A coin has .48 probability to be heads and .52 to be tails. Heads is considered a "win" (+1) and tails is considered a "loss" (-1). Each win or loss is added to a cumulative score. How do I figure out the probability of the score reaching +10 before it reaches -10 and vice versa?
I was thinking over this and wondered why you specifically picked 6 tails and 4 heads... it could have very well been
- 2 tails (-2)
- 1 tail, 1 head, 2 tails (-1+1-2 = -2)
- 1 tail, 2 heads, 3 tails (-1+2-3 = -2)
- 1 tail, 2 heads, 2 tails, 1 head, 2 tails (-1+2-2+1-2 = -2)
and so on...
What I see is that there are limitless ways to get -2, not just 4 heads and 6 tails
Also, the way you put it, you are also including instances where you can get 4 heads successively then 6 tails, which means it already gets to +2 before getting to -2...
Well of course it matters. It takes at least ten tosses to reach either 10 or -10.
If we toss the coin only nine times, then the most we can have is +9 not +10.
If we get ten heads in a row, then we get +10 with the probability of .
Suppose we have tossed it ten times and find we now have only +8: nine heads and one tail with the probability of .
After the next toss we will be at either +9 or +7.
You were absolutely correct: this is a very complicated problem.
Let me take a crack at this. I don't think it is that complicated, especially when you use my time honored trick of simplifying thinking by going to an extreme limit.
As an aside - I think Plato's probabilities may be overstated as they do not divide by the number of different ways of getting the same result. Sort of like saying the prob of winning the 649 lottery is 49!/44!. Since order does not matter and there are 6! ways of getting the same line you have to divide by the 720 permutations.
First off, I would tell to you to just get a fair coin and this would be a lot easier - lol! That would make the answer 0.5.
I think the question is complicated by the wording. "reach +10 before -10" but I think this will make itself clear soon.
What if the question were "reach +1 before -1" This only takes one flip right? There are two possible outcomes and only one satisfies the desired target score and the other satisfies the score to be avoided. So it is easy to see the answer to this would 0.48. Correct?
Now let's examine the question "reach +2 before -2". Initially it seems as if there are four possible outcomes (HH, TH, HT, TT) resulting in scores (+2, 0, 0, -2). But this is only IF you restrict it to 2 tosses. But if you are at the 0 score, you are not done yet and you continue to a 3rd toss, which will not get you to either of the target scores, so you toss a 4th time. This will get you to the target or put you back at 0 again where you would continue. This implies that there are not just 4 possible outcomes, instead; there are an infinite number of outcomes!
Going on to higher targets will not help this any - you still have an infinite number of outcomes to consider. As a result, I conclude that the way the question is posed makes it impossible to answer. QED
As soon as you set a limit on the number of tosses, the question is solvable. "What is the probability of reaching +10 before -10 after 15 tosses"
What do you think?
You toss a coin ten times. On each toss if a head appears then add one to the total otherwise subtract a one.
The string “ ” gives sum .
Thus the number of ways to rearrange the string “ ” is the number of ways to get .
That number is .
Now I was guilty of over simplifying the question.
I was trying to get around some deep counting problems.
I now don’t think that is possible.
To solve this we need to know about Markov chains and/or random walk theory.
The reason for that is following. Suppose we have tossed the coin eight times and have six heads and two tails. That gives us a total 4. After the next coin toss, the ninth, the new sum is either 5 or 3.
That is a random walk. At each stage the total changes by one up or down.
The question is, “What is the probability of getting a sum of 10 before getting a sum of -10?”
This is the Gambler's Ruin Problem. See
The Gambler's Ruin