91Ó°ÊÓ

Gambler\(A\)and\(B\)both have the initial fortune of\(50\)dollars. The gambler\(A\)will win on any single game play against the gambler\(B\)with probability\(p\).

Also, one can win one dollar from the other each play in the game, or they can double the stakes, and one can win two dollars from the other on each play of the game.

It will be solved by the conditional probability method. Here we calculate the initial probability.

Assume the conditional probability to calculate the winning probability.

For \(p < \frac{1}{2}\) it is an unfavourable case. So, in unfavourable cases, it is defined by some conditions. Let's take k dollars and will stop if it reaches either \(0\) and \(n\) .

Let \({{\rm A}_1}\) denote the event that a gambler \({\rm A}\) wins two dollars on the game's first play. And also, let \({{\rm B}_1}\) denote the event that the gambler \({\rm A}\) loses one dollar on the first play of the game, and let denote the event that the gambler's fortune \({\rm A}\) ultimately reaches \(k\) dollars(say) before it reaches \(0\) dollars.

Then it is given by

\(\begin{array}{c}pr\left( W \right) = pr\left( {{{\rm A}_1}} \right)pr\left( {W|{A_1}} \right) + pr\left( {{{\rm B}_1}} \right)pr\left( {W\left| {{{\rm B}_1}} \right.} \right)\\ = p \times pr\left( {W\left| {{{\rm A}_1}} \right.} \right) + \left( {1 - p} \right) \times pr\left( {W\left| {{{\rm B}_1}} \right.} \right)\end{array}\)

Here denote that\(p\)= Number of wins

\(1 - p\) = Number of loss

If the initial fortune of the gambler\({\rm A}\)is\(i\)dollars\(\left( {i = 1, \ldots ,k - 1} \right)\), then\(pr\left( W \right) = {a_i}\).

If a gambler\({\rm A}\)wins two dollars on the first play of the game, then fortune increases to\(\left( {i + 1} \right)\)dollars, and the conditional probability\(pr\left( {W|{{\rm A}_1}} \right)\)that the fortune increases\(k\)dollars are\({a_{i + 1}}\). If\({\rm A}\)one loses one dollar on the first play of the game, then the fortune decreases to\(\left( {i - 1} \right)\)dollars and the conditional probability\(pr\left( {W|{{\rm B}_1}} \right)\)that the fortune decreases\(k\)to dollars are\({a_{i - 1}}\). Hence it is given by\({a_i} = p{a_{i + 1}} + \left( {1 - p} \right){a_{i - 1}}\). Here is a take\(p = 0.5\)and also\(1 - p = 0.5\)

Unfair cases it is defined as

\(1 - {a_i} = {a_i} \times \frac{{{{\left( {\frac{{1 - p}}{p}} \right)}^k} - \left( {\frac{{1 - p}}{p}} \right)}}{{{{\left( {\frac{{1 - p}}{p}} \right)}^k} - 1}}\)

Then

\({a_i} = \frac{{{{\left( {\frac{{1 - p}}{p}} \right)}^i} - 1}}{{{{\left( {\frac{{1 - p}}{p}} \right)}^k} - 1}}\)

Here

\(p = 0.5\)

\(1 - p = 0.5\)

\(i = 50\)

\(k = 100\)

Therefore

\(\begin{aligned}{l}{a_{50}}& = \frac{{{{\left( 1 \right)}^{50}} - 1}}{{{{\left( 1 \right)}^{100}} - 1}}\\ &= 0\end{aligned}\)

In this condition to not say the greatest winning probability

\({\rm A}\) .

For \(p > \frac{1}{2}\) it is a favourable case. Define the theorem in the same state in step 3. But for, favourable cases \({a_i}\) defined as \(\left( {1 - {a_1}} \right) = {a_1}\sum\limits_{i = 1}^{k - 1} {{{\left( {\frac{{1 - p}}{p}} \right)}^i}} \)

Here\(i = 50\)and\(k = 100\)it is given by

\(\begin{aligned}{c}\left( {1 - {a_1}} \right)& = {a_1}\sum\limits_{i = 1}^{99} {{{\left( {0.5} \right)}^{50}}} \\\left( {\frac{{1 - {a_1}}}{{{a_1}}}} \right) &= \sum\limits_{i = 1}^{99} {8.89} \\\left( {\frac{{1 - {a_1}}}{{{a_1}}}} \right) &= 43.961\\{a_1} &= 0.022\end{aligned}\)

For favourable cases \(0.022\) probability.

For \(p = \frac{1}{2}\) it is equally favourable cases. Define the same theorem which is defined in step-3, step-4. Then for equally favourable cases \({a_i}\) defined as \({a_i} = \frac{i}{k}\) .

Here\(i = 50\)and\(k = 100\)it is given by

\(\begin{aligned}{c}{a_i} &= \frac{i}{k}\\{a_{50}} &= \frac{{50}}{{100}}\\ &= 0.5\end{aligned}\)

Hence in condition to say that it has the greatest probability.