Fair game problem

[SGS]

Here is a extra credit problem for my stats class that I am struggling with. Any help? A fair game is when the EV of either bet is 0.

Jay and Ted decide toss a fair coin until either five consecutive heads or two consecutive tails appears. If two consecutive tails appears first Jay has agreed to pay Ted $15.00. How much should Ted agree to pay Jay if five consecutive heads appears first in order for this to be a fair game?

SGS

[Yads]

Think, how much more likely is getting 2 consecutive tails compared to 5 consecutive heads?

[LetYouDown]

Let me be the first to say, I highly doubt it's as simple as 1/4 vs. 1/32. [img]/images/graemlins/smile.gif[/img]

[AaronBrown]

Five consecutive heads has probability 1/32, two tails has probability 1/4. If you know one of these events has happened, Bayes Theorem tells you that (1/32)/(1/32 + 1/4) = 1/9 of the time it will be five heads and (1/4)/(1/32 + 1/4) = 8/9 of the time it will be two tails.

In nine games, Jay pays 8*$15 = $120. So Ted must pay $120 when five heads appear to make the game fair.

[LetYouDown]

On first glance, that looks right.

[Siegmund]

It's a challenging game to solve. Not as easy as it looks.

Jay's chance of winning is not 1/8. His chance of *immediately* winning is 1/32 vs. Ted's 1/4, but the remaining 23/32 of the time they play on, and Ted is helped more by that than Jay is.

Here is how I approached it:

The first flip is equally likely to be heads or tails.

If the first flip is heads, or a sequence of less than 2 tails is ended by one head, we are in "State H": 1/16 of the time this leads to a win for Jay, 15/16 of the time this leads to state T.

If the first flip is tails, or a sequence of less than 5 heads is ended by one tail, we are in "State T": 1/2 of the time this leads to a win for Ted, 1/2 of the time this leads to state H.

We can jump back and forth between states any number of times before the game ends, and we add up an infinite series to see how often each player wins.

Ted wins: 1/4 + 15/64 + 15/128 + 225/2048 + 225/4096 + ... = = 31/34.

Jay wins: 1/32 + 1/64 + 15/1024 + 15/2048 + 225/32768 + ... = 3/34.

And if the chances of winning are 31/34 to 3/34, the payoff to make it a fair game is (31/3)(15) = $155.

To get the extra credit, you will, of course, have to be able to show where the terms of those series came from and how you added them up. I wouldn't want to spoil ALL the fun for you.

[AaronBrown]

This is correct, my answer was wrong. In effect, I assumed that if either player did not win immediately that the bet was off. But, as Seigmund points out, if there is not an immediate win, Ted's odd's improve.

You can do this without infinite series. Let X be the number of heads since the last tail, and T(X) be the probability of Ted winning given X. We know:

T(4) = T(0)/2
T(3) = [T(4) + T(0)]/2
T(2) = [T(3) + T(0)]/2
T(1) = [T(2) + T(0)]/2
T(0) = [T(1) + 1]/2

That's easily solved to:

T(4) = 8/17
T(3) = 12/17
T(2) = 14/17
T(1) = 15/17
T(0) = 16/17

The first flip can be either heads, in which case we get T(1) or tails, in which case we get T(0). So Ted's probability of winning is (15/17 + 16/17)/2 = 31/34. Since he wins $15 31 times out of 34, that's a total of $465, he has to make that back by getting $465/3 = $155 the other three times.

[Weatherhead03]

I wish I could be so smart [img]/images/graemlins/frown.gif[/img]

[diebitter]

Take one from the other for 50% probability, and then work out the probabilities for the remainder, and multiply the fee.

eg 5 consecs - 2 consecs = 3 consecs

3 consecutive heads (or tails) = 8/1

therefore he should be paid 8 * 15 bucks = 120 bucks

[irchans]

Very nice proof Aaron.