I'm new to learning probability and am a bit stummped on this simple scenario: I wondered if I could walk you through all my different reasoning, so you can point out what I've misunderstood about the different techniques of probability that thread into working out something like this?
Two players are taking it in turns to flip a coin, scoring 1 point if it lands on head, 0 if tail – thus an equal chance each player will score a point, per go. First one to score 5 points wins the round.
As it stands, Player 1 has 4 points, and Player 2 has 3 – it's P1's go: what is the probability that P1 will go on to win the tournament, and likewise for P2?
Here are my thoughts and where I'm stuck:
1) Clearly, P1 has 50% chance of scoring the 1 point he needs to win the round on his next go, but 50% surely isn't the answer – what if the round played to 100, P1 had 99 points and P2 just 1 – still it's 50/50 P1 will get the winning head on the next go, but overall it's far more likely P1 will go onto win without P2 ever catching up, than just 50%?
2) I tried next to think of it like this – the only way for P2 to win is to score two consecutive heads, and P1 of course to score a tail in between, leading (with P1 going first) to the vector: THTH. Now, the probability of that exact combination is (0.5)^4 = P(Player 2 Wins). Hence, P(Player 1 Wins) = 1 – (0.5^4).
3) But then, I again realised that this isn't correct – rather than THTH being the only vector of scores leading to P2 winning, you could of course have any of an infinite number, each more unlikely than the last, leading to P2 winning – TTTTTHTH for example.
So, some sort of integration would be required to sum the probability that P2 would win, over all infinite cases of diminishing probability – subtract that from 1, and you have P(P1). Or do it the other way around to find the probability of each, of course.
Is my assessment that 3 is the needed strategy, correct – i.e. some sort of integration between 0 tosses to inifinity tosses is required?
Allow me to just return to 2) and modify the scenario a bit – say instead of each flipping a coin, with a tail meaning no one scores a point that go, say a referee flips the coin, and heads is a point for P1 tails a point for P2.
This totally changes it now, and should make it simpler – now every go, one of the players is gauranteed to score a point. Now we have a finite number of paths to victory (and probability vectors) for each player.
1) But still I'm a bit confused: with the scores at P1: 4 & P2: 3, the only way for P2 to win is by scoring two straight tails: (TT). The probabiltiy the ref tosses two tails is (0.5)^2 = 0.25. Hence P(P2) = 0.25, and P(P1) = 1-0.25 = 0.75.
But, this simple scenario has a sample space of these outcomes: (H), (TH), (TT). I.e. those are the only sequences in which the ref could toss the coin and end the game.
2) But of those outcomes, 2 of them mean victory for P1 – so looking at it like this, P(P1) = 2/3??
Here are my thoughts on this, and what I believe is the final correct answer: I forgot that those 3 possible scenarios in the sample space are not equally likely. (H) at 0.5 is twice as likely as (TH) or (TT). If then you account for the weighting of their likeliness, and instead have a new pseudo-Sample Space of (H), (H), (TH), (TT) and consider each of those to be equally likely, then indeed 3/4 scenarios mean victory for P1, which agrees with 1-P(P2) = 1-(0.5)^2 = 0.75.
Is this final interpretation correct?
Thanks very much for reading through my long question! Really appreciate it, as I wanted to lay out all my thoughts on the different probabilty ideas so you can point out what I'm understanding correctly, and where I'm going wrong.
Many thanks indeed! Really appreciate it.
Best Answer
Your reasoning looks sound, but the possibility of very long strings of tails makes it undesirable to try to solve this through enumeration. Let's do it through states instead. Let $P_+(n,m)$ be the probability of $A$ winning if $A$ needs $n$ more wins and $B$ needs $m$ more win and it is $A's$ turn. Let $P_-(n,m)$ be the same thing if it is $B's$ turn. We note that symmetry tells us that $$P_-(n,m)=1-P_+(m,n)$$
Of course, the problem is asking for $P_+(1,2)$. As a quick remark, we note that, no matter what, $A$ will get at least $2$ tries for a win and, of course, $A$ might win even without winning one of those two...so $$P_+(1,2)>.75$$
It helps to solve for $P_+(1,1)$ first. In that situation, we see that either $A$ wins on the first try or the situation is reversed. Thus $$P_+(1,1)=\frac 12\times 1 +\frac 12\times (1-P_+(1,1))\implies P_+(1,1)=\frac 23$$
Now, consider $P_+(1,2)$. Considering $A's$ first toss we see that $$P_+(1,2)=\frac 12\times 1 +\frac 12\times P_-(1,2)$$ Now, considering $B's$ first turn : $$P_-(1,2)=\frac 12\times P_+(1,1)+\frac 12\times P_+(1,2)=\frac 12\times \frac 23+\frac 12\times P_+(1,2)=\frac 13+\frac 12\times P_+(1,2)$$
It follows that $$P_+(1,2)=\frac 12\times 1 + \frac 16+ \frac 14\times P_+(1,2)\implies 12P_+(1,2)=6+2+3\times P_+(1,2)\implies$$$$\implies \boxed { P_+(1,2)=\frac 89}$$