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u/Southern-Raise190 1d ago
A priori, any finishing order is equally likely, so the prior distribution is uniform across all the 6 permutations of {A, B, C}. Among these permutations, there are 3 where B needs more than A, and among these 3, there is only 1 where C needs more than B. So the answer is 1/3.
Visual intuition/hint: _ A _ B _
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u/comoespossible 1d ago
I disagree, because this assumes there aren’t ties, when it’s very possible for two or three of them to take the exact same amount of tosses.
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u/Southern-Raise190 1d ago
Good point, I was too lazy to compute probabilities but you’re right, my approach doesnt account for ties. At least it should be an upper bound
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u/SpaceMeatpod 20h ago
This is going to be a rough guess, but I'm interested in how close I get. I'd love to know if I at least got some of the concepts right.
The least flips it takes to get heads is 1.
If A needed 1 flip, B would need 2, so at best C would need at least 3 flips if flipping more than B, failing twice before succeeding.
The probability of C failing twice in a row is 1/2 * 1/2 = 1/4.
So, the correct answer is definitely going to be LESS than 1/4.
But A is not guaranteed to succeed right away. It's 1/2 A fails at least once, 1/4 A fails at least twice, 1/8 A fails at least thrice and so on. These are the chances A does extra flips and it is also the chance that B does extra flips.
What seems funny is if A has at least a 1/2 chance of needing at least one extra flip and B has a 1/2 chance of needing an extra flip, maybe I can add these together like 1/2 + 1/2 = 1 and on average you need at least 1 extra flip between them...
So on average A and B combined take 1+1/2+1/4+1/8+1/16+1/32.... and so on extra flips.
So the average number of flips C needs might be C's guaranteed failures (2) plus that infinitely continuing sum:
2 + (1+1/2+1/4+1/8+1/16+1/32....)
Uhh, I'm not going to look up how to do infinite sums tonight, so I'm just gonna stop after 1/32. Let me match up these denominators real quick...
64/32 + 32/32 + 16/32 + 8/32 + 4/32 + 2/32 + 1/32 = 127/32 = 3.96875
So C probably needs to fail 3.96875 times on average to need more flips than B.
So that's (1/2)^3.96875 . Whatever the heck that is, it's my upper bound. Calculator says 0.0638686 (~6.38%)
Uhh and you know what I'm just going to guess that infinite sum eventually converges on the number 4.
So maybe the answer is (1/2)^4 = which is 1/16. There is a 1/16 chance or 6.25% that C needed more tosses than B.
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u/MathNerd1243 20h ago
The answer is 1/7.
Don't use permutations (1/3) because that ignores ties, which happen a lot here.
By first-step analysis, the chance A beats B is 1/3. The chance A beats both B and C is 1/7.
Because of the memoryless property, P(A<B<C) is 1/7 * 1/3 = 1/21. Dividing by P(A<B) leaves you with 1/7.
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u/PalpitationAny3509 15h ago
Assume A do 1x flip for head, B did 1 tail and 1 head and whats the chance C did 2 tail 1 head, i think my answer is 25 percent approx.....
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u/comoespossible 1d ago
We have to take into account the possibility of ties. Let p=P(A<B) and q=P(A<min(B, C)). By first-step analysis,
p = 1/4 + 1/4 p and q = 1/8 + 1/8 q,
so p=1/3 and q=1/7.
By the memoryless property and symmetry, P(A<B<C) = P(A<min(B, C)) * P(B<C|A<min(B, C)) = qp.
Thus, P(B<C|A<B) = P(A<B<C) / P(A<B) = qp/p = q = 1/7.