Pierfrancesco PierQR Aiello
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 Jan9 comment Pick a card from a set with the help of a dice Yep, the "whole process" is the key that i missed in your earlier posts. Is a powerful shrewdness. About your biased coin problem, i'll think about it another time :) . Jan9 revised Pick a card from a set with the help of a dice added 204 characters in body Jan9 awarded Supporter Jan9 comment Pick a card from a set with the help of a dice Why is splitting a bad idea? Using your concept of "repeat the whole process to model a bigger dice": given two rows, first with three and the second with four cards, then do a first roll to pick a row and then reroll to pick a card. If the second roll shows a number that is greater than the number of cards in the row, repeat the whole process. So each card has 1/12 of prob, while if i repeat only the second roll (to pick the card in the row) i'll loss the equal probability. Jan9 comment Pick a card from a set with the help of a dice Right! I was stuck because if r>7 then i would reroll the dice only to choose again from 7 to 12 , and this doesn't work! This shrewdness will work even with the method exposed in the question (split cards in different rows). Jan9 revised Pick a card from a set with the help of a dice added 73 characters in body Jan9 accepted Pick a card from a set with the help of a dice Jan9 comment Pick a card from a set with the help of a dice Ok, your explanation is quite clear. So why am i stuck with the following thought (as above, i pick the example with 7 cards)? With your method the first roll choose the first six sides or the latter six sides of the 12-sided dice with probability of 1/2, right? If yes, then we roll again to choose one of the six sides previously chosen. Each side has, then, 1/12 of prob. , this is ok for me; but if i consider the condition of "usefull throw" i have: $P(a \mod 2 = 0)P(b \in 1,...,6 | b \ \text{can be} \in 1,...,6 ) = 1/12$ and $P(a \mod 2 = 1)P(b = 7 | b \in {7} )$ no ok it's 1/12 too. Jan9 awarded Commentator Jan9 comment Pick a card from a set with the help of a dice i'm thinking about it (i'm unsure), thanks again. Jan9 comment Pick a card from a set with the help of a dice In the example written above the positions from one to six have a probability of $\frac{1}{6} \cdot \frac{1}{2}$ while the seventh position has $1 \cdot \frac{1}{2}$ , since we don't consider others positions. Then there isn't an equal probability for each card. If i'm wrong, please tell me where. Thanks in advance! Jan9 comment Pick a card from a set with the help of a dice Ok after a night i must review my position. Even if your answer seems beautiful to me doesn't solve the problem. In fact your method it's equal to a particular arrangement of cards with the method exposed in the question. For example, with 7 cards (so in the case $6 < n \le 12$), we have: one row with six cards and one with only one (instead of one row with three cards and the other with four). But in this manner it's true that all twelve position has the same probability, but these aren't all covered! So kicks in a truncated probability! Jan8 accepted Conventions for notation of function exponentation. Jan8 awarded Scholar Jan8 comment Pick a card from a set with the help of a dice @dtldarek maybe i didn't get your point sorry. if i understand you said: first throw choose if we consider only the odd or even cards [given seven cards]. Is it right? If yes, then is the same of splitting seven cards in two rows, one with three and the other with four cards. Jan8 comment Pick a card from a set with the help of a dice yes, i already think about it after your answer. Thanks! Jan8 comment Pick a card from a set with the help of a dice Nice! Anyway it seems to solve the problem with 7 cards, but with others odds numbers? 9,11,and so on? If i apply the same idea, that is: pick each 6-elements subset of 9 cards, there are $\binom{9}{6}$ possible subsets (IIRC), that is quite a lot. Jan8 asked Pick a card from a set with the help of a dice Oct7 comment Conventions for notation of function exponentation. thanks for answers. Filmor brought a good argument (even if i'm a Gauss fan here xD ) and now i'm reading the link about arcsin. Oct7 awarded Cleanup