Probabilities/Expectations Question....Intuitive...?

Interesting question and a surprising result!

A fairly well known result is that the average number of flips to get a single head is 1 + 1/2 + 1/4 + 1/8 ... = 2

Now, once you get a head, the next tail will inevitably form an HT sequence. Inevitably. Whether it goes HT, HHT, HHHT, it doesn't matter - the next tail forms an HT sequence. So the av number of flips to HT = av number of flips to H + av number of flips to next T = 4.

But after the first head, the next head does NOT inevitably form an HH sequence. So while av number of flips to H + av number of flips to next H = 4, that does not guarantee that you will get an HH sequence. So for instance, if the sequence is HTH, we have got our second head but we don't yet have our first HH sequence. So that is why av number of flips to HH > 4.

Another way of looking at it is that, even though the HH and HT sequences are equally likely, the HH sequences are more clustered (you can get two in a row with the sequence HHH), but the HT sequences must be more spread out (the closest you can get two HT sequences is HTHT), so you're more likely to get HT first. But I like the first explanation better.

ok wow this became stressful. so we would desire to discover E(Xi), it somewhat is the envisioned form of tosses it's going to take to hit an i'th bin when I-a million packing containers are already occupied. permit's inspect what happens once you're throwing the balls: once you throw a ball, the prob of hitting an unoccupied bin are (b - (i - a million))/b, and the prob of hitting an occupied bin are (i-a million)/b. you will save throwing balls till you hit an unoccupied bin, so which you additionally could make a table with the prob of it taking diverse numbers of throws: # | prob. a million | (b - (i - a million))/b 2 | (i-a million)/b * (b - i + a million)/b 3 | ((i-a million)/b)^2 * (b - i + a million)/b, and generally you have got a geometrical series the place the 1st term is c = (b - (i - a million))/b, and the ratio of consecutive words is r = (i-a million)/b. i'm basically utilising the letters c and r to make the equations look less demanding. in this notation, P(n tosses) = c*r^(n-a million). provided that we are computing envisioned fee, we would desire to sum n*P(n tosses) = nc*r^(n-a million). there's a formula for sums like this; it somewhat is sum(nc*r^n) = c/(a million-r)^2. so plug the expressions for r and c into this formula, and after an incredible form of simplifying, you will get b/(b-i+a million). so as that's what E(Xi) is. to get the envisioned form of tosses to fill each and all of the packing containers, you basically sum E(Xi) as i is going from a million to n. i'm no longer there's a extreme high quality formula for expressing this sum, so which you will probable basically bypass away your answer in sigma notation.