to the power of
to the power of
It also means XOR.***(For binary and such)
Working as fast as I can on Fusion 3
From what you describe the boxes are randomly shuffled and the user gets to pick a box to see if a red ball is inside. Therefore the user has a 1% chance of picking the right box. On the second turn the boxes are randomly shuffled and the user picks a box to see if a red ball is inside. Again, since these are mutually exclusive events, in order words, the chance of picking the right box is the "same" each turn.
However if you are asking what is the chance of picking the correct box on the third turn then the answer is 99% * 99% * 1% = .9801%. In other words what is the chance of failing the first time, failing the second time, and successing the third time.
A simplier example...flipping a coin. Each try is mutually exclusive. So the chance of getting a heads on a single flip is 50%. On three tries it would be .5 * .5 * .5= .125 * 100 = 12.5%
Really what you are asking is what is the chance of getting the event of (tails, tails, heads) given three turns.
The chance that you will get the ball after x number of times would be the same as the chance that you will not get the ball at all subtracted from 100%.
(1 - 0.99 to the x power) * 100
40 tries will give you 33%.
You do not need to get the ball on the very last try; you could have it any of the 40 times you select a box. If, however, you want the program to tell you the chance you will succeed at the very last try, read the post above this.
40 tries will not give you 60% because the boxes are reset every time. 100 tries does not guarantee a red ball; nothing does.
This is what Im looking forOriginally Posted by Root
Are you able to explain this in an MMF event using Counter?? I have really no idea how to type in "to the power of" here, as ^ seems to make a synthax error.
Try the advanced math object. There you can raise a number to a power.
STeve
The calculation is not correct. Let's assume two turns. What is the chance of getting guess correct if you have 100 boxes and you can only select one each turn.
1% chance of success of picking the box "with" the ball
99% chance of picking the box "without" the ball.
Randomly shuffle the boxes.
What is the chance of getting the guess correct on this second try? (These are mutually exclusive events meaning it does not matter what happened on the first guess, it does not effect the second guess percentage of picking a box. Each time the user successes or fails.)
1% chance of success of picking the box with the ball
99% chance of success of picking a box without the ball
Now what is the chance the user guesses correctly on the second try? This is just another way of saying what is the chance of the user failing on the first try and succedes on the second try. Lets consider what possible things could happen:
Given two turns what are the possible outcomes:
(Fail, Fail) user fails both time to pick correct box
(Fail, Success) user fails on first try but successes on second
(Success, Fail) user picks correctly on first and fails on second
(Success, Success) user picks correctly both times
What is the chance of picking the ball on the second try. This means the user fails on the first try and successes on the second try. So it is represened by (Fail, Success).
Lets figure the possible chances of each possible outcome.
(Fail, Fail)
Here the user fails on both trys so percentage would be
(.99 * .99)*100 = 98.01%
(Fail, Success)
Here the user fails on the first try and successes on the second
(.99 * .01)*100 = .99%
(Success, Fail)
Here the user succedes on the first try and fails on the second
(.01 * .99)*100 = .99%
(Success, Success)
Here the user is successful on both trys of pick the right box
(.01 * .01)* 100 = .01%
These are all of the possible outcomes. So the total of them all has to add up to 100%.
.9801 + .0099 + .0099 +.0001 = 1 * 100 = 100%
So if the question is what is the chance on any turn of picking the correct box the answer is 1% because each turn you have always a 1% chance.
What is the chance of the user winning on the second turn, meaning they fail on the first turn and win on the second turn it is .99%.
I dont think this is exactly what I ment. What Im looking for is that you check a box X times... Lets say 10. My question then is, what are the chances, in total, of getting a ball out of these 10 attempts?? It does not have to be first, it does not have to be last, just one of those 10 attempts.
While the above analysis is correct, I'm not sure that it helps answer the original question. I think the best way of looking at it is to calculate the failure rate for each turn.
Assuming that there are ten boxes and one ball, the failure rate, the probability of not picking the ball would be 9 out of 10 = 90% or 0.9
Failure rate for turn two (AND turn 1) would be 0.9 x 0.9 = 0.81
Failure rate for turn three (AND turns 1 and 2) would be 0.9 x 0.9 x 0.9 = 0.729
So the failure rate for any number of turns n would be:
0.9 to the nth power
The success rate would be (1 - 0.9 to the nth power)
So the final formula to use to give the percentage "chance" of picking the ball for any n number of turns would be:
(1 - 0.9 to the nth power)x 100
Of course, you could substitute any other probability for 0.9, let's call it P based on the number of boxes and balls. The more general formula would be:
(1 - P to the nth power) x 100
It's amazing how such a simple question leads to such a complex answer.
Trying to directly calculate the success rate is more complex.
The success rate for turn 1 (ONLY) would be 0.1
The success rate for turn 2 (ONLY) would be 0.9 x 0.1
The success rate for turn 3 (ONLY) would be 0.9 x 0.9 x0.1
Then these would need to be added up, which would make a messy formula.
Steve
In MMF2 , you can just do...just like that. No need for any extension...Code:NumA pow NumB
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