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.
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:
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!
March 6th 2010, 09:53 AM
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.
March 6th 2010, 10:35 AM
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.
March 6th 2010, 10:54 AM
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 $90). If you continue the calculation, you should find the best strategy at each try.
Let me know if you have any more trouble.
March 6th 2010, 11:23 AM
When you put it that way, that's actually a great idea.
I got my solution another way already though. I just wrote down the expected return for each amount of flips. (0.5^n*$)
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!
March 7th 2010, 12:05 AM
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%.