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#### What is the probability of this problem?

63 viewed last edited 2 years ago
Anshul Malik
1

A and B roll a dice each. A wins if his dice number is greater than B. B wins if his dice number is greater than or equal to A.

If A gets a 1, then B will win for sure. So B makes up a rule that A will play the same game again if A rolls a 1 in the first throw. The winner of the second game (if it reaches that stage) will be the final winner even if A rolls a 1 again.

What is the probability that A will win this game?

Mahesh Godavarti
3

I think the problem statement can be worded better for easier understanding. However, I think I got it.

Probability of A winning = 1 - Probability of B winning.

Probability of B winning =

Probability of B throwing an equal or greater than when A throws a number greater than 1 in the first throw +

Probability that A throws a 1 in the first throw X Probability of B getting an equal or higher number in the second throw than A =

\frac{1}{6 } \times (\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6}) + \frac{1}{6} \times \frac{1}{6} \times (\frac{6}{6} + \frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6}) =

\frac{1}{6} \times \frac{1}{6} \times 5 \times 6 \times \frac{1}{2} + \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times 6 \times 7 \times 1/2 = 1/6 \times 1/6 \times 1/2 \times (30 + 7) = \frac{37}{72}

Therefore, probability of A winning = 35/72.

Earlier, if B had not changed the rule then the probability of B winning would be

1/6 \times (6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6) = 1/6 \times 1/6 \times 6 \times 7 \times 1/2 = 7/12 .

And the probability of A winning would be 5/12.

Essentially, B leveled the playing field a little bit by changing the rules of the game.

Vivekanand Vellanki
0
It would be good to explain how the probability of B throwing a number greater than or equal to A is determined.
Anshul Malik
0
But the cases in which A rolls a 1 in the first roll should not be counted as sample cases right? Also, the probability that the game is decided in the first roll is 5/6? If I'm understanding conditional probability correctly, then this should also be factored in right?
Mahesh Godavarti
0
I just swt up the sample space and added up all the cases that end up in B's victory. You could set it up as a conditional as well, with A rolling a 1 or not rolling a 1. Or the game finishing in one throw or finishing in two throws, you will get the same answer. In fact, I will set it up and show you that it is the same. Look for my second answer.
Mahesh Godavarti
0

Let's do it another way. Let's condition it first on the event that the game will finish in one throw or two throws.

Then P(B wins) = P(one throw) P(B wins | one throw) + P(two throws) P(B wins | two throws).

P(one throw) = P(A throws a number other than 1) = 5/6

P(two throws) = P(A throws a 1) = 1/6.

P(B wins | one throw) = P(A throws a 2 | A threw a number from 2 to 5) X P(B throws 2 or higher) + P(A throws a 3 | A threw a number from 2 to 5) X P(B throws 3 or higher) + and so on till A throws a 6 = 1/5 * 5/6 + 1/5 * 4/6 + 1/5 * 3/6 + 1/5 *2/6 + 1/5 * 1/6.

Therefore, P(one throw) P (B wins | one throw) = 5/6 * [1/5 * 5/6 + 1/5 * 4/6 + 1/5 *3/6 + 1/5 *2/6 + 1/5 * 1/6 ] = 1/6 * 5/6 + 1/6 *4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 *1/6 = 1/6 * [5/6 + 4/6 + 3/6 + 2/6 + 1/6].

P(B wins | two throws) = P(A throws a 1 in the second throw | A threw a number from 2 to 5 in the first throw) X P(B throws 1 or higher in the second throw) + P(A throws a 2 in the second throw | A threw a number from 2 to 5 in the first throw) X P(B throws 2 or higher in the second throw) + and so on till A throws a 6 in the second throw

Since, the throws are independent what A throws in the second is independent of what is thrown in the first. Therefore, the above expression is the same as:

P(B wins | two throws) = P(A throws a 1 in the second throw) X P(B throws 1 or higher in the second throw) + P(A throws a 2 in the second throw) X P(B throws 2 or higher in the second throw) + and so on till A throws a 6 in the second throw = 1/6 * 6/6 + 1/6 *5/6 + 1/6 * 4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 * 1/6

Therefore, P (two throws) P(B wins | two throws) = 1/6 * [1/6 * 6/6 + 1/6 *5/6 + 1/6 * 4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 * 1/6] = 1/6 * 1/6 * [6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6]

Therefore, P(B wins) = 1/6 * [5/6 + 4/6 + 3/6 + 2/6 + 1/6] + 1/6 * 1/6 * [6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6].

Which is the same as what we had in the other answer.