Mixing
Pick 5 numbers - calculate probability that respective 4-number combinations appear for $n-th$ time (given...
$begingroup$
We pick 5 unique (distinct) numbers out of [1..36] numbers (like playing a lottery). Each time we pick 5 unique numbers ("play one ticket") thus we select 5 distinct combinations of 4 numbers out of these 5 (let call such a combination - "tuple4").
We repeat picking 5 unique (distinct) numbers ("play one ticket") many times. After that we average how often certain tuple4 appeared.
It turns out that in N plays there were
- 21322 (66%) distinct tuple4 that appeared exactly once (1 time)
- 8335 (26%) distinct tuple4 that appeared exactly 2 times (none of tuples from item above included here, same applies below)
- 2171 (7%) distinct tuple4 that appeared exactly 3 times
- 481 (1%) - distinct tuple4 that appeared exactly 4 times
Percentage is taken from total number of tuple4 appearances (that is: $N*5$).
Now we "play" again (pick 5 unique (distinct) numbers) - what is the probability that among 5 distinct tuple4 that sach a play generates:
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - we don't care (also for the first time or for any $k-th$ time)- Exactly 2 would appear for the first time and any of the remaining 3 would appear for second or third time (but not for first or 4th+ time)
How can I construct such formulas?
P.S. In all 376'992 possible combinations of 5 numbers from [1..36] there are 58'905 possible distinct tuple4. But I think this information and number of plays (N) is not needed. In my case N = 9400.
My thinking was like
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - we don't care (also for the first time or for any $k-th$ time): $(0.66)*(0.66)*(1-0.66)*((1-0.66)+0.66)*((1-0.66)+0.66) =$ 15%
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - only for the second time or more (but not for the first time): $(0.66)*(0.66)*(1-0.66)*((1-0.66))*((1-0.66)) = $ 1,7%
But these results don't seem to reflect what happens really (I did a good random-number computer simulation), so I think I made a mistake somewhere.
probability combinatorics probability-theory probability-distributions conditional-probability
asked Jan 10 at 12:09
user10777718user10777718
1464
$endgroup$
add a comment |
$begingroup$
We pick 5 unique (distinct) numbers out of [1..36] numbers (like playing a lottery). Each time we pick 5 unique numbers ("play one ticket") thus we select 5 distinct combinations of 4 numbers out of these 5 (let call such a combination - "tuple4").
We repeat picking 5 unique (distinct) numbers ("play one ticket") many times. After that we average how often certain tuple4 appeared.
It turns out that in N plays there were
- 21322 (66%) distinct tuple4 that appeared exactly once (1 time)
- 8335 (26%) distinct tuple4 that appeared exactly 2 times (none of tuples from item above included here, same applies below)
- 2171 (7%) distinct tuple4 that appeared exactly 3 times
- 481 (1%) - distinct tuple4 that appeared exactly 4 times
Percentage is taken from total number of tuple4 appearances (that is: $N*5$).
Now we "play" again (pick 5 unique (distinct) numbers) - what is the probability that among 5 distinct tuple4 that sach a play generates:
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - we don't care (also for the first time or for any $k-th$ time)- Exactly 2 would appear for the first time and any of the remaining 3 would appear for second or third time (but not for first or 4th+ time)
How can I construct such formulas?
P.S. In all 376'992 possible combinations of 5 numbers from [1..36] there are 58'905 possible distinct tuple4. But I think this information and number of plays (N) is not needed. In my case N = 9400.
My thinking was like
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - we don't care (also for the first time or for any $k-th$ time): $(0.66)*(0.66)*(1-0.66)*((1-0.66)+0.66)*((1-0.66)+0.66) =$ 15%
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - only for the second time or more (but not for the first time): $(0.66)*(0.66)*(1-0.66)*((1-0.66))*((1-0.66)) = $ 1,7%
But these results don't seem to reflect what happens really (I did a good random-number computer simulation), so I think I made a mistake somewhere.
probability combinatorics probability-theory probability-distributions conditional-probability
asked Jan 10 at 12:09
user10777718user10777718
1464
$endgroup$
$begingroup$
Not sure I got question number 1. Do you mean: what is the probability that at least 3 of the 5 new 4-tuples never appeared in the previous last $N$ game sessions?
