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How much money would it cost to “buy out” every possible number in the 6-digit Lottery?
Estimating the number of tickets bought in a lotteryDetermine if this lottery system is profitableFind the minimum number of tickets to guarantee the win of a n-bit binary lottery?How many 6 digit numbers are possible with no digit appearing more than thrice?Expected value lottery (4 digits)The Number Of Possible Four Digit Numbers With Ordered DigitsProbability of finding a random number in a random interval vs. random samplescombinatorics. how many options to make a ticket?Number of $10$ digit numbers such that every digit that appears appears exactly twiceHow many powerball tickets will guarantee a 3/5 match?
$begingroup$
So, the constraints are as follows:
- First 5 numbers are digits 1 ; 69
- The 6th number is a number 1 - 26
- Order does not matter
- Numbers do not repeat
- Tickets cost $2 each
My reasoning is as follows: for the first 5 digits, there are clearly $$69 times 68 times 67 times 66 times 65$$ possibilities.
For the 6th digit, there are 26 possible choices, but we must account for the fact that the previous digits could've been within the range of $[1, 26]$ (or we end up overcounting). This is where I'm a little less certain: clearly I can't just multiply the previous number by 26. The "worst" case is where all 5 previous numbers were also in the range of $[1, 26]$; the "best" case is where all of them were in the range $[27, 69]$. My thought process was to just multiply my previous number by $21$ (because $26 - 5 = 21$), but I think that this may be undercounting because there are $$43 times 42 times 41 times 40 times 39$$ possible numbers that they could take that are greater than 26.
It's easy to account for the $2 tickets because if I want to calculate how much money it would take to buy enough tickets to guarantee a win, I'd just have to multiply my previous result by 2.
So, my proposed solution is that it would cost $$69 times 68 times 67 times 66 times 65 times 21 times 2$$
dollars to buy enough Lottery tickets to guarantee that you would win.
Can someone help me with the number of possible values for the final digits? Is $21$ correct, or are there more?
combinatorics proof-verification combinations gambling lotteries
asked Mar 28 at 17:56
EJoshuaSEJoshuaS
444719
$endgroup$
add a comment |
$begingroup$
So, the constraints are as follows:
- First 5 numbers are digits 1 ; 69
- The 6th number is a number 1 - 26
- Order does not matter
- Numbers do not repeat
- Tickets cost $2 each
My reasoning is as follows: for the first 5 digits, there are clearly $$69 times 68 times 67 times 66 times 65$$ possibilities.
For the 6th digit, there are 26 possible choices, but we must account for the fact that the previous digits could've been within the range of $[1, 26]$ (or we end up overcounting). This is where I'm a little less certain: clearly I can't just multiply the previous number by 26. The "worst" case is where all 5 previous numbers were also in the range of $[1, 26]$; the "best" case is where all of them were in the range $[27, 69]$. My thought process was to just multiply my previous number by $21$ (because $26 - 5 = 21$), but I think that this may be undercounting because there are $$43 times 42 times 41 times 40 times 39$$ possible numbers that they could take that are greater than 26.
It's easy to account for the $2 tickets because if I want to calculate how much money it would take to buy enough tickets to guarantee a win, I'd just have to multiply my previous result by 2.
So, my proposed solution is that it would cost $$69 times 68 times 67 times 66 times 65 times 21 times 2$$
dollars to buy enough Lottery tickets to guarantee that you would win.
Can someone help me with the number of possible values for the final digits? Is $21$ correct, or are there more?
combinatorics proof-verification combinations gambling lotteries
asked Mar 28 at 17:56
EJoshuaSEJoshuaS
444719
$endgroup$
$begingroup$
Try choosing the 6th number first.
$endgroup$
– FredH
Mar 28 at 18:11
add a comment |
$begingroup$
So, the constraints are as follows:
- First 5 numbers are digits 1 ; 69
- The 6th number is a number 1 - 26
- Order does not matter
- Numbers do not repeat
- Tickets cost $2 each
My reasoning is as follows: for the first 5 digits, there are clearly $$69 times 68 times 67 times 66 times 65$$ possibilities.
For the 6th digit, there are 26 possible choices, but we must account for the fact that the previous digits could've been within the range of $[1, 26]$ (or we end up overcounting). This is where I'm a little less certain: clearly I can't just multiply the previous number by 26. The "worst" case is where all 5 previous numbers were also in the range of $[1, 26]$; the "best" case is where all of them were in the range $[27, 69]$. My thought process was to just multiply my previous number by $21$ (because $26 - 5 = 21$), but I think that this may be undercounting because there are $$43 times 42 times 41 times 40 times 39$$ possible numbers that they could take that are greater than 26.
It's easy to account for the $2 tickets because if I want to calculate how much money it would take to buy enough tickets to guarantee a win, I'd just have to multiply my previous result by 2.
So, my proposed solution is that it would cost $$69 times 68 times 67 times 66 times 65 times 21 times 2$$
dollars to buy enough Lottery tickets to guarantee that you would win.
Can someone help me with the number of possible values for the final digits? Is $21$ correct, or are there more?
combinatorics proof-verification combinations gambling lotteries
asked Mar 28 at 17:56
EJoshuaSEJoshuaS
444719
$endgroup$
So, the constraints are as follows:
- First 5 numbers are digits 1 ; 69
- The 6th number is a number 1 - 26
- Order does not matter
- Numbers do not repeat
- Tickets cost $2 each
My reasoning is as follows: for the first 5 digits, there are clearly $$69 times 68 times 67 times 66 times 65$$ possibilities.
