Mixing
How to calculate scratch card probability
$begingroup$
A scratch ticket has 15 cells.
How do you calculate the probability of scratching 6 winning images before scratching 9 losing images?
I'm trying to build a simple online game and struggling with the probabilities :(
probability combinatorics discrete-mathematics
edited Jan 30 at 14:16
Scientifica
6,80941335
asked Jan 30 at 14:08
xtremetomxtremetom
31
$endgroup$
migrated from mathematica.stackexchange.com Jan 30 at 14:10
This question came from our site for users of Wolfram Mathematica.
add a comment |
$begingroup$
A scratch ticket has 15 cells.
How do you calculate the probability of scratching 6 winning images before scratching 9 losing images?
I'm trying to build a simple online game and struggling with the probabilities :(
probability combinatorics discrete-mathematics
edited Jan 30 at 14:16
Scientifica
6,80941335
asked Jan 30 at 14:08
xtremetomxtremetom
31
$endgroup$
migrated from mathematica.stackexchange.com Jan 30 at 14:10
This question came from our site for users of Wolfram Mathematica.
add a comment |
$begingroup$
A scratch ticket has 15 cells.
How do you calculate the probability of scratching 6 winning images before scratching 9 losing images?
I'm trying to build a simple online game and struggling with the probabilities :(
probability combinatorics discrete-mathematics
edited Jan 30 at 14:16
Scientifica
6,80941335
asked Jan 30 at 14:08
xtremetomxtremetom
31
$endgroup$
A scratch ticket has 15 cells.
How do you calculate the probability of scratching 6 winning images before scratching 9 losing images?
I'm trying to build a simple online game and struggling with the probabilities :(
probability combinatorics discrete-mathematics
probability combinatorics discrete-mathematics
edited Jan 30 at 14:16
Scientifica
6,80941335
asked Jan 30 at 14:08
xtremetomxtremetom
31
edited Jan 30 at 14:16
Scientifica
6,80941335
asked Jan 30 at 14:08
xtremetomxtremetom
31
edited Jan 30 at 14:16
Scientifica
6,80941335
edited Jan 30 at 14:16
Scientifica
6,80941335
edited Jan 30 at 14:16
Scientifica
6,80941335
6,80941335
asked Jan 30 at 14:08
xtremetomxtremetom
31
asked Jan 30 at 14:08
xtremetomxtremetom
31
asked Jan 30 at 14:08
xtremetomxtremetom
31
31
migrated from mathematica.stackexchange.com Jan 30 at 14:10
This question came from our site for users of Wolfram Mathematica.
migrated from mathematica.stackexchange.com Jan 30 at 14:10
This question came from our site for users of Wolfram Mathematica.
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
I think you can formulate your problem in terms of the negative hypergeometric distribution, by asking the question as: What is the probability of having all six winning images revealed when we hit all nine losing images? Since then it is clear that the winning scenario (revealing all six winning images) happened first, regardless if after that we continue to reveal the remaining (losing) images. But if we have less than all six winning images revealed when revealing the nineth losing image we lose. In terms of the negative hypergeometric distribution with pmf
$$P(X=k)=frac{{{k+r-1}choose{k}}{{N-r-k}choose{K-k}}}{N choose K}$$
$N$ is the total number of images on the scratch ticket, $K$ is the total number of winning images, $r$ is the number of revealed losing images after which we stop and $k$ is the number of winning images we see when we have stopped.
Now your question becomes what is the value of
$$P(k=6,N=15,K=6,r=9)=frac{{{6+9-1}choose{6}}{{15-9-6}choose{6-6}}}{15 choose 6}=frac{3003*1}{5005}=frac{3}{5}$$
answered Jan 30 at 15:02
Thies HeideckeThies Heidecke
1186
$endgroup$
$begingroup$
This does the trick nicely :)
$endgroup$
– xtremetom
Jan 30 at 16:02
add a comment |
$begingroup$
I assume that we know beforehand that the card will have exactly 6 winning images amongst the 15 on the card.
There are 6 winning images in the 15 images. When choosing the first image to reveal, there is a $frac{6}{15}$ probability of it being a winning one.
There are now 5 winning images left in the remaining 14 images. When choosing the second image to reveal, there is a $frac{5}{14}$ probability of it being a winning one.
There are now 4 winning images left in the remaining 13 images. etc.
