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Player P plays a game whereby he bets on an event E occuring.

Event E occurs with probability q. So ¬E occurs with probability (1 - q)

The payoff per round is 1:1 -> For every 1 unit P bets on E , if E occurs, E collects 1 unit as the payoff, and keeps the 1 unit he bet; and if ¬E occurs, P loses the 1 unit he bet.

He adopts a strategy of playing agressive when he is winning, and playing safe when he is losing:

  1. P plays exactly 10 rounds of the game
  2. P always bets on E in every round
  3. P will start the first round with a bet of 1 unit
  4. After each and every loss, P will bet 1 unit on the next round, if there is a next round
  5. If P wins the i^th round, he will increase his bet for the next round ( (i + 1)^th round ) by 2 times of the amount he bet in the i^th round.

ie, P bets 1 unit in i^th round and won. He will bet 2 units in the (i+1)^th round. If he wins again, he will bet 4 units in the (i+2)^th round.

  1. After a triple consecutive win, P returns to betting 1 unit on the following round.
  2. Rules 1 to 6 applies to every round of the game.

An example of a game,

Outcomes E E E E ¬E ¬E E E ¬E ¬E
Bets 1 2 4 1 2 1 1 2 4 1

Question: What is the expectation of the payoff if P adopts this strategy?

Note: No use of markov chain

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I need help with this question. I was thinking of using the usual way of calculating expectation, until I realised that the payoffs for games with the same number of wins and losses differs with the sequence of occurances of the wins and losses.

I tried to calculate the expectation of playing 1 round, 2 rounds, 3 rounds ... individually to try to find a relationship between them, but to no avail. I could not find any perceivable relation. In fact, the expectation became too tedious to calculate beyond 3 rounds.

Is there any intuitive way of solving this problem?

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This solution was written before the poster modified the question and added rule 6. The following is only valid if P continues to double indefinitely after each win.

Let $E(n)$ be the expected payoff on the next $n$ rounds, assuming P stakes 1 unit on the first of those rounds. So we want $E(10)$.

The first thing to note is that with n rounds to play, if P stakes $x$ units on the first of those rounds, the expected payoff will be $xE(n)$ (the expectation scales up).

With $n+1$ rounds left to play, with P staking 1 unit on the first of those:

  • with probability $q$, P will gain 1 and (because they will then stake 2) expect to gain $2E(n)$ on their remaining rounds
  • with probability $1-q$, P will lose 1 and then expect to gain $E(n)$ on their remaining rounds.

So

$E(n+1) = q(1+2E(n)) + (1-q)(-1+E(n))$

$E(n+1) = 2q-1+(q+1)E(n)$

This is a recurrence relation that we can solve, using the initial condition that:

$E(1) = q - (1-q) = 2q - 1$

By considering a solution of the form $E(n) = A\alpha^n+B$ the recurrence relation leads to $\alpha=(q+1)$ and $B=\frac{1-2q}{q}$. The initial condition leads to $A = \frac{2q-1}{q}$. So

$E(n) = \frac{(2q-1)((q+1)^n-1)}{q}$

Substitute $n=10$ to answer the question.

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  • $\begingroup$ Thanks MilesB! Sorry about this but I've realised that I just omitted some important information in the strategy and have updated the question to reflect that. They are in points 6 and 7 $\endgroup$ Mar 13 at 16:49
  • $\begingroup$ That invalidates my solution! $\endgroup$
    – MilesB
    Mar 13 at 16:50
  • $\begingroup$ I find the "reset after 3 wins" thing to be pretty artificial. I like this solution. I am hoping you might clarify: "The first thing to note is that with n rounds to play, if P stakes 𝑥 units on the first of those rounds, the expected payoff will be 𝑥𝐸(𝑛) (the expectation scales up)." This isn't totally clear to me since you reset your bets to 1, not x, if you lose. So this scaling rule is not yet clear to me. $\endgroup$
    – Jake Mirra
    Mar 13 at 17:22
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Suppose we have $n$ rounds to go and P's stake is going to be $x$ on the first of those rounds. $x$ can be 1, 2 or 4. If it's 1 or 2 and they win they double their stake, but if it's 4 and they win, that signals three wins and their stake returns to 1.

Let $E(n,x)$ be the expected payout on the $n$ rounds assuming the stake of $x$.

The following three equations then follow (by considering the probability of a win then the probability of a loss):

$E(n+1,1) = q(1+E(n,2))+(1-q)(-1+E(n,1)) = 2q-1 + (1-q)E(n,1)+qE(n,2)$

$E(n+1,2) = q(1+E(n,4)) + (1-q)(-2+E(n,1))=2(2q-1)+(1-q)E(n,1)+qE(n,4)$

$E(n+1,4) = q(1+E(n,1)) + (1-q)(-4+E(n,1))=4(2q-1)+E(n,1)$

Initial conditions for these recurrence relations are:

$E(1,1) = q + (1-q)(-1) = 2q - 1$

Similarly, $E(1,2) = 2(2q-1)$

and $E(1,4) = 4(2q-1)$

If we express this using matrices, then:

Define $E_n = \begin{pmatrix} E(n,1) \\ E(n,2) \\ E(n,4) \\ \end{pmatrix}$

Define $u = (2q-1)\begin{pmatrix} 1 \\ 2 \\ 4 \\ \end{pmatrix}$

Define $Q = \begin{pmatrix} 1-q & q & 0 \\ 1-q & 0 & q \\ 1 & 0 & 0 \\ \end{pmatrix}$

Then $E_1 = u$

and $E_{n+1} = u + Q E_n$

The solution to this takes the form

$E_n=Q^n a + b$

where $b = (I-Q)^{-1} u$ and $a = Q^{-1} (u-b)$

Now all that remains is to calculate $Q^{-1}$ and $Q^{10}$, use those to calculate $a$ and $b$ and so calculate $E_{10}$ from which $E(10,1)$ can be deduced.

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