= (1/6) * SUM(n=1 to 5) {[(n+1)/6] * 5Cn * (1/6)^n * (5/6)^(5-n)}= 0.0509

<t>N = no of tossesP(N=n) = Probability of getting AT LEAST 8 consecutive tails ANYTIME before getting (the final) 8 consecutive heads, in n tossesP(N<=15) = 0P(N=16) = (1/2)^16...P(N=n) = { ((n-8)-8+1) / (n-8)C8 } * (1/2)^16Here, the first term is the fraction of selection of 8 consecutive position...

<t>I get the generalized formula as:P(X >= i) = [2 * 2^(n-i) / 2^n] * [(n-i+1)/(nCi)]Term 1: Probability of getting the same face (=2) for i tosses and getting any face for the rest (2^(n-i) ). These i tosses need not be sequential.Term 2: This is the fraction of i SEQUENTIAL tosses in a process of ...

<t>Bhutes is right.@noexpert:One way is to calculate P(X>=i) for all i and then use P(X=i) = P(X>=i) - P(X>=(i+1))eg, P(X>=2) = (1/5) * {prob that player 2 & 3 are assigned nos less than that of player 1}= (1/5) * { 0 + 0 + [2/4 * 1/3] + [3/4 * 2/3] + [4/4 * 3/3] } (ith term corresponds to the c...

<t>QuoteIt will need to write a long article to illustrate the point.Well, please do that. As for me, I have come to a point where it is nearly impossible to explain my argument in any different way than how it has been explained in my recent posts. So, I would expect any counter-argument to quote a...

<t>umm, I do not see any supporting argument for the post that had introduced 4.6. For your info, I am quoting that post below...QuoteIt is not so difficult to come up with a strategy that would allow the swimmer to get away if the ratio (pursuer's speed) / (swimmer's speed) = 4 (or, say, ).As for f...

<t>I do not know how you get this number 4.6 (I never got to know since the time it was quoted early on, in this thread)Since my argument has been that R is forced to run 'behind' (ie, same angular direction as) S, and if S changes his direction and comes head-on (which is suicidal), there is no rea...

<t>Quoteit depends on when R discovers that he is fooled, i.e when he finds out chasing on course is the best.Like I said in my previous post, irrespective of whether R (or C) knows S's max speed (and hence knowing when R crosses the critical circle), R should never change his direction outside the ...

<t>QuoteIf he knows this important information, of course he knows that staying on course will be the best way as he knows that his angular speed is faster than that of the swimmer. Chaser (C) will definitely know that his angular speed is greater than the swimmer's (S) since S is outside the critic...

<t>After S crosses the critical circle, the max angular speed of C is always greater than the max angular speed of S. In fact, critical circle is defined by this only - the circle below which the max angular velocity of S is greater than that of C (and vice versa above this circle). So, till S reach...

I am not sure what you mean by 'S can fool C by swimming at a reduced speed.' And, I am also not sure what S's speed has to do with C's direction. Just for clarity, can you tell me the direction (left/right) that C would choose for different ranges of S's speed and why so?

<t>If the chaser (C) changes his direction, he is just going to lose time and let the swimmer (S) get closer to the edge (with no real advantage for C) - your zigzag method is an extreme case of this change of direction strategy. So, C has to pick a direction and run even if he does not know S's max...

<t>QuoteOriginally posted by: panpsi get min=0.1, max=0.7...Argued as follows:A=event rains on sat. P(A)=0.6B=event rains on sun. P(B)=0.3P(dry)=1- (P(A)+P(B)-P(AB))=0.1 + P(AB)Three cases:AB=empty set -> P(dry)=0.1AB=A -> P(dry)=0.7AB=B -> P(dry)=0.4P(AB) = P(B/A).P(A) -(1)P(AB)= P(A/B).P(B) -(2)In...

<t>QuoteOriginally posted by: PaolosThe table below (basically the same table of Vit2007) partitions the sample space---------------------------------------------------------------------------------------....Sat - Sun.............perfect dependence............ mutual exclusivity---------------------...

QuoteOriginally posted by: Tochiro...and the question is how you get a right guess probability strictly above 0.5 in any case...Sorry, my mistake. I was thinking about strategies that would give a right guess probability strictly above 0.5 in the long run (and not in each case). Thanks.