2
$\begingroup$

While reading the answer of a previous question Binomial Distribution Question (Exactly/At Least $x$ Trials for Success), it got me thinking a little. I know the reasoning must be flawed somewhere, but I can't tell where. So here it goes:

Given

I have a biased coin, 5% chance of H, 95% chance of T. I toss the coin and I win when the coin turns up H. I can play indefinitely, until I win.

Question

What is the probability of getting an H after at most 5 throws?

Reasoning so far

$$ \Pr(\text{5 or fewer tosses}) = \Pr(\text{exactly 1 toss}) + \Pr(\text{exactly 2 tosses}) + \Pr(\text{exactly 3 tosses}) + \Pr(\text{exactly 4 tosses}) + \Pr(\text{exactly 5 tosses})\\ =0.05^1\cdot0.95^0+0.05^1\cdot0.95^1+0.05^1\cdot0.95^2+0.05^1\cdot0.95^3+0.05^1\cdot0.95^4\\ =0.226 $$ This could eventually be interpreted that if you've had a long enough string of T, you're bound to eventually have a H... which you shouldn't! At every toss you have exactly 5% probability of getting one!

EDIT: Thinking about it a little more, the result doesn't say that after 4 T tosses you've got 22.6% proba of getting a H. But rather that, given the conditions, in 5 throws you've got a 22.6% proba of getting a H... Which still doesn't sound right!

Say you watch someone play the game, and he's 13 tosses in, all of them T. Knowing that in 14 tosses he's got a 51.2% proba of getting a H, you'd be inclined to bet the next toss would result in a H, wouldn't you?

That's my question: this flawed reasoning suggests you should take the bet, however we know if you take it you've got only a 5% proba of winning... Why is this happening?

$\endgroup$
1
$\begingroup$

you are mixing two different things :

  • probability of an event $A_n$ : $\mathbb P(A_n)$
  • probability of an event $A_n$ knowing $B$ : $\mathbb P_B(A_n)$

Here, the probability of winning in less than $n$ tosses will reach $1$ as $n$ goes to infinity because you will eventually get at least one head sometimes. This is $\mathbb P(A_n)$.

But if you observe a game, and you see the guy tossing the coin and getting 10 tails, you know that the first ten tosses are tails, so the next toss as probability $\mathbb P_B(A_{11})$ where $B$ is "the first 10 tosses are tails". And this is equal to $\mathbb P(A_{1})$

The geometric distribution is memoryless.

$\endgroup$
1
$\begingroup$

Hint: Just compute $1-0.95^5$. Besides that your'e talking about two different events. The probability that the fifth toss yields $T$ is $0.95$. The other event is to get at least one $H$ in five tosses: this is achieved when not all tosses are $T$. Now $P(\text{all tosses are $T$})=0.95^5$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.