The probability of going through all 6 reveals successfully is therefore $frac{6}{15}cdot frac{5}{14}cdot frac{4}{13}cdot frac{3}{12}cdot frac{2}{11}cdot frac{1}{10}$.
answered Jan 30 at 14:19
Jaap ScherphuisJaap Scherphuis
4,232717
$endgroup$
$begingroup$
But this applies only, if you need to find the 6 winning tickets in the first six scratches, right? If i understand the question correctly, it's only required to reveal the six winning images before all nine losing images are revealed.
$endgroup$
– Thies Heidecke
Jan 30 at 14:22
$begingroup$
@ThiesHeidecke It is ambiguous. In your interpretation it would simply be equivalent to choosing an order in which to reveal the images such that the 15th is not a winning image. That is simply $9/15$.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:38
$begingroup$
I think it's a bit more subtle than that. It's essentially a race between revealing all six winning images before all nine losing images. There are many possible realizations of both scenarios, and you want the probability of the winning scenario happening before the losing scenario.
$endgroup$
– Thies Heidecke
Jan 30 at 14:47
1
$begingroup$
@ThiesHeidecke The losing scenario always takes 14 reveals (which only finds 5 of the winning images so the unscratched box must be the missing one). It does not matter that you can win earlier, because the only way to lose is to go on to the bitter end and fail. The probability of losing in this way is the probability of the 15th box having a winning image, or $6/15$. Winning therefore has probability $9/15$. If you do it the hard way and add up probabilities of winning in 6,7,8...14 reveals, you'll get the same answer.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:59
$begingroup$
Thanks for the explanation, you are right, now i get it! I wrote up the explanation via neg. hypergeometric distribution up, anyway, because i thought it still might be interesting.
$endgroup$
– Thies Heidecke
Jan 30 at 15:04
|
show 1 more comment
$begingroup$
One more observation that I'm surprised no one has mentioned is that one scratches six winners before nine losers if and only if the last panel scratched is a loser. Since nine out of the fifteen are losers, the probability that one scratches six winners first is $9/15 = 3/5$.
Perhaps I'm not understanding the set-up correctly, if this reasoning doesn't apply.
answered Jan 31 at 0:23
Brian TungBrian Tung
26.2k32656
$endgroup$
add a comment |
$begingroup$
Direct enumeration is another possibility. All possible drawings are obtained by the following Mathematica codes:
ticket = Join[Table[w, 6], Table[l, 9]];
ords = Permutations[ticket];
Length[ords] (* output 5005 *)
The count is $5005$.
Drawing 6 w's before the 9 l's requires that the first 14 scratches contain 6 w's. The number of drawings which satisfy this is
Length[ Select[ #[[1 ;; 14]] & /@ ords, Count[#, w]==6 &]]
and this gives $3003$. The required probability is $3003/5005$.
edited Jan 31 at 4:09
Lee David Chung Lin
4,47841242
answered Jan 30 at 23:16
Tomas GarzaTomas Garza
1
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think you can formulate your problem in terms of the negative hypergeometric distribution, by asking the question as: What is the probability of having all six winning images revealed when we hit all nine losing images? Since then it is clear that the winning scenario (revealing all six winning images) happened first, regardless if after that we continue to reveal the remaining (losing) images. But if we have less than all six winning images revealed when revealing the nineth losing image we lose. In terms of the negative hypergeometric distribution with pmf
$$P(X=k)=frac{{{k+r-1}choose{k}}{{N-r-k}choose{K-k}}}{N choose K}$$
$N$ is the total number of images on the scratch ticket, $K$ is the total number of winning images, $r$ is the number of revealed losing images after which we stop and $k$ is the number of winning images we see when we have stopped.
Now your question becomes what is the value of
$$P(k=6,N=15,K=6,r=9)=frac{{{6+9-1}choose{6}}{{15-9-6}choose{6-6}}}{15 choose 6}=frac{3003*1}{5005}=frac{3}{5}$$
answered Jan 30 at 15:02
Thies HeideckeThies Heidecke
1186
$endgroup$
$begingroup$
This does the trick nicely :)
$endgroup$
– xtremetom
Jan 30 at 16:02
add a comment |
$begingroup$
I think you can formulate your problem in terms of the negative hypergeometric distribution, by asking the question as: What is the probability of having all six winning images revealed when we hit all nine losing images? Since then it is clear that the winning scenario (revealing all six winning images) happened first, regardless if after that we continue to reveal the remaining (losing) images. But if we have less than all six winning images revealed when revealing the nineth losing image we lose. In terms of the negative hypergeometric distribution with pmf
$$P(X=k)=frac{{{k+r-1}choose{k}}{{N-r-k}choose{K-k}}}{N choose K}$$
$N$ is the total number of images on the scratch ticket, $K$ is the total number of winning images, $r$ is the number of revealed losing images after which we stop and $k$ is the number of winning images we see when we have stopped.
