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Ask the Wizard #328
Ace2
The answer could be approximated as expressed 1  (prob(no 1's) + prob(no 2's) + ... + prob(no 6's)) = 1  6*(5/6)^20 = apx. 0.84349568.
However that would doublesubtract the situations where two different sides never got rolled. There are combin(6,2)=15 ways to choose two sides out of six. The probability that any two given sides never get rolled is (4/6)^20. We need to add those to the probability, because they got subtracted twice in the previous step. So, now we're at 1  6*(5/6)^20 + 15*(4/6)^20 = apx. 0.84800661.
This question is asked and discussed in my forum at Wizard of Vegas.
However, if any group of three sides that had never been rolled would have been triplesubtracted in the first step and tripleadded in the second step. We need to subtract them back out as a state where not all six sides were rolled. There are combin(6,3) = 20 ways to choose three sides out of six. The probability that any specific three sides are never rolled is (3/6)^20. So, now we're at 1  6*(5/6)^20 + 15*(4/6)^20  20*(3/6)^20= apx. 0.847987537.
However, if any group of four sides that had never been rolled would have been quadruplesubtracted in the first step, quadrupleadded in the second step, and quadruple subtracted in the third step. We need to add them back in, because each such state was already subtracted out twice. There are combin(6,4) = 15 ways to choose four sides out of six. The probability that any specific four sides are never rolled is (2/6)^20. So, now we're at 1  6*(5/6)^20 + 15*(4/6)^20  20*(3/6)^20 + 15*(2/6)^20 = apx. 0.84798754089.
However, if all 20 rolls were the same numbers, this situation would have been quintuplesubtracted in the first step, quintupleadded in the first step, quintuplesubtracted in the third step, and quintupleadded in the fourth step. We need to subtracted them back out. So, now we're at 1  6*(5/6)^20 + 15*(4/6)^20  20*(3/6)^20 + 15*(2/6)^20  6*(1/6)^20 = apx. 0.84798754089.
So the answer is 16*(5/6)^20+COMBIN(6,4)*(4/6)^20COMBIN(6,3)*(3/6)^20+COMBIN(6,2)*(2/6)^206*(1/6)^20 = apx. 0.84798754089.
A sign has ten light bulb sockets, each with a light bulb in it. Each socket takes a different size light bulb. Besides the light bulb already in each socket, there is one spare per socket. The life of each light bulb is distributed exponentially*, with a mean lifetime of one day. As soon as a light bulb dies, the spare will immediately replace it, if there still is a spare for that socket.
What is the expected time until the last light bulb burns out?
Ace2
Here is my solution (PDF).
This question is asked and discussed in my forum at Wizard of Vegas.
A casino dealer is working on a new Three Card Poker variant. She takes all the face cards from a standard deck and thoroughly shuffles them. She then deals 3 cards to Player #1, three cards to Player #2, three cards to Player #3, and the final three cards to Player #4. What is the probability that all four hands will contain a straight (JQK of any suit)?
Gialmere
The probability the first hand is AKQ is 1*(8/11)*(4/10) = 29.09%.
The probability the second hand is AKQ, given the first already is, equals 1*(6/8)*(3/7) = 32.14%.
The probability the third hand is AKQ, given the first and second already are, equals 1*(4/5)*(2/4) = 40.00%
The cards left must be AKQ, given the first three hands are. Thus, the probability is the product of the three above probabilities, which is 216/5775 = apx. 0.037402597.
This question is asked and discussed in my forum at Wizard of Vegas.
Lawrence
You can expect to be down 6000/22 = 272.73 bets.
The standard deviation of 6000 bets is sqrt(6000)*0.954545 = 73.93877.
Thus, you are 272.73/73.94 = 3.688556 standard deviations above expectations. Using the Gaussian curve, the probability of being up this many standard deviations or more is apx. 0.000112765 = apx. 1 in 8868.