Gambler's Chance

Probability Level 5

Vishal a beginner to gambling enters a casino at midnight. There are only four members playing dice at a table in that lonely place. They introduce him to Arjun Dixit a very successful and rich gambler. Vishal wants to challenge him.

Arjun says, "I shall roll two dices and we will bet on two outcomes," whilst picking up two dices on the corner of the table, Arjun continues, "I will give you two choices, what will you bet on the roll of the dice? A sum of 12 in one go, or back to back rolls with the sum of seven ". Vishal thinks for a moment and says, "Sum of Seven!"

Arjun smiles sheepishly and rolls over the dice. They put together a wager of $10 . After a few rolls Arjun keeps the money in his pocket much to the surprise of Vishal.

What was the probability of Arjun's winning ?

The answer is 0.5384.

1 solution

Satyen Nabar
Apr 12, 2015

Let the probability of Arjun winning by throwing a $12$ before two consecutive sums of $7's$ be $p$

Arjun can win the game in the following ways - - -

1) He rolls a $12$ at first throw with probability of $\frac{1}{36}$ and the game is over.

2) He rolls a $7$ and then he rolls a $12$ and wins with a probability of $\frac{1}{216}$

3) He rolls a neutral roll with probability $\frac{29}{36}$ and the game restarts with original probability of winning $p$ .

4) He rolls a $7$ and then a neutral number with probability $\frac{29}{216}$ and the game restarts with original probability of winning $p$ .

Taken together we have,

$p = \frac{1}{36}\ + \frac{1}{216}\ + \frac{29}{36}\times(p) +\frac{29}{216}\times (p)$

Solving $p = 0.538$

Gambler's Chance

The answer is 0.5384.

1 solution

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