how to calculate probability of winning exactly one number at lottery
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There are 38 numbers, I pick 7. What are probabilities I win exactly one number, no more and no less.
Is this possible to calculate.
It would be simple to calculate what are the chances I don't win any number, but how can you calculate chances of winning exactly one?
probability
asked Jan 16 at 19:28
StengaStenga
296
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add a comment |
$begingroup$
There are 38 numbers, I pick 7. What are probabilities I win exactly one number, no more and no less.
Is this possible to calculate.
It would be simple to calculate what are the chances I don't win any number, but how can you calculate chances of winning exactly one?
probability
asked Jan 16 at 19:28
StengaStenga
296
$endgroup$
$begingroup$
To be clear, are you saying that you choose 7 numbers between 1 and 38, and there are 7 winning numbers, and there is exactly one match? If so, are repetitions allowed i.e. can you choose (1,1,1,1,1,1,1)?
$endgroup$
– pwerth
Jan 16 at 19:30
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@pwerth: yes, 7 numbers between 1 and 38 and no repetions are allowed. If 16 is selected, it's eliminated.
$endgroup$
– Stenga
Jan 16 at 19:32
add a comment |
$begingroup$
There are 38 numbers, I pick 7. What are probabilities I win exactly one number, no more and no less.
Is this possible to calculate.
It would be simple to calculate what are the chances I don't win any number, but how can you calculate chances of winning exactly one?
probability
asked Jan 16 at 19:28
StengaStenga
296
$endgroup$
There are 38 numbers, I pick 7. What are probabilities I win exactly one number, no more and no less.
Is this possible to calculate.
It would be simple to calculate what are the chances I don't win any number, but how can you calculate chances of winning exactly one?
probability
probability
asked Jan 16 at 19:28
StengaStenga
296
asked Jan 16 at 19:28
StengaStenga
296
asked Jan 16 at 19:28
StengaStenga
296
asked Jan 16 at 19:28
StengaStenga
296
asked Jan 16 at 19:28
StengaStenga
296
296
$begingroup$
To be clear, are you saying that you choose 7 numbers between 1 and 38, and there are 7 winning numbers, and there is exactly one match? If so, are repetitions allowed i.e. can you choose (1,1,1,1,1,1,1)?
$endgroup$
– pwerth
Jan 16 at 19:30
$begingroup$
@pwerth: yes, 7 numbers between 1 and 38 and no repetions are allowed. If 16 is selected, it's eliminated.
$endgroup$
– Stenga
Jan 16 at 19:32
add a comment |
$begingroup$
To be clear, are you saying that you choose 7 numbers between 1 and 38, and there are 7 winning numbers, and there is exactly one match? If so, are repetitions allowed i.e. can you choose (1,1,1,1,1,1,1)?
$endgroup$
– pwerth
Jan 16 at 19:30
$begingroup$
@pwerth: yes, 7 numbers between 1 and 38 and no repetions are allowed. If 16 is selected, it's eliminated.
$endgroup$
– Stenga
Jan 16 at 19:32
$begingroup$
To be clear, are you saying that you choose 7 numbers between 1 and 38, and there are 7 winning numbers, and there is exactly one match? If so, are repetitions allowed i.e. can you choose (1,1,1,1,1,1,1)?
$endgroup$
– pwerth
Jan 16 at 19:30
$begingroup$
To be clear, are you saying that you choose 7 numbers between 1 and 38, and there are 7 winning numbers, and there is exactly one match? If so, are repetitions allowed i.e. can you choose (1,1,1,1,1,1,1)?
$endgroup$
– pwerth
Jan 16 at 19:30
$begingroup$
@pwerth: yes, 7 numbers between 1 and 38 and no repetions are allowed. If 16 is selected, it's eliminated.
$endgroup$
– Stenga
Jan 16 at 19:32
$begingroup$
@pwerth: yes, 7 numbers between 1 and 38 and no repetions are allowed. If 16 is selected, it's eliminated.
$endgroup$
– Stenga
Jan 16 at 19:32
add a comment |
1 Answer
1
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You have chosen $7$ numbers. In order for you to match exactly one number, the winning numbers must include $6$ of the $31$ numbers you did not choose.
$$frac{7cdotbinom{31}{6}}{binom{38}{7}}approx40.839%$$
answered Jan 16 at 19:38
Daniel MathiasDaniel Mathias
1,40518
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add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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active
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votes
$begingroup$
You have chosen $7$ numbers. In order for you to match exactly one number, the winning numbers must include $6$ of the $31$ numbers you did not choose.
$$frac{7cdotbinom{31}{6}}{binom{38}{7}}approx40.839%$$
answered Jan 16 at 19:38
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
add a comment |
$begingroup$
You have chosen $7$ numbers. In order for you to match exactly one number, the winning numbers must include $6$ of the $31$ numbers you did not choose.
$$frac{7cdotbinom{31}{6}}{binom{38}{7}}approx40.839%$$
answered Jan 16 at 19:38
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
add a comment |
$begingroup$
You have chosen $7$ numbers. In order for you to match exactly one number, the winning numbers must include $6$ of the $31$ numbers you did not choose.
$$frac{7cdotbinom{31}{6}}{binom{38}{7}}approx40.839%$$
answered Jan 16 at 19:38
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
You have chosen $7$ numbers. In order for you to match exactly one number, the winning numbers must include $6$ of the $31$ numbers you did not choose.
$$frac{7cdotbinom{31}{6}}{binom{38}{7}}approx40.839%$$
answered Jan 16 at 19:38
Daniel MathiasDaniel Mathias
1,40518
answered Jan 16 at 19:38
Daniel MathiasDaniel Mathias
1,40518
answered Jan 16 at 19:38
Daniel MathiasDaniel Mathias
1,40518
answered Jan 16 at 19:38
Daniel MathiasDaniel Mathias
1,40518
1,40518
add a comment |
add a comment |
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$begingroup$
To be clear, are you saying that you choose 7 numbers between 1 and 38, and there are 7 winning numbers, and there is exactly one match? If so, are repetitions allowed i.e. can you choose (1,1,1,1,1,1,1)?
$endgroup$
– pwerth
Jan 16 at 19:30
$begingroup$
@pwerth: yes, 7 numbers between 1 and 38 and no repetions are allowed. If 16 is selected, it's eliminated.
$endgroup$
– Stenga
Jan 16 at 19:32