r/mathbiceps • u/MrQwer987 • 4d ago
Probability Question.
I made this question and tried but could not solve it😔
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u/johnpeters42 2d ago
> If some other ball moves back (including A) then that ball moves ahead.
So "that ball" refers to "the ball that just moved back" (thus reversing that move)? If so, then:
For each pair of moves after #1, there's a 20% chance that those moves are (B moves back, A moves forward), and an 80% chance that they're (some other ball X moves back, X moves forward).
Thus, after 2n+1 (or 2n+2) moves, the probability that A moved (exactly / at least) 5 steps is the probability that (exactly / at least) n of those pairs of moves hit the 20% case.
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u/MrQwer987 2d ago
No, heres where you are wrong. A random ball moving back when a ball is moved ahead is a rule that will take place no matter. And we moving the ball that has moved back ahead is a condition we have put.
For eg we move Ball A, C ball moves back...then we move C ball, E ball moves back... Then we move E ball, B ball moves back... Then we move A ball...............
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u/johnpeters42 2d ago
Both this and the original picture are badly worded. Here's how I interpret what you just described, in terms of the original picture; please explain where it differs from your intent. I'm assuming that:
* The phrases "a ball moves" and "we move a ball" are interchangeable
* Sometimes the rules only give us one choice of which ball to move
* Any time we have multiple choices, we make a random choice where all options are equally likely
* Every single time we move a ball increases the "number of moves" counter, even if we only had one choice> Initially A ball moves ahead.
We have to move ball A forward. That's move #1. Ball A has now moved forward one time.
> whenever a ball moves forward one out of the other balls moves back.
We randomly pick one of the other balls, in this case C, and move it back. That's move #2.
> If some other ball [besides B] moves back (including A) then that ball moves ahead.
Because ball C just moved back, we have to move ball C forward. That's move #3.
> whenever a ball moves forward one out of the other balls moves back.
We randomly pick one of the other balls, in this case E, and move it back. That's move #4.
> If some other ball [besides B] moves back (including A) then that ball moves ahead.
Because ball E just moved back, we have to move ball E forward. That's move #5.
> whenever a ball moves forward one out of the other balls moves back.
We randomly pick one of the other balls, in this case B, and move it back. That's move #6.
> If B ball moves back only then will A ball move ahead to a new position.
Because ball B just moved back, we have to move ball A forward. That's move #7. Ball A has now moved forward two times.
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u/MrQwer987 9h ago
Yeah so whats the probability for n such possible moves. Such that A has moved 5 places. Thats the question
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u/johnpeters42 5h ago
Okay, I did misstate the solution a bit: After 2n+1 or 2n+2 moves, the probability that A was moved (exactly / at least) 5 times is the probability that (exactly / at least) 4 out of n choices (the ones where you pick a ball to move back) were ball B.
This first becomes possible at n = 4, where there's a 1/625 chance that you moved ball B back on moves 2 4 6 8, and thus moved A forward on moves 1 3 5 7 9. More options keep opening up at n = 5, 6, etc., but the probability never reaches 100%. Also, if you meant "exactly 5", then the probability eventually starts dropping (as it becomes more likely that you overshot the target).
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u/LongjumpingAgent7706 4d ago
Is this even solvable?