For the 6th digit, there are 26 possible choices, but we must account for the fact that the previous digits could've been within the range of $[1, 26]$ (or we end up overcounting). This is where I'm a little less certain: clearly I can't just multiply the previous number by 26. The "worst" case is where all 5 previous numbers were also in the range of $[1, 26]$; the "best" case is where all of them were in the range $[27, 69]$. My thought process was to just multiply my previous number by $21$ (because $26 - 5 = 21$), but I think that this may be undercounting because there are $$43 times 42 times 41 times 40 times 39$$ possible numbers that they could take that are greater than 26.
It's easy to account for the $2 tickets because if I want to calculate how much money it would take to buy enough tickets to guarantee a win, I'd just have to multiply my previous result by 2.
So, my proposed solution is that it would cost $$69 times 68 times 67 times 66 times 65 times 21 times 2$$
dollars to buy enough Lottery tickets to guarantee that you would win.
Can someone help me with the number of possible values for the final digits? Is $21$ correct, or are there more?
combinatorics proof-verification combinations gambling lotteries
combinatorics proof-verification combinations gambling lotteries
asked Mar 28 at 17:56
EJoshuaSEJoshuaS
444719
asked Mar 28 at 17:56
EJoshuaSEJoshuaS
444719
edited Mar 28 at 18:09
EJoshuaS
asked Mar 28 at 17:56
EJoshuaSEJoshuaS
444719
asked Mar 28 at 17:56
EJoshuaSEJoshuaS
444719
asked Mar 28 at 17:56
EJoshuaSEJoshuaS
444719
444719
$begingroup$
Try choosing the 6th number first.
$endgroup$
– FredH
Mar 28 at 18:11
add a comment |
$begingroup$
Try choosing the 6th number first.
$endgroup$
– FredH
Mar 28 at 18:11
$begingroup$
Try choosing the 6th number first.
$endgroup$
– FredH
Mar 28 at 18:11
$begingroup$
Try choosing the 6th number first.
$endgroup$
– FredH
Mar 28 at 18:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Start with the sixth digit, then work your way backwards. If you choose the sixth number first, then there are only 68 numbers remaining for the other choices.
$$dbinom261dbinom685$$
By using Binomial Coefficients, we are not taking order into account (because you say in the beginning that order does not matter).
So, this gives the total number of ways of picking the six numbers.
answered Mar 28 at 18:12
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
$begingroup$
Thanks, that's way simpler than what I was trying.
$endgroup$
– EJoshuaS
Mar 28 at 18:18
add a comment |
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1 Answer
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$begingroup$
Start with the sixth digit, then work your way backwards. If you choose the sixth number first, then there are only 68 numbers remaining for the other choices.
$$dbinom261dbinom685$$
By using Binomial Coefficients, we are not taking order into account (because you say in the beginning that order does not matter).
So, this gives the total number of ways of picking the six numbers.
answered Mar 28 at 18:12
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
$begingroup$
Thanks, that's way simpler than what I was trying.
$endgroup$
– EJoshuaS
Mar 28 at 18:18
add a comment |
$begingroup$
Start with the sixth digit, then work your way backwards. If you choose the sixth number first, then there are only 68 numbers remaining for the other choices.
$$dbinom261dbinom685$$
By using Binomial Coefficients, we are not taking order into account (because you say in the beginning that order does not matter).
So, this gives the total number of ways of picking the six numbers.
answered Mar 28 at 18:12
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
$begingroup$
Thanks, that's way simpler than what I was trying.
$endgroup$
– EJoshuaS
Mar 28 at 18:18
add a comment |
$begingroup$
Start with the sixth digit, then work your way backwards. If you choose the sixth number first, then there are only 68 numbers remaining for the other choices.
$$dbinom261dbinom685$$
By using Binomial Coefficients, we are not taking order into account (because you say in the beginning that order does not matter).
So, this gives the total number of ways of picking the six numbers.
answered Mar 28 at 18:12
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
Start with the sixth digit, then work your way backwards. If you choose the sixth number first, then there are only 68 numbers remaining for the other choices.
$$dbinom261dbinom685$$
By using Binomial Coefficients, we are not taking order into account (because you say in the beginning that order does not matter).
So, this gives the total number of ways of picking the six numbers.
answered Mar 28 at 18:12
InterstellarProbeInterstellarProbe
3,154728
answered Mar 28 at 18:12
InterstellarProbeInterstellarProbe
3,154728
answered Mar 28 at 18:12
InterstellarProbeInterstellarProbe
3,154728
answered Mar 28 at 18:12
InterstellarProbeInterstellarProbe
3,154728
3,154728
$begingroup$
Thanks, that's way simpler than what I was trying.
$endgroup$
– EJoshuaS
Mar 28 at 18:18
add a comment |
$begingroup$
Thanks, that's way simpler than what I was trying.
$endgroup$
– EJoshuaS
Mar 28 at 18:18
$begingroup$
Thanks, that's way simpler than what I was trying.
$endgroup$
– EJoshuaS
Mar 28 at 18:18
$begingroup$
Thanks, that's way simpler than what I was trying.
$endgroup$
– EJoshuaS
Mar 28 at 18:18
add a comment |
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$begingroup$
Try choosing the 6th number first.
$endgroup$
– FredH
Mar 28 at 18:11