So, only add an even number of moves to the last known good "Total moves". The "Right Moves" and "Left Moves" formulae will result in non-integers, which will result in a wrong (often >1) "Binomial Probability".
Therefore, you cannot have an od distance from the start with an even number of total steps.īug 1: An odd "Final Position" causes fractional (impossible) moves and a jaggy (wrong) graph.įix: Change the "Total moves" formula from =C8+2*$C$6 to =C8+MROUND($C$6,2).Įxplanation: You can't do an odd number of moves and end up in the same position. If S is even, then subtracting any multiple of 2 will keep it even. If S is odd, then subtracting any multiple of 2 will keep it odd. Since R and S have to be whole numbers, to get all the possible results for D, you take the total number of steps minus all multiples of 2. the distance from the start is the total number of steps minus 2 times the number of steps right. Substitute this into the distance equation:ĭ = |S-R-R| or, to simplify it, D = |S-2R|
Now, total distance from the start (D) = |L-R|Īnd since S = L+R, solve for L and you get L = S-R Since you have an even number of steps (10, 20, 30,etc.), each one being exactly one unit in directly opposing directions (so that we could say that 1 step right is equivalent to -1 steps left), you cannot finish with an odd number of units away from the center, given an even number of total steps. 25:29, the convergence chart seems to have an error whenever you input the probability of finishing an odd number of steps away from the start point.
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