In many state lotteries you can pay 50 cents to buy a single lottery ticket consisting of 6 two-digit numbers between 01 and 60, inclusive. Based on the results of a random drawing, the ticket may win any of the following prizes (including no prize), with the probabilities indicated.
X $0 $3 $30 $1000 Jackpot
P(X) 0.9623 0.0357 0.00194 0.0000369 0.000000143
Your chance of winning $3 is about 3.6%. Your chance of winning the jackpot is close to nonexistent. Your chance of winning nothing is overwhelming. Suppose that the jackpot is worth $1,000,000 this week. What is the expected value of the ticket? (That is, what is the mean of the discrete random variable X?) The result of your calculation will be a decimal number. Round this to the nearest whole number of cents, and give an answer that is a whole number. Do not include units or money symbols in your answer.
+330 Alan’s valuation for the probabilities listed above are correct.
However, the probabilities listed do not correlate to the values of the R of N lottery described. There is no value of N where R=6 that distributes the probabilities depicted above.
The probabilities for R of N, and K of R (where K is a sub set of R) lotteries are calculated by hypergeometric distributions and the probabilities are discreet. The general formula is
$$P(k)=\frac{\binom{R}{K}*\binom{N-R}{R-K}}
{\binom{N}{R}}$$
Where N is the pool of numbers, R is the quantity drawn, and K is the quantity correctly matching R. The special case where K=R P(k) is equal to 1/(C(N,R)).
The hypergeometric discreet probability distributions are:
For N =70 and R =6 and K = 6 to 0
K=6: 0.0000000199744886
K=5: 0.0000064717343010
K=4: 0.0004287523974380
K=3: 0.0099089442963447
K=2: 0.0947542798337962 ----->
K=1: 0.3790171193351850 ----->
K=0: 0.5158844124284460 -----> 0.989655811597427 cumulative probability for “loosing tickets” (k<4)
-------------These values are close to the low end.
For N =38 and R =6 and K = 6 to 0
K=6: 0.000000362229464
K=5: 0.000069548057164
K=4: 0.002694987215111
K=3: 0.035933162868147 -------> Near value
K=2: 0.195386573095551 ------->
K=1: 0.437665923734035 ------->
K=0: 0.328249442800526 -------> 0.961301939630113 cumulative probability for “loosing tickets”(k<4)
Note:In many cases of R of N lotteries the probability of selecting one correct number is higher than selecting none.
~~D~~
+33616 $${\mathtt{ExpectedValue}} = {\mathtt{0}}{\mathtt{\,\times\,}}{\mathtt{0.962\: \!3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{0.035\: \!7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,000}}{\mathtt{\,\times\,}}{\mathtt{0.000\: \!036\: \!9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,000\,000}}{\mathtt{\,\times\,}}{\mathtt{0.000\: \!000\: \!143}} \Rightarrow {\mathtt{ExpectedValue}} = {\mathtt{0.287}}$$
Expected value = 0.287 dollars = 28.7cents.
To the nearest whole number of cents: Expected value = 29
+330 Alan’s valuation for the probabilities listed above are correct.
However, the probabilities listed do not correlate to the values of the R of N lottery described. There is no value of N where R=6 that distributes the probabilities depicted above.
The probabilities for R of N, and K of R (where K is a sub set of R) lotteries are calculated by hypergeometric distributions and the probabilities are discreet. The general formula is
$$P(k)=\frac{\binom{R}{K}*\binom{N-R}{R-K}}
{\binom{N}{R}}$$
Where N is the pool of numbers, R is the quantity drawn, and K is the quantity correctly matching R. The special case where K=R P(k) is equal to 1/(C(N,R)).
The hypergeometric discreet probability distributions are:
For N =70 and R =6 and K = 6 to 0
K=6: 0.0000000199744886
K=5: 0.0000064717343010
K=4: 0.0004287523974380
K=3: 0.0099089442963447
K=2: 0.0947542798337962 ----->
K=1: 0.3790171193351850 ----->
K=0: 0.5158844124284460 -----> 0.989655811597427 cumulative probability for “loosing tickets” (k<4)
-------------These values are close to the low end.
For N =38 and R =6 and K = 6 to 0
K=6: 0.000000362229464
K=5: 0.000069548057164
K=4: 0.002694987215111
K=3: 0.035933162868147 -------> Near value
K=2: 0.195386573095551 ------->
K=1: 0.437665923734035 ------->
K=0: 0.328249442800526 -------> 0.961301939630113 cumulative probability for “loosing tickets”(k<4)
Note:In many cases of R of N lotteries the probability of selecting one correct number is higher than selecting none.
~~D~~
