# Math Help - dice game

1. ## dice game

Players A and B roll a pair of dice in turn, with player A rolling first. A's objective is to obtain a sum of 6 (on one roll), while B's ojective is to obtain a sum of 7. The winner is the first player to reach his or her objective. What's the probability that player A wins?

Does my approach make any sense? I'm probably making this way too hard.

I let $A_{W}$ be the event that A wins, and $A_{i}$ be the event that player A rolls a sum of i (i=2,3,...,12) on his first roll.

so $P(A_{W}) = P(A_{W}|A_{2})P(A_{2})+ P(A_{W}|A_{3})P(A_{3}) + P(A_{W}|A_{4})P(A_{4})$ $+ P(A_{W}|A_{5})P(A_{5}) + P(A_{W}|A_{6})P(A_{6}) + P(A_{W}|A_{7})P(A_{7}) + P(A_{W}|A_{8})P(A_{8})$ $+ P(A_{W}|A_{9})P(A_{9}) + P(A_{W}|A_{10})P(A_{10}) + P(A_{W}|A_{11})P(A_{11}) + P(A_{W}|A_{12})P(A_{12})$

when $i \not= 6$, $P(A_{W}|A_{i}) = \frac {30}{36}P(A_{W})$, which is the probability that player B does not win on his first roll times the probability that A wins (assuming independence of course)

then $P(A_{W}) = \frac {30}{36}P(A_{W})\frac{1}{36} + \frac {30}{36}P(A_{W})\frac{2}{36}$ $+ \frac {30}{36}P(A_{W})\frac{3}{36} + \frac {30}{36}P(A_{W})\frac{4}{36}$ $+ \frac{5}{36} + \frac {30}{36}P(A_{W})\frac{6}{36} + \frac {30}{36}P(A_{W})\frac{5}{36}$ $+ \frac {30}{36}P(A_{W})\frac{4}{36} + \frac {30}{36}P(A_{W})\frac{3}{36}$ $+ \frac {30}{36}P(A_{W})\frac{2}{36} + \frac {30}{36}P(A_{W})\frac{1}{36}$

and then solve for $P(A_{W})$

I get 30/61, or about 49%

2. I would add the probability of winning on his/her first attempt plus his/her second....

To win on the first attempt would be ${5\over 36}$.

To win on the second attempt would be $\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr) \biggl( {5\over 36}\biggr)$.

The third would be $\biggl( {31\over 36}\biggr)^2\biggl( {5\over 6}\biggr)^2 \biggl( {5\over 36}\biggr)$.

Sum this via a geometric series should produce the answer.

${5\over 36}\sum_{k=0}^{\infty}\biggl( {31\over 36}\biggr)^k\biggl( {5\over 6}\biggr)^k={5\over 36}\sum_{k=0}^{\infty}\Biggl(\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr)\Biggr)^k$.

I just did the math and I got 31/60 too.

${5/36\over 1- (31/36)(5/6)}=30/61$

3. Yeah, I made the problem way, way too complicated. I was conditioning on the outcome of the first roll.

4. Hello, Random Variable!

You got the right answer . . . Great work!
There are, as you suspected, other approaches.

Players $A$ and $B$ roll a pair of dice in turn, with player $A$ rolling first.
$A$'s objective is to obtain a sum of 6, while $B$'s ojective is to obtain a sum of 7.
The winner is the first player to reach his or her objective.
What's the probability that player $A$ wins?
We have: . $\begin{array}{ccc}P(\text{sum 6}) \:=\: \frac{5}{36} & & P(\text{sum 7}) \:=\:\frac{1}{6} \\ \\[-4mm]
P(\text{not 6}) \:=\: \frac{31}{36} & & P(\text{not 7}) \:=\:\frac{5}{6} \end{array}$

Consider the ways that $A$ can win . . .

$A$ wins on his first roll.
$P(\text{A, 1st roll}) \:=\:\tfrac{5}{36}$

$A$ wins on his second roll.
. . $A$ does not get 6 on his first roll: . $\tfrac{31}{36}$
. . $B$ does not get 7 on his first roll: . $\tfrac{5}{6}$
. . $A$ gets 6 on his second roll: . $\tfrac{5}{36}$
$P(\text{A, 2nd roll}) \:=\:(\tfrac{31}{36})(\tfrac{5}{6})\tfrac{5}{36} \;=\;(\tfrac{155}{216})\tfrac{5}{36}$

$A$ wins on his third roll.
. . $A$ does not get 6 on his first roll: . $\tfrac{31}{36}$
. . B does not get 7 on his first roll: . $\tfrac{5}{6}$
. . A does not get 6 on his second roll: . $\tfrac{31}{36}$
. . B does not et 7 on his second roll: . $\tfrac{5}{6}$
. . A get 6 on his third roll: . $\tfrac{5}{36}$
$P(\text{A, 3rd roll}) \:=\:(\tfrac{31}{36})(\tfrac{5}{6})(\tfrac{31}{36} )(\tfrac{5}{6})\tfrac{5}{36} \:=\:(\tfrac{155}{216})^2\tfrac{5}{36}$

And we see the pattern . . .

