Introducing Scooby Snacks!

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It took me a while to find the solution to your probability problem but now I can give you a pretty realistic solution.The math of probability problems is generally simple, simple arithmetic, but the difficulty lies with the definitions of the probabilities. In this case we are talking of so called 'conditional probabilities'. This means that the probability of a certain outcome is conditioned by another probability. To obtain the final probabilty of any outcome or event you have to multiply the two probabilities. And there is one more important aspect. If multiple outcomes constitute the event we are looking for, in our case, that one of the four 'second cards'- I call them 'picks' - is in near of one of the 'first cards' - I call them 'targets' - then the probabilities have to be summed up (the probability of the first pick OR the probability of the second pick OR ... third...OR...fourth pick).
Now the essential part: The number of probabiblities of the picks to be near the targets is first conditioned by the probalities of the targets not to overlay each other: If the targets lay nicely far from each other than each has 2 places on each side for the probability of the picks to fall there. So totally we have 4x4=16 places where one of the 4 picks can end up. So we can define the total probability that one or the other of the 4 picks to fall near the 4 targets as:
P=P(0)x16C+P(1)x15C+P(2)x14C+P(3)x13C+...+P(10)x6C+P(11)x5C+P(12)x4C, where:
P(0), P(1), P(2),...P(12) are the probabilities of the targets to be placed with 0,1,2,...11 or max 12 overlapping target places for the picks and
C=1/48+1/47+1/46+1/45=0.0859 This are the probabilities of the picks to fall near one of the targets because out of the total 52 places, 4 are accupied by the targets so the first pick has a chance of 1/52-4=1/48, the next has 1/52-5=1/47... to fall near one of the target places. That's why if the target areas have no overlaps there are 4x4=16 places near the targets, with one overlap we have 16-1=15 target places, with 2 overlaps there are 16-2=14 places, a.s.o. The max target areas overlaps, when all 4 targets are side by side then there are only 2+2=4 target places left for the picks so the sum of probabilities ends with P(12)x4C.
Now, we have to define the overlap probabilities for the targets as follows:
Each target has 4 places on each side where no other target should fall in order to avoid any overlap areas for the picks. That's why the target no-overlap probability is: P(0)=[(51-8)/51]x[(50-16)/50]x[(49-24)/49]=0.2925 because the second target has 51-8 places out of the available 51 to avoid the overlap with the first target, the third target has 50-16 places to avoid the first two target areas and the fourth target has 49-24 places for the same reason.
Now, there is one more important thing we have to define. P(1), P(2), P(3) and P(4) - these are target areas overlaps of 1, 2, 3 or 4 places (at 4 overlap two targets will lay side by side) we define as follows:
P(1)=P(2)=P(3)=P(4)=[(51-8)/51]x[(50-16)/50]x(6/49)=0.0702 Why 6/49, because for these configurations one of the targets has to be placed in one of the 6 places which are around the already 3 existing no-overlaping targets in order to have 1, 2, 3 or 4 overlaps (each of the first 3 no-overlaping targets has only 2 places, one on each side, where the fourth target has to be placed in order to achieve these configurations).
The rest of the 8 target overlaping probabilities we define differently: We start with the extreme case that all the four targets are side by side:
P(12)=(6/51)x(2/50)x(1/49) because the second target has to fall max 3 places around the first, the third target has only 2 places available and the fourth one has only 1 place left to fall into.
P(11)=(8/51)x(3/50)x(2/49), P(10)=(10/51)x(4/50)x(3/49), P(9)=(12/51)x(5/50)x(4/49), ...until P(5)=(20/51)x(9/50)x(8/49). These probalilities are extremely small and for the purpose of our problem I practically ignore them (the 'largest', P(5)=0.0115 and the rest, P(6,7,8,9,10 and 11) are even way smaller).
So, now we can sum up the final probabilty for our task:
P = P(0)x16C + P(1)x15C + P(2)x14C + P(3)x13C + P(4)x12C = 0.4020 + 0.0904 + 0.0844 + 0.0783 + 0.0723 = 0.7274 or approx 73%. Which looks like a correct probality according to your practical experience.
So, you can see, you definetely don't need 'Chaos Theory' to solve the problem !
I hope this will be satisfactory explanation of your practical experience.
Regards,
Gabe.-