flipping furniture
Whilst doing experimental probabilty, Ragknot ended up with ten coins under his couch. He decided it was now worth the effort to retrieve his accrued fortune. Amazingly the first nine coins he found were heads up. What is the probability the last one will be heads up as well?
Clarification - As I might have outsmarted myself, I want the problem to be interpreted as:
If you flip 10 coins and at least 9 of them land heads up, what is the probability that all 10 will be heads up?
Clarification - As I might have outsmarted myself, I want the problem to be interpreted as:
If you flip 10 coins and at least 9 of them land heads up, what is the probability that all 10 will be heads up?
Labels: Probability
posted by Chris at
5:41 AM
16 Comments:
50/50
Hi Ragknot. I hope that you're wrong :) Have you tried it?
50% probability of heads
Because this is my brthday weekend, there is a chance that my brain is a litle more fried than usual, I have just added a clarification to the original problem. 50% or 1/2 is not the answer I'm seeking.
I wrote the problem in the original form in homage to Ragknot's wonderfully funny answer to the "Coin tossing (yet another)" blog.
1/11
Cry Wolf. You got it. Do you think that the first form of the question is correct? I am intellectually challenged at the moment (that's an understatement).
This post has been removed by the author.
Are you drinking?
I don't see any difference in the outcome, unless you found my coin
collection that are heads/heads.
Hi Ragknot. Because my brain is not up to scratch, I'm unable to determine whether or not the two statements of the problem are the same. It seemed right when I psted it, but I decided to play safe and give the less contentious version.
Do you agree with 1/11 in either/both cases?
Alright I'll try to explain how I got this...
If you consider the case where there are exactly 9 heads:
H H H H H H H H H T
Since order does not matter, there are really 10 ways that I can arrange the coins:
H H H H H H H H T H
H H H H H H H T H H
etc...
On the other hand, there is only one way that all 10 can be heads:
H H H H H H H H H H
Therefore, it is 10 times as likely that the final coin will be tails, making the probability 1/11.
To answer your question, Chris, the only thing about the original question that leaves room for interpretation is this:
"Amazingly the first nine coins..."
When you say "the first 9 coins", it implies that the order in which they are arranged does matter. For example, in the revised edition, he can find a tails coin as say the third coin, and as long as all the others are heads, there's no problem.
However, when you say "the first nine", this is not the case. So in fact saying it like that eliminates 9 out of the 10 possibilities of having 9 heads and 1 tails, making the probability 50-50.
I suppose phrasing it as "Amazingly, nine of the coins he found were heads up."
Hi Cry Wolf. You've hit the part that worried me. At present, I'm quite incapable of deciding whether or not the two versions of the problem are equivalent. The "amazingly..." was my attempt at replacing the safer "at least..." as you have supposed.
I meant each coin toss is independent of any other toss, even if it has not been tossed yet.
If you say that there will be 11 heads, then that's not independent.
It does not matter if you say, you've tossed 10 coins before or you will toss 10 coins after the 11th coin. The 11th coin is independent.
What, you had 11 coins under your couch! I'm sure it was only 10 ;)
I've had a kip, so my brain is much better today. The answer to the second form of the question definitely is 1/11.
I'm going to re-post the original problem (or a variation on it) as "flipping furniture 2", as anyone looking here, will assume that this blog is finished. But I may be a while as I'm not sure what the question actually is!
Chris,
Ok, I'll look for flippen 2.
I can't understand what your looking for, but I know if you mean flipping a single coin, no matter how many after or how many before, the chance is 50/50.
Let's say, I move my couch and see ten heads, and I am looking for the eleventh one, cause I know there were eleven. I see it leaning against the wall. What is the probability that the heads is facing away from the wall so I can see it? Well it will be 50/50!
While writing this, I saw something. Can you take 2 away from 11 and make it even?
Yeah, easy, remove the "el".
Sometimes I read a ToM, and it hits me the wrong way, and I can't even try to figure it out, because the situation throws me off track.
Like your cows eating all the grass. I was so shocked I couldn't think. Or more recently, the military car running out of gas in the desert. I can't think of a solution, because I can't get passed the absurdity. Maybe it it had been me driving my truck, not the military.
Hi Ragknot, 2 from 11, can be done in any odd based counting system. e.g. 11 (base 3) = 4 (base 10), knock of 2 (base 3 or 10) = 2 (base 3 or 10).
I doubt I'd ask a coin flipping problem where the answer was a half, unless that was counter-intuitive. As you see, in this problem the answer is 1/11 (I'm staying with the 10 coins for now).
In the case of the situation that you just proposed, I'm not sure what the answer is - that's why I haen't posted flipping 2 yet. The exact phraseology makes a big difference. In your case with 11 coins, if the 11th is a tails, the probability of you finding the 10 heads first is 1/11. That suggests to me that the probability that the remaining coin is heads, is nearer to 1 than 1/2, I cannot be more precise than that at the moment.
Although I'm not sure, I believe that the answer to the original question is a 1/2.
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