Probability Question

[X Abstract X]

Hey guys, I need some help with a probability question.

- We have 4 10-sided dice
- Success level starts at 0
- If a 9 or 10 is rolled, ++success
- If a 1 is rolled, --success

We roll the 4 dice and after all 4 rolls, if success < 0, we get a "fail"

How can I calculate the probability of getting a "fail"?

Edit: I think I have an answer but I might be crazy. Can anyone confirm this?

P(fail) = 1/10 * 8/10 +
2/10 * 1/10 * 16/10 * 8/10 +
3/10 * 2/10 * 1/10 * 6/10 * 4/10 * 2/10 +
3/10 * 2/10 * 1/10 * 24/10 * 16/10 * 8/10 +
4/10 * 3/10 * 2/10 * 1/10 * 8/10 * 6/10 * 4/10 * 2/10 +
4/10 * 3/10 * 2/10 * 1/10 * 32/10 * 24/10 * 16/10 * 8/10

[Ginto8]

well for it to fail, 1 must be rolled at least 1 more time than 9 or 10, so we have these 2 possible cases:
-0*(9 or 10) and (1 or 2 or 3 or 4)*(1), where P = 343/1000*1/10+49/100*1/100+7/10*1/1000+1/10000
-1*(9 or 10) and (2 or 3)*(1), where P = 2/10*7/10*1/100+2/10*1/1000
so P(fail) = 343/1000*1/10+49/100*1/100+7/10*1/1000+1/10000+2/10*7/10*1/100+2/10*1/1000 = 0.0416

Someone should check my work on this one though

[N64vSNES]

well for it to fail, 1 must be rolled at least 1 more time than 9 or 10, so we have these 2 possible cases:
-0*(9 or 10) and (1 or 2 or 3 or 4)*(1), where P = 343/1000*1/10+49/100*1/100+7/10*1/1000+1/10000
-1*(9 or 10) and (2 or 3)*(1), where P = 2/10*7/10*1/100+2/10*1/1000
so P(fail) = 343/1000*1/10+49/100*1/100+7/10*1/1000+1/10000+2/10*7/10*1/100+2/10*1/1000 = 0.0416

Someone should check my work on this one though

....Fucking 12 year olds are way to smart these days

[eatcomics]

well for it to fail, 1 must be rolled at least 1 more time than 9 or 10, so we have these 2 possible cases:
-0*(9 or 10) and (1 or 2 or 3 or 4)*(1), where P = 343/1000*1/10+49/100*1/100+7/10*1/1000+1/10000
-1*(9 or 10) and (2 or 3)*(1), where P = 2/10*7/10*1/100+2/10*1/1000
so P(fail) = 343/1000*1/10+49/100*1/100+7/10*1/1000+1/10000+2/10*7/10*1/100+2/10*1/1000 = 0.0416

Someone should check my work on this one though

....Fucking 12 year olds are way to smart these days

Well you see... no they aren't. Ginto is just a straight up bad ass.

[bnpph]

Here is how I solved it - I am sure there is a better way.
I first found probability of success incrementing or decrementing.
Then I found chance of rolling 4 dice (for example, rolling 2 + and 2 -)
Then I found number of times the result could be found (for example, 3+ 1- can be: +++-, ++-+, +-++, -+++)
Then I added up the values for -1, -2, -3 , -4 to find probability.

Code: Select all

_ = 7/10 chance
+ = 2/10 chance
- = 1/10 chance

+4 = (+,+,+,+)  1 *   16/10000                                                      =    16/10000
+3 = (+,+,+,_)  4 *   56/10000                                                      =   224/10000
+2 = (+,+,_,_)  6 *  196/10000 + (+,+,+,-)  4 *   8/10000                           =  1208/10000
+1 = (+,_,_,_)  4 *  686/10000 + (+,+,-,_) 12 *  28/10000                           =  3080/10000
+0 = (_,_,_,_)  1 * 2401/10000 + (+,-,_,_) 12 *  98/10000 + (+,+,-,-) 6 *   4/10000 =  3601/10000
-1 = (-,_,_,_)  4 *  343/10000 + (-,-,+,_) 12 *  14/10000                           =  1540/10000
-2 = (-,-,_,_)  6 *   49/10000 + (-,-,-,+)  4 *   2/10000                           =   302/10000
-3 = (-,-,-,_)  4 *    7/10000                                                      =    28/10000
-4 = (-,-,-,-)  1 *    1/10000                                                      =     1/10000

(1540 + 302 + 28 + 1) / 10000 = 18.71% chance

So you will have 18.71% of success being < 0

I have tested this with a quick program, and got ~19%, and so I am sure this is the correct.

[Ginto8]

So you will have 18.71% of success being < 0

I have tested this with a quick program, and got ~19%, and so I am sure this is the correct.

Ah, yeah, I forgot to factor in the different combinations (ie ++-- and +-+- etc.).

[eatcomics]

So you will have 18.71% of success being < 0

I have tested this with a quick program, and got ~19%, and so I am sure this is the correct.

Ah, yeah, I forgot to factor in the different combinations (ie ++-- and +-+- etc.).

Its alright we still <3 you ginot :D