[–] [S] 2 points 1 year ago* (last edited 1 year ago) (1 child)

This makes sense. And I can see how to apply when you add more dice with different probability.

... however, when I did that math, I ended up with 87.04%

1-((1-.64) x (1-.4) x (1-.4)) = x

1-(.36 x .6 x .6) = x

1-.1296 = .8704

87.04% chance to get 2 winning dice

Hold up. Something seems off with this.

I tried calculating the probability of landing 2 losing side by using the same method and it doesn't add up

.2 x .2 = .04

.2 x .5 = .1

.2 x .5 = .1

1-((1-.04) x (1-.1) x (1-.1) = x

1-(.96 x .9 x .9) = x

1-.7776 = .2224

22.24% chance to have 2 losing dice.

But 87.04+22.24 = 109.28%

You would think that adding the winning combinations to the losing combinations would be 100%

What am I missing

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    I can't seem to wrap my head around how to calculate this, nor the reasoning behind it.

    Say you have 3 10-sided dice. 2 dice have 8 "winning" faces and 2 "losing" faces. So an 80% chance of winning. The third dice has 5 "winning" and 5 "losing" sides, so a 50% chance of winning.

    If you roll all three of these dice at the same time:

    -What is the probability of at least 1 dice winning.

    -What is the probability of at least 2 dice winning.

    -What is the probability of all 3 winning.

    From what I've seen I can calculate the probability of all 3 winning by doing:

    0.8 x 0.8 x 0.5=0.32

    A 32% chance of all 3 winning at the same time.

    But what about the others, I can't seem to figure it out.

    [–] 16 points 1 year ago* (2 children)

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