Now your question becomes what is the value of
$$P(k=6,N=15,K=6,r=9)=frac{{{6+9-1}choose{6}}{{15-9-6}choose{6-6}}}{15 choose 6}=frac{3003*1}{5005}=frac{3}{5}$$
answered Jan 30 at 15:02
Thies HeideckeThies Heidecke
1186
$endgroup$
$begingroup$
This does the trick nicely :)
$endgroup$
– xtremetom
Jan 30 at 16:02
add a comment |
$begingroup$
I think you can formulate your problem in terms of the negative hypergeometric distribution, by asking the question as: What is the probability of having all six winning images revealed when we hit all nine losing images? Since then it is clear that the winning scenario (revealing all six winning images) happened first, regardless if after that we continue to reveal the remaining (losing) images. But if we have less than all six winning images revealed when revealing the nineth losing image we lose. In terms of the negative hypergeometric distribution with pmf
$$P(X=k)=frac{{{k+r-1}choose{k}}{{N-r-k}choose{K-k}}}{N choose K}$$
$N$ is the total number of images on the scratch ticket, $K$ is the total number of winning images, $r$ is the number of revealed losing images after which we stop and $k$ is the number of winning images we see when we have stopped.
Now your question becomes what is the value of
$$P(k=6,N=15,K=6,r=9)=frac{{{6+9-1}choose{6}}{{15-9-6}choose{6-6}}}{15 choose 6}=frac{3003*1}{5005}=frac{3}{5}$$
answered Jan 30 at 15:02
Thies HeideckeThies Heidecke
1186
$endgroup$
I think you can formulate your problem in terms of the negative hypergeometric distribution, by asking the question as: What is the probability of having all six winning images revealed when we hit all nine losing images? Since then it is clear that the winning scenario (revealing all six winning images) happened first, regardless if after that we continue to reveal the remaining (losing) images. But if we have less than all six winning images revealed when revealing the nineth losing image we lose. In terms of the negative hypergeometric distribution with pmf
$$P(X=k)=frac{{{k+r-1}choose{k}}{{N-r-k}choose{K-k}}}{N choose K}$$
$N$ is the total number of images on the scratch ticket, $K$ is the total number of winning images, $r$ is the number of revealed losing images after which we stop and $k$ is the number of winning images we see when we have stopped.
Now your question becomes what is the value of
$$P(k=6,N=15,K=6,r=9)=frac{{{6+9-1}choose{6}}{{15-9-6}choose{6-6}}}{15 choose 6}=frac{3003*1}{5005}=frac{3}{5}$$
answered Jan 30 at 15:02
Thies HeideckeThies Heidecke
1186
answered Jan 30 at 15:02
Thies HeideckeThies Heidecke
1186
answered Jan 30 at 15:02
Thies HeideckeThies Heidecke
1186
answered Jan 30 at 15:02
Thies HeideckeThies Heidecke
1186
1186
$begingroup$
This does the trick nicely :)
$endgroup$
– xtremetom
Jan 30 at 16:02
add a comment |
$begingroup$
This does the trick nicely :)
$endgroup$
– xtremetom
Jan 30 at 16:02
$begingroup$
This does the trick nicely :)
$endgroup$
– xtremetom
Jan 30 at 16:02
$begingroup$
This does the trick nicely :)
$endgroup$
– xtremetom
Jan 30 at 16:02
add a comment |
$begingroup$
I assume that we know beforehand that the card will have exactly 6 winning images amongst the 15 on the card.
There are 6 winning images in the 15 images. When choosing the first image to reveal, there is a $frac{6}{15}$ probability of it being a winning one.
There are now 5 winning images left in the remaining 14 images. When choosing the second image to reveal, there is a $frac{5}{14}$ probability of it being a winning one.
