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Coin Flip

On Monday, you flip a coin all day. You start flipping it until you see the pattern Head, Tail, Head. You record the number of flips required to reach this pattern, and start flipping again (and counting up from 1 again) until you see that pattern again, you record the second number, and start again. At the end of the day you average all of the numbers you’ve recorded. On Tuesday you do the EXACT same thing except you flip until you see the pattern Head, Tail, Tail.

Will Monday’s number be higher than Tuesday’s, equal to Tuesday’s, or lower than Tuesday’s?

Tuesday’s pattern will require, on average, fewer flips to be achieved. The two patterns differ only in their final required elements (Heads on Monday and Tails on Tuesday). So let’s look at success and failure AFTER successfully flipping the first two elements (HT). When you flip HTH on Monday, you stop counting. But suppose you fail and flip HTT on Monday, you would have to wait until you see the next H to start hoping for the pattern again. On Tuesday, however, if you flip HTT, you stop counting. But if you flip HTH, you have failed to achieve your goal of HTT, but you are one-third of the way to achieving it again (because you got the last H which can be used as the first element of the pattern).

On average, HTH will occur in about 10 flips and HTT will occur in about 8 flips.

Comments


R. Crabb

I like how the straightforward intuition doesn't quite work. For a sequence of 3 flips, any of the patterns are equally likely, so why wouldn't they be the same? Obviously the trick is that you start over each time you hit the sequence. If you just looked at the overall sequence of flips throughout the day and just counted how many times each sequence appeared, the numbers should be equal.

  

Aakash

Exactly, I find this arises so often with statistics/probability questions...intuition isn't worth a whole lot. After figuring it out or understanding the solution, you start to wonder how many other things you haven't thought about correctly.


AJB

You should really be more careful how you ask the question. You're asking for an exact answer to a question about a future event. That is impossible (mathematically/logically). Instead you mean to be asking: "Is Monday's number more likely to be higher, lower, or equal to Tuesday's number?"

Also, I get 9 and 7 (not 10 and 8) using a simple Monte Carlo simulation implemented in PARI/GP with the following lines. (Note in PARI/GP each command prompt starts with "?" and each echoed result starts with "%[ans #]":

? nx=0; ny=0; totalx=0; totaly=0;
? x=[2,2,2];
? y=[2,2,2];
? xcounter=0; ycounter=0;
? for(a=1,1000000,x[1]=x[2]; x[2]=x[3]; x[3]=random(2); y[1]=y[2]; y[2]=y[3]; y[3]=random(2); if(x==[0,1,0],totalx=totalx+nx; xcounter=xcounter+1; nx=0; x=[2,2,2], nx=nx+1); if(y==[0,1,1],totaly=totaly+ny; ycounter=ycounter+1; ny=0; y=[2,2,2], ny=ny+1);)
? totalx/xcounter*1.
%50 = 8.9860094468688522953095197675231428315
? totaly/ycounter*1.
%51 = 6.9901880098759118838541625050937652314
  

AJB

Oh no, I think I should have initialized nx=1 and ny=1, which would have given me 10 and 8, so I think you were correct!


AJB

Corrected code :)

? nx=1; ny=1; totalx=0; totaly=0;
? x=[2,2,2];
? y=[2,2,2];
? xcounter=0; ycounter=0;
? for(a=1,1000000,x[1]=x[2]; x[2]=x[3]; x[3]=random(2); y[1]=y[2]; y[2]=y[3]; y[3]=random(2); if(x==[0,1,0],totalx=totalx+nx; xcounter=xcounter+1; nx=1; x=[2,2,2], nx=nx+1); if(y==[0,1,1],totaly=totaly+ny; ycounter=ycounter+1; ny=1; y=[2,2,2], ny=ny+1);)
? totalx/xcounter*1.
%9 = 9.9747538727020637786400406970364976609
? totaly/ycounter*1.
%10 = 8.0048669591111395728602990618295923922
  

Aakash

Whoa - thanks for sharing!


Ravi

Wow, you even went as far as the Monte Carlo simulation. Impressive.

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Submitted by
Aakash
almost 2 years ago
Likes
Difficulty 6.0 ?

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Math statistics


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