$Geometric distribution$

Question

$Geometric distribution$

Two palyers battle in a competition and they each have 50% of winning 1 point at each round. The winner of the tournament is the first who is
(a) two points ahead
(b) three points ahead
Let X be the number of rounds played. Calcuate E[X], the distributuion of P(X=n), and Var[X] in the cases of (a) and (b)

Here is what I thought for part (a), but I am not quite sure about (b).

1st and 2nd rounds can be seen as a mini match. In the same way, 3rd and 4th is also a mini match. 5th and 6th is also a mini match and etc. The game ends when the first mini match is not a draw. There are two situations to end the game. 1. Player A wins. If player A wins, the probability of wining that last mini match is 1/4 (1/2 * 1/2). 2. Player A loses, it is the same probability (1/4). Therefore, the total probability of final mini match is 1/2 (1/4 + 1/4). Based on this pattern, the number of rounds X ~ 2Geo(1/2). E(X) = 4 Var(X) = 8.

Probability Probability Distribution

Lanx1594

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Alessandro Iraci

0

Hi, I have a nice way to give you P(X=n) for case (b), but probability is not my field and to get E[X] and Var[X] I need more time than what the offered amount is worth. Is it okay if I just give you P(X=n)? It seems you know by heart how to get E[X] and Var[X] yourself. Otherwise I can do that if you increase the offered amount.
Alessandro Iraci

0

Honestly also just finding P(X=n) would be worth more than the value of the question, but I thought it was an interesting one to answer and so I computed that regardless. Let me know your preference!
Daniel90

0

I can answer this question, but your offered amount is too low.
Professormo

0

What level is this? Are we allowed to use Markov Chains?
Alessandro Iraci

0

There is no need, I have a very elementary solution, I just don't want to spend time computing expected value and variance for this little money.
Lanx1594

0

Yes, you can use Markov Chain. It is my first time using this website, I did not know how much is low amount and how much is high amount. I I put most of the money on another question (card shuffling). If you want to get more, you can have a look at that one. I have worked out part a in this question (2 points ahead), so I just need (b). To be more specific, I just need the distribution for b. Thank you very much for your help!
Alessandro Iraci

0

I have the distribution for b, if you're ok with just that I will answer.
Alessandro Iraci

0

But please confirm!
Lanx1594

0

Sure.you can tell me the distribution. I just wrote my answer my part b. X ~ 2Geo(1/4) +1. E(X)= 9, Var(X)= 48. I guess this makes your situation a bit easier.
Lanx1594

0

So I can see if I am correct or not.
Alessandro Iraci

0

I get a different answer. I'll post it.
Lanx1594

0

I don't think this make sense to me. In the three points situation, we can regard 2nd&3rd, 4th&5th, 6th&7th ..... rounds as mini-match, which means that 1st round is seprerated and not being included in the mini match. Therefore, there are two situations happening in the game: 1. the points difference is either 1point. 2. the game ends becasue the person who led the 1 point wins the final two mini match (p=1/4). Therefore, all the mini matches follows Geo(1/4), so X~2Geo(1/4)+1.
Lanx1594

0

that plus 1 means we plus the first round that did not get included in the mini matches.
Alessandro Iraci

0

I think we're saying the same thing, except that I don't know the notation for the distribution (which I already admitted). The probability of the match ending after 2n+1 rounds is still (3/4)^(n-1) * 1/4, which you counted as "not finishing in the first n-1 mini-matches times finishing in the n-th", which is the same thing I said (but counted in a different way).
Alessandro Iraci

0

I checked the definition, and it is indeed 2 Geo(1/4) + 1, which gives P(X=2n+1) = (3/4)^(n-1) * 1/4. My bad for not checking before, but the answer I gave is also correct.
Lanx1594

0

Ok, cool! Thank you very much for confirming this!
Alessandro Iraci

0

No worries! Also your solution is more elegant than mine, I have to say!

Answer

The answer is accepted.

