Unexpected Probability problem!! 2

Probability Level 5

You have n coins with you numbered 1 to n . But, they are biased. r t h r^{th} coin in the set has a probability of 1 r + 1 \frac{1}{r+1} of showing t a i l s \color{#CEBB00}{tails} and r r + 1 \frac{r}{r+1} of showing h e a d s \color{#EC7300}{heads} . Example - Coin numbered 7 t h 7^{th} has 1 8 \frac{1}{8} probability of showing t a i l s \color{#CEBB00}{tails} and 7 8 \frac{7}{8} probability of showing h e a d s \color{#EC7300}{heads} .

You perform they following tiring process with them, starting from coin numbered 1

  • Step-1: You flip the coin you have and note its outcome. Let it be O r O_r where r r is the number of the coin you flipped.

  • Step-2: You put away your coin and pick coin numbered ( r + 1 ) (r + 1) where r r is your current coin number. If r + 1 = n r + 1 = n , you jump to Step-4 . Else, you go to Step-1 again with current coin.

  • Step-3: If O r = t a i l s O_r = \color{#CEBB00}{tails} , you pick coin ( r − 1 ) t h (r - 1)^{th} after process of ( r + 1 ) t h (r + 1)^{th} coin is over. Else, you pick coin r t h r^{th} after process of ( r + 1 ) t h (r + 1)^{th} coin is over. If index of coin picked = ( r − 1 ) = 0 = (r - 1) = 0 , you terminate the process. If coin picked = ( r − 1 ) ≠ 0 = (r - 1) \neq 0 , you repeat Step-3 with that coin. Else, you go to Step - 1 with the current coin(coin r r ).

  • Step-4: For coin numbered n n , you flip it till you encounter a t a i l \color{#CEBB00}{tail} . Once you encounter it, you go to Step-3 with coin ( n − 1 ) (n - 1) .

Find the expected value of the number of times you'll flip the 6 t h coin \color{#3D99F6}{6^{th} \text{ coin}} .

All of my problems are original

Try easier version Unexpected Probability problem!!

Difficulty: † † † † † \dagger \dagger \dagger \dagger \dagger


The answer is 5040.

Aryan Sanghi by Aryan Sanghi , 17, India

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1 solution

Aryan Sanghi
Jun 29, 2020

Let E r / p i c k e d E_{r/picked} be the expected value of number of times coin r r is flipped once it is picked.

E r / p i c k e d = ∑ (number of times coin is flipped) × (probability that it is flipped that number of times) E_{r/picked} = \sum\text{(number of times coin is flipped) × (probability that it is flipped that number of times)}

E r / p i c k e d = 1 × 1 r + 1 + 2 × ( r r + 1 ) 1 × 1 r + 1 + 3 × ( r r + 1 ) 2 × 1 r + 1 … ∞ E_{r/picked} = 1 × \frac{1}{r + 1} + 2 × (\frac{r}{r+1}) ^ 1 × \frac{1}{r + 1} + 3 × (\frac{r}{r+1}) ^ 2 × \frac{1}{r + 1} \ldots \infty

E r / p i c k e d = r + 1 \color{#3D99F6}{\boxed{E_{r/picked} = r + 1}}

Let E r E_r be the expected value of number of times a coin is flipped in total

Now, logically E 1 = E 1 / p i c k e d = 2 E_1 = E_{1/picked} = 2

Also, we get a reccurence relationship

E r = E r / p i c k e d × E r − 1 \boxed{E_r = E_{r/picked} × E_{r-1}}

E r = ( r + 1 ) × E r − 1 = ( r + 1 ) × r × E r − 2 = ( r + 1 ) × r × ( r − 1 ) × E r − 3 = … = ( r + 1 ) × r × ( r − 1 ) × … × 3 × 2 E_r = (r + 1) × E_{r - 1} = (r + 1) × r × E_{r - 2} = (r + 1) × r × (r - 1) × E_{r - 3} = \ldots = (r + 1) × r × (r - 1) × \ldots × 3 × 2

E r = ( r + 1 ) ! where ! denotes factorial symbol. \boxed{E_r = (r + 1)!} \text{ where \boxed{!} denotes factorial symbol.}

E 6 = 7 ! = 5040 \color{#3D99F6}{\boxed{E_6 = 7! = 5040}}

2 Helpful 2 Interesting 4 Brilliant 0 Confused

I shall delete my solution now as you already posted. I would suggest next time you post a little later when a lot of people have tried to solved or you see some solutions.

Siddharth Chakravarty - 11 months, 2 weeks ago

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I'll definitely follow your suggestion. Thanku. :)

Aryan Sanghi - 11 months, 2 weeks ago

I wondered if my question was confusing. But, now since 2 people have solved, I am relieved. :)

Aryan Sanghi - 11 months, 2 weeks ago

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The wording was a bit confusing, I might say therefore I wasted my 1 try interpreting something else. But second time I got it right.

Siddharth Chakravarty - 11 months, 2 weeks ago

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@Siddharth Chakravarty – Actually when I made it, my first confusion was how would I word it. This was the best way I found. Even I felt wording confusing although. But, that is the real fun, isn't it?

Aryan Sanghi - 11 months, 2 weeks ago

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@Aryan Sanghi – Haha! If this had been my exam question, where obviously we had 1 try I would leave it as an option because I want full marks, hehe...

Siddharth Chakravarty - 11 months, 2 weeks ago

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@Siddharth Chakravarty – Exactly 😂.

Aryan Sanghi - 11 months, 2 weeks ago

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