There are now 4 winning images left in the remaining 13 images. etc.
The probability of going through all 6 reveals successfully is therefore $frac{6}{15}cdot frac{5}{14}cdot frac{4}{13}cdot frac{3}{12}cdot frac{2}{11}cdot frac{1}{10}$.
answered Jan 30 at 14:19
Jaap ScherphuisJaap Scherphuis
4,232717
$endgroup$
$begingroup$
But this applies only, if you need to find the 6 winning tickets in the first six scratches, right? If i understand the question correctly, it's only required to reveal the six winning images before all nine losing images are revealed.
$endgroup$
– Thies Heidecke
Jan 30 at 14:22
$begingroup$
@ThiesHeidecke It is ambiguous. In your interpretation it would simply be equivalent to choosing an order in which to reveal the images such that the 15th is not a winning image. That is simply $9/15$.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:38
$begingroup$
I think it's a bit more subtle than that. It's essentially a race between revealing all six winning images before all nine losing images. There are many possible realizations of both scenarios, and you want the probability of the winning scenario happening before the losing scenario.
$endgroup$
– Thies Heidecke
Jan 30 at 14:47
1
$begingroup$
@ThiesHeidecke The losing scenario always takes 14 reveals (which only finds 5 of the winning images so the unscratched box must be the missing one). It does not matter that you can win earlier, because the only way to lose is to go on to the bitter end and fail. The probability of losing in this way is the probability of the 15th box having a winning image, or $6/15$. Winning therefore has probability $9/15$. If you do it the hard way and add up probabilities of winning in 6,7,8...14 reveals, you'll get the same answer.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:59
$begingroup$
Thanks for the explanation, you are right, now i get it! I wrote up the explanation via neg. hypergeometric distribution up, anyway, because i thought it still might be interesting.
$endgroup$
– Thies Heidecke
Jan 30 at 15:04
|
show 1 more comment
$begingroup$
I assume that we know beforehand that the card will have exactly 6 winning images amongst the 15 on the card.
There are 6 winning images in the 15 images. When choosing the first image to reveal, there is a $frac{6}{15}$ probability of it being a winning one.
There are now 5 winning images left in the remaining 14 images. When choosing the second image to reveal, there is a $frac{5}{14}$ probability of it being a winning one.
There are now 4 winning images left in the remaining 13 images. etc.
The probability of going through all 6 reveals successfully is therefore $frac{6}{15}cdot frac{5}{14}cdot frac{4}{13}cdot frac{3}{12}cdot frac{2}{11}cdot frac{1}{10}$.
answered Jan 30 at 14:19
Jaap ScherphuisJaap Scherphuis
4,232717
$endgroup$
$begingroup$
But this applies only, if you need to find the 6 winning tickets in the first six scratches, right? If i understand the question correctly, it's only required to reveal the six winning images before all nine losing images are revealed.
$endgroup$
– Thies Heidecke
Jan 30 at 14:22
$begingroup$
@ThiesHeidecke It is ambiguous. In your interpretation it would simply be equivalent to choosing an order in which to reveal the images such that the 15th is not a winning image. That is simply $9/15$.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:38
$begingroup$
I think it's a bit more subtle than that. It's essentially a race between revealing all six winning images before all nine losing images. There are many possible realizations of both scenarios, and you want the probability of the winning scenario happening before the losing scenario.
$endgroup$
– Thies Heidecke
Jan 30 at 14:47
1
$begingroup$
@ThiesHeidecke The losing scenario always takes 14 reveals (which only finds 5 of the winning images so the unscratched box must be the missing one). It does not matter that you can win earlier, because the only way to lose is to go on to the bitter end and fail. The probability of losing in this way is the probability of the 15th box having a winning image, or $6/15$. Winning therefore has probability $9/15$. If you do it the hard way and add up probabilities of winning in 6,7,8...14 reveals, you'll get the same answer.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:59
$begingroup$
Thanks for the explanation, you are right, now i get it! I wrote up the explanation via neg. hypergeometric distribution up, anyway, because i thought it still might be interesting.
$endgroup$
– Thies Heidecke
Jan 30 at 15:04
|
show 1 more comment
$begingroup$
I assume that we know beforehand that the card will have exactly 6 winning images amongst the 15 on the card.
There are 6 winning images in the 15 images. When choosing the first image to reveal, there is a $frac{6}{15}$ probability of it being a winning one.
