# Binomial distribution false reasoning

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?

• 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 $\mathbb P(A_{1})$
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$})=0.95^5$.