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Help with a multiple simultaneous probability problem. (lemm.ee)

submitted 4 days ago* (last edited 4 days ago) by Chefdano3@lemm.ee to c/askmath

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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.

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[–] owenfromcanada@lemmy.ca 2 points 4 days ago* (last edited 4 days ago) (1 children)

For at least one winning, we can calculate the probability of all three losing, then take the inverse.

So for at least one die winning (that is, not all three losing), we have:
1 - ( (1 - 0.8) * (1 - 0.8) * (1 - 0.5) ) = 1 - 0.02 = 98%

The probability of at least two winning is harder, but we can do it by combining multiple possibilities.

The probability of dice 1 and 2 winning is:
0.8 * 0.8 = 64%

The probability of dice 1 and 3 winning is:
0.5 * 0.8 = 40%

The probability of dice 2 and 3 winning is:
0.5 * 0.8 = 40%

Similar to before, we can calculate the probability of none of those things happening, then take the inverse:
1 - ( (1 - 0.64) * (1 - 0.4) * (1 - 0.4) ) = ~~76.96%~~ 87.04%

[–] Chefdano3@lemm.ee 2 points 4 days ago* (last edited 4 days ago) (1 children)

~~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

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

[–] owenfromcanada@lemmy.ca 2 points 4 days ago

You're right, I flubbed my calculation. The probability of at least two winning is 87.04%.

For your other question, the key words are at least. The probability of at least one die winning, which includes the probability of more dice also winning (two or three).

I'm on mobile so I can't do it now, but if you're interested you could create a table of every possibility and their respective probabilities. That table should add up to exactly 100%, and you can add up all the valid combinations for a given criteria to verify the shorter math we did before.