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**4 members have voted**

January 4th, 2020 at 2:14:36 AM
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I think I'm going to call 2020 the year of math appreciation.

In celebration, here is your next math puzzle.

What is the expected value of this game?

As usual, a beer to the first correct answer and solution. Previous winners of a beer are asked to not post their own answers for 24 hours from this posting. To get the beer, an expression of the answer is need, not an estimate or result of a simulation.

Please use spoiler tags for answers and solutions.

In celebration, here is your next math puzzle.

- A deck begins with one white and one black card
- You draw a card at random
- If you draw a black card, the game ends and you're paid $1 for each black card in the deck, including the one you drew.
- If you drew a white card, it is added back to the deck along with another black card. Then go back to rule 2.

What is the expected value of this game?

As usual, a beer to the first correct answer and solution. Previous winners of a beer are asked to not post their own answers for 24 hours from this posting. To get the beer, an expression of the answer is need, not an estimate or result of a simulation.

Please use spoiler tags for answers and solutions.

Last edited by: Wizard on Jan 4, 2020

It's not whether you win or lose; it's whether or not you had a good bet.

January 4th, 2020 at 3:05:43 AM
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Did you mean the number of black cards drawn, the number of white cards drawn, or the number of black cards that landed up in the deck?

If it's the number of black cards drawn then "3) if you draw a black card, the game ends." so you can only ever draw one black card.

If it's how long you keep going then the problem is essentially that you win N-1 units if you can keep going until the Nth round. The chances of getting through each round get lower and lower, so does the sum of (N-1) / p (picking N-1 white cards, then a black card) converge.

If it's the number of black cards drawn then "3) if you draw a black card, the game ends." so you can only ever draw one black card.

If it's how long you keep going then the problem is essentially that you win N-1 units if you can keep going until the Nth round. The chances of getting through each round get lower and lower, so does the sum of (N-1) / p (picking N-1 white cards, then a black card) converge.

January 4th, 2020 at 4:27:28 AM
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Quote:WizardI think I'm going to call 2020 the year of math appreciation.

In celebration, here is your next math puzzle.

- A deck begins with one white and one black card
- You draw a card at random
- If you draw a black card, the game ends and you're paid $1 for each black card you drew
- If you drew a white card, a black card is added to the deck. Then go back to rule 2.

What is the expected value of this game?

As usual, a beer to the first correct answer and solution. Previous winners of a beer are asked to not post their own answers for 24 hours from this posting. To get the beer, an expression of the answer is need, not an estimate or result of a simulation.

Please use spoiler tags for answers and solutions.

I think you misdescribed the game somewhere. You can only ever draw one black card, so triggering the payment of $1 per black card drawn just means that each game is worth exactly $1.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

January 4th, 2020 at 5:42:45 AM
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I think it's easier to assume the payout depends on something like how many cards you manage to draw or rounds you keep going. Depending on your assumptions for starting conditions (e.g. do you get paid $0 if your first card is black, and $1 if your first two cards are white then black, etc. ) the answer will be something like x or x-1.Quote:unJonI think you misdescribed the game...

I had already sent a PM to Mike but it's now overnight in the US

January 4th, 2020 at 7:22:02 AM
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I did explain the problem incorrectly. Step 3 should say you're paid $1 for each black card in the deck, including the one you drew.

For example, there is a 50% chance you pick the black card immediately and win $1.

There is a (1/2)*(2/3) chance you pick a white card first and then a black one, winning $2.

There is a (1/2)*(1/3)*(3/4) chance you pick a white card first, then another white card, and then a black one, winning $3.

I apologize for the typo.

For example, there is a 50% chance you pick the black card immediately and win $1.

There is a (1/2)*(2/3) chance you pick a white card first and then a black one, winning $2.

There is a (1/2)*(1/3)*(3/4) chance you pick a white card first, then another white card, and then a black one, winning $3.

I apologize for the typo.

It's not whether you win or lose; it's whether or not you had a good bet.

January 4th, 2020 at 8:52:56 AM
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I assume that if you draw the white card, it is put back into the deck along with an additional black card - otherwise, the most you can win is $2, since the second draw will be from a deck with 2 black and no white cards

January 4th, 2020 at 11:12:24 AM
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Assume the game costs $1 to play. The return on a “for-1” basis will be:

1st draw: 1/2 * $2 = 1 / 1!

2nd draw: 1/2 * 2/3 * $3 = 2 / 2!

3rd draw: 1/2 * 1/3 * 3/4 * $4 = 3 / 3!

This is clearly the series 1/1! + 2/2! + 3/3! .... which equals e for a return of 2.72 for 1 or 1.72 to 1.

So the value of the game is e - 1 = $1.72

1st draw: 1/2 * $2 = 1 / 1!

2nd draw: 1/2 * 2/3 * $3 = 2 / 2!

3rd draw: 1/2 * 1/3 * 3/4 * $4 = 3 / 3!

This is clearly the series 1/1! + 2/2! + 3/3! .... which equals e for a return of 2.72 for 1 or 1.72 to 1.

