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Bux in a Box (Posted on 2007-02-14) Difficulty: 2 of 5
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)

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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      0.891167849395690640
10      0.915927144870069729
15      0.980514734136347864
20      0.984046834351722546
25      0.995941826992179986
30      0.996620044712462552
35      0.999105127524373599
40      0.999248488018559448
45      0.999796490311090862
50      0.999828254816081767
55      0.999952808903541517
60      0.999960045562275520
65      0.999988908932061224
70      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
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