A and B throw alternatively a pair of dice, A wins if he throws a sum

Question

A and B throw alternatively a pair of dice, A wins if he throws a sum of 8 before B throws a sum of 9 and B wins if he throws sum of 9 before A throws a sum of 8. If A throws first, what is the probability of A winning the game?

A. 31/36
B. 45/76
C. 31/76
D. 5/36
E. 1/9

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A. 31/36
B. 45/76
C. 31/76
D. 5/36
E. 1/9

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ScottTargetTestPrep · Accepted Answer

Solution:

Recall that there are 36 possible outcomes when a pair of dice is thrown. The outcomes yielding a sum of 8 are: (2,6), (3,5), (4,4), (5,3), and (6, 2). Thus, the probability that A wins on any given throw is 5/36, and the probability that A doesn’t win is 31/36.

For B, there are 4 outcomes yielding a sum of 9: (3, 6), (4, 5), (5, 4), and (6, 3), and so the probability that B wins on any given throw is 4/36 = 1/9. The probability that B doesn’t win on any given throw is 8/9.

The experiment describes a geometric random variable, in which the game stops when A rolls a sum of 8 (and B has not yet won). Let’s determine the first several trials of this experiment to understand the pattern that develops:

First throw: 
P(A wins on first throw) = 5/36

Second throw: 
P(A wins on second throw) = P(A wins on second throw) x P(A didn’t win on first throw and B didn’t win on first throw)
P(A wins on second throw) = 5/36 x

Third throw: 
P(A wins on third throw) = P(A wins on third roll) x P(A didn’t win on first throw and B didn’t win on first throw) x P(A didn’t win on second throw and B didn’t win on second throw)
P(A wins on third throw) = 5/36 x   x  
P(A wins on third throw) = 5/36 x  
 
We want to answer the question of calculating the probability that A wins the game. This happens if A wins on the first throw OR A wins on the second throw OR A wins on the third throw, or fourth, or fifth, etc. Because the number of throws needed until A wins is unknown, let’s sum the probabilities already calculated and extend the sum infinitely.

P (A wins game) = 5/36 + {5/36 x  } + {5/36 x   x  } + …

We can see that we are generating the sum of a geometric series, with first term a = 5/36 and multiplier r =  . The formula for the sum of a geometric series a / (1 – r) can be used to calculate the probability that A wins the game:

a / (1 – r) = 5/36 /   = 5/36 / (1 – 248/324) = (5/36) / (76/324) = 45/76

Answer: B

anjanita · Answer

The answer is B. Please find below my explanation:

P(Sum8)= 5/36

(2,6),(3,5),(4,4),(6,2),(5,3)

P(Sum9)= 4/36 = 1/9

(3,6),(6,3),(4,5),(5,4)

∴P(Awins) = (5/36)+ (31/36)*(8/9)*(5/36) + (31/36)*(8/9)*(31/36)*(8/9)*(5/36)
= {(5/36)}/{1- ((31/36)*(8/9))} = 45/76

acf840 · Answer

I am seeing the explanation here and realizing that I didn't do any calculations. I'm not sure whether I used correct approach.

I just thought, since the probability of sum of 8 is 5/36 and sum of 9 is 4/36, that means that probability of A winning is just a little higher than P(B) winning. Only option C is such an option. All other options are a bit extreme.

I would again like to clarify, I don't think my assumption is 100% Correct here.

Regor60 · Answer

I would say that your intuition is correct but that you didn't proceed correctly from there.

Since there are no ties, the difference from 1 for each of the answer choices should represent B's probability of winning.

As you say, your intuition suggests that B's probability calculated as above should be slightly less than A's.

Answer A suggests B's probability is 1-31/36 = 5/36. Well that's much less than A's, so that doesn't seem right.

Answer C suggests B's probability is higher which definitely doesn't seem correct.

Answer D suggests B's probability is much higher, again doesn't make sense.

Answer E is higher still for B than even D, so not correct.

Answer B suggests B's is 31/76 versus A's 45/76, which seems consistent with your initial observation of A's being slightly higher than B's, so that answer makes the most sense going on intuition alone.

I would say your approach if done correctly is the fastest way to solve this problem.

Posted from my mobile device

pappal · Answer

given: two dices thrown ;A wins when he throws 8 before B throws 9; A starts. P(A wins)=?
P(8)=5/36, P(not 8)=(1-5/36)=31/36 ,P(not 9)=(1-4/36)=32/36
A wins means: P(8) or P(not 8)*P(not 9)*P(8) or..........so on
which forms an infinite G.P. of the form; 
5/36 + 31/36*32/36*5/36 + (31/36)^2 *(32/36)^2 * 5/36.........................where a=5/36 and r=(31*32)/36^2 <1
required P=a/1-r=45/76
B

A and B throw alternatively a pair of dice, A wins if he throws a sum

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A and B throw alternatively a pair of dice, A wins if he throws a sum

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