Thread: Tossing a coin

Hey there!
I did a course in probability about a year ago, but things have gotten rusty.
Although I normally hate to have to bother other people with my problems, I need to turn somewhere to get some help. I kind of googled this site up.

My problem:
I have to toss a coin, and keep getting heads every time. For each time that I continue I will gain a better prize, but if I fail, I loose everything. I can stop at any given point and leave with my prize. So it comes down to me having a 0.5^n chance to get the n:th prize.
Here are the prizes for n successes in a row:

1. $1
2. $3
3. $5
4. $10
5. $25
6. $120
7. $180
8. $200
9. $400
10. $1400

Now let's say that I have a pretty large amount of tries available, say 100 tries. What amount of tosses would be the best to aim for, if I want the highest probability to earn the most.

I would appreciate not only a solution, but also an explanation on why. My math knowledge is pre-university. (2nd grade at Swedish gymnasium)

Thanks in advance for any help!

Let's say you're at the $400 level (you tossed 9 heads in 9 tries). Should you toss again? If you do, you have a 50-50 chance getting the $1400 or nothing, so your expected return is 0.5*($0) + 0.5*($1400) = $700, which is greater than $400, so the odds say to flip the coin. I suppose the real answer depends on how many tries you have and how much of a gambler you are, but all we can do here is calculate the expected return.

Now suppose you have tossed 8 heads in 8 tries. If you toss again, you either get $0 or you're into the situation in the previous paragraph, which we decided has an expected return of $700. So you should toss again, because your expected return is $350 and you only get $200 if you don't toss.

I think you can calculate the rest.

Let us know if you're still stuck.

Reasoning like that, going past the second coin would not be worth it, since the expected return would be $2.5, and I already secured $3 if I stop.
However, it might still be worth going on if further steps have a high enough expected return.
Counting expected returns for each single scenario could be done, but would take forever, and that's why I'm looking for a more mathematical way. Might very well be that there's not a better way of doing it though.

Your intuition is correct - it would be better to go on past the $3 level. We need to get a number that shows us this, though. That's why we need to work backward from the $1400 level. Each time we move back a level, we've accounted for all the possiblities from that level forward.

So the expected return after 8 tries is $350. After 7 tries, we can take the $180 or flip the coin with an expected value of $350/2 = $175. So the best strategy is to take the money. After 6 tries, we should also take the money ($120 instead of $90). If you continue the calculation, you should find the best strategy at each try.

Let me know if you have any more trouble.

When you put it that way, that's actually a great idea.
I got my solution another way already though. I just wrote down the expected return for each amount of flips. (0.5^n*$)
Eg:
1. 0,5$
2. 0,75$
etc.

My conclusion; I should aim for the 6th option.
To add, the fee for each time playing is $1. So, chances to win at least once in the time I spend the money earned from it once, are 82%

Anyway, thanks a lot for your help, I appreciate it a lot!

Well done. I get the same answer.

You actually have only 10 strategies to choose from - flip the coin x number of times and if they're all heads, take the money (x is between 1 and 10 inclusive). So your expected return from a strategy is the winning amount times the chances of winning, and the largest expected return is for x=6: $120/(2^6) or $1.875.

In 120 tries, your chances of losing 120 times in a row are (63/64)^120 or about 15.1%, so your chances of winning at least once are 84.9% (how did you get 82%, by the way?). This method also gives x=6 for the best strategy. For x=1 to 10, I get 50.0%, 57.8%, 48.7%, 47.6%, 54.8%, 84.9%, 75.6%, 54.3%, 54.3%, and 74.5%.