Probability question

Hi, can someone explain to me why the answer to the following question is not simply (4/15)...

Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always 4/5 and the probability of Geri hitting the target is always 2/3 , what is the probability that Tom wins the competition?

(edited 7 years ago)

Reply 1

chazwomaq

Original post by staircase

If the probability of Tom hitting the target is always 5/4 and the probability of Geri hitting the target is always 3/2

I do not understand those probabilities. Can you rephrase or type the exact question in full?

Reply 2

NotNotBatman

Original post by staircase

Hi, can someone explain to me why the answer to the following question is not simply (4/15)...

Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always 5/4 and the probability of Geri hitting the target is always 3/2 , what is the probability that Tom wins the competition?

How can the probabilities be more than 1?

Reply 3

staircase

Original post by chazwomaq

I do not understand those probabilities. Can you rephrase or type the exact question in full?

sorry must ave been a copy and past issue, Ill edit it now

Reply 4

Kiritsugu

Original post by staircase

For Tom to win, he needs to hit and then Geri needs to fail. They could keep
going forever if they both either keep hitting it or keep failing.

For Tom to win, he needs to hit = 4/5 AND
Geri needs to not hit = 1/3.

4/5 * 1/3 = 4/15. However this alone is not the answer.

There's also the probability that Tom could win later on.
Let's say he wins on the 2nd round... the probability is that 1/5*1/3 (both lose) * 4/15 = 1/15 * 4/15 = 4/225.
or they both win = 4/5*2/3 = 8/15 then *4/15 = 32/225
Add these two together... 36/225

I actually don't even know where I'm going with this....... or maybe....

3rd round = 1/15*^how many times both lost (2 in this case) * 4/15 = 1/225 * 4/15 = 4/3375

I'm probably wrong, but I think it is ((1/15)^n * 4/15)+((8/15)^y * 4/15) where n is the number of times both have not hit and y is where they've both hit. Someone correct me...

(edited 7 years ago)

Reply 5

59wlb

The probability Tom wins in the first round is 4/15.
The probability the game goes on to a second round is (4/5 * 2/3) + (1/3 * 1/5) = 9/15
Once they are in the second round it is the same probability of winning as he did before the first round, say p. It is as if the game has reset.
So p = 4/15 + (9/15)*p.
Rearranging give (6/15)*p = 4/15
and the answer is p = 2/3.

(edited 7 years ago)

Reply 6

TheConfusedMedic

Is there a limit to how many attempts they can have because this could go on forever?

Reply 7

staircase

Original post by 59wlb

why do you do (9/15)p??

PS: I understand 9/15 but why * it by p?

Reply 8

59wlb

Original post by staircase

why do you do (9/15)p??

PS: I understand 9/15 but why * it by p?

It uses something called the law of total probability.
P(A) = P(A|B)*P(B) + P(A|C)*P(C) + P(A|D)*P(D)
where B, C, and D cover all possible outcomes. (i.e. P(B)+P(C)+P(D)=1. It doesn't have to be three terms, but I am using three for this example).

In this case A is the probability that Tom wins, B is the probability that he wins in the first round, C is the probability that no one wins in the first round, and D is the probability that Geri wins in the first round.

Then we have:
P(B) = 4/15
P(C) = 9/15
P(D) = 2/15
P(A|D) = 0 since Tom can't win if the other person has already won.
P(A|B) = 1 since he has already won in the first round.
P(A|C) = P(A) since the first round doesn't affect the second round.

Then P(A) = 1*4/15 + P(A)*9/15 + 0*2/15 is the equation I had before.
I hope that's a bit clearer.

(edited 7 years ago)

Reply 9

59wlb

Original post by staircase

why do you do (9/15)p??

PS: I understand 9/15 but why * it by p?

I think I've thought of an easier way of thinking about it, so you might want to ignore my last reply.
9/15 is the probability the game goes to a second round.
But we also want Tom to win.
So we need P(second round AND tom wins)=P(second round)*P(Tom wins)

Reply 10

Starlight2000

Original post by staircase

Is this GCSE?
Have you tried probability tree diagram?

Reply 11

Forum User

Tom wins on the first round with probability 4/5 * 1/3 = 4/15
Geri wins on the first round with probability 1/5 * 2/3 = 2/15

Thus Tom is twice as likely to win on the first round as Geri. The same is true on any subsequent round, should they reach it. So p(Tom wins) = 2 x p(Geri wins). Someone has to win. So p(Tom wins) + p(Geri wins) = 1.

Those two equations give the answer.