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

 Bux in a Box (Posted on 2007-02-14)
You are playing the following game. You are offered two closed boxes, each having a random amount of money between \$0 and \$100 inside. After picking one and noting the amount of money inside, you are asked to state whether the other box contains more or less money. If you are right, you win \$1. If wrong, you lose \$1. If you play a large number of times, is there a strategy where you can be almost certain to leave with more money than you started with?

 See The Solution Submitted by Kenny M Rating: 3.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: solution -- overall chances of coming out ahead | Comment 2 of 3 |
(In reply to solution by Charlie)

With the given probability of a win, the probability that one will come out ahead, net, for given numbers of times played, is as follows:

` 5      0.89116784939569064010      0.91592714487006972915      0.98051473413634786420      0.98404683435172254625      0.99594182699217998630      0.99662004471246255235      0.99910512752437359940      0.99924848801855944845      0.99979649031109086250      0.99982825481608176755      0.99995280890354151760      0.99996004556227552065      0.99998890893206122470      0.999990588091779791`

10   for I=0 to 100
20    if I<50 then P=(100-I)//101:else P=I//101
30    T=T+P
40   next
50   Op=T//101
60   print Op,T/101
70   P=T/101
100   for N=5 to 70 step 5
110   T=0
120   for I=0 to int(N/2)
130    T=T+combi(N,I)*P^I*(1-P)^(N-I)
140   next
150   print N,1-T
200   next N

 Posted by Charlie on 2007-02-14 11:23:53

 Search: Search body:
Forums (0)