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Alessandro Iraci

0

Hi, I have a nice way to give you P(X=n) for case (b), but probability is not my field and to get E[X] and Var[X] I need more time than what the offered amount is worth. Is it okay if I just give you P(X=n)? It seems you know by heart how to get E[X] and Var[X] yourself. Otherwise I can do that if you increase the offered amount.
Alessandro Iraci

0

Honestly also just finding P(X=n) would be worth more than the value of the question, but I thought it was an interesting one to answer and so I computed that regardless. Let me know your preference!
Daniel90

0

I can answer this question, but your offered amount is too low.
Professormo

0

What level is this? Are we allowed to use Markov Chains?
Alessandro Iraci

0

There is no need, I have a very elementary solution, I just don't want to spend time computing expected value and variance for this little money.
Lanx1594

0

Yes, you can use Markov Chain. It is my first time using this website, I did not know how much is low amount and how much is high amount. I I put most of the money on another question (card shuffling). If you want to get more, you can have a look at that one. I have worked out part a in this question (2 points ahead), so I just need (b). To be more specific, I just need the distribution for b. Thank you very much for your help!
Alessandro Iraci

0

I have the distribution for b, if you're ok with just that I will answer.
Lanx1594

0

Sure.you can tell me the distribution. I just wrote my answer my part b. X ~ 2Geo(1/4) +1. E(X)= 9, Var(X)= 48. I guess this makes your situation a bit easier.
Alessandro Iraci

0

I get a different answer. I'll post it.
Lanx1594

0

I don't think this make sense to me. In the three points situation, we can regard 2nd&3rd, 4th&5th, 6th&7th ..... rounds as mini-match, which means that 1st round is seprerated and not being included in the mini match. Therefore, there are two situations happening in the game: 1. the points difference is either 1point. 2. the game ends becasue the person who led the 1 point wins the final two mini match (p=1/4). Therefore, all the mini matches follows Geo(1/4), so X~2Geo(1/4)+1.
Lanx1594

0

that plus 1 means we plus the first round that did not get included in the mini matches.
Alessandro Iraci

0

I think we're saying the same thing, except that I don't know the notation for the distribution (which I already admitted). The probability of the match ending after 2n+1 rounds is still (3/4)^(n-1) * 1/4, which you counted as "not finishing in the first n-1 mini-matches times finishing in the n-th", which is the same thing I said (but counted in a different way).
Alessandro Iraci

0

I checked the definition, and it is indeed 2 Geo(1/4) + 1, which gives P(X=2n+1) = (3/4)^(n-1) * 1/4. My bad for not checking before, but the answer I gave is also correct.
Lanx1594

0

Ok, cool! Thank you very much for confirming this!
Alessandro Iraci

0

No worries! Also your solution is more elegant than mine, I have to say!

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

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Let's call $P_0(n)$ the probability that the game is not finished and the players are even after $n$ rounds, $P_2(n)$ the probability that the game is not finished and either of the players is 2 points ahead after $n$ rounds, and $P(n)$ as a shorthand for $P(X=n)$, i.e. one player went 3 points ahead on the $n$-th round.

We clearly have $P_0(2n+1) = P_2(2n+1) = 0$ for every $n$, $P_0(0) = 1$, $P_2(0) = 0$, for $n \geq 1$ \[ P_0(2n) = P_2(2n) = \frac{1}{2} P_0(2n-2) + \frac{1}{4} P_2(2n-2), \] and finally $P(2n) = 0$ for every $n$, and $P(2n+1) = \frac{1}{2} P_2(2n)$.

In fact, if the players are even, two rounds later they're even again with probability $1/2$, and they have a two points distance (either direction) with probability $1/2$; similarly, if either player is $2$ points ahead, two rounds later they're even with probability $1/4$, that player is still two points ahead with probability $1/4$, and the game ends the next round with probability $1/2$. Notice in particular that $P_0(2n) = P_2(2n)$ for $n\geq1$, so $P_2(2n+2) = \frac{3}{4} P_2(2n)$ for $n \geq 1$.

We can then deduce \[ P(2n+1) = \frac{1}{2} P_2(2n) = \frac{1}{2} \left( \frac{3}{4} \right)^{n-1} P_2(2) = \frac{1}{4} \left( \frac{3}{4} \right)^{n-1} = \frac{1}{3} \left( \frac{3}{4} \right)^{n} \] for $n \geq 1$ (and $P(1) = P(2n) = 0$), so, if I understand the notation, the distribution would be $2 \cdot \mathsf{Geo}(3/4) + 3$ (but I am not sure about the notation - anyway, you have the distribution now).

Alessandro Iraci

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$Geometric distribution$

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