probability question for the maths geniuses??

Question

What are the odds (approximate will do) of getting 10 correct choices in a play your cards right higher lower scenario?
Assume normal deck of cards with unknown first card and sensible guessing (i.e. always lower on an 8 or above, always higher on a 6 or below and random on a seven) and same card loses. The cards weren't replaced but if its easier you can assume they had no bearing on following guesses.

This was a bonus round at my local pub quiz and i wondered if it was ridiculously improbable. I realize its a pretty complicated problem so any kind of 'near enough' calculating will be good enough.

Thanks

Answer 1

It is fairly improbable -- the probability of winning is somewhere around 2%.

In order to make this problem tractable, I modeled it with replacement, so that the probability of winning depended only on the card showing and not on the cards previously played. This enabled me to use markov chains to model the system. The model contained 14 states, the first 13 corresponding to the various ranks of cards that might be showing and a 14th indicating that the player had already lost. A 14×14 transition matrix was then constructed, with the number in the mth row and nth column indicating the probability that you would transition into the mth state if you were currently in the nth. More precisely, the matrix was as follows:

A_(m, n) = {1/13 if n<7 ∧ n 1/13 if n>7 ∧ m 1/26 if n=7 ∧ m≠n ∧ m≠14
0 if n<14 ∧ m=n
0 if n=14 ∧ m≠n
n/13 if m=14 ∧ n≤7
(14-n)/13 if m=14 ∧ 7 1 if n=14 ∧ n=14}

In general, for all (n, m) from 1 to 13, this matrix represents the probability that you will continue play with a card of rank m if your current cards is of rank m. for m=14 and n≠14, it represents the probability that you will lose if your current card is of rank n. For n=14 and m≠14, it represents the probability that you will continue play with a card of rank m given that you have already lost (which is obviously zero), and for n=14 and m=14, it represents the probability that you will lose given that you have already lost.

Now, given a vector 14×1 vector x representing the probability that your showing card is of rank n (or for n=14, the probability that you have already lost), Ax represents the probability that your showing card is of rank n at the beginning of the next round. It therefore follows that if x is initialized to [1/13, 1/13, 1/13, 1/13, 1/13, 1/13, 1/13, 1/13, 1/13, 1/13, 1/13, 1/13, 1/13, 0], (which is the probability distribution for the first card), that the 14th value of A^10 x will represent the probability that you lose during any of 10 rounds.

Plugging A and x into maxima (a free computer algebra system, available at http://maxima.sourceforge.net ), we find that the probability that you lose during any of the 10 rounds is 1751544548118 / 1792160394037. Your probability of winning is 1 - 1751544548118 / 1792160394037, which is 40615845919 / 1792160394037, or about 0.022663064117553.

Answer 2

the sector is countless. what number workers are there in total? 10/x = share of workers chosen. As a million/2 commuters holiday practice, a million/2 get to artwork in different concepts. Random pattern of 10 workers might want to produce all walkers/all practice riders, and so on. that's no longer in simple terms 50%*10=5. possibility says interior of ninety 5% self belief, that's in all possibility 5 of 10 are practice riders, notwithstanding, as that's the most in all possibility effect given the parameters.