Coin Tossing

davegage, for n+1 coins the analysis is much simpler than the one you gave:

Bill must toss more heads OR more tails, but cannot toss more heads AND tails. Since the situation is perfectly symmetric with respect to heads and tails, it must be the case that there is precisely a 50% chance that Bill will toss more heads than Ben.

And for n coins the argument is more complicated than the one you never gave. 😵

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Originally posted by THUDandBLUNDER
davegage, for n+1 coins the analysis is much simpler than the one you gave:

Bill must toss more heads OR more tails, but cannot toss more heads AND tails. Since the situation is perfectly symmetric with respect to heads and tails, it must be the case that there is precisely a 50% chance that Bill will toss more heads than Ben.

And for n coins the argument is more complicated than the one you never gave. 😵

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my analysis was about as simple as it gets...it was nothing more than brute force...i am sure there are more elegant solutions, and i suppose there is beauty in simplicity as opposed to brute force.

i'm still not sure why you think i am wrong for n coins...your question as i understood it was what is the probability that bill will toss more heads if each person has n coins. i still submit the answer is n/(2n+2), and the proof is again very simple by brute force -- consider all possibilities and all cases where bill throws more.

i mean, consider a specific case of n=3 for example:

there are sixteen possibilities (i,j) where i is the number of heads bill throws and j is the number of heads ben throws:

(0,0); (1,0); (2,0); (3,0); (0,1); (1,1); (2,1); (3,1); (0,2); (1,2); (2,2); (3,2); (0,3); (1,3); (2,3); (3,3)

there are clearly 6 cases where i > j. so the probability is 6/16 or 3/8. or generally, n/(2n+2).

maybe it's more complicated, as you say, but i don't see how.

THUD,

is it that i also need to normalize by the probability of throwing a certain outcome (i,j)? after all, (3,3) is not very likely (they both have to throw all heads). i think that's probably why it's more complicated like you say...i think you're right about that.

sorry, i always see these things more clearly in hindsight 🙂

Originally posted by davegage
THUD,

is it that i also need to normalize by the probability of throwing a certain outcome (i,j)? after all, (3,3) is not very likely (they both have to throw all heads). i think that's probably why it's more complicated like you say...i think you're right about that.

sorry, i always see these things more clearly in hindsight 🙂

davegage, it is actually easier to first work out the probability of them both throwing the same number of heads.
If this probability = P(s) then we require (1 - P(s)) /2

Suppose they throw 2n coins and get the same number of heads.
Now flip all Bill's coins over.
There will now be n heads and n tails showing.
So P(s) equals probability of throwing n heads from 2n coins.
This equals (2n Choose n) / 2^2n

Hence required probability = (1/2) - [(2n choose n) / 2^(2n+1)]

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Originally posted by davegage
there are sixteen possibilities (i,j) where i is the number of heads bill throws and j is the number of heads ben throws:

(0,0); (1,0); (2,0); (3,0); (0,1); (1,1); (2,1); (3,1); (0,2); (1,2); (2,2); (3,2); (0,3); (1,3); (2,3); (3,3)

there are clearly 6 cases where i > j. so the probability is 6/16 or 3/8. or generally, n/(2n+2).

maybe it's more complicated, as you say, but i don't see how.

You are not including the order of the heads/tails.
There are (N Choose n) ways of throwing n heads from N throws
where (N Choose n) = N! / [n!*(N-n)!]
The total number of ways of making N throws = 2^N

Therefore the probability of throwing n heads from N throws is
P = (N Choose n) / 2^N
eg, if N = 6 we get (6 Choose 3) / 2^6 = 5/16
Hence required probability = (1 - 5/16) / 2 = 11/32, slightly more than 1/3

Originally posted by THUDandBLUNDER
davegage, it is actually easier to first work out the probability of them both throwing the same number of heads.
If this probability = P(s) then we require (1 - P(s)) /2

Suppose they throw 2n coins and get the same number of heads.
Now flip all Bill's coins over.
There will now be n heads and n tails showing.
So P(s) equals probability of thr ...[text shortened]... als (2n Choose n) / 2^2n

Hence required probability = (1/2) - [(2n choose n) / 2^(2n+1)]

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right...I was considering all outcomes, but i also need to consider that not all outcomes are equally probable...

actually, i like your solution above best...much simpler and rather crafty

Originally posted by THUDandBLUNDER
Actually, Bowmann was right for n+1 coins but he gave no reasoning.

When they both have n coins the calculation is a bit more complicated.

Hey! I was right without a reason before Bowmann....😀