shallow probability (1 Viewer)

[OLDMAN]

Bob tosses 11 coins, Penny tosses 10. (i) What is the probability that Bob has more heads than Penny. (ii) Generalize to n+1,n respectively.

 

[Affinity]

(i)50%
(ii)50%

was almost tempted into summing up binomials..

 

[Jumbo Cactuar]

Well I hate probability but please .. is that question for real. I shouldn't talk I'm failing hard. Back to work for me!

 

[Affinity]

it's a real question, and.. not that hard(long) if you think about it

 

[McLake]

Nice Question

(i) 1/2
Why?: Same probability for first 10 tosses, 50-50 for Bob's last toss
(ii) 1/2
Why?: Same probability for first n tosses, 50-50 for n+1 toss

 

[underthesun]

Just curious: since i don't understand this anyways.

Let's say Jim throws 10 coins, and Bob throws 10 coins.

At this point doesn't jim have 50% chance of getting more heads than Bob?

then you throw the coin another time.. Jim should have more than 50% chance of getting more heads than bob.. shouldn't he?

Or I'm just confused?

 

[ND]

Yeh i'm thinking the same as you underthesun (although your use of their names is kinda confusing me ).
For the 1st 10 (or n) tosses, the prob is the same. But for the final toss, Bob has a 1/2 prob, where Penny has 0.

Meh i dunno...

 

[flyin']

I can't see how it's 50-50. *shrugs* And to think I'm studyin' so much Probability and Statistics at Uni.

 

[underthesun]

Hey, if it really is 50-50, next time you want to toss a coin with someone, toss twice and point to this thread

Im thinking the chance of getting more should be in the ratio of n+1:n..

 

[OLDMAN]

Let M=Outcome Bob has more Heads than Penny.

P(M)+P(not M)=1 mutually exclusive and complementary (of course)

But P(not M)=P(Bob has more Tails than Penny) -check this out assuming Bob has n+1 and Penny n coins.

By symmetry P(M)=P(not M)=1/2 P(more heads)= P(more tails)

 

[Archman]

ye its 1/2
the probability of bob getting k heads is (n+1)Ck/2^(n+1) and he'll win if penny gets less than k heads.
say penny has the probability of P(k) doing that.
(aCb is a choose b)
the probabilty of bob getting (n+1)-k heads is (n+1)C(n+1-k)/2^(n+1) = (n+1)Ck/2^(n+1) and he'll win if penny gets n-k heads or less, in other words, k tails or more.
since the chance of getting k tails is equivalent, the chance of getting k tails or more is really, 1-P(k)

notice how (n+1)Ck/2^(n+1) * P(k) + (n+1)C(n+1-k)/2^(n+1) * (1-P(k)) = (n+1)Ck/2^(n+1) = (n+1)C(n+1-k)/2^(n+1)

therefore, chance of bob winnin:
sum(i = 0 to n+1) (n+1)Ci/2^(n+1) * P(i)
= 1/2 sum (n+1)Ci/2^(n+1)
= 2^(n+1)/2^(n+2)
= 1/2

ah oops, didnt c that post by oldman, ah well

 

[underthesun]

So, let Bob throw a coin twice, and penny once.

There is a 1/2 chance of bob getting more heads than penny?

I still don't get it..

 

[OLDMAN]

_________________________________________________
Quote
So, let Bob throw a coin twice, and penny once.

There is a 1/2 chance of bob getting more heads than penny?

I still don't get it..
____________________________________________________

Underthesun:

Required probability is

P(Bob HH)+P(Bob HT or TH)*P(Penny TT)= 1/4+1/2*1/2=1/2

Sorry to edit a continuation, but underthesun has given me another proof.
Above proves for n=1.
Assume true for n=k, ie. Even chance for Bob to have more heads if he has k+1, and Penny k coins. Now give each of them another coin...

 

[Affinity]

I was thinking:

consider only the first n coins for each person.

it's either
BOB has more heads or
PENNY has more or
EQUAL

case 1: the last coin bob tosses does not affect the outcome, bob will have more heads
case 2: the last coin of bob does not change the outcome, bob will not have more heads than penny (at most the last coin will make them equal)
case 3: the last coin is the 'decider' bob will have more heads if this last coin turns out to be a head, which will happen half of the time.

now, case 1 is just as likely as case 2, (symmetrical)

so.. it's 50-50 overall

 

[underthesun]

Ah i got it now. thanks