How to calculate EV of a dice? (Layman's explanation)

Factual Questions
1

The opening paragraphs of this article give an excellent layman’s explanation on how to calculate the EV of a standard dice, and then gives another example of how to calculate EV for a weighted dice.

Let’s say I’m playing some sort of game where I’m using a standard dice, where each number has a 1/6 chance of appearing. However, in this game, any time I roll a 1, 2 or 3, it doesn’t count, and I get to roll again (in other words, only rolls of 4, 5, 6 “count” in this particular game).

I know the EV of my dice for this game, when the rolls count, is 5, however, what’s the formula to calculate that? I’m looking for a formula that closely mimics the formula in my link above.

I’m not a mathematician, please keep explanations in layman’s terms.

Thanks!

2

In layman’s terms the only values that count are 4, 5 and 6, and they come up 1/3 of the time.
41/3 + 51/3 + 6*1/3 = 15/3 = 5

To include the rerolls in your formula you’d have to do something like this:

The EV of a reroll is the same as the EV of any roll, and you’ll have a reroll half the time:

EV = EV1/2 + 41/6 + 51/6 + 61/6

EV1/2 = 41/6 + 51/6 + 61/6

EV = 5

3

Thanks naita,

I think this is something close to what I’m looking for.

In the dice example, it’s quite easy to “recalculate” that 4/5/6 will only come up 1/3 of the time in my game, so adjusting the numbers is easy.

But what about a situation where a simple readjustment isn’t so easy? Eg, a 64 sided dice numbered 1-64, with weighted biases on certain numbers? How would you do it then?

4

Ahh ok, I think I see what you’ve done here. You’ve used the same method described in my link, but you’ve then taken the result and divided it by the fraction of rolls that are getting ignored/rerolled, which in this case, is 1/2.

2.5 / (1/2) = 5.

Is that right?

5

The EV will be the sum, over all possible results, of that result’s value times the proportion of times that result comes up.

In your case, there are four possible results: ultimately getting a 4, ultimately getting a 5, ultimately getting a 6, and rerolling forever.

It’s clear that 4, 5, and 6 come up the same proportion of the time as each other. [And were your die weighted, the proportion of the time they came up would be in proportion to those weights].

If you can deduce or stipulate that rerolling forever occurs only a negligible proportion of the time, then this means 4, 5, and 6 each have proportion 1/3 of coming up, and the EV is 1/3 * 4 + 1/3 * 5 + 1/3 * 6 = 5. If you can’t conclude that rerolling forever occurs only a negligible proportion of the time, then you’re in trouble, because you haven’t even defined what the resulting value should be in that case, so the EV will not be well defined.

Of course, standardly, you can conclude that rerolling forever occurs only a negligible proportion of the time, by the reasoning that to reroll forever requires first rolling a 1, 2, or 3 (as happens on half of die rolls), then rerolling forever again after that, so that if p is the proportion of the time one rerolls forever, we have p = 1/2 * p, and thus p = 0.

6

Well that depends on how they are weighted.

Yes, except I started by justifying that division in a more general sense. Let’s stick with a six sided die and use the bias from your link. The half in the version of my formula you quote is the sum of the probabilities of rolling 4, 5 or 6, so with the 1-5 being equally likely and 6 coming up half the time the expected value becomes:

EV = (41/10+51/10+6*1/2)/(1/10+1/10+1/2) = 39/10 / 7/10 = 39/7 = 5.57

7

I think I’ve got it. Thank you.

8

For future reference – one is a die, more than one are dice.

9

However much my inner pedant agrees with you, dictionaries now list dice as the singular as well as plural.

10

Yeah, the singular usage grates on me, too, but it’s considered standard nowadays.

11

That still doesn’t make it ‘right’.

…but I fear we’re fighting a losing battle on this one…

12

It actually does. Like I said, it sounds fricking weird to my ears, but it’s become standard so far as I could tell, and so far as the dictionaries tell me. I’m fine with that.