help me figure some percentages for a game for FS

[Murmur 3,941]

1 hour ago, blaird22 said:

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%

That's not correct. The event that one gets 18-20 on at least one roll, is the complement of the event that one gets 1-17 on both rolls. The latter event has (following your correct calculation for the first case) 17/20 * 17/20 = 289/400 probability, so its complementary event has 1-289/400=111/400=27.75% probability. It's actually a bit less than 30%.

[briansommers 74]

I'm using dice. I decided to go with 100 sided dice.

this of course is easy in a spreadsheet with =RANDBETWEEN(1,100)

Then I can set my target number in an easy percentage.

I use it to generate a customer job to be flown.

You get more dice the closer you move toward a larger airport representing more people.

The chance of getting a customer is 25% to begin with. Then as you gain more experience and become more known in the community your chance of getting work becomes easier.

This is represented by increasing your chance of getting a customer. So for every 10 jobs you do you percentage goes up by 1% and max at 80%

 

then in addition the more "money" I put in "advertising" that will also get me more dice, or a larger chance of getting a customer, but still not guaranteed. It's a gamey mechanic but works well and gives a very realistic feel.

[blaird22 35]

 

6 hours ago, Murmur said:

That's not correct. The event that one gets 18-20 on at least one roll, is the complement of the event that one gets 1-17 on both rolls. The latter event has (following your correct calculation for the first case) 17/20 * 17/20 = 289/400 probability, so its complementary event has 1-289/400=111/400=27.75% probability. It's actually a bit less than 30%.

You know what, I was actually thinking I was probably wrong about that part omw home from work. Thanks for the correction and explanation

[mabe54 17]

On 12/19/2017 at 0:10 PM, 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)?

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.

Yes

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.

Yes

Now, this is where I'm confused.

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

Yes

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

30%

Cheers,

[Murmur 3,941]

2 hours ago, mabe54 said:

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.

Yes

Now, this is where I'm confused.

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

Yes

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

30%

Uhm... No. See the example I made in the previous post. I have 50% chance of getting a head flipping a coin. But I don't have a 100% chance of getting at least a head flipping two coins.

 

[odourboy 1,289]

3 hours ago, mabe54 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.

Yes

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.

Yes

Now, this is where I'm confused.

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

Yes  NO  

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

30%

Cheers,

With 2 X 20 sided dice, there are 400 possible outcomes. 111 outcomes have an 18 or greater on one or BOTH dice. 111/400 = 27.75%

[mabe54 17]

With 2 X 20 sided dice, there are 400 possible outcomes. 111 outcomes have an 18 or greater on one or BOTH dice. 111/400 = 27.75%

Yeah right. You have 400 combinations but the OP just want one of any particular three in either dice. The two dices are independent from one another so his chances just double.

No cheers,

 

[odourboy 1,289]

Apparently one of us is not getting it, so let's simplify and go to a universe where a 3-side die can exist...

Say you 'win' if a 3 comes up but you lose if a 1 or a 2 comes up. Odds of winning with a single die is 1 in 3 (33.3%) - correct? So we add a second, now the odds of winning are now 2 in 3 (66.6%) because the dice are independent, right?

WRONG!

Let's enumerate all the outcomes and I'll '*' any winning combinations:

(1,1), (1,2), (1,3)*, (2,1), (2,2), (2,3)*, (3,1)*, (3,2)*, (3,3)*

Count the wins. 5 in 9 (55.5%), not 2 in 3. Extend this to the 20 sided dice and you'll see it's 27.75%, not 30%

This is my interpretation of the problem. I may have it wrong and if so, my apologies.

[briansommers 74]

Yep, that's crazy how that works, but that does prove it.