$endgroup$
– Matteo
Jan 10 at 14:14
$begingroup$
$21322+8335cdot 2+2171cdot 3+481cdot 4=46429$ occurrences. The total number of occurrences is $Ncdot 5=9400cdot 5=47000$, so $571$ occurrences are missing. These are the 4-tuples that occurred more than 4 times, but we do not know how many distinct 4-tuples this represents. As such, we cannot determine how many distinct 4-tuples have not yet appeared or find the probability that some of these will for the first time with the next pick.
$endgroup$
– Daniel Mathias
Jan 10 at 14:40
add a comment |
$begingroup$
We pick 5 unique (distinct) numbers out of [1..36] numbers (like playing a lottery). Each time we pick 5 unique numbers ("play one ticket") thus we select 5 distinct combinations of 4 numbers out of these 5 (let call such a combination - "tuple4").
We repeat picking 5 unique (distinct) numbers ("play one ticket") many times. After that we average how often certain tuple4 appeared.
It turns out that in N plays there were
- 21322 (66%) distinct tuple4 that appeared exactly once (1 time)
- 8335 (26%) distinct tuple4 that appeared exactly 2 times (none of tuples from item above included here, same applies below)
- 2171 (7%) distinct tuple4 that appeared exactly 3 times
- 481 (1%) - distinct tuple4 that appeared exactly 4 times
Percentage is taken from total number of tuple4 appearances (that is: $N*5$).
Now we "play" again (pick 5 unique (distinct) numbers) - what is the probability that among 5 distinct tuple4 that sach a play generates:
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - we don't care (also for the first time or for any $k-th$ time)- Exactly 2 would appear for the first time and any of the remaining 3 would appear for second or third time (but not for first or 4th+ time)
How can I construct such formulas?
P.S. In all 376'992 possible combinations of 5 numbers from [1..36] there are 58'905 possible distinct tuple4. But I think this information and number of plays (N) is not needed. In my case N = 9400.
My thinking was like
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - we don't care (also for the first time or for any $k-th$ time): $(0.66)*(0.66)*(1-0.66)*((1-0.66)+0.66)*((1-0.66)+0.66) =$ 15%
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - only for the second time or more (but not for the first time): $(0.66)*(0.66)*(1-0.66)*((1-0.66))*((1-0.66)) = $ 1,7%
But these results don't seem to reflect what happens really (I did a good random-number computer simulation), so I think I made a mistake somewhere.
probability combinatorics probability-theory probability-distributions conditional-probability
asked Jan 10 at 12:09
user10777718user10777718
1464
$endgroup$
We pick 5 unique (distinct) numbers out of [1..36] numbers (like playing a lottery). Each time we pick 5 unique numbers ("play one ticket") thus we select 5 distinct combinations of 4 numbers out of these 5 (let call such a combination - "tuple4").
We repeat picking 5 unique (distinct) numbers ("play one ticket") many times. After that we average how often certain tuple4 appeared.
It turns out that in N plays there were
- 21322 (66%) distinct tuple4 that appeared exactly once (1 time)
- 8335 (26%) distinct tuple4 that appeared exactly 2 times (none of tuples from item above included here, same applies below)
- 2171 (7%) distinct tuple4 that appeared exactly 3 times
- 481 (1%) - distinct tuple4 that appeared exactly 4 times
Percentage is taken from total number of tuple4 appearances (that is: $N*5$).
Now we "play" again (pick 5 unique (distinct) numbers) - what is the probability that among 5 distinct tuple4 that sach a play generates:
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - we don't care (also for the first time or for any $k-th$ time)- Exactly 2 would appear for the first time and any of the remaining 3 would appear for second or third time (but not for first or 4th+ time)
How can I construct such formulas?
P.S. In all 376'992 possible combinations of 5 numbers from [1..36] there are 58'905 possible distinct tuple4. But I think this information and number of plays (N) is not needed. In my case N = 9400.