$P(\text{A wins}) \;=\;\tfrac{5}{36} + (\tfrac{155}{216})\tfrac{5}{36} + (\tfrac{155}{216})^2\tfrac{5}{36} + (\tfrac{155}{216})^3\tfrac{5}{36} + \hdots$

. . . . . . . $= \;\tfrac{5}{36}\underbrace{\left[1 + \tfrac{155}{216} + (\tfrac{155}{216})^2 + (\tfrac{155}{216})^3 + \hdots \right]}_{\text{geometric series}}$

The geometric series has: . $a = 1,\;r = \tfrac{155}{216}$
. . Its sum is: . $\frac{1}{1-\frac{155}{216}} \:=\:\frac{1}{\frac{61}{216}} \:=\:\tfrac{216}{61}$

Therefore: . $P(\text{A wins}) \;=\;\frac{5}{36}\cdot\frac{216}{61} \;=\;\frac{30}{61}$

Edit: I was almost ready to post this long long disseration
. . . and I was called away for supper.
When I returned, I posted it without checking for earlier replies . . . *blush*

Nice work, matheagle!
.

5. The second part of the problem asks for the expected number of rolls until there is a winner. To answer that you would probably have to use conditioning. Otherwise you would have an infinite series without an obvious sum (I think).

6. Sorobanned, I gave you over an hour to solve this.
Seeing your past work I was waiting for you to do this.
I had dinner, watched a movie and still was surprised I didn't see your geometric sum.

$P(X=k)= \biggl( {5\over 36}\biggr) \Biggl(\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr)\Biggr)^{k-1}$ for k=1,2,...

So $EX={5\over 36}\sum_{k=1}^{\infty}k\Biggl(\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr)\Biggr)^{k-1}$

The trick here is to notice the derivative of $x^k$ and after differentiating

letting $x=\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr)$

7. But $P(X=2) = (\frac {31}{36}) (\frac{6}{36})$, no?

8. This is what I came up with in the past 30 minutes:

E(X) = E[X|A wins on his first roll]P(Awins on his first roll) + E[X|B wins on his first roll]P(B wins on his first roll) + E[X|neither A or B win on their first roll]P(neither A or B win on their first roll)

$E[X] = 1* \frac{5}{36} + 2* (\frac {31}{36}) (\frac {1}{6}) + (2+E[X])(\frac {31}{36}) (\frac {5}{6})$

solving for E(X) I get 402/61 or about 6.6

9. Originally Posted by Random Variable
But $P(X=2) = (\frac {31}{36}) (\frac{6}{36})$, no?

Nope, I worked it out above, you're missing that B must fail once too.

For A to win on the second attempt that would be $\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr) \biggl( {5\over 36}\biggr)$.

That's A fails on first try AND B FAILS ON THE FIRST TRY and then A wins on second try.

10. The second question asks for the expected number of rolls until there is a winner. So P(X=2) is the probability B wins on his first roll.

11. Originally Posted by Random Variable
The second question asks for the expected number of rolls until there is a winner. So P(X=2) is the probability B wins on his first roll.
My rv X was the trial on which A won.

LET W be the trial in which there is a winner.
For W to be 1 we need either {A wins immediately} OR {A losses and B wins}

So $P(W=1)=(5/36)+(31/36)(1/6)$.

and the word 'rolls' is confusing, there are 2 rolls per play.
Actually 4, 2 by each player.
So I figure you mean how many times they play this game.

12. I thought you were responding to my second question. Sorry.

13. If player A wins on his first roll, then there is a winner after 1 roll.

If player B wins on his first roll, then there is a winner after 2 rolls.

If player A wins on his second roll, then there is a winner after 3 rolls.

And so on.

14. Originally Posted by Random Variable
If player A wins on his first roll, then there is a winner after 1 roll.

If player B wins on his first roll, then there is a winner after 2 rolls.

If player A wins on his second roll, then there is a winner after 3 rolls.

And so on.
That's not how I intrepreted it.
But that's easier.
So W=1 means A won immediately on the FIRST round
W=2 means A didn't win but B won immediately
W=3 means neither won on first round, but A won on his second try.

It's 2 am, bedtime

15. I'm so stupid. I just realized that all I had to do was condition on whether or not player A rolls a sum of 6 on his first roll.

Let $A_{W}$ be the event that player A wins

Let $A_{6}$ be the event that player A rolls a sum of 6 on his first roll.

Let $\bar{A}_{6}$ be the event that player A does not roll a sum of 6 on his first roll.

$P(A_{W}) = P(A_{W}|A_{6})P(A_{6}) + P(A_{W}|\bar{A}_{6})P(\bar{A}_{6})$

$P(A_{W}) = 1* \frac {5}{36} + \frac {5}{6} P(A_{W}) \frac {31}{36}$

$P(A_{W}) = 30/61$

And I'm sticking with my answer to the second question.

Let $X$ be the total number of rolls until there is a winner.

Let $A_{6}$ be the event that A rolls a sum of 6 on his first roll.

Let $B_{7}$ be the event that B rolls a sum of 7 on his first roll (which means A didn't roll a 6 on his first roll).

Let $\bar{B}_{7}$ be the event that B doesn't roll a 7 on his first roll.

$E[X] = E[X|A_{6}]P(A_{6}) + E[X|B_{7}]P(B_{7}) + E[X|\bar{B}_{7}]P(\bar{B}_{7})$

$E[X] = 1* \frac{5}{36} + 2* (\frac {31}{36}) (\frac {1}{6}) + (2+E[X])(\frac {31}{36}) (\frac {5}{6})$

EDIT: $E[X] = \frac {402}{61}$

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