deeper probability (1 Viewer)

[OLDMAN]

Two players bet on the outcome of a toss of two coins. Bob bets that double heads will be tossed first. Penny bets that two consecutive single head will be tossed first (that is, exactly one head and one tail, Penny wins if this happens twice, one after the other). They keep tossing until one player wins. What is the probability that Bob wins?

Two solution approaches wanted. On second thoughts, change that to three!

 

[ND]

Ok consider two tosses. Bob can win by getting 2 heads either on the 1st or 2nd toss: 1/4*2=1/2. Penny has a 1/2 prob of getting a single head on 1st and 1/2 on second: 1/2^2=1/4. Probability of Bob winning is 2/3?

I dunno, i can't do probability....

 

[OLDMAN]

ND : for a solution approach, check out 1999 HSC 7b.

 

[McLake]

1) ND's way

2) Finite State Machine with probablities

3) ???

 

[flyin']

I'm not sure about this solution but here's my attempt:

Approach 1:

P(B wins in 1 attempt) = P(HH) = 1/4
P(P wins in 1 attempt) = P(H)P(H) = (1/2)(1/2) = 1/4
P(Neither win in 1 attempt) = 1 - (1/4 + 1/4) = 1/2

So the probability that Bob wins is given by the probability he wins in the first attempt, or in the second attempt (that is - Neither win in attempt 1 and Bob wins attempt 2), or in the third attempt (that is - Neither win in attempt 1 and 2, and Bob wins attempt 3)... etc.

P(Bob wins)
= (1/4) + (1/4)(1/2) + (1/4)(1/2)^2 + (1/4)(1/2)^3 ...
= (1/4) { (1/2)^0 + (1/2)^1 + (1/2)^2 + (1/2)^3 ...}
= (1/4) {[1 - (1/2)^n] / [1 - 1/2]}, where n is infinity
= 1/2.

 

[flyin']

Approach 2:

Since the probability that both (Bob and Penny) will win are equally likely in one attempt (in this case, 1/4) and in subsequent attempts, it is expected that the probability that one wins is 1/2 (and the other consequently is also a half).

Is this a valid approach?


As for another approach, I'm out of ideas.

 

[OLDMAN]

__________________________________________________
Quote from flyin'
Since the probability that both (Bob and Penny) will win are equally likely in one attempt ...
__________________________________________________

Just to clarify that Bob can win in just one attempt, but Penny requires two successful consecutive attempts. I have edited the problem statement correspondingly.

 

[flyin']

Should've guessed, I misread the question!


Before, I attempt it... could you clarify this bit: "(that is, exactly one head and one tail, Penny wins if this happens twice, one after the other)."

 

[OLDMAN]

An example of how Penny can win : first toss HT, second toss TH - exactly one head happened each toss.

 

[Archman]

1. one of them has to win, but whenever TT occur, either of them has as good chance of winnin as they have at the start. look at cases where theres no TT involved
HH: 1/3, B wins
HT or TH: 2/3, wait for next toss
if HH, 2/3*1/3 = 2/9: B wins, otherwise, P wins
so its 1/3 + 2/9 = 5/9

 

[Archman]

2. i dunno if u can call this a different approach, instead of starting from the start, u work backwards, look at the last two tosses of the coins. the working out will be exactly the same as above.

 

[spice girl]

Look, I dun think anybody has the solution yet, and btw, i can only find one, so i shall not be too greedy.

2 throws (or less):
P(bob) = 1/4 + 3/4 * 1/4 = 7/16
P(penny) = 1/2 * 1/2 = 1/4
P(neither) = 5/16

Now of this 5/16, lets see the possible combinations:
HT TT
TH TT
TT TT
TT HT
TT TH

3 throws:
Since if noone wins, there's a 3/5 chance of TT being the last throw, P(penny) = 2/5 * 1/2 = 1/5
P(bob) = 1/4 (since Bob doesn't care wot turned up last)

Now if noone wins on the 3rd throw:
if 2nd throw was TT, possible throws are TT, TH, HH
if 2nd throw was TH, HT, only TT is possible.

Thus if noone wins on the 3rd throw, there REMAINS a 3/5 chance of TT being the last throw, with 1/5 being HT, and 1/5 being TH (i.e. the outlook on the last throw REMAINS CONSTANT)

So we can say that if the game drags onto 3 throws or larger, P(bob) = 1/4 and P(penny) = 1/5 on every go.

