All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Pick a card, any card.. (Posted on 2008-03-11)
You sit down with a well mixed deck containing A cards marked "+" and B cards marked "—". You may draw cards from this deck as long as you want, i.e., you can stop playing at any point. Each time you draw a + card you are given \$1 and each time you draw a — card you have to pay \$1. Cards are not replaced after having been drawn.

What would be a fair amount to pay for the right to play (i.e., what is the expected payoff) and under what circumstance should a player cease drawing?

 No Solution Yet Submitted by FrankM Rating: 2.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 A fair amount? | Comment 27 of 37 |

After Charlie's initial post, I had been working on this problem looking for an equation that might provide the fair amount if given A and B. During my initial foray, I had arrived at the same results as had Leming, e.g., recognizing that there may be positive return even with A < B.

I have not found the single equation, yet I do not believe FrankM's equation, given in his soluton is correct:
A(A+1-B)/(A+1) when A >= B.  It provides a close approximation, but it is not precise.

I am not guaranteeing the following will work for all non-negative integer values of A given B where A >= B, but the following seem to provide the values in Leming's tables for the given values of B:

Let D = (A+B)!/(A!*B!)
Where B = 0: (A^1)/D
Where B = 1: (A^2)/D
Where B = 2: (A^3 + A^2 - 2*A)/(2*D)

Edited on March 14, 2008, 12:19 am
 Posted by Dej Mar on 2008-03-13 17:49:43

 Search: Search body:
Forums (0)