What is the EV of this game?

[Go Blue]

You play a game with me. I flip a coin. These are the possible outcomes:

Flip 1:
Heads: You get $1 and I flip again.
Tails: You get nothing and the game ends.

Flip 2:
Heads: You get another $2 and I flip again.
Tails: You get nothing and the game ends (you still keep the $1).

Flip 3:
Heads: You get another $4 and I flip again.
Tails: You get nothing and the game ends (you still keep the $3 from the first 2 flips).

This countinues for a total of 10 flips, where the payout for heads doubles each time. Obviously, if at any time tails comes up, the game ends and you keep anything you made from the prior flips, if any.

My question is: What is the EV of this game? I was told that it is $5 because the EV of each flip is 50 cents. Is this correct? If so, please explain and elaborate on this.

[AaronBrown]

$5.00 is correct.

You have a 50% chance of winning the $1 prize.
You have a 25% chance of winning the $2 prize.
You have (1/2)^k chance of winning the 2^(k-1) prize.

Each of these has an expected value of $0.50. There are ten of them, so the total expected value is $5.00.

[Go Blue]

[ QUOTE ]
$5.00 is correct.

You have a 50% chance of winning the $1 prize.
You have a 25% chance of winning the $2 prize.
You have (1/2)^k chance of winning the 2^(k-1) prize.

Each of these has an expected value of $0.50. There are ten of them, so the total expected value is $5.00.

[/ QUOTE ]

I see. So the fact that the winnings are cumulative doesn't affect this? As in, you have a 12.5% chance of winning the $4 prize, but if you win that, you also have won $3 from the first two flips?

[ThinkQuick]

[ QUOTE ]
[ QUOTE ]
$5.00 is correct.

You have a 50% chance of winning the $1 prize.
You have a 25% chance of winning the $2 prize.
You have (1/2)^k chance of winning the 2^(k-1) prize.

Each of these has an expected value of $0.50. There are ten of them, so the total expected value is $5.00.

[/ QUOTE ]

I see. So the fact that the winnings are cumulative doesn't affect this? As in, you have a 12.5% chance of winning the $4 prize, but if you win that, you also have won $3 from the first two flips?

[/ QUOTE ]

You're right to think like that, but that's why the game is worth $5 and not $.50. Do you see why?

[BruceZ]

[ QUOTE ]
[ QUOTE ]
$5.00 is correct.

You have a 50% chance of winning the $1 prize.
You have a 25% chance of winning the $2 prize.
You have (1/2)^k chance of winning the 2^(k-1) prize.

Each of these has an expected value of $0.50. There are ten of them, so the total expected value is $5.00.

[/ QUOTE ]

I see. So the fact that the winnings are cumulative doesn't affect this? As in, you have a 12.5% chance of winning the $4 prize, but if you win that, you also have won $3 from the first two flips?

[/ QUOTE ]

There are 2 ways that we can compute the EV here, and it is a good idea to keep both of these methods in your toolbox. You have suggested using the definition of EV by considering the total amounts that we can win for the whole game, multiplying each of these possible outcomes by the probability of winning that amount, and summing as follows:

WL: $1 * 1/4
WWL: $3 * 1/8
WWWL: $7 * 1/16
WWWWL: $15 * 1/32
WWWWWL: $31 * 1/64
WWWWWWL: $63 * 1/128
WWWWWWWL: $127 * 1/256
WWWWWWWWL: $255 * 1/512
WWWWWWWWWL: $511 * 1/1024
WWWWWWWWWW: $1023 * 1/1024
---------------------------------------------------
total EV = $5

An easier way in this case is to use the property that the EV of a sum is the sum of EVs. If we can figure out the contribution to the EV of each flip, then we can add the contribution for the 10 flips to get the total EV. On the nth flip we win 2^(n-1) dollars, provided that we win the first n flips in a row, and that has probability (1/2)^n. So before we make any flips, the nth flip has an EV of 2^(n-1)*(1/2)^n = 1/2 dollars or 50 cents. Adding 50 cents for each of the 10 flips gives an overall EV of $5.

flip 1: $1 * 1/2 = $0.50
flip 2: $2 * 1/4 = $0.50
flip 3: $4 * 1/8 = $0.50
...
flip 10: $512 * 1/1024 = $0.50
---------------------------------------------------
total EV = $5

[Go Blue]

Yeah, I think I see it now (I think). I'm typically good with these kinds of things but I wasn't (still kind of am not) sure about this one. Thanks for the explanations.