Raffle Odds for 1:125000, 3 winners, bought 5 tickets
If I bought 5 raffle tickets with a 1:125000 odds rating- what are my odds at winning if there are 3 winners, with winning tickets NOT returned to the bucket each time?
probability
asked Nov 28 '18 at 17:40
SandraSandra
11
add a comment |
If I bought 5 raffle tickets with a 1:125000 odds rating- what are my odds at winning if there are 3 winners, with winning tickets NOT returned to the bucket each time?
probability
asked Nov 28 '18 at 17:40
SandraSandra
11
add a comment |
If I bought 5 raffle tickets with a 1:125000 odds rating- what are my odds at winning if there are 3 winners, with winning tickets NOT returned to the bucket each time?
probability
asked Nov 28 '18 at 17:40
SandraSandra
11
If I bought 5 raffle tickets with a 1:125000 odds rating- what are my odds at winning if there are 3 winners, with winning tickets NOT returned to the bucket each time?
probability
probability
asked Nov 28 '18 at 17:40
SandraSandra
11
asked Nov 28 '18 at 17:40
SandraSandra
11
asked Nov 28 '18 at 17:40
SandraSandra
11
asked Nov 28 '18 at 17:40
SandraSandra
11
asked Nov 28 '18 at 17:40
SandraSandra
11
11
add a comment |
add a comment |
1 Answer
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Let $N$ be the total number of tickets (in your case $N=125000$)
There are $binom{N}{3}$ ways in which the $3$ winning tickets can be chosen.
There are $binom{N-5}{3}$ ways in which the $3$ winning tickets can be chosen so none of your tickets are selected.
Hence the probability you lose is:
$$frac{binom{N-5}{3}}{binom{N}{3}}=frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$$
So the probability you have at least one winning ticket is $1-frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$
answered Nov 28 '18 at 17:49
Jorge FernándezJorge Fernández
75.1k1190191
Thank you for taking the time to reply. How are the odds effected by purchasing 5 tickets? The total number of tickets for sale is 375000. Since there are 3 winners, the company lists the odds as 1:125000. I don't know how many tickets I will have purchased by the time the drawing is done, but currently I have 5. :) Have a great weekend!!! Thanks again. :)
– Sandra
Dec 1 '18 at 2:52
Oh I see, my bad. Ill modify the answer
– Jorge Fernández
Dec 1 '18 at 3:59
add a comment |
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1 Answer
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1 Answer
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Let $N$ be the total number of tickets (in your case $N=125000$)
There are $binom{N}{3}$ ways in which the $3$ winning tickets can be chosen.
There are $binom{N-5}{3}$ ways in which the $3$ winning tickets can be chosen so none of your tickets are selected.
Hence the probability you lose is:
$$frac{binom{N-5}{3}}{binom{N}{3}}=frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$$
So the probability you have at least one winning ticket is $1-frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$
answered Nov 28 '18 at 17:49
Jorge FernándezJorge Fernández
75.1k1190191
Thank you for taking the time to reply. How are the odds effected by purchasing 5 tickets? The total number of tickets for sale is 375000. Since there are 3 winners, the company lists the odds as 1:125000. I don't know how many tickets I will have purchased by the time the drawing is done, but currently I have 5. :) Have a great weekend!!! Thanks again. :)
– Sandra
Dec 1 '18 at 2:52
Oh I see, my bad. Ill modify the answer
– Jorge Fernández
Dec 1 '18 at 3:59
add a comment |
Let $N$ be the total number of tickets (in your case $N=125000$)
There are $binom{N}{3}$ ways in which the $3$ winning tickets can be chosen.
There are $binom{N-5}{3}$ ways in which the $3$ winning tickets can be chosen so none of your tickets are selected.
