190 result in wins for you (20*19/2), and thus 210 result in not-wins.

If you have four dice, the probability of all of them being not-wins for you is then (210/400)[sup]4[/sup]., which is 0.07597. The probability of at least one of them being a win for you is one minus this, or 0.92403.

If you do know your opponent’s roll beforehand, the analysis is the same, but the caluculation of the probability of wins or not-wins with one roll is easier.

If your opponent rolls D20+5, that changes the probability of wins or not-wins with one roll. In this case, there are still 400 possible combinations (20[sup]2[/sup]), but here only 105 result in wins for you [(20-5)*(19-5)/2], and thus 295 result in not-wins. Continuing the calculation, if you have four dice, the probability of all of them being not-wins for you is then (295/400)[sup]4[/sup]., which is 0.2958, and the probability of at least one of them being a win for you is one minus this, or 0.7042.

Let’s say it’s a N-sided dice.

probability your friend rolls an x = P(X = x) = 1/N
probability you roll higher than x = P(X > x) = (N-x)/N
probability you roll lower than x = 1 - P(X > x) = x/N
probability you roll lower than x on all 4 rolls = (x/N)^4
probability you roll higher than x on at least one roll = 1 - (x/N)^4

So now for any x and N you can figure out the probability.

It’s been years since statistics for me too, but if I read your OP correctly you are not adding your 4 dice together, so we need not concern ourselves with a bell-curve distribution of the sums.

If his die is a 1 (1/20 chance), then each of your dice has a 19/20 chance to be higher.

If his die is a 2 (1/20 chance), then each of your dice has an 18/20 chance to be higher.
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If his die is a 19 (1/20 chance), then each of your dice has a 1/20 chance to be higher.

If his die is a 20 (1/20 chance), then each of your dice has a 0/20 chance to be higher.

Once you have done this for all possible single-die rolls, multiply each of the final probabilities resulting from his throw by 1/20, then add all of them together for the total probability that at least one of your dice will be higher than any throw of his. I get 0.475, or 9.5/20.

If I’m full of it, a more prob-oriented Doper will no doubt inform us. If I’m not, they will no doubt tell you of a much simpler formula that has been lost to my memory. But I think this works.

In the case where you add 5 to his result, it’s just a matter of accounting for that in the calculations.

If his total is a 6 (1+5), each of your dice has a 14/20 chance to exceed it, etc.

Anything he rolls over a 14 has no chance of being beaten.

I get 0.2625, or 5.25/20.

Yech! I knew there was some way to incorporate the presence of 4 dice into this that I was forgetting. Carterba has it right.

For each x, there is a 1-((x/20)^4) chance you will beat it.

Figure out the results for all 20 possible values of x, multiply each by 0.05, and add them together.

I get 0.774.

I think you’re all making this too difficult, there’s no need to invoke statistical distribution to answer this. When all five dice are rolled (one of his, four of yours), arrange them in ascending order. What are the odds that the one die of his is the highest? One in 5. What are the odds it ain’t? 4 in 5.

The only fine point is around ties. Note that your two interpretations of the problem deal with ties in different ways. [“How do I calculate the odds of at least one of my rolls being higher than his one roll” is not the same as “(or the probability that his one roll is better than all 4 of mine)?”]

So to make the math easy, assume a tie result in a do over. Then the odds are 80% you win, 20% he does. End of story.
If he gets a +5, then you do indeed need to get into the statistics.