{"word_count": 561, "character_count": 3700}
This evening, while I was about to wash the dishes, I heard my elders starting a game (call them Him and Her)Him: "I have picked - in my head - a number, lower than 50. Try to guess..."Her: "No way, too difficult..."Him: "You can try five different numbers..."Her: ".,. um ... No, no way..."Me: "Wait... each time we suggest a number, you tell us if yours is either above, or below ?"You can see me coming clearly, can't you ? Using a simple subdivision rule, we have a fast algorithm (and indeed, if I have to choose between washing the dishes and playing with the kids...)Him: "um.... ok"Her: "Daddy, are you sure we will win ?"Me: "Well... I cannot promise that we will win... but I am rather sure [sic] that we will win quite frequently: more gains than losses..." (I guess).Her: "Great ! I am playing with daddy..."Him: "um.... wait, is it one of you trick, again ? I don't to play anymore... Do you want to see the books we've chosen at the library ?"Her: "Sure..."Me: "What ? no one wants to see if I was right ? that we have indeed more than 50% chances to win..."Him and her: "No !"The point of that story ? If we listen to kids, science will not go forward, trust me. But I am curious... I want to see if my intuition was correct. Actually, the intuition was based on the fact that&#62; 2^5[1] 32 &#62; 2^6[1] 64so in 5 or 6 steps the algorithm of subdivision should converge. I guess... I mean, I do not know for sure, since 50 is not a power of 2, so it might be difficult, each time, to split in two: we have to deal only with integers here...To be sure, let us substitute my laptop to my son... to pick up numbers, randomly (yes, sometimes I feel like I am Doctor Tenma, 天馬博士). The algorithm is simple: there are bounds, and at each stop I should suggest the middle of the interval. If the middle is not an integer, I suggest either the integer below or the integer above (with equal probabilities).cutinhalf=function(a,b){m=(a+b)/2if(m %% 1 == 0){m=m}if(m %% 1 != 0){m=sample(c(m-.5,m+.5),size=1)}return(round(m))}The following functions runs 10,000 simulations, and tells us how many times, out of 5 numbers suggested, we got the good one.winning=function(lower=1,upper=50,tries=5,NS=100000){SIM=rep(NA,NS)for(simul in 1:NS){interval=c(lower,upper)(unknownnumber=sample(lower:upper,size=1))success=FALSEfor(i in 1:tries){picknumber=cutinhalf(interval[1],interval[2])if(picknumber==unknownnumber){success=TRUE}if(picknumber&#62;unknownnumber){interval[2]=picknumber}if(picknumber&#60;unknownnumber){interval[1]=picknumber}#print(c(unknownnumber,picknumber,success,interval))};SIM[simul]=success};return(mean(SIM))}It looks like the probability that we got the good number is higher than 60%,&#62; winning()[1] 0.61801Which is not bad. And if the upper limit was not 50, but something else, the probability of winning would have been the following.VWN=function(n){winning(upper=n)}V=Vectorize(VWN)(seq(25,100,by=5))plot(seq(25,100,by=5),V,type="b",col="red",ylim=c(0,1))Actually, after losing a couple of times, I am rather sure that my son would have to us that we can suggest only four numbers. In that case, the probability would have been close to 30%, as shown on the blue curve below (where four numbers only can be suggested)Anyway, as intuited, with five possible suggestions, we were quite likely to win frequently. Actually with a probability of almost 2 out of 3...and 1 out of 3 if my son had decided to pick an number between 1 and 100, or only 4 possible suggestions... Those are quite large actually, when we think about it. It reminds me that McGyver story I mentioned a few months ago... Anyway, calculating probabilities is nice, but I still have to wash the dishes...