Gambling challenge...

[Browen]

Hypothetical situation...would be interesting to hear the responses!

Easy easy one first:
Someone offers you a game where a single die is rolled and you are paid £1 if a 1 comes, £2 if a 2 comes all the way up to 6. How much would you be willing to pay to play the game?

Now the tough one:
Someone offers you a game based on the toss of a fair coin.
If heads comes up, he pays you £2 and the game ends.
If tails comes up, he tosses the coin again. If heads comes up this time, he pays you £4 and the game ends.
The game continues until heads comes up, with the payout doubling each time tails comes up (£8, then £16, then £32 etc).

What are you willing to pay to play this game?

(Example: H = £2, TH = £4, TTH = £8, TTTH = £16, TTTTTTTTTTH = £2,048.)

[The Incompetent-Donkey]

Well the first one's just the EV of the dice roll
which is (1 + 2 + 3 + 4 + 5 + 6)/6= $3.5

The second one I have to think about

[Browen]

First one: you're right of course, it just sets up the second problem quite nicely!

[The Incompetent-Donkey]

After some thought I think the EV is potentially infinite on the second. Therefore at first glance it would be +EV to make any bet on the second game but I will do more thinking

Well Ive done up to £8 worth of bets and the results are consistent

N=amount bet

EV= (2-N)/2 + (4-N)/4 + (8-N)/6...etc

when done for $2

EV= 0 + 0.5 + 0.75 +0.875 +0.9375....

for $4

EV= -1 + 0 + 0.5 +0.75 + 0.875 + 0.9375....

for $8

EV= -3 - 1 + 0 + 0.5 + 0.75 + 0.875.....

so Im extrapolating that they all add to infinite

So trying to simplify the first formula

EV= 1-(n/2) + 1-(n/4) ......

ok Im stuck maybe someone who's good at maths can show that this will always lead to a positive result

[bridgey321]

I'd go and play poker :p

[Browen]

Thinking about the EV...
50% chance of getting £2 = £1 EV
25% of getting £4 = £1 EV
12.5% of getting £8 = £1 EV
etc.

Sum them up and you get £1+£1+£1+...... which goes to infinity.
Hence it should be profitable (in the long term) to bet £1000000 a go.

The problem lies when you have a finite bankroll...how many times do you have to be able to play for £10 a game to be profitable most of the time? How many times at £20 a game?

[Pates]

This game is getting trickier. @ Your last post, are you talking about £10 profit a game or £10 bet in each game?

[Browen]

This game is getting trickier. @ Your last post, are you talking about £10 profit a game or £10 bet in each game?

I'm talking about a £10 bet in each game...

[Pates]

If you're playing £10 a game then you're going to lose money most of the time and make money just under 12.5% of the time. But everytime you lose money 87.5% of the time you need to make back that loss by playing more. I think the answer to this question is a hell of a lot of games at £10 and verging on a 100k bankroll for £20 a game. I'm sure if I sat down for a while with pen and paper I could work out just how much money you'll need to turn a profit on average.

[Browen]

If you played for £10 a game, you'd need a £10k bankroll to be confident of making a profit.
If you played for £20 a game, you'd need £20m.

If Bill Gates wanted to use his entire fortune (~£35bn) to try and make a profit, he could only play up to £30.12 a game, otherwise he'd most likely lose.

Even though the EV is infinite, the average return increases very very slowly, and the decreasing utility of money (£20 means less to someone who has £2m than to someone who has £50) makes it even less appealing to play above a certain level.
http://en.wikipedia.org/wiki/St._Petersburg_paradox

This leads to the other question...if you could only play this game once, how much would you be willing to bet on it?