help me figure some percentages for a game for FS

[briansommers]

I'm working on a financial game to be used with flight sim.

If I have a 20 sided die I have 1/20=5% chance of any one number coming up.

In order to be successful lets say I need to roll an 18 or higher, (18,19,20) 
if my math is correct I have a 15% chance of rolling that on one 20 sided die.

Now, this is where I'm confused.

If I use two dice, do I have 30% chance? 

then is that equivalent of rolling 1 die with a target number of >14 (15-20)?

[Bert Pieke]

Winning probability is calculated as favorable outcomes / total combinations.

So in your example:

Favorable outcomes:  120

Total combinations: 400

120/400 = 30%  I would think..

 

[BlueStar]

2 hours ago, briansommers said:

If I use two dice, do I have 30% chance? 

1.5 percent (6/400) based on the way the question was stated. 

blaustern

[Murmur]

4 hours ago, briansommers said:

I'm working on a financial game to be used with flight sim.

If I have a 20 sided die I have 1/20=5% chance of any one number coming up.

In order to be successful lets say I need to roll an 18 or higher, (18,19,20) 
if my math is correct I have a 15% chance of rolling that on one 20 sided die.

Now, this is where I'm confused.

If I use two dice, do I have 30% chance? 

then is that equivalent of rolling 1 die with a target number of >14 (15-20)?

 

The probability of rolling 18 or higher on at least one of the dices, is: 111/400=27.75%.

Infact the possible winning combinations are 3*17 (18, 19 or 20 on the first dice, and between 1 and 17 on the second one) + 3*17 (same as before but with swapped dices) + 3*3 (18, 19 or 20 on both dices) = 111.

 

[briansommers]

how are you figuring your numbers?

this is one 20 sided die

so I can generate a number from 1-20

so one number is 5%

there is only one single die.

[Bert Pieke]

then is that equivalent of rolling 1 die with a target number of >14 (15-20)?

That would be 6 / 20.. 30%

 

[Bert Pieke]

1 hour ago, Murmur said:

 

The probability of rolling 18 or higher on at least one of the dices, is: 111/400=27.75%.

Infact the possible winning combinations are 3*17 (18, 19 or 20 on the first dice, and between 1 and 17 on the second one) + 3*17 (same as before but with swapped dices) + 3*3 (18, 19 or 20 on both dices) = 111.

 

These things are always more complicated than they seem... but since the two dice are independent, is there a difference between rolling one die twice and rolling two dice side by side ? 

[Murmur]

37 minutes ago, briansommers said:

how are you figuring your numbers?

this is one 20 sided die

so I can generate a number from 1-20

so one number is 5%

there is only one single die.

In your first post, you mentioned two dices. In this case, the probability of having 18 or greater on at least one dice (so that means 18 or greater on the first dice, on the second dice, or on both) is not 30%, but 27.75%.

So that is not equivalent to rolling the same dice with a target >14. An equivalence would be rolling a 400-sided dice with a target >289 (infact from 290 to 400 there are 111 numbers).

 

19 minutes ago, Bert Pieke said:

These things are always more complicated than they seem... but since the two dice are independent, is there a difference between rolling one die twice and rolling two dice side by side ? 

No, they are equivalent of course. :-) The probability of having 18 or greater on at least one of the two rolls, is again 111/400=27.75%.

 

[Bert Pieke]

18 minutes ago, Murmur said:

No, they are equivalent of course. :-) The probability of having 18 or greater on at least one of the two rolls, is again 111/400=27.75%.

 

OK,  I roll once, the probability is 3/20.

I roll one more time, the probability is 3/20.

The combined probability is not 3/20 + 3/20? 

[Murmur]

1 minute ago, Bert Pieke said:

OK,  I roll once, the probability is 3/20.

I roll one more time, the probability is 3/20.

The combined probability is not 3/20 + 3/20? 

No it isn't. :-) A simpler example: if I flip a coin, the probability of getting head is 50%. If I flip it a second time, the probability is again 50%. But the probability of getting head in at least one of the two flips is not 50%+50%=100%, but it's 75%.

 

[Bert Pieke]

4 minutes ago, Murmur said:

No it isn't. :-) A simpler example: if I flip a coin, the probability of getting head is 50%. If I flip it a second time, the probability is again 50%. But the probability of getting head in at least one of the two flips is not 50%+50%=100%, but it's 75%.

 

That is a very convincing example

Reminds me of why I hated Statistics classes while studying Engineering.

We used slide rules, not fancy log tables etc..

[DaveCT2003]

2 hours ago, Murmur said:

No it isn't. :-) A simpler example: if I flip a coin, the probability of getting head is 50%. If I flip it a second time, the probability is again 50%. But the probability of getting head in at least one of the two flips is not 50%+50%=100%, but it's 75%.

 

Excellent.

 

For the OP.... There are a varying array of dice / probability calculators available online (Google is your friend).  But you don't have to use dice to figure what you're looking for - and there are both probability and scripts available online as well. 

Best wishes.

[mabe54]

It remains 1/20 = 0.05 for each face no matter what his preferences are because these are not part of the dice or its outcomes.

Cheers,

 

[BlueStar]

35 minutes ago, mabe54 said:

It remains 1/20 = 0.05 for each face no matter what his preferences are because these are not part of the dice or its outcomes.

Cheers,

 

If he rolls both at the same time then they are no longer independent and thus my 1.5%. 

blaustern

[blaird22]

Die (that's the singular of dice, correct?) 1 is 15%

2nd die is also 15%

So, if you want them BOTH to be 18-20, then it would be 15% of 15% (.15 X .15 = .0225, or 2.25%)

Or: 3/20 X 3/20=.0225 or 2.25%

Or, if you only need one, it would be 6 (total positive outcomes) / 20 (total possible digits since they're the same on each die) for 30%