So the value of the game is e - 1 = $1.72

It’s all about making that GTA

January 4th, 2020 at 12:02:20 PM
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Quote:ThatDonGuyI assume that if you draw the white card, it is put back into the deck along with an additional black card - otherwise, the most you can win is $2, since the second draw will be from a deck with 2 black and no white cards

Yes, I just reworded again to make that clear.

It's not whether you win or lose; it's whether or not you had a good bet.

January 4th, 2020 at 12:03:12 PM
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Quote:Ace2Assume the game costs $1 to play. The return on a “for-1” basis will be:

1st draw: 1/2 * $2 = 1 / 1!

2nd draw: 1/2 * 2/3 * $3 = 2 / 2!

3rd draw: 1/2 * 1/3 * 3/4 * $4 = 3 / 3!

This is clearly the series 1/1! + 2/2! + 3/3! .... which equals e for a return of 2.72 for 1 or 1.72 to 1.

So the value of the game is e - 1 = $1.72

I agree! I owe you a beer.

It's not whether you win or lose; it's whether or not you had a good bet.

January 4th, 2020 at 12:49:54 PM
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Quote:WizardQuote:Ace2Assume the game costs $1 to play. The return on a “for-1” basis will be:

1st draw: 1/2 * $2 = 1 / 1!

2nd draw: 1/2 * 2/3 * $3 = 2 / 2!

3rd draw: 1/2 * 1/3 * 3/4 * $4 = 3 / 3!

This is clearly the series 1/1! + 2/2! + 3/3! .... which equals e for a return of 2.72 for 1 or 1.72 to 1.

So the value of the game is e - 1 = $1.72

I agree! I owe you a beer.

Since somebody solved this, here's a more rigorous proof:

The probability of stopping after N black cards are in the deck is 1/2 x 1/3 x ... x 1/N x N/(N + 1) = N / (N + 1)!, so the EV for N black cards is N x N / (N + 1)! = N{sup]2 / (N + 1)!

The total EV = 1

^{2}/ 2! + 2

^{2}/ 3! + 3

^{2}/ 4! + ...

Start with 1 / 0! + (1 / 1! - 1 / 1!) + (1 / 2! - 1 / 2!) + (1 / 3! - 1 / 3!) + ... = 1 / 0! = 1

(1 / 0! + 1 / 1! + 1 / 2! + 1 / 3! + ...) - (1 / 1! + 1 / 2! + 1 / 3! + ...) = 1

(1 x 1) / (0! x 1) + (1 x 2) / (1! x 2) + (1 x 3) / (2! x 3) + (1 x 4) / (3! x 4) + ...) - (1 / 1! + 1 / 2! + 1 / 3! + ...) = 1

(1 / 1! + 2 / 2! + 3 / 3! + 4 / 4! + ...) - (1 / 1! + 1 / 2! + 1 / 3! + ...) = 1

(1 - 1) / 1! + (2 - 1) / 2! + (3 - 1) / 3! + (4 - 1) / 4! + ... = 1

1 / 2! + 2 / 3! + 3 / 4! + 4 / 5! + ... = 1

(1 x 3) / (2! x 3) + (2 x 4) / (3! x 4) + (3 x 5) / (4! x 5) + (4 x 6) / (5! x 6) + ... = 1

(1 x 3) / 3! + (2 x 4) / 4! + (3 x 5) / 5! + (4 x 6) / 6! + ... = 1

((1 x 3) / 3! + (2 x 4) / 4! + (3 x 5) / 5! + (4 x 6) / 6! + ...) + (1 / 3! + 1 / 4! + 1 / 5! + 1 / 6! + ...) = 1 + (1 / 3! + 1 / 4! + 1 / 5! + 1 / 6! + ...)

(1 x 3 + 1) / 3! + (2 x 4 + 1) / 4! + (3 x 5 + 1) / 5! + (4 x 6 + 1) / 6! + ... = 1 + (1 / 3! + 1 / 4! + 1 / 5! + 1 / 6! + ...)

Substitute n

^{2}for (n - 1)(n + 1) + 1:

2

^{2}/ 3! + 3

^{2}/ 4! + 4

^{2}/ 5! + ... = 1 + 1 / 3! + 1 / 4! + 1 / 5! + 1 / 6! + ...

Add 1/2 (= 1

^{2}/ 2! = 1 / 2!) to both sides:

1

^{2}/ 2 + 2

^{2}/ 3! + 3

^{2}/ 4! + 4

^{2}/ 5! +... = 1 + 1 / 2! + 1 / 3! + 1 / 4! + 1 / 5! + 1 / 6! + ...

The left side of the equation is the EV; the right side is (1 / 0! + 1 / 1! + 1 / 2! + 1 / 3! + 1 / 4! + 1 / 5! + 1 / 6! + ...) - 1

1 / 0! + 1 / 1! + 1 / 2! + 1 / 3! + 1 / 4! + 1 / 5! + 1 / 6! + ... = e by Taylor's theorem, so the EV = e - 1