A Lotto bet is picking 6 numbers out of 49 -- if you pick the correct combination, you get the jackpot!
If N persons play, there will be many repeats, since it's highly probable that some combinations will be chosen by two persons or more. (This is known as the "birthday paradox".)
What's the expected number of DIFFERENT combinations that will be chosen, if N persons play? (Assume these persons pick their combinations totally randomly.)
(In reply to
my solution by Ady TZIDON)
If p=(s-1)*(s-2)*(s-3)*...(s-k )...*(s-N+1)/(S^(N-1)) and the expected number of different combinations is N*p, then, when N=s, p=(s-1)!/(s^(s-1)) and the expected number of different combinations would be (s-1)!/(s^(s-2)).
In the simple case that I gave originally, for s=10, that would be 9!/(10^8) = 0.0036288. Clearly that expected value is larger.
In the case of s=C(49,6)=13,983,816, (s-1)! is 5.09456697 x 10^93,850,017, which is much less than 13,983,816^13,983,814, so the expected number of different values would be only an infinitesimal fraction.
Here's an evaluation of your formula for various N, showing N, p and N*p:
2000 0.8667906644994577999 1733.5813289989155999357
3000 0.7249038095932000619 2174.7114287796001859441
4000 0.5643947166555643515 2257.5788666222574062672
5000 0.4090907651277387745 2045.4538256386938725305
6000 0.2760503920116564839 1656.302352069938903711
7000 0.1734150535030699466 1213.9053745214896268831
8000 0.1014174603352282436 811.3396826818259494062
9000 0.0552158744998651441 496.9428704987862977751
10000 0.0279858332744194512 279.858332744194512995
11000 0.0132048468890099776 145.2533157791097544468
12000 0.0058002553429323945 69.603064115188734498
13000 0.0023717955320933506 30.8333419172135586344
14000 0.0009028623571569044 12.6400730001966617394
15000 0.0003199468119638217 4.7992021794573269536
16000 0.0001055464773157981 1.688743637052770273
17000 0.0000324128476834766 0.5510184106191037302
18000 0.0000092660768321063 0.1667893829779141167
19000 0.0000024659123875676 0.0468523353637848158
20000 0.0000006108860356235 0.0122177207124703028
21000 0.0000001408774025805 0.0029584254541911454
22000 0.0000000302425793817 0.0006653367463985149
Notice how it goes down after a while.
The program uses logarithms to avoid overflow:
5 S=combi(49,6)
10 for N=2000 to 22000 step 1000
20 LNum=0
30 for I=S-N+1 to S-1
40 LNum=LNum+log(I)
50 next
55 Lp=LNum-(N-1)*log(S):P=exp(Lp):E=N*P
60 print N,P,E
70 next
The results of my formula (after the typo was corrected) are:
2000 1999.8570558440459625842
3000 2999.6783296844847674345
4000 3999.4281076214049976602
5000 4999.1063947673916414075
6000 5998.7131962340449614474
7000 6998.2485171335580267427
8000 7997.7123625785567609058
9000 8997.1047376785105082852
10000 9996.4256475453432250585
11000 10995.6750972870725626615
12000 11994.8530920177215569979
13000 12993.9596368426030864938
14000 13992.9947368750768808918
15000 14991.9583972229180110886
16000 15990.8506229925446130908
17000 16989.6714192930303217428
18000 17988.4207912338353846232
19000 18987.0987439216326473044
20000 19985.705282462452874995
21000 20984.2404119610722366784
22000 21982.7041375289795596902
which seems reasonable.
That segment of the program is:
100 for N=2000 to 22000 step 1000
110 print N,S*(1-((S-1)/S)^N)
120 next
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Posted by Charlie
on 2005-01-15 15:45:14 |
re: my solution
