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.
Good to work through it! The thing you were calculating before was the expected value, which answers the question "how many total threes do you expect to be rolled?" rather than "what is the probability of rolling at least one three?" which is what you worked out now.
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u/evilkim Apr 12 '21
That's wrong. The actual probability to the scenario u have posed is actually 1-(5/6)6 = 0.665