Sunday, April 24, 2022

The Problem of points - Without Replacement

Consider an Urn with $A$ green balls and $B$ red balls. Two players A and B play a game where they draw balls from this Urn one at a time (without replacement). If the drawn ball is green, Player A gets a point. Else, Player B gets a point.

If Player A needs $a$ points to win the game whereas Player B needs $b$ points, the question that we address in this post concerns the probability that Player A wins the match.

As the title says, this is just the Problem of points but without replacement. We've already encountered the problem in our blog in the post titled A note on Partial fractions where we saw how Pascal and Fermat independently solved the problem with different reasoning.

Interestingly, Pascal reasoning involves the (CDF of) Neg. Binomial distribution whereas Fermat's idea involves the (CDF of) Binomial distribution. The central insight in both their ideas is in realising that we'll need a maximum of $a+b-1$ games to conclude the match.

Let's consider the original Problem for points for a moment. If we let $p$ and $q$ represent the probability of A and B winning a single game, then using Pascal's idea,

$\displaystyle \mathbb{P}(\text{A wins the match})=\sum_{k=0}^{b-1}\binom{k+a-1}{k}p^aq^k$

where the individual terms are probabilities from the Neg. Binomial distribution, $NB(a,p)$.

Using Fermat's idea,

$\displaystyle \mathbb{P}(\text{B wins the match})=\sum_{k=b}^{a+b-1}\binom{a+b-1}{k}q^kp^{a+b-1-k}$

where the individual terms are probabilities from the Binomial distribution, $B(a+b-1,p)$.

Now using our discussion about the Neg. Hypergeometric distribution in the previous case and Pascal's idea, we could solve the Without-Replacement case of the Problem of points. Or we could go for simplicity and use the Hypergeometric distribution and Fermat's idea.

Either way, let $P(A,a,B,b)$ be the probability Player A winning the game in the setup described at the start of the post and $M$ be a very large number. It should be clear the classical case can be derived out of this setup as $P(pM,a,qM,b)$ with $p+q=1$.

Obviously, when $A=B$ and $a=b$, either player has an equal chance of winning because of symmetry. Similarly, when $A=B$ and $a<b$, Player A has an advantage. Same goes when $a=b$ and $A>B$.

But consider $p(50,25,40,20)$. Player A has 10 more balls than Player B but needs 5 more points to win the game. It isn't immediately clear whether this is advantageous to Player A or not? In this case, the probability of Player A winning the match is only (approx.) 49.07%.

Naturally, we might think as we break symmetry, the match skews towards one way. Perhaps the most surprising result about the Without-Replacement case of Problem of points is the one family of parameters that form an exception.

Consider $p(21, 11, 19, 10)$. Player A has two more balls than Player B but needs one extra point to win the match. Surprisingly, this match is evenly poised. Not just this, for positive integer $n$,

$\displaystyle p(2n+1,n+1,2n-1,n)=\frac{1}{2}$

Clearly, the parameters that define the game lacks symmetry. Still the game is evenly poised between the two players. I could neither come up with a reason nor a simple explanation of why this should be the case? If you can, please post it in the comments below.


Clear["Global`*"];
pp[A_, a_, B_, b_] := (A / (A + B)) Binomial[A - 1, a - 1] Sum[Binomial[B, j] / Binomial[A + B - 1, j + a - 1], {j, 0, b - 1}];
p[A_, a_, B_, b_] := Sum[Binomial[j + a - 1, j] Binomial[A - a + B - j, B - j], {j, 0, b - 1}] / Binomial[A + B, B];
k = 2;
ListPlot[Table[p[2n + k, n + k, 2n - k, n], {n, 100}]]
p[50, 25, 40, 20]
N[%]
p[21, 11, 19, 10]
N[%]
p[26, 13, 22, 12]
N[%]


Until then
Yours Aye
Me

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