Probability Question

EPGFoley

Can anybody help me with this:

Two people, Tom and Gerry, play a game in which each of them tosses a coin in succession until one of them gets a head, who then becomes the winner. If Tom starts with the first toss (so the tosses go Tom - Gerry - Tom - Gerry - Tom etc.. ) what is the conditional probability that Tom is the winner, given that he does not win on the third toss?

I though I had it by just taking the prob of Tom winning on 3rd toss away from prob of him winning in total, but now im just confused...

pkr_ennis

afaik it's still 50/50 as the coin has no memory.

RoundTower

EPGFoley wrote: »

I though I had it by just taking the prob of Tom winning on 3rd toss away from prob of him winning in total, but now im just confused...

it is the probability that he wins, but not on the third toss, divided by the probability that he does not win on the third toss.

Fremen

pkr_ennis wrote: »

afaik it's still 50/50 as the coin has no memory.

You're thinking of "given that he doesn't win by the third toss". Besides, it's not 50/50 since whoever goes first has an advantage.

RoundTower

pkr_ennis wrote: »

afaik it's still 50/50 as the coin has no memory.

this answer can suit pretty much any probability problem though

blackbody

The question is "what is the probability that Tom is the winner, given that he does not win on the third toss?" the next time Tom has a chance of winning is on his fifth toss. At this point 4 tails must have been tossed and the probability of Tom winning at the fifth toss is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32 given that the probability of either a head or a tail is 50:50 i.e. 1/2. so the probability of 4 tails followed by a head is 1/32.

Reillyman

By winning on the third toss, does that mean Tom's third toss or the third overall toss?

blackbody

Given that Tom has lost the third toss we can disregard all previous tosses. Now I think we find the probability of Tom winning just after his loss and before gerrys go. If Tom is to win, Gerry must toss a tail followed by Tom tossing a head. So at this point the odds of Tom winning are*1/2 x 1/2 = 1/4.*

Mellor

blackbody wrote: »

Given that Tom has lost the third toss we can disregard all previous tosses. Now I think we find the probability of Tom winning just after his loss and before gerrys go. If Tom is to win, Gerry must toss a tail followed by Tom tossing a head. So at this point the odds of Tom winning are*1/2 x 1/2 = 1/4.*

He might not win on the 5th, but still win if Jerry loses the 6th and so on.
Basically, if he losses the 3rd, then it's the equivalent of Jerry going first starting all over again. And his chance of winning is the same as jerrys chance on not winning.

So the conditional prob of winning, given that you don't win by the third is equal to "1 - (probability of winning when going first)"

Easily solved with a geometric series

frido

The question seems a bit ambiguous. And you really can't be in probability - it causes tears. - Three different answers.

I think you might mean what is the summed probability of Tom winning on the 1st, 5th, 7th, 2n+1th toss.
Which would be 0.5(1+0.5^2+0.5^4+0.5^6...) - 0.5^2
which is 0.5(1+r+r^2 etc) - r, where r is 0.25 with solution
0.5(4/3 - 1/4) = 2/3-1/8
= 13/24

Alternatively it means what are the odds of tom winning on the 5th, 7th, 2n+1th toss, given there are no winners by or on the third toss. I.e. probability of a win on the 1st, 2nd, 3rd =0, which is equivalent to a non conditional scenario with George going first.
P(Wg) = 1/2(1 + 1/2^2....) = 2/3
P(Tom winning) = 1/3

Alternatively it means that Tom's 3rd toss is invalid in which case it simply isn't counted and can be completely ignored. i.e George throws twice.
And the probabilty of tom winning on the 1st is 0.5, on the 3rd is 0. Now ignoring the third toss, letting George throw twice and reconstructing your probability tree I think that it should wiggle out with a bit of messing around as
0.25+0.25(1/1-0.25) = 1/4 + 1/3 = 7/12.

Like I said - don't be ambiguous. Sprechen English.

Mellor

frido wrote: »

The question seems a bit ambiguous. And you really can't be in probability - it causes tears. - Three different answers.

I think you might mean what is the summed probability of Tom winning on the 1st, 5th, 7th, 2n+1th toss.
Which would be 0.5(1+0.5^2+0.5^4+0.5^6...) - 0.5^2
which is 0.5(1+r+r^2 etc) - r, where r is 0.25 with solution
0.5(4/3 - 1/4) = 2/3-1/8
= 13/24

Alternatively it means what are the odds of tom winning on the 5th, 7th, 2n+1th toss, given there are no winners by or on the third toss. I.e. probability of a win on the 1st, 2nd, 3rd =0, which is equivalent to a non conditional scenario with George going first.
P(Wg) = 1/2(1 + 1/2^2....) = 2/3
P(Tom winning) = 1/3

Alternatively it means that Tom's 3rd toss is invalid in which case it simply isn't counted and can be completely ignored. i.e George throws twice.
And the probabilty of tom winning on the 1st is 0.5, on the 3rd is 0. Now ignoring the third toss, letting George throw twice and reconstructing your probability tree I think that it should wiggle out with a bit of messing around as
0.25+0.25(1/1-0.25) = 1/4 + 1/3 = 7/12.

Like I said - don't be ambiguous. Sprechen English.

It's the second one. The fact that the questions says conditional probability ands not just probability that clarifies. The conditional part means that the 3 toss occured and he didn't win, hence nobody won previously


Also, your first and third situations are the same, phrased differently, in each case you are ignoring the 3 toss, ot assigning it a 0% chance.

The reason your answers are different is because you made a mistake in the 3rd.
The extra toss isn't 1/4 chance, 1/4 is the second toss, the extra isn't this as it might not be needed (if he won on the previous toss)

frido

Mellor wrote: »

It's the second one. The fact that the questions says conditional probability ands not just probability that clarifies. The conditional part means that the 3 toss occured and he didn't win, hence nobody won previously


Also, your first and third situations are the same, phrased differently, in each case you are ignoring the 3 toss, ot assigning it a 0% chance.

The reason your answers are different is because you made a mistake in the 3rd.
The extra toss isn't 1/4 chance, 1/4 is the second toss, the extra isn't this as it might not be needed (if he won on the previous toss)

Hi Mellor,
Admittedly it wasn't what the question asked at all, but in the first situation I meant that I was just summing Tom's probability of winning on the 1st, 5th, 7th, .. etc go which I had as
1/2 + 1/2^3 + 1/2^5 + 1/2 ^7 ..... - 1/2^3 = 1/2(1+1/4+1/4^2... - 1/4)

and in the third I was ignoring the third throw completely and solving for
Tw + Tl.Gl.Gl.Tw + TL.Gl.Gl.Tl.Gl.Tw....
=_____ 1/2 + 1/2^4 + 1/2^6 + 1/2^8...
= 1/4 + 1/4 + 1/4^2 + 1/4^3 + 1/4^4...
Wasn't very clear with what I was doing.

Thanks for clarifying conditional, yikes - things I'm suppose to know.