Editing Talk:2659: Unreliable Connection
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::The maths above is indeed correct enough. The 2^11 relates to the total number of unique paths it can take (assuming a bounce left/right just enough to strike the nearest offset pin below to force a new left/right bounce choice) from the first divider through to any of the 11 final left-right pin-bounces (and onto the 12 switches, at which point we're not bothered with the bouncing - diagram suggests the balls leap outwards and don't hit any other switches). | ::The maths above is indeed correct enough. The 2^11 relates to the total number of unique paths it can take (assuming a bounce left/right just enough to strike the nearest offset pin below to force a new left/right bounce choice) from the first divider through to any of the 11 final left-right pin-bounces (and onto the 12 switches, at which point we're not bothered with the bouncing - diagram suggests the balls leap outwards and don't hit any other switches). | ||
::"11 choose 3" is a way how to ask, given 11 items (possible bounces), how many unique and unordered combinations of exactly 3 of these must exist (leftward-bounces, the rest being right-bounces) to filter onto the off-connected switch. (This is the same as "11 choose 8", if you decide to ask how many right-bounces are necessary, the rest being left-bounces.) That could be layer 1 (the 1-pin), 2 (the 2-pins) and 3 (...), before going consistently right to the final strike of the switch, or layers 9+10+11 (after being pure-right 1..8), but with many intermediate tracks across the pin-spacs (165 in total, as it happens; and it would be 55 to hit switch 10. Or 2, instead of 3, if you orientate things the other way round). | ::"11 choose 3" is a way how to ask, given 11 items (possible bounces), how many unique and unordered combinations of exactly 3 of these must exist (leftward-bounces, the rest being right-bounces) to filter onto the off-connected switch. (This is the same as "11 choose 8", if you decide to ask how many right-bounces are necessary, the rest being left-bounces.) That could be layer 1 (the 1-pin), 2 (the 2-pins) and 3 (...), before going consistently right to the final strike of the switch, or layers 9+10+11 (after being pure-right 1..8), but with many intermediate tracks across the pin-spacs (165 in total, as it happens; and it would be 55 to hit switch 10. Or 2, instead of 3, if you orientate things the other way round). | ||
β | :: 165/2048 (paths hitting the off-switch (at #9) divided by all paths that might happen) is a tad over 8%. On the assumption that it's fair and unbiased and you don't get more rattling around than a simple (single half-step) left/right distribution. | + | :: 165/2048 (paths hitting the off-switch (at #9) divided by all paths that might happen) is a tad over 8%. On the assumption that it's fair and unbiased and you don't get more rattling around than a simple (single half-step) left/right distribution. |