@markr,

I think I solved it for the most of it, but i need help with b), the numbers don't add up to a equivalent verification. Look after the explanations for your last post below.

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In your b^i example, the player also didn't have a 1, but you called it a 6-ball match.

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You're right, the player can't ever match all 6, I ignored the 2, my bad

lottery units: 1-2 4 6 8 8

player's units: 1 4 6 8 8 8

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How do you figure? There are four zeros. You can't draw five or six zeros. You're not replacing; you just happen to have multiples of each digit.

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Right again, there can only be 4 repeated zeroes and 5 of other units from 0 to 9, that was a quick example though. This means we must proportion the probability of b) extracting 1/5 from 1/10 of the whole, because 0 (zero) lacks a fifth appearance in the sample space.

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Correct me if I'm wrong please; I've yanked standard approach by the hair a bit here

the solution for a) should be:

a) because we're having sampling w/o replacement for the distinct units part, e.g.

lottery units: 1-2 4 6 8

player units: 1 4 6 8

we can apply the hypergeometric distribution:

Player) because it picks with 5 distinct units out of which 4 match 4 out of 5 distinct units of the lottery, we have C(5,4) way to pick success, C(10-5, 1) way to fail (that's the unit that always fails actually), and total sample space is C(10,5), thus the probability is:

C(5,4)*C(10-5,1)/C(10,5)=25/252

Lottery) because it picks with 5 distinct units out of which 4 match 4 out of 5 distinct units of the player, we have C(5,4) way to pick success, C(10-5, 1) way to fail (that's the unit that always fails actually), and total sample space is C(10,5), thus the probability is:

C(5,4)*C(10-5,1)/C(10,5)=25/252

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Now for b^i and b^ii, they are one and the same case (separating them is rather irrelevant to the problem I think now), that is from the remaining two units out of the 6, only one is able to match with the repeated unit of the lottery, or both.

*Because we're sampling with replacement in this case, we can use normal probability math, and corroborate with the binomial distribution afterwards.

Now the problem is that the player has one more unit, that was used to help the probability of the sampling w/o replacement at a). Should this unit contribute to the probability of the sampling w/ replacement b)? Since it fails to be a distinct unit to be more exact in a) I think yes, it should contribute to the probability of b) because it can only be a repeated unit in accord with the rule that the player set of units contain no adjacent or sequenced units.

The question is then, what is the probability that 2 random units will hit 1 unit picked out of 4 random units from 0 to 9. Probability is (4/10) + (4/10)=8/10=4/5?

4/10 is from one of the two player units that will be repeated with a unit from the player set of units that matches the lottery units (they can't really be repeated with anything else but that set in actuality), and has 1/10 chance of being any of the units from 0 to 9, times 4 units that match between player and lottery (it can match in 4 ways a matching unit between the lottery and the player), which is in total: 4/10 chance times 2 tries = 4/5.

Viceversa, the probability that the repeated unit in the lottery set hits only the matching units (and also repeated units of the player) between the lottery and the player is 4/10 for one of the player's repeated units, 4/5 for both, and 1/10 chance that it will match the unit that doesn't match with the player. This means 4/5 of the time the player can match 5 units (4 distinct 1 repeated) with the lottery, and 1/5 of the time only 4 distinct units.

The respective probabilities are 4/5 probability *4/5 of the time = 16/25 and 1/5 of the time * 1/10 probability = 1/50, where the total probability of success is (80*16/25 + 20*1/50)/100=0.516

Another approximation is the binomial distribution:

C(10,5)[80*(4/5)^5] *[20*(5/5-4/5)^5]=42.27858432/100=0.423... (rather far apart from the mean approximation). A little help here with verification?

The first attempt before verifying it with the player side, and I thought the probability was 4/10 not 4/5, the average was 0.34 and the binomial was 3.2105299968/10?=0.321... like this:

C(10,5)[80*(4/10)^5] *[20*(10/10-4/10)^5]

Seems to me that the lottery's timing with whether it repeats the 6th unit with the adjacent unit that doesn't match with the player is an absolute matter in the account.

Verification from player's point of view:

4/5 chance times 4/5 of the time success = 16/25 chance for success

1/10 times 1/5 of the time success = 1/50 chance for failure

(80*16/25 + 20*1/50)/100=

=(16*16/5 + 2/5)/100=0.516

I could adjust the above probability with minus 2% here, for zero not making a fifth appearance, which is 10th part (1/10) times a fifth (1/5) = 1/50, and because we're talking continuous distribution here on the long run, that would mean minus the fiftieth part of the whole, which is 2% I think.

For now, I'm interested in matching 5 units between player and lottery, which is optimal. Combining the probability w/o replacements with the one w/ replacements is irrelevant because the repeated units can't be avoided for once, and the ratio I need can be extracted from the probabilities w/o replacement for sequenced units alone, which was the first problem. Not to mention it would be hard to prove the combination of probabilities, even if correct

Total probability would be:

(25/252) *0.516=0.05112... (1 sequenced units and one repeated units, matching)

and the verification would be:

(3)(22)and(2)repeated units occur:

96% of the time with an ~0.2326... probability=~0.223296...

times

?% of the time sequenced units with a ? probability

(haven't done sequenced units calculations yet)

~0.223296 * x = 0.05112

x=0.05112/~0.223296 =0.2289... probability of 2 sequenced units ?% of the time (perhaps about 96% too)

which looks good to me at this point, though not verified yet thoroughly and scientifically.