Hi. It took me most of the day to read this from the start of the thread with a lot of skimming. How did I get here? I thought of a possible experiment, and looked on the challenge site – suddenly what I have done last week becomes relevant.
To calculate the odds you need to use the binomial distribution probability (BDF) formula. Odds are the reciprocal of the BDF.
BDF= {N!/((N-X)!*X!)} * (p**X) * {(1-p)**(N-X)}
You can find this on an Exel spreadsheet under BINOMDIST(X, N, p, FALSE). X is the number of hits, N is the number of attempts, p is the probability per choice (=1/number of choices), and D is set to FALSE.
If you remove the first card, then the second card, and so on, then the formula does not work, and you need another formula. Either put the chosen card back and shuffle , or accept the BDF is an approximation for a low number of attempts, or use the correct formula.
If a 52 card deck is used and the card must be fully identified eg Ace of Hearts, Two of Spades, then p is 1/52. If there are three tries then N = 3. The following odds apply
None correct ( X = 0) the odds are 1.06 to 1 against a chance of success
One correct ( X = 1) the odds are 18 to 1 against a chance of success
Two correct ( X = 2) the odds are 919 to 1 against a chance of success
Three correct ( X = 3) the odds are 140,000 to 1 against a chance of success
If flaccon gets none correct (the most probable result) and asks for another test, then one could start with a “clean slate” and accept the same odds.
Strictly speaking a new test means 6 attempts (N=6) with results summed ie 0+0, 0+1,0+2, and 0+3.
These odds are: 1.12; 9.5; 194; and 7,452.