There are now 5 winning images left in the remaining 14 images. When choosing the second image to reveal, there is a $frac{5}{14}$ probability of it being a winning one.
There are now 4 winning images left in the remaining 13 images. etc.
The probability of going through all 6 reveals successfully is therefore $frac{6}{15}cdot frac{5}{14}cdot frac{4}{13}cdot frac{3}{12}cdot frac{2}{11}cdot frac{1}{10}$.
answered Jan 30 at 14:19
Jaap ScherphuisJaap Scherphuis
4,232717
$endgroup$
I assume that we know beforehand that the card will have exactly 6 winning images amongst the 15 on the card.
There are 6 winning images in the 15 images. When choosing the first image to reveal, there is a $frac{6}{15}$ probability of it being a winning one.
There are now 5 winning images left in the remaining 14 images. When choosing the second image to reveal, there is a $frac{5}{14}$ probability of it being a winning one.
There are now 4 winning images left in the remaining 13 images. etc.
The probability of going through all 6 reveals successfully is therefore $frac{6}{15}cdot frac{5}{14}cdot frac{4}{13}cdot frac{3}{12}cdot frac{2}{11}cdot frac{1}{10}$.
answered Jan 30 at 14:19
Jaap ScherphuisJaap Scherphuis
4,232717
answered Jan 30 at 14:19
Jaap ScherphuisJaap Scherphuis
4,232717
answered Jan 30 at 14:19
Jaap ScherphuisJaap Scherphuis
4,232717
answered Jan 30 at 14:19
Jaap ScherphuisJaap Scherphuis
4,232717
4,232717
$begingroup$
But this applies only, if you need to find the 6 winning tickets in the first six scratches, right? If i understand the question correctly, it's only required to reveal the six winning images before all nine losing images are revealed.
$endgroup$
– Thies Heidecke
Jan 30 at 14:22
$begingroup$
@ThiesHeidecke It is ambiguous. In your interpretation it would simply be equivalent to choosing an order in which to reveal the images such that the 15th is not a winning image. That is simply $9/15$.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:38
$begingroup$
I think it's a bit more subtle than that. It's essentially a race between revealing all six winning images before all nine losing images. There are many possible realizations of both scenarios, and you want the probability of the winning scenario happening before the losing scenario.
$endgroup$
– Thies Heidecke
Jan 30 at 14:47
1
$begingroup$
@ThiesHeidecke The losing scenario always takes 14 reveals (which only finds 5 of the winning images so the unscratched box must be the missing one). It does not matter that you can win earlier, because the only way to lose is to go on to the bitter end and fail. The probability of losing in this way is the probability of the 15th box having a winning image, or $6/15$. Winning therefore has probability $9/15$. If you do it the hard way and add up probabilities of winning in 6,7,8...14 reveals, you'll get the same answer.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:59
$begingroup$
Thanks for the explanation, you are right, now i get it! I wrote up the explanation via neg. hypergeometric distribution up, anyway, because i thought it still might be interesting.
$endgroup$
– Thies Heidecke
Jan 30 at 15:04
|
show 1 more comment
$begingroup$
But this applies only, if you need to find the 6 winning tickets in the first six scratches, right? If i understand the question correctly, it's only required to reveal the six winning images before all nine losing images are revealed.
$endgroup$
– Thies Heidecke
Jan 30 at 14:22
$begingroup$
@ThiesHeidecke It is ambiguous. In your interpretation it would simply be equivalent to choosing an order in which to reveal the images such that the 15th is not a winning image. That is simply $9/15$.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:38
$begingroup$
I think it's a bit more subtle than that. It's essentially a race between revealing all six winning images before all nine losing images. There are many possible realizations of both scenarios, and you want the probability of the winning scenario happening before the losing scenario.
$endgroup$
– Thies Heidecke
Jan 30 at 14:47
1
$begingroup$
@ThiesHeidecke The losing scenario always takes 14 reveals (which only finds 5 of the winning images so the unscratched box must be the missing one). It does not matter that you can win earlier, because the only way to lose is to go on to the bitter end and fail. The probability of losing in this way is the probability of the 15th box having a winning image, or $6/15$. Winning therefore has probability $9/15$. If you do it the hard way and add up probabilities of winning in 6,7,8...14 reveals, you'll get the same answer.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:59
$begingroup$
Thanks for the explanation, you are right, now i get it! I wrote up the explanation via neg. hypergeometric distribution up, anyway, because i thought it still might be interesting.
