# Probability(dice problem)

• January 19th 2010, 11:51 AM
donsmith
Probability(dice problem)
Hello guys,

i need help with this problem

The probability of winning on a single toss of the dice is p. A starts, and
if he fails, he passes the dice to B, who then attempts to win on her toss. They continue tossing the dice back and forth until one of them wins. What are their respective probabilities of winning?

I have used conditional probability rule and i get the same probability for person A and B. (Wondering)

Thanks
• January 19th 2010, 12:28 PM
novice
Quote:

Originally Posted by donsmith
Hello guys,

i need help with this problem

The probability of winning on a single toss of the dice is p. A starts, and
if he fails, he passes the dice to B, who then attempts to win on her toss. They continue tossing the dice back and forth until one of them wins. What are their respective probabilities of winning?

I have used conditional probability rule and i get the same probability for person A and B. (Wondering)

Thanks

The question doesn't look right. Without being given the winning criteria, you will have infinite tosses.

Suppose $A$ and $B$ takes turn in tossing, and the first person who gets a head or three tails will win, then it will be like this:

Suppose $A$ goes first, then $p_A(H) = 0.5$.
If A got a tail on his first toss, then the probability of B winning will be depend on A getting a tail, then p_B(TH) = (0.5)(0.5) = 0.25

Suppose that $B$ got a tail, then $A's$ probability of winning will be $p_A(\{TTT\} \cup \{TTH\}) = (0.5)(0.5)(0.5)+(0.5)(0.5)(0.5)=0.25$
$A$, in this case, has the advantage for getting the first toss. In this game of toss, $\Sigma p_A = p_A(H) + p_A(TH)+p_A(\{TTT\}\cup \{TTH\})= 0.5+0.25 = 0.75$

whereas $\Sigma p_B = p_B(TH) = 0.25$ only.
• January 19th 2010, 12:40 PM
donsmith
I agree the question doesn't look right.

We are tossing a dice not a coin but i see your point.

thanks
• January 19th 2010, 12:54 PM
novice
Quote:

Originally Posted by donsmith
I agree the question doesn't look right.

We are tossing a dice not a coin but i see your point.

thanks

I am on a semester break. Rolling a dice is easy, but writing it for homework is too much work, and Dice rolling is worse. (Rofl)
• January 19th 2010, 01:31 PM
donsmith

but I'll get it after few hours of sleep
• January 9th 2011, 08:10 PM
howl457
P(A wins) = Sum P(A wins on (2n + 1) roll)
= Sum (1 - p)^(2n) * p
= Sum [(1-p)^2]^n
= p * 1/(1 - (1 - p)^2) --> |(1 - p)^2| < 1
= p/(2p - p^2)
= 1/(2-p)

P(B wins) = 1 - P(A wins) = (1-p) / (2-p)
• January 10th 2011, 05:18 PM
matheagle
IF I undersatnd this, A win via the 1,3,5,7,...tosses, which are

$p+(1-p)^2p+(1-p)^4p+\cdots$

$=p\sum_{n=0}^{\infty}\left((1-p)^2\right)^n={p\over 1-(1-p)^2}={1\over 2-p}$

Which is what the howling did
• January 11th 2011, 07:05 PM
awkward
I like the previous solution(s), but here is an alternative approach which avoids the use of infinite series.

Let A's probability of winning be t, and condition on the outcome of the first toss. A wins on the first toss with probability p. Otherwise, it's B's turn and B will win with probability t, i.e. A will win with probability 1-t. So

P(A wins) = P(A wins | A wins first toss) * P(A wins first toss) + P(A wins | A doesn't win first toss) * P(A doesn't win first toss)

I.e.,

t = 1 (p) + (1-t) (1-p)

Now solve for t.