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 The Magic Cafe Forum Index » » Magical equations » » A math question for you... (0 Likes) NJJ Inner circle 6439 Posts  Posted: Apr 24, 2009 05:24 am 0 I have 64 numbered balls. The chances of picking the number 1 is 1 in 64. If I repeat the exercise 64 times (returning the selected ball after each selection) what are the chances of picking the number 1 at some point? rgranville Elite user Boston area 462 Posts  Posted: Apr 24, 2009 07:52 am 0 The probability of picking the desired number with one trial is 1/64, as you observed. The probability of NOT picking the number is 63/64 (which is 1 - 1/64). You'll see why we care about that. The probability of picking the desired number AT LEAST once in two trials is 1 minus the probability of NOT picking the number either time in the two trials. The probability of not picking the number in two trials is 63/64 * 63/64, which is 3969/4096. So the probability of getting the desired number in at least one of the trials is 1 - 3969/4096, which is 127/4096, about 0.03. If you play this game of two trials 100 times, you'll get the desired number at least once in about 3 games (on average). If you allow three trials, the probability of getting the desired number at least once is 1 - (63/64 * 63/64 * 63/64) = 1 - 250047/262144, about 0.046. The more trials you allow, the greater the probability that you will pick the desired number at least once. So when you say, "what are the chances of picking the number 1 at some point," it matters what that "some point" is. With 44 trials, the probability is about .5 - the odds are about even that you will or won't pick the desired number at least once. With 88 trials, the probability goes to about .75 - you WILL pick the desired number at least once 3 out of 4 times. If you allow 146 trials (and anyone's left awake to witness) the probability is about .9. :banana: stanalger Special user St. Louis, MO 996 Posts  Posted: Apr 24, 2009 07:53 am 0 Prob(picking 1 at some point) = 1 - Prob(never picking 1) =1-(63/64)^64 which approx. equals .635 Angelo the Magician Loyal user Vienna(Austria/Europe) 217 Posts  Posted: May 4, 2009 07:30 am 0 Antoher question to this problem: If you try, perhaps you are lucky and get the ball with the "1" early or if you are unlucky you need thousands trials. What is the average? How many trials do you need in average to get the ball with the "1". Solution: <- ruofytxis <- (in german it is called "geometrische Verteilung" which means: geometrical distribution) Angelo www.zauberschloss.at sean_mh Loyal user 229 Posts  Posted: May 5, 2009 09:53 am 0 Picking the number one at least once can be modelled by the situation "keep picking balls and returning them to the bag until you have chosen the '1' for the first time. This is an example of the geometric probability distribution. If X represents the number of trials (ball selections) until the first success (the number '1'), then P(X=x) = p*(1-p)^(x-1), where p is the probability of a success. In this example p= 1/64. So, for example, the probability that 5 trials are needed before seeing the first ball with a '1' is P(X=5) =(1/64)*(1-1/64)^4 = 0.015. The expected value of the geometric distribution is 1/p. So the expected number of trials before seeing a ball with a '1' on it is 1/(1/64) = 64. Sean The Magic Cafe Forum Index » » Magical equations » » A math question for you... (0 Likes)
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