Almintas2l

Answered

2022-07-18

Geometric distribution where failure probability is not $1-p$

The typical geometric distribution is defined from the success probability p, i.e., a r.v. G~Geometric(p), would have PMF... $P[G=g]=(1-p{)}^{g-1}p$

Fischer and Spassky play a chess match in which the first played to win a game wins the match. After 10 successive draws. the match is declared drawn. Each game is won by Fischer with probability 0.4 is won by Spassky with probability 0.3. and is a draw with probability 0.3. independent of previous games.

a) What is the probability that Fischer wins the match?

For part a, the answer is obviously ${\mathrm{\Sigma}}_{k=1}^{10}(0.3{)}^{k-1}(0.4)\approx 0.571$.

But we could also say that r.v. L~CustomGeometric($p=0.4$, $q=0.3$)...

$P[L=l]={q}^{l-1}p=(0.3{)}^{l-1}(0.4)$

Where the answer to part a would just come from evaluating the CDF of L, i.e., $P[L\le 10]$. Is there a name for this more generic version of the Geometric distribution?

The typical geometric distribution is defined from the success probability p, i.e., a r.v. G~Geometric(p), would have PMF... $P[G=g]=(1-p{)}^{g-1}p$

Fischer and Spassky play a chess match in which the first played to win a game wins the match. After 10 successive draws. the match is declared drawn. Each game is won by Fischer with probability 0.4 is won by Spassky with probability 0.3. and is a draw with probability 0.3. independent of previous games.

a) What is the probability that Fischer wins the match?

For part a, the answer is obviously ${\mathrm{\Sigma}}_{k=1}^{10}(0.3{)}^{k-1}(0.4)\approx 0.571$.

But we could also say that r.v. L~CustomGeometric($p=0.4$, $q=0.3$)...

$P[L=l]={q}^{l-1}p=(0.3{)}^{l-1}(0.4)$

Where the answer to part a would just come from evaluating the CDF of L, i.e., $P[L\le 10]$. Is there a name for this more generic version of the Geometric distribution?

Answer & Explanation

phravincegrln2

Expert

2022-07-19Added 19 answers

Step 1

As was already pointed out, if L is the number of games played until Fischer wins the match, then it is not a well-defined random variable, because Spassky can win with nonzero probability. In such a case, how is L defined? For instance, there could be 3 ties, and the fourth game is won by Spassky. You can't say $L=4$, yet the match has stopped: no more games are played.

A modification to this thinking may be possible. Suppose we call G the random number of games played until one of the two players wins. We do not impose a limit to the total number of games played. Then clearly $G\sim \mathrm{Geometric}(p=0.4+0.3).$.

Then, given G, we know that every game until the final game played was a tie. The final game, having been won decisively, was won by Fischer with probability $\frac{0.4}{0.4+0.3}=\frac{4}{7}.$.

Step 2

In fact, this is the unconditional probability of Fischer winning the match, if there is no restriction on how many games are played. If we do restrict the maximum number of games to $n=10$, then there is a nonzero probability the match results in a tie, which is $(0.3{)}^{10}=\mathrm{0.0000059049.}$.

But if the match is decisive, then Fischer still wins with odds 4:3, i.e. the probability is $\frac{4}{4+3}(1-(0.3{)}^{10})=\frac{1428562993}{2500000000}.$

As was already pointed out, if L is the number of games played until Fischer wins the match, then it is not a well-defined random variable, because Spassky can win with nonzero probability. In such a case, how is L defined? For instance, there could be 3 ties, and the fourth game is won by Spassky. You can't say $L=4$, yet the match has stopped: no more games are played.

A modification to this thinking may be possible. Suppose we call G the random number of games played until one of the two players wins. We do not impose a limit to the total number of games played. Then clearly $G\sim \mathrm{Geometric}(p=0.4+0.3).$.

Then, given G, we know that every game until the final game played was a tie. The final game, having been won decisively, was won by Fischer with probability $\frac{0.4}{0.4+0.3}=\frac{4}{7}.$.

Step 2

In fact, this is the unconditional probability of Fischer winning the match, if there is no restriction on how many games are played. If we do restrict the maximum number of games to $n=10$, then there is a nonzero probability the match results in a tie, which is $(0.3{)}^{10}=\mathrm{0.0000059049.}$.

But if the match is decisive, then Fischer still wins with odds 4:3, i.e. the probability is $\frac{4}{4+3}(1-(0.3{)}^{10})=\frac{1428562993}{2500000000}.$

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