Card sharp

Originally posted by mtthw
There's an additional complication to this. You can improve your chances by using an intelligent strategy.

For an extreme case, consider what happens if the first four cards are all Kings (and assume you haven't lost yet). You now know there are no more Kings. So if you keep saying "King" you win.

I agree, but I thought you had to say the standard ranking order. (which is basically equivalent odds-wise to any prefixed order that you choose)

Originally posted by Palynka
I agree, but I thought you had to say the standard ranking order. (which is basically equivalent odds-wise to any prefixed order that you choose)

Yes, you're right. The order of the ranks you name must go in order, which is equivalent to choosing a fixed order of any kind.

I also agree that the odds posted on the website sound wrong after this discussion. Have to look into this further.

Originally posted by Palynka
I agree, but I thought you had to say the standard ranking order. (which is basically equivalent odds-wise to any prefixed order that you choose)

Yes, you're right. Sorry. Might make a nice extension to the problem, though. What are the odds if you can state any order, and can use the optimal strategy?

Originally posted by mtthw
Yes, you're right. Sorry. Might make a nice extension to the problem, though. What are the odds if you can state any order, and can use the optimal strategy?

First one is random, 12/13, second optimally you choose the card that came out 48/51. Then for the third card is:

Prob(2nd card equal to 1st)*48/50+Prob(2nd different 1st)*47/50=
= 3/51*48/50+48/51*47/50

And so on. The logic is there, but calculating all the tree needs either a lot of patience or some recursive programming...

PS: I'm not sure if I have to factor the fact that the 2nd is different than the first, given that I didn't lose yet. I think so, so replace 48/51 with 44/47 and 3/51 with 3/47... Or not. What do you think?