So, that is how I spent my first hours of the New Year!!
I was wondering how to start the year. Figured, let’s do something different.
So, I sat down with the last puzzle I had given and after two hours of putting pen to the paper, finally solved the probability problem. Whew!!! hopefully that would be some mental exercise for the little of whatever remains of the grey cells. Here is the answer…..
(BTW, anybody who cares to actually go thru the solution, please let me know if I have made any calculation errors. I am sure about the logic. And I have double checked the LCMs, GCMs and all that… just in case, I still made a silly mistake, point it out please)
The three people A,B,C get a random sequence, of which there can be 6. Let’s look at those 6 cases one by one.
CASE 1. Sequence: ABC.
Note that A, who goes first, would ether shoot B or C (random). If he shoots B, B is dead (since A is a sure shot). Prob of B surviving is 0. C then shoots at A. Chance of A being shot is 1/2 which means C survives. But if C misses (prob 1/2), A will for sure shoot him in his chance next. So, A, B and C prob are 1/2, 0, 1/2
If A shoots at C, C is dead, B then shoots at A. Prob of A surviving is 1/5 (since B is 80% accurate) but after that A will, for sure, kill B. So, the prob of A, B and C surviving are 1/5, 4/5, 0
Since the target for A is chosen in random, both the cases are equally likely. Therefore weighting the prob equally between the two events, the Case 1 prob for A, B and C surviving are 7/20, 2/5, 1/4. [7/20 equals half of 1/2 plus half of 1/5 etc etc etc)
CASE 2. Sequence: ACB.
You will note that this is same as Case 1. A either shoots B dead first and C shoots at A or A shoots C dead first and then B shoots at A. If A survives in either case, he finishes off C or B as the case might be. Exact situation as above. So the prob for A, B and C are 7/20, 2/5, 1/4.
Before I go to next cases, let us look at the pattern where A is dead (unlike the above case). So we have B and C and one of them will go first.
B and C are there alive and B shoots first. The prob of B surviving is when
(*) B shoots C dead with first shot – 4/5 OR
(*) B misses (1/5), but then C misses too (1/2) and then in the second chance, B succeeds (4/5) OR
(*) the same sequence as before except in the last step B misses (1/5) and then again C misses (1/2) and in the third chance B succeeds (4/5) and so on.
That prob would be 4/5 + 1/5×1/2×4/5 + 1/5×1/2×1/5×1/2×4/5 + …. which equals to (4/5) x (1+1/10 + 1/100+…) which equasl to (4/5) x 1.11111… which equals to 0.888888… which is 8/9. So B is 8/9 and C is 1/9.
B and C are there alive but this time C shoots first. The prob of C surviving is when
(*) C succeeds with first shot (1/2) OR
(*) C misses (1/2) but then B also misses (1/5) and then C succeeds in second shot (1/2) OR
(*) same sequence as above except join the last step C misses (1/2) and then B misses again (1/5) and C succeeds in the now third shot (1/2).
That prob would work out to be 1/2 + 1/2×1/5×1/2 + 1/2*1/5*1/2*1/5*1/2+….. which is equal to (1/2) x (1 + 1/10 + 1/100 +…) which is equal to 1/2 x 1.1111… which is equal to 0.55555… which is 5/9. So B and C’s prob are 4/9 and 5/9
Let’s get back to our original cases.
CASE 3. Sequence: BAC.
B can shoot at A first. If he misses (1/5), then now it becomes ABC which is CASE 1. So A, B, C prob are 1/5 of CASE 1 which is 7/100, 2/25 and 1/20.
If however, he did kill A (4/5) now it is he and C left, which C going first – in other words, PATTERN Y. Multiplying PATTERN Y by 4/5, now we have A, B and C as 0, 16/45 and 4/9.
Adding these two cases, we get A, B and C as 7/100, 98/225 and 89/180
Or, B could have shot at C. If he did succeed (4/5), next A is going to shoot him dead. So the prob of surviving (after factoring is the 4/5 prob) are 4/5, 0, 0.
But if he failed in shooting C (1/5), now we have CASE 2!!! Therefore the prob are (1/5 of CASE 2) = 7/100, 2/25 and 1/20.
Adding the above two, we have 87/100, 2/25 and 1/20.
Since B would have picked A or C in random, the probability for CASE 3 is the average of the above two which would work out to be 47/100, 58/225, 49/180
CASE 4. Sequence BCA
I will come to this case later.
CASE 5. Sequence CAB
If C shoots at A then it could either be he succeeds (1/2) and then we have PATTERN X. Thus prob = 1/2 of (0, 8/9 and 1/9) = 0, 4/9 and 1/18
or C could have missed A (1/2) and then we have CASE 1!! Thus prob would be 7/40, 1/5 and 1/8.
Adding the above two, we have 7/40, 29/45 and 13/72
Or, if C shot at B first then it could have been that he either succeeded (1/2) in which case A would kill him next making the prob of A, B and C being 1/2, 0, 0
If C missed (1/2) then we again have CASE 1. Thus as calculated before, the prob are 7/40, 1/5 and 1/8.
Adding the above two, we have 27/40, 1/5 and 1/8.
Since C would have picked A or B in random, we weigh the above two cases equally and now have this Case prob as 17/40, 19/45 and 11/72.
CASE 6. Sequence CBA
If C shoots at B first and succeeds (1/2), A will shoot him dead for sure next. Thus we have the prob as 1/2, 0, and 0
If C however missed B (1/2), we have CASE 3!!!. That would be 1/2 of CASE 3 which is 47/200, 29/225 and 49/360
Combined, the prob are 147/200, 29/225 and 49/360
On the other hand, if C shot at A first,
If C succeeded (1/2), then we have PATTERN X which means the prob is 1/2 of PATTERN X making it 0, 4/9 and 1/18
If C however failed (1/2), we have now CASE 3!! Thus the prob would be 47/200, 29/225 and 49/360
Combined, the prob are 47/200, 129/225 and 69/360
Since C would have picked A or B at random, we average the above two to get the rolling prob: 97/200, 79/225 and 59/360
Now let’s go back to
CASE 4. Sequence BCA
If B shoots at A first and succeeds (4/5), we have PATTERN Y. Prob then is 0, 16/45 and 4/9
But if B failed in killing A (1/5), then we have CASE 5!! Prob then is 1/5 of CASE 5 which is 17/200, 19/225 and 11/360
Together that would be 17/200, 99/225 and 171/360
If B shoots at C first and succeeds (4/5), then next A will for sure kill him making the prob 4/5, 0, 0
But if B failed in shooting C dead (1/5), we have CASE 5 again! Prob, as above is 17/200, 19/225 and 11/360
Together that would be 177/200, 19/225 and 11/360
Since whether B shot at A first or C first was random, averaging the above two, we get the CASE 4 prob as 97/200, 59/225 and 91/360
So finally (phew) we have the following:
Sequence and prob of A or B or C surviving as:
ABC: 7/20, 2/5, 1/4
ACB: 7/20, 2/5, 1/4
BAC: 47/100, 58/225, 49/180
BCA: 97/200, 59/225, 91/360
CAB: 17/40, 19/45, 11/72
CBA: 97/200, 79/225, 59/360
Since that above are all equally likely, we finally take the average of those six cases to get to the final answer of A, B and C as: 513/1200 , 471/1350 and 483/2160 !!
Phew!!! Ok! Now I am ready for the New Year.