Thus P(bob on 3rd go or later) = 5/16 * (1/4) / (1/4 + 1/5) = 5/16 * 5/9
P(penny on 3rd go or later) = 5/16 * (1/5) / (1/4 + 1/5) = 5/16 * 4/9

thus P(bob overall) = 7/16 + 25/144
P(penny overall) = 1/4 + 20/144

*phew* it all adds up to 1

 

[Affinity]

Would have to dispute that result I got 5/9 for Bob and 4/9 for Penny, and seems to reflect the results from a simulation better than 11/18 and 7/18

the results for bob from the simulation were (100000 trials)
55.518%
55.516%
55.579%
55.674%

did it this way:

ok.. let the probability of Penny winning in exactly n+2 moves be P(n)

we know that P(0)= (1/2)*(1/2) = 1/4
and P(1) = 1/16

now:

lets call HT/ TH event A and TT event B

P(k+2) = (1/4)*P(k+1) + (1/8)*P(k)

because all sequences of events that lead to P(k+2)
can be constructed by preceding each sequence in p(k) by "AB"
and by preceding each sequence for P(k+1) by B
(sorry I know this isn't clearly expressed, Mods please feel free to edit)

now this is a second order linear homogeneous recurrence relation

and it has solution in the form P(n) = C*(-1/4)^n + D*(1/2)^n
solving for C,D by using values for n=0 and n=1 gives:

C = 1/12 D = 1/6

now we find sum of the infinite series P(n)
that gives 4/9
therefore the probability of penny winning is 4/9
and that of bob winning is 5/9

 

[deyveed]

Whoa. Would a question like this be in the exam? It seems very hard and something that takes ages to learn.
My initial guess was Bob = 1/2 and Penny = 1/2 since chucking 1 coin twice or 2 coins once seems to have the same probability.
Oh well... Seems like i'm off by heaps.

btw, Affinity, did you have to learn something outside 4u to solve this because you mentioned "second order linear homogeneous recurrence relation"?

 

[Lazarus]

"second order linear homogeneous recurrence relation" - sounds suspiciously similar to first-year uni calculus.

 

[turtle_2468]

Discrete actually.

 

[Archman]

i thought it was "second order linear homogeneous recurrence relation with constant coefficients" lol

 

[Affinity]

hmm OLDMAN mentioned three ways.. so there must be some 4U way to do this

Archman, let's expand that to "Second order linear homogenous non-trivial reccurence relation with constant coefficients"

 

[OLDMAN]

___________________________________________________
Quote deyveed
Whoa. Would a question like this be in the exam? It seems very hard and something that takes ages to learn.
___________________________________________________

The truth is "no holds barred" in these last questions, can't hold the examiner to the syllabus. But despite this, the examiners have been quite remarkably fair, as I believe this question is.

This question is similar to 1999 Q7b.

Method 1.

Draw a probability tree tracing how Bob can win. Start with a node called "square one". Emanating from "square one" are : a toss of HH with probability 1/4, Bob wins; a toss of TT with probability 1/4, back to "square one" ;a toss of a TH or HT, followed by HH with probability 1/2*1/4- Bob wins, or followed by TT with probability 1/2*1/4 back to "square one", or followed by TH or HT with prob 1/2*1/2 -Bob loses . All these are mutually exclusive.

Probability (Bob wins w/o going back to "square one")= 1/4+1/2*1/4=3/8
Probability(back to "square one")= 1/4+1/2*1/4=3/8

P(Bob wins)= 3/8+3/8*3/8+[(3/8)^2](3/8)+[(3/8)^3](3/8)+...
=3/5.

 

[Archman]

Originally posted by Affinity
...
and it has solution in the form P(n) = C*(-1/4)^n + D*(1/2)^n
solving for C,D by using values for n=0 and n=1 gives:

C = 1/12 D = 1/6

now we find sum of the infinite series P(n)
that gives 4/9
therefore the probability of penny winning is 4/9
and that of bob winning is 5/9

ah i c... i was way off.
anyway, affinity had the correct method, except that there was a slight calculation mistake:

(sum 0 to infinity) 1/12*(-1/4)^n + 1/6*(1/2)^n
= (1/12)/(5/4) + (1/6)/(1/2)
= 1/15 + 1/3
= 6/15 = 2/5

thats pennys chance, so bobs chance will be 1 - 2/5 = 3/5

p.s. in case anyone who dunno how the -1/4 and 1/2 came about, they are the roots of x^2=1/4*x+1/8