"Probability spread question " Topic

Working on my travel cube based D&D game but need a probability question asked.

Mission – Find the relic

I have 6 locations it could be at. So would I have a better spread of finding it on which of these two systems.

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Roll D6
Roll at the first location find it on a 6+
Roll at the second location find it on a 5+
Roll at the third location find it on a 4+
Etc.
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or
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Roll 2d6 at each location and on a double you have found it.

I don't want it to always be the last place but also don't want it so probable it is the first place they look.

Any help or suggestions would be great!
Thanks
DZT

Along the "any help or suggestions" line…

Finding it on the first go (no matter the odds) may spoil the game.

Make the odds of finding the relic separate from successfully recovering it.

Increase the odds of successfully recovering it related to the number of clues/aides discovered -which would be in different locations, thus making it worthwhile extending the game even if its location were found early on (eg: "We know where it is, but our odds of getting it are only 1 in 6…let's look for some more clues/aides to help.").

Roll 2d6 at each location and on a double you have found it.

The probability of finding it using that mechanic is exactly the same as rolling 1D6 and finding it on a 6. Your first suggestion seems much better:

Roll D6
Roll at the first location find it on a 6+
Roll at the second location find it on a 5+
Roll at the third location find it on a 4+
Etc.

Obviously the 2nd has the greater spread of probability because it includes not finding it at all wheras using the first it is certain to be found within 6 locations.

I like the first system better. There is always a chance you will find the dingus in the first room but not very good. It might be in the second room. It is much more likely in the third.

The probability of finding it in the first room is 1/6.
The probability of finding it in the second room is 2/6*5/6 (the 5/6 is the chance you will not find it in the first room).

The third room would be 3/6*4/6*5/6

And so forth.

With that system, you WILL find the dingus by the last room…which is the 6th room. You may find it in room's 1 through 5.

If you want a bigger dungeon, use a bigger die and do the same thing, adding one to the chance each time.

John

The first thing is what Gildas Facit said. Your first method ends at 100% probability on the sixth trial; the second method converges, but never actually reaches 100%.

The second thing is, as Major Bumsore says, rolling doubles on 2d6 (6/36) is the same probability as rolling any given number on a d6 (1/6), barring some reroll mechanic in your system that affects single dice. Also, rolling doubles has a psychological effect that is different that rolling a "6" on a d6; I think that even affects people who know the probabilities cold.

The probability of finding the thing by (not on) the nth location is:

Progressive : .16, .44, .72, .90, .98, 1.00
Doubles:      .16, .30, .42, .51, .59, .66, .72, .76, .80, …

So, it depends on what you are looking for. Even though the doubles technically never ends (though is easy enough to artificially terminate by not rolling for the last one), it grows a little slower and allows for more than six locations.

A real good way to do something like this, if you are not wedded to dice, is with a set of playing cards. Five clubs and the ace of spades (your target item) gives you a similar probability distro as the progressive method. And it has a "feel" that is more like the reality of the situation – the objective is already "out there" when you start and you're trying to find it.

The double isn't good as the odds never improve from 1 in 6. For 5 out of 6 games they won't find anything at all. How about, using two dice:
1st attempt – 12, 1 in 36 chance
2 nd- 10, 11, 12, 1 in 6 chance
3rd, 8 or above, 5 in 12, a little under 50%
4th – 6 Or above, 7 in 12, a little over 50%
5th – 5 or above, 25 in 36, about 70%
6th – automatic
Due to the bell curve nature of the probabilities, most games would result in result 3 or 4, with a reasonable chance of a 5.

Why not just pick chits out of a hat?

I like etotheipi's card idea.
It's also easy to modify for size or numbers of Items.

A little graphics work, and they don't need to be plain playing cards either.

Since this is sampling without replacement, shouldn't the denominator be the number of sites left? If it's not in room 1, there are five remaining rooms, so it's 1/5. Oooh, we can do a probability tree. I loved those in grad school.

Since this is sampling without replacement, shouldn't the denominator be the number of sites left?

Good point, you are absolutely right. Assuming of course that the relic is actually somewhere in this dungeon in the first place …

Oh, Major, you are truly an evil genius!

If it's not in room 1, there are five remaining rooms, so it's 1/5. Oooh, we can do a probability tree. I loved those in grad school.

Yeah, but then you need a d5, d3, and d2. I have those (including a pipped d10 that became pipped d5!), but most people don't. I think he's looking for a simple progression that approximates the {.16, .33, .50, .67, .83, 1.0} progression. Or maybe one weighted a little more to the right, so the discovery happens generally a little later.

A little graphics work, and they don't need to be plain playing cards either.

I love sanding down and drawing on playing cards, or just drawing over them (… or buying blanks, if you want to cheat!). With printers, microperf business card sheets, and word processor templates, it's pretty easy to make a decent looking card set.

If its a simple 16% increment the OP is looking for, then it's very easy. Turn 1, you need a 1, turn 2, 2 or less, turn 3, 3 or less etc. I think he wants it more weighted to the latter portion though to make the game last longer. I'm struggling to see where the probability of rolling a double (or any combination) increases depending on the previous rolls. The dice have no memory. If this were the case, the Martingdale system in a casino would be a guaranteed winner and Vegas would go bust. Every roll is a separate event.

Roll 2d6 at each location and on a double you have found it.

Uh, if the entire mission is to find the relic, and they only find it on a set of dice rolls, you obviously have a problem because the dice roll determines the winner--and that could happen on any turn. So, now you want the dice rolls to have less chance of rolling the first or early turns. What's the point of playing the game? It's just a set of dice rolls.

You might want to rethink what mechanisms 'discover' the relic.

Since this is sampling without replacement, shouldn't the denominator be the number of sites left? If it's not in room 1, there are five remaining rooms, so it's 1/5. Oooh, we can do a probability tree. I loved those in grad school.

Outstanding Glenn! +1 to you!

I'm struggling to see where the probability of rolling a double (or any combination) increases depending on the previous rolls.

It doesn't. The people above are talking about the cumulative probability function not the instantaneous probability weight.