Share Price, page-7449

[Apols for the above badly presented and explained post - hope some got the gist of it. Also was a reply to @florafilan.]

Assuming independence of trials, and with all trials given the same probability of success/fail, the possible permutations and their probability outcomes for W = 0.5 are correct:
I.e. 0 wins = 0.0625; 1 win = 0.2500; 2 wins = 0.375; 3 wins = 0.2500; 4 wins = 0.0625 - which necessarily sum to 1.

These figure change dramatically if W is assigned 0.8
I.e. 0 wins = 0.0016; 1 win = 0.0256; 2 wins = 0.1536; 3 wins = 0.4096; 4wins = 0. 4096 - which necessarily sum to 1. [Interesting to note that for this case (W = 0 8) pr W=3 and pr W=4 are identical and sum to just over 80%.]

Importantly, these simple stats assume total independence of probable outcomes between trials - result of future trial(s) is unaffected by result of previous trial(s).

As intimated by HT's brilliant post of yesterday a strong case can be made for non independence of results which would add complexity to any statistical analysis. Briefly, previous trial result(s) would create adjustment to initial future trial probability assumptions, dependent on the degree of intra-independence assigned.

Also, in a commercial sense, the strength of efficacy results would play a significant valuation role.