Hence the probability you lose is:
$$frac{binom{N-5}{3}}{binom{N}{3}}=frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$$
So the probability you have at least one winning ticket is $1-frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$
answered Nov 28 '18 at 17:49
Jorge FernándezJorge Fernández
75.1k1190191
Thank you for taking the time to reply. How are the odds effected by purchasing 5 tickets? The total number of tickets for sale is 375000. Since there are 3 winners, the company lists the odds as 1:125000. I don't know how many tickets I will have purchased by the time the drawing is done, but currently I have 5. :) Have a great weekend!!! Thanks again. :)
– Sandra
Dec 1 '18 at 2:52
Oh I see, my bad. Ill modify the answer
– Jorge Fernández
Dec 1 '18 at 3:59
add a comment |
Let $N$ be the total number of tickets (in your case $N=125000$)
There are $binom{N}{3}$ ways in which the $3$ winning tickets can be chosen.
There are $binom{N-5}{3}$ ways in which the $3$ winning tickets can be chosen so none of your tickets are selected.
Hence the probability you lose is:
$$frac{binom{N-5}{3}}{binom{N}{3}}=frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$$
So the probability you have at least one winning ticket is $1-frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$
answered Nov 28 '18 at 17:49
Jorge FernándezJorge Fernández
75.1k1190191
Let $N$ be the total number of tickets (in your case $N=125000$)
There are $binom{N}{3}$ ways in which the $3$ winning tickets can be chosen.
There are $binom{N-5}{3}$ ways in which the $3$ winning tickets can be chosen so none of your tickets are selected.
Hence the probability you lose is:
$$frac{binom{N-5}{3}}{binom{N}{3}}=frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$$
So the probability you have at least one winning ticket is $1-frac{(N-5)(N-6)(N-7)}{N(N-1)(N-2)}$
answered Nov 28 '18 at 17:49
Jorge FernándezJorge Fernández
75.1k1190191
edited Dec 1 '18 at 4:01
answered Nov 28 '18 at 17:49
Jorge FernándezJorge Fernández
75.1k1190191
answered Nov 28 '18 at 17:49
Jorge FernándezJorge Fernández
75.1k1190191
answered Nov 28 '18 at 17:49
Jorge FernándezJorge Fernández
75.1k1190191
75.1k1190191
Thank you for taking the time to reply. How are the odds effected by purchasing 5 tickets? The total number of tickets for sale is 375000. Since there are 3 winners, the company lists the odds as 1:125000. I don't know how many tickets I will have purchased by the time the drawing is done, but currently I have 5. :) Have a great weekend!!! Thanks again. :)
– Sandra
Dec 1 '18 at 2:52
Oh I see, my bad. Ill modify the answer
– Jorge Fernández
Dec 1 '18 at 3:59
add a comment |
Thank you for taking the time to reply. How are the odds effected by purchasing 5 tickets? The total number of tickets for sale is 375000. Since there are 3 winners, the company lists the odds as 1:125000. I don't know how many tickets I will have purchased by the time the drawing is done, but currently I have 5. :) Have a great weekend!!! Thanks again. :)
– Sandra
Dec 1 '18 at 2:52
Oh I see, my bad. Ill modify the answer
– Jorge Fernández
Dec 1 '18 at 3:59
Thank you for taking the time to reply. How are the odds effected by purchasing 5 tickets? The total number of tickets for sale is 375000. Since there are 3 winners, the company lists the odds as 1:125000. I don't know how many tickets I will have purchased by the time the drawing is done, but currently I have 5. :) Have a great weekend!!! Thanks again. :)
– Sandra
Dec 1 '18 at 2:52
Thank you for taking the time to reply. How are the odds effected by purchasing 5 tickets? The total number of tickets for sale is 375000. Since there are 3 winners, the company lists the odds as 1:125000. I don't know how many tickets I will have purchased by the time the drawing is done, but currently I have 5. :) Have a great weekend!!! Thanks again. :)
– Sandra
Dec 1 '18 at 2:52
Oh I see, my bad. Ill modify the answer
– Jorge Fernández
Dec 1 '18 at 3:59
Oh I see, my bad. Ill modify the answer
– Jorge Fernández
Dec 1 '18 at 3:59
add a comment |
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