$endgroup$
– Thies Heidecke
Jan 30 at 15:04
$begingroup$
But this applies only, if you need to find the 6 winning tickets in the first six scratches, right? If i understand the question correctly, it's only required to reveal the six winning images before all nine losing images are revealed.
$endgroup$
– Thies Heidecke
Jan 30 at 14:22
$begingroup$
But this applies only, if you need to find the 6 winning tickets in the first six scratches, right? If i understand the question correctly, it's only required to reveal the six winning images before all nine losing images are revealed.
$endgroup$
– Thies Heidecke
Jan 30 at 14:22
$begingroup$
@ThiesHeidecke It is ambiguous. In your interpretation it would simply be equivalent to choosing an order in which to reveal the images such that the 15th is not a winning image. That is simply $9/15$.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:38
$begingroup$
@ThiesHeidecke It is ambiguous. In your interpretation it would simply be equivalent to choosing an order in which to reveal the images such that the 15th is not a winning image. That is simply $9/15$.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:38
$begingroup$
I think it's a bit more subtle than that. It's essentially a race between revealing all six winning images before all nine losing images. There are many possible realizations of both scenarios, and you want the probability of the winning scenario happening before the losing scenario.
$endgroup$
– Thies Heidecke
Jan 30 at 14:47
$begingroup$
I think it's a bit more subtle than that. It's essentially a race between revealing all six winning images before all nine losing images. There are many possible realizations of both scenarios, and you want the probability of the winning scenario happening before the losing scenario.
$endgroup$
– Thies Heidecke
Jan 30 at 14:47
1
1
$begingroup$
@ThiesHeidecke The losing scenario always takes 14 reveals (which only finds 5 of the winning images so the unscratched box must be the missing one). It does not matter that you can win earlier, because the only way to lose is to go on to the bitter end and fail. The probability of losing in this way is the probability of the 15th box having a winning image, or $6/15$. Winning therefore has probability $9/15$. If you do it the hard way and add up probabilities of winning in 6,7,8...14 reveals, you'll get the same answer.
$endgroup$
– Jaap Scherphuis
Jan 30 at 14:59
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@ThiesHeidecke The losing scenario always takes 14 reveals (which only finds 5 of the winning images so the unscratched box must be the missing one). It does not matter that you can win earlier, because the only way to lose is to go on to the bitter end and fail. The probability of losing in this way is the probability of the 15th box having a winning image, or $6/15$. Winning therefore has probability $9/15$. If you do it the hard way and add up probabilities of winning in 6,7,8...14 reveals, you'll get the same answer.
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– Jaap Scherphuis
Jan 30 at 14:59
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Thanks for the explanation, you are right, now i get it! I wrote up the explanation via neg. hypergeometric distribution up, anyway, because i thought it still might be interesting.
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– Thies Heidecke
Jan 30 at 15:04
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Thanks for the explanation, you are right, now i get it! I wrote up the explanation via neg. hypergeometric distribution up, anyway, because i thought it still might be interesting.
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– Thies Heidecke
Jan 30 at 15:04
|
show 1 more comment
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One more observation that I'm surprised no one has mentioned is that one scratches six winners before nine losers if and only if the last panel scratched is a loser. Since nine out of the fifteen are losers, the probability that one scratches six winners first is $9/15 = 3/5$.
Perhaps I'm not understanding the set-up correctly, if this reasoning doesn't apply.
answered Jan 31 at 0:23
Brian TungBrian Tung
26.2k32656
$endgroup$
add a comment |
$begingroup$
One more observation that I'm surprised no one has mentioned is that one scratches six winners before nine losers if and only if the last panel scratched is a loser. Since nine out of the fifteen are losers, the probability that one scratches six winners first is $9/15 = 3/5$.
Perhaps I'm not understanding the set-up correctly, if this reasoning doesn't apply.
answered Jan 31 at 0:23
Brian TungBrian Tung
26.2k32656
$endgroup$
add a comment |
$begingroup$
One more observation that I'm surprised no one has mentioned is that one scratches six winners before nine losers if and only if the last panel scratched is a loser. Since nine out of the fifteen are losers, the probability that one scratches six winners first is $9/15 = 3/5$.