My thinking was like
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - we don't care (also for the first time or for any $k-th$ time): $(0.66)*(0.66)*(1-0.66)*((1-0.66)+0.66)*((1-0.66)+0.66) =$ 15%
At least 3 would appear for the first time (thinking of all last N plays), remaining 2 - only for the second time or more (but not for the first time): $(0.66)*(0.66)*(1-0.66)*((1-0.66))*((1-0.66)) = $ 1,7%
But these results don't seem to reflect what happens really (I did a good random-number computer simulation), so I think I made a mistake somewhere.
probability combinatorics probability-theory probability-distributions conditional-probability
probability combinatorics probability-theory probability-distributions conditional-probability
asked Jan 10 at 12:09
user10777718user10777718
1464
asked Jan 10 at 12:09
user10777718user10777718
1464
asked Jan 10 at 12:09
user10777718user10777718
1464
asked Jan 10 at 12:09
user10777718user10777718
1464
asked Jan 10 at 12:09
user10777718user10777718
1464
1464
$begingroup$
Not sure I got question number 1. Do you mean: what is the probability that at least 3 of the 5 new 4-tuples never appeared in the previous last $N$ game sessions?
$endgroup$
– Matteo
Jan 10 at 14:14
$begingroup$
$21322+8335cdot 2+2171cdot 3+481cdot 4=46429$ occurrences. The total number of occurrences is $Ncdot 5=9400cdot 5=47000$, so $571$ occurrences are missing. These are the 4-tuples that occurred more than 4 times, but we do not know how many distinct 4-tuples this represents. As such, we cannot determine how many distinct 4-tuples have not yet appeared or find the probability that some of these will for the first time with the next pick.
$endgroup$
– Daniel Mathias
Jan 10 at 14:40
add a comment |
$begingroup$
Not sure I got question number 1. Do you mean: what is the probability that at least 3 of the 5 new 4-tuples never appeared in the previous last $N$ game sessions?
$endgroup$
– Matteo
Jan 10 at 14:14
$begingroup$
$21322+8335cdot 2+2171cdot 3+481cdot 4=46429$ occurrences. The total number of occurrences is $Ncdot 5=9400cdot 5=47000$, so $571$ occurrences are missing. These are the 4-tuples that occurred more than 4 times, but we do not know how many distinct 4-tuples this represents. As such, we cannot determine how many distinct 4-tuples have not yet appeared or find the probability that some of these will for the first time with the next pick.
$endgroup$
– Daniel Mathias
Jan 10 at 14:40
$begingroup$
Not sure I got question number 1. Do you mean: what is the probability that at least 3 of the 5 new 4-tuples never appeared in the previous last $N$ game sessions?
$endgroup$
– Matteo
Jan 10 at 14:14
$begingroup$
Not sure I got question number 1. Do you mean: what is the probability that at least 3 of the 5 new 4-tuples never appeared in the previous last $N$ game sessions?
$endgroup$
– Matteo
Jan 10 at 14:14
$begingroup$
$21322+8335cdot 2+2171cdot 3+481cdot 4=46429$ occurrences. The total number of occurrences is $Ncdot 5=9400cdot 5=47000$, so $571$ occurrences are missing. These are the 4-tuples that occurred more than 4 times, but we do not know how many distinct 4-tuples this represents. As such, we cannot determine how many distinct 4-tuples have not yet appeared or find the probability that some of these will for the first time with the next pick.
$endgroup$
– Daniel Mathias
Jan 10 at 14:40
$begingroup$
$21322+8335cdot 2+2171cdot 3+481cdot 4=46429$ occurrences. The total number of occurrences is $Ncdot 5=9400cdot 5=47000$, so $571$ occurrences are missing. These are the 4-tuples that occurred more than 4 times, but we do not know how many distinct 4-tuples this represents. As such, we cannot determine how many distinct 4-tuples have not yet appeared or find the probability that some of these will for the first time with the next pick.
$endgroup$
– Daniel Mathias
Jan 10 at 14:40
add a comment |
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$begingroup$
Not sure I got question number 1. Do you mean: what is the probability that at least 3 of the 5 new 4-tuples never appeared in the previous last $N$ game sessions?
$endgroup$
– Matteo
Jan 10 at 14:14
$begingroup$
$21322+8335cdot 2+2171cdot 3+481cdot 4=46429$ occurrences. The total number of occurrences is $Ncdot 5=9400cdot 5=47000$, so $571$ occurrences are missing. These are the 4-tuples that occurred more than 4 times, but we do not know how many distinct 4-tuples this represents. As such, we cannot determine how many distinct 4-tuples have not yet appeared or find the probability that some of these will for the first time with the next pick.
$endgroup$
– Daniel Mathias
Jan 10 at 14:40