av84fun said:
You should be careful about allowing the opponent to flip. Just as is true with knife throwers, it is entirely possible, with practice, to control the number of rotations over a given distance.
You do get some bouncing with coin tosses that you don't get with knife throwing and of course, 80% is WAY too large of an objective. But I know people who can flip for a given side of a coin WAY more than 50% of the time.
In addition, consider a similar bet. What odds would require if I were to bet you that you cannot flip one side exactly 50 times out of 100 chances?
The probabilities are 1 in 2^100 so, according to the odds there is an extreme liklihood of a 50/50 split...BUT...
There is a problem. Quote me the odds on the money and I'll respond with what the problem is.
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Jim
Oh, alright. I'll tell ya what the problem is anyway.
In betting on the exact probabilities, there are 3 potential outcomes and 2 are unfavorable.
1. The coins fall 50/50 so you win.
2. The distribution goes less than the odds and you lose.
3. The distribution goes greater than the odds and you lose.
But, wait...there's more.
Even if the distribution is perfect, you can't get to 100 tosses without completing 99 and at 99 even if the distribution is falling perfectly the "score" will be H45-T44 (assuming the player is calling heads).
So, on the last toss, the coin doesn't care what the last 99 outcomes were. All it knows is that there is a head and a tail and the odds are 50/50 on the last toss.
Therefore, the odds can NEVER be better than even money even if the number of coins tossed is 1 million because it will ALWAYS come down to that last coin toss.
In any event, since there are 3 possible outcomes...each of which carries a 1 in 2 probability,/...even money... your odds of success are 1 in 3.
But wait...it gets worse...MUCH worse.
The standard deviation for 100 coin tosses is +/-5. So, out of 11 outcomes, only 1 wins the bet
But it gets worse. 1 standard deviation is expected only about 68% of the time so now you're talking about roughly a 15-1 shot of ending up with a 50/50 distribution.
So, you can lay 10-1 odds all day against that outcome....I THINK!
It has been a LONG time since I've studied binominal probabilities!
But try it with 20 tosses which is a bet you can easily make in the pool hall. The standard deviation would be +/-1 so there are 3 likely outcomes...only 1 of which is a winner for the mark...but even then, he only has a 68% chance that the deviation will be as small as +/-1.
In any event you can lay 2-1 odds all night and win over time.
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