Specialty Dice Probability Question

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I’ve looked high and low to try to figure out the answer to this, but I’m not sure I’m knowledgeable enough in this area to do so.

Say you have 6-sided dice that are normal in every way except that the side that would be the number 6 says “roll again with additional die”. So if you rolled the sixth side, you would reroll it and roll another die as well. I’ve seen this phenomena called “exploding dice”.

My question is.

If you rolled 5 of these dice with the goal of getting a value of 4 or higher (a 4 or 5) on three of them, considering you might ultimately end up rolling more than 5 dice (if you the “roll again with additional die” side on any of them), what is the probability of this?

Is there a formula that expresses this that I can use for other scenarios?

Also, is there a formula that could explain the average roll of one of these dice? I know with a normal die the average roll is 3.5, but how about with this special die?

Thanks a million to anyone who’s able to help steer me in the right direction!

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@frostyO,

The only descriptions of exploding dice that I've seen have you count the 6, then re-roll it and add the new roll to it - repeating as necessary.

Are you saying that if you roll a 6, you then roll two dice and repeat as necessary? If so, do you count the original (and any subsequent) 6 in the total?

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@markr,

Yes, in this case there would be no "6", only a roll-over, plus the addition of another die in the new roll. It's a bit different than other exploding dice in that way.

So if you rolled a 4, you have one four.

But if you rolled the special side (the 6th side), you would reroll that die, not counting the original roll, and also roll a new die, and then end up with two dice rolled. (and it's possible that you might roll the special side again, and end up rolling 3+ dice in total)

Hope that makes sense.

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@frostyO,

So, at the end of this process (starting with 5 dice), you'll have 5 or more dice showing a number between 1 and 5. You want the probability that at least 3 of them are showing 4 or 5, correct?

For the expected value of one die, I get 3.75.

V = (5/6 * 3) + (1/6 * 2V)

2 Replies

@markr,

That's exactly right, thanks!

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@frostyO,

Second answer is in previous post.

The first problem is rather complicated. The probability of 3 or more depends on how many dice end up getting rolled. I can compute the probability of a given die resulting in 1, 2, 3, ..., n dice because of sixes that are rolled, but with five dice to start with, the combinations get out of hand rather quickly. The easiest way to deal with this is via simulation. After one billion trials, I get that the probability of 3 or more 4s and 5s is:
0.468255028

Does a bit under 50% coincide with your empirical observations?

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@markr,

Quote:

For the expected value of one die, I get 3.75.

V = (5/6 * 3) + (1/6 * 2V)

I have a problem with this equation because with an ordinary die the expected value is 3.5 but in this exploding scenario a 6 is never "scored"(unlike in more usual re-rolling games where points are scored on each roll and re-roll). No matter how many exploding sides are rolled the (eventual) results are always 1,2,3,4 or 5. The 1,2,3,4,5 are all equally likely so I'm thinking the expectation is 3.

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@nacredambition,

There's a 5/6 chance of stopping after the first roll. The expected value in that case is 3, as you point out.
However, there is a 1/6 chance that the first roll will be replaced by two rolls, and so on. Clearly, the expected value must be greater than 3.

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@markr,

Yes, I thought it would be under 50%, but what I'm really trying to do is find an equation that allows me to plug in any number of dice, with any number of sides, and any target numbers to hit and see the probability.

In the simplest example, if you're trying to roll a 5 with one die, then you have a 1/6 chance of rolling the '5' side, but you also have a 1/6 chance of rolling the side that explodes the dice, in which case you now have a 1/6 chance on each of two dice to hit the 5, but also two additional 1/6 chances to roll the exploding dice again. Is there an equation that can explain this more simple example perhaps?

I know simulation can provide the answers to specific questions, but I'm hoping to eventually use this info in a game that will have dice that act this way with 4, 8, 10, 12 and 20 sides as well, and to figure out the rules, I need to know how difficult hitting different target numbers will be.

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@frostyO,

I might be able to help you out with trying to roll at least one of one or more numbers with a single exploding die of any size. Let me think about that...

1 Reply

@markr,

Ok, that would be SUPER helpful! I really appreciate anything you can come up with.

1 Reply

@frostyO,

I think this will do it for you:

S = number of sides on the die (S >= E+N)
E = number of sides that cause the die to explode (E <= S)
N = number of sides that you're trying to roll one of (N <= S)
P = probability of getting at least one of the sides you're trying to roll

For E > 0:
P = [2*E - S + sqrt((2*E - S)^2 + 4*E*N)] / [2 * E]

For E = 0:
P = N / S

For your original example with a single die:
S = 6 because you're using a six-sided die
E = 1 because the number six is the only number that causes the die to explode
N = 2 because you're trying to roll a four or a five

1 Reply

@markr,

Thanks so much, Mark. That's SO helpful. I appreciate your expertise very much.

Best

J

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@frostyO,

If you're still looking for an answer to this, I think I can provide you with a program that will compute probabilities for multiple dice.

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Specialty Dice Probability Question

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