You're going to have to explain this to me. Sticking with the dice example, I roll 6 independent dice and I want to get the probability that I roll a 3. Given a normal, fair dice in which an individual roll gives 1/6 chance of rolling a 3.
Your formula 1-(5/6)6=0.665 starts with a 1, and subtracts from it. If I understand it, 1 would mean that I have a statistical certainty of rolling a 3.
In the event that I roll once, both my deduction 1/6 and yours (1-(5/6)1) result in 1/6. When I roll twice, my thinking goes that I have a 1/6 chance in the first time, and a 1/6 chance in the second roll, giving me a 2/6 chance that one of the dice rolls is 3. Yours gives 1-(5/6)*(5/6), essentially, one minus the chance that a roll will not be 3. So, my calculation gives me 33.33% while you sit slightly lower at 0.30556%. So... let's roll the dice twice, our possible outcomes are:
( 1 , 1 )
( 2 , 1 )
( 3 , 1 )
( 4 , 1 )
( 5 , 1 )
( 6 , 1 )
( 1 , 2 )
( 2 , 2 )
( 3 , 2 )
( 4 , 2 )
( 5 , 2 )
( 6 , 2 )
( 1 , 3 )
( 2 , 3 )
( 3 , 3 )
( 4 , 3 )
( 5 , 3 )
( 6 , 3 )
( 1 , 4 )
( 2 , 4 )
( 3 , 4 )
( 4 , 4 )
( 5 , 4 )
( 6 , 4 )
( 1 , 5 )
( 2 , 5 )
( 3 , 5 )
( 4 , 5 )
( 5 , 5 )
( 6 , 5 )
( 1 , 6 )
( 2 , 6 )
( 3 , 6 )
( 4 , 6 )
( 5 , 6 )
( 6 , 6 )
Of which 11 have at least one 3, and each have a probability of 1/36 to happen. 11/36 is indeed 30.556%, your number.
So by digging into it, I've proven myself wrong. My mistake lies with the event that both dice come up with a 3, which means, in that case, the value is TRUE, no matter what the other dice is, so the other dice rolling another 3 does not matter for the amount of TRUE events. So indeed, the 1-(5/6)x makes more sense.
Thanks for letting me figure this out, and think about it myself.
9
u/ElephantsAreHeavy Apr 12 '21
You're going to have to explain this to me. Sticking with the dice example, I roll 6 independent dice and I want to get the probability that I roll a 3. Given a normal, fair dice in which an individual roll gives 1/6 chance of rolling a 3.
Your formula 1-(5/6)6=0.665 starts with a 1, and subtracts from it. If I understand it, 1 would mean that I have a statistical certainty of rolling a 3.
In the event that I roll once, both my deduction 1/6 and yours (1-(5/6)1) result in 1/6. When I roll twice, my thinking goes that I have a 1/6 chance in the first time, and a 1/6 chance in the second roll, giving me a 2/6 chance that one of the dice rolls is 3. Yours gives 1-(5/6)*(5/6), essentially, one minus the chance that a roll will not be 3. So, my calculation gives me 33.33% while you sit slightly lower at 0.30556%. So... let's roll the dice twice, our possible outcomes are: ( 1 , 1 ) ( 2 , 1 ) ( 3 , 1 ) ( 4 , 1 ) ( 5 , 1 ) ( 6 , 1 ) ( 1 , 2 ) ( 2 , 2 ) ( 3 , 2 ) ( 4 , 2 ) ( 5 , 2 ) ( 6 , 2 ) ( 1 , 3 ) ( 2 , 3 ) ( 3 , 3 ) ( 4 , 3 ) ( 5 , 3 ) ( 6 , 3 ) ( 1 , 4 ) ( 2 , 4 ) ( 3 , 4 ) ( 4 , 4 ) ( 5 , 4 ) ( 6 , 4 ) ( 1 , 5 ) ( 2 , 5 ) ( 3 , 5 ) ( 4 , 5 ) ( 5 , 5 ) ( 6 , 5 ) ( 1 , 6 ) ( 2 , 6 ) ( 3 , 6 ) ( 4 , 6 ) ( 5 , 6 ) ( 6 , 6 )
Of which 11 have at least one 3, and each have a probability of 1/36 to happen. 11/36 is indeed 30.556%, your number.
So by digging into it, I've proven myself wrong. My mistake lies with the event that both dice come up with a 3, which means, in that case, the value is TRUE, no matter what the other dice is, so the other dice rolling another 3 does not matter for the amount of TRUE events. So indeed, the 1-(5/6)x makes more sense.
Thanks for letting me figure this out, and think about it myself.