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What is the expected number of tosses such that there are 3 consecutive sixes.

Denote the event $T$=number of tosses until first non-six

Let $X$ be the event of getting $3$ consecutive sixes in $n$ trials.

We need to find $\mathbb E [n]$

My reasoning: Probability of getting 3 consecutive sixes in $n$ tosses is:

$\mathbb [X|T=1]P[T=1]+\mathbb P [X|T=2]\mathbb P[T=2]+\mathbb P [X|T=3]\mathbb P[T=3]+\mathbb P [X|T>3]\mathbb P[T>3]$

$\mathbb P [X|T>3]\mathbb P[T>3]=0$ since $T>3$ means that we already had 3 sixes rolled.

So in $n$ trials we have a $\mathbb P$ of getting 3 sixes, so the $\mathbb E[n]=1/\mathbb P$

I know the answer should be 258

Community
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GRS
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1 Answers1

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METHOD I: Let $E$ denote the answer. Let $E_1$ denote the answer if you have one $6$ before you start (so you win if your first two tosses are $6$ but otherwise you need three in a row). Similarly let $E_2$ denote the answer if you have two $6's$ before you start (so you win if the first toss is a $6$ but otherwise you need three in a row). Now, starting from scratch imagine the first roll. Either you get a $6$ (probability $\frac 16$) or you don't. Thus $$E=\frac 16\,(E_1+1)+\frac 56 \,(E+1)\implies E=E_1+6$$

Similarly, we see: $$E_1=\frac 16\,(E_2+1)+\frac 56\,(E+1)$$ $$E_2=\frac 16\,(1)+\frac 56\,(E+1)=1+\frac 56\,E$$

This system is easily solved to yield the desired answer.

METHOD II: Consider the first three tosses. There are $4$ cases.

case 1. The first toss is a non-six (probability $\frac 56$). You wind up at the start after one toss.

case 2. The first toss is a six, but the second is a non-six (probability $\frac 16*\frac 56=\frac 5{36}$). You wind up back at the start after two tosses.

case 3. The first two tosses are sixes, but the third is a non-six (probability $\frac 16*\frac 16*\frac 56=\frac 5{216}$). You wind up back at the start after three tosses.

case 4. The first three tosses are sixes (probability $\frac 1{216}$). You win!

Thus we have $$E=\frac 5{6}(E+1)+\frac 5{36}(E+2)+\frac 5{216}(E+3)+\frac 1{216}(3)\implies E=258$$

lulu
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  • Thanks, but I have to answer the question using conditions on non-6 rolls – GRS Jan 18 '16 at 21:23
  • Not sure what you mean by that. I see the calculation you began, but I don't understand what you are doing. What is $\mathbb P$? – lulu Jan 18 '16 at 21:26
  • Ok...I have a guess. I'll write it up that way too. – lulu Jan 18 '16 at 21:30
  • it's the $P$ of getting 3 sixes in $n$ trials. Which is given by the equation above. There are only 3 possible ways: I get 1st non-six with 1 trial, get 1st non-six with 2 trials or I get 1st non-six with 3 trials. If I get 1st non-six after 3+ trials, it means that I already got 3 consecutive sixes, and the event already happened. – GRS Jan 18 '16 at 21:34
  • Take a look at my second solution. Is that close enough? – lulu Jan 18 '16 at 21:37
  • perfect, thanks, I'll have a read through it right now – GRS Jan 18 '16 at 21:43
  • Is there not a deeper reason why it turns out to be: $6+6^2+6^3=258?$ It seems to generalize. – DinkyDoe Mar 22 '23 at 16:24
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    @DinkyDoe Don't know about deeper, but it's easy to prove the generalized form of that inductively. Say that we know that the expected number of tosses needed to see $n$ consecutive sixes was $E_n=\sum_{i=1}^n6^i$. Then, to see $n+1$ sixes we need to first see $n$ sixes and then we need another $6$. Thus $E_{n+1}=6(E_n+1)$ which implies what you want. – lulu Mar 22 '23 at 16:40
  • Yes, that proves it inductively. Nice. I guess the result could have been proven right away with this recurrence without that tedious work ;) – DinkyDoe Mar 22 '23 at 16:52