Perhaps I'm not understanding the set-up correctly, if this reasoning doesn't apply.
answered Jan 31 at 0:23
Brian TungBrian Tung
26.2k32656
$endgroup$
One more observation that I'm surprised no one has mentioned is that one scratches six winners before nine losers if and only if the last panel scratched is a loser. Since nine out of the fifteen are losers, the probability that one scratches six winners first is $9/15 = 3/5$.
Perhaps I'm not understanding the set-up correctly, if this reasoning doesn't apply.
answered Jan 31 at 0:23
Brian TungBrian Tung
26.2k32656
answered Jan 31 at 0:23
Brian TungBrian Tung
26.2k32656
answered Jan 31 at 0:23
Brian TungBrian Tung
26.2k32656
answered Jan 31 at 0:23
Brian TungBrian Tung
26.2k32656
26.2k32656
add a comment |
add a comment |
$begingroup$
Direct enumeration is another possibility. All possible drawings are obtained by the following Mathematica codes:
ticket = Join[Table[w, 6], Table[l, 9]];
ords = Permutations[ticket];
Length[ords] (* output 5005 *)
The count is $5005$.
Drawing 6 w's before the 9 l's requires that the first 14 scratches contain 6 w's. The number of drawings which satisfy this is
Length[ Select[ #[[1 ;; 14]] & /@ ords, Count[#, w]==6 &]]
and this gives $3003$. The required probability is $3003/5005$.
edited Jan 31 at 4:09
Lee David Chung Lin
4,47841242
answered Jan 30 at 23:16
Tomas GarzaTomas Garza
1
$endgroup$
add a comment |
$begingroup$
Direct enumeration is another possibility. All possible drawings are obtained by the following Mathematica codes:
ticket = Join[Table[w, 6], Table[l, 9]];
ords = Permutations[ticket];
Length[ords] (* output 5005 *)
The count is $5005$.
Drawing 6 w's before the 9 l's requires that the first 14 scratches contain 6 w's. The number of drawings which satisfy this is
Length[ Select[ #[[1 ;; 14]] & /@ ords, Count[#, w]==6 &]]
and this gives $3003$. The required probability is $3003/5005$.
edited Jan 31 at 4:09
Lee David Chung Lin
4,47841242
answered Jan 30 at 23:16
Tomas GarzaTomas Garza
1
$endgroup$
add a comment |
$begingroup$
Direct enumeration is another possibility. All possible drawings are obtained by the following Mathematica codes:
ticket = Join[Table[w, 6], Table[l, 9]];
ords = Permutations[ticket];
Length[ords] (* output 5005 *)
The count is $5005$.
Drawing 6 w's before the 9 l's requires that the first 14 scratches contain 6 w's. The number of drawings which satisfy this is
Length[ Select[ #[[1 ;; 14]] & /@ ords, Count[#, w]==6 &]]
and this gives $3003$. The required probability is $3003/5005$.
edited Jan 31 at 4:09
Lee David Chung Lin
4,47841242
answered Jan 30 at 23:16
Tomas GarzaTomas Garza
1
$endgroup$
Direct enumeration is another possibility. All possible drawings are obtained by the following Mathematica codes:
ticket = Join[Table[w, 6], Table[l, 9]];
ords = Permutations[ticket];
Length[ords] (* output 5005 *)
The count is $5005$.
Drawing 6 w's before the 9 l's requires that the first 14 scratches contain 6 w's. The number of drawings which satisfy this is
Length[ Select[ #[[1 ;; 14]] & /@ ords, Count[#, w]==6 &]]
and this gives $3003$. The required probability is $3003/5005$.
edited Jan 31 at 4:09
Lee David Chung Lin
4,47841242
answered Jan 30 at 23:16
Tomas GarzaTomas Garza
1
edited Jan 31 at 4:09
Lee David Chung Lin
4,47841242
edited Jan 31 at 4:09
Lee David Chung Lin
4,47841242
edited Jan 31 at 4:09
Lee David Chung Lin
4,47841242
4,47841242
answered Jan 30 at 23:16
Tomas GarzaTomas Garza
1
answered Jan 30 at 23:16
Tomas GarzaTomas Garza
1
answered Jan 30 at 23:16
Tomas GarzaTomas Garza
1
1
add a comment |
add a comment |
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