# calling all math nerds...

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Okay, this happened on Friday night:

I'm fighting a monster with 4 hit points.  I get to roll 6 dice, and I have an item that gives me the "add one to the result of one die" ability in effect, as well as another item that allows me to "reroll as many dice as I want on one test per round" in effect.

I roll 2 successes, a 4, and a bunch of misses.  Essentially a very average result on 6 dice for a non-blessed person.  That's when it get's tricky.

A) My immediate reaction was to add a pip to the 4, making it a hit, then reroll the other three dice.

B) My friend told me I should reroll all 4 of the misses and save the added pip for after the reroll.

So, assuming all I care about is defeating this monster (i.e. doing at least 2 more wounds to it on this encounter) and further assuming that saving up the reroll for a different test in a later part of the turn isn't required, which strategy above (A or B) gives me the greatest chance of killing the monster?

Strategy A is an easy enough math problem.  You have 3 hits in your pocket and need one more, the odds of failing to hit one more time on 3 dice is 8/27, meaning that the odds of success my way are 19/27, or roughly 70%.

For the life of me I cannot figure out how to correctly count it up my friend's way (B).  Every time I do, I get a slightly different result.  That said, his way usually wins.  But then that's not reliable given that I might be adding it wrong every way I've tried so far.

Help?

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I don't know how the math works out exactly but my instinct is for method A. If you end up needing to use the pip on the reroll, then why not just have done it in the first place? Seems like option B just gives you more opportunities to fail the rolls. Also, you said 4 hit points(toughness) but not how much damage you are trying to prevent. Since each hit prevents damage as well as dealing it.

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If my math is correct so is rolling all 4 gives 55% of getting 2 success with 0-1 pip.

(1/3)^4+4*(1/3)^3*1/6+4*(1/3)^3*1/2+12*(1/3)^2*1/6*1/2+6*(1/3)^2*(1/2)^2+12*1/3*1/6*(1/2)^2

4 without pip + 4 with 1 pip + 3 without pip + 3 with 1 pip + 2 without pip + 2 with 1 pip

Edit: Which seems logical to me. To reroll a success can't be good.

Edit2: Realised that I had forgotten the instances with more than 1 pip rolled.

So above + 2 with 2 pips (6*(1/3)^2*(1/6)^2) + 1 with 2 pips (12*(1/3)*(1/6)^2*(1/2)) + 1 with 3 pips (4*(1/6)^3*(1/3)) which gives 63.58% and since I did a backwards check also I am quite confident about this number which don't fit with the other calculation here.

Edited by StLemon

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ArkhamCalc app shows that you have a 70.4% chance for success in Option 1, and a 68.8% chance in Option 2.

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My friend whom I mentioned in the OP told me he ran a simulation on his computer, rolled 4 dice 1 million times and came up with something like 65% chance of success using option B.  I still can't figure out where my math is wrong though.

For the record, I was playing as Jenny Barnes, I was fighting a Shoggoth sitting on the gate that had just spawned it, and wanted to close the gate, so I had to defeat the Shoggoth to do that.  At the time I had the Flamethrower, the Lucky Cigarette Case, the Camera, the Magnifying Glass, and one clue.  I think I forgot about Jenny's gun that she comes with (the .25 Automatic).  That would have given me another die, actually.