Confused about a probability question

gerry87

Was just asked the following question, and i can't make sense of it.

You're playing a dice game, you get €1 for rolling 1, €2 for rolling 2 and so on... what is the fair price you would pay to play this game?

€3.5, done.

Then they asked, if you get another go of the game (pay another €3.5), what outcome of the first game would induce you to go again?

I can't remember what they said the answer was, but the question didn't make sense to me. Surely the first game shouldn't factor into your decision making?

Am i thinking about this wrong?

seamus

No, the outcome of the first game shouldn't factor into your thinking, if you're being asked after having thrown the first dice.

Are you sure that's the question that was asked?

The only thing I can think of is that if you win €6 on the first go, then you're effectively up by €2.50. If you then roll a 1 on the second go, you're on €3.50 - you've broken even.

So if you do roll a 6 on the first go, then you have a 5 in 6 chance of being in profit after the second roll. Does that sound right?

gerry87

seamus wrote: »

No, the outcome of the first game shouldn't factor into your thinking, if you're being asked after having thrown the first dice.

Are you sure that's the question that was asked?

The only thing I can think of is that if you win €6 on the first go, then you're effectively up by €2.50. If you then roll a 1 on the second go, you're on €3.50 - you've broken even.

So if you do roll a 6 on the first go, then you have a 5 in 6 chance of being in profit after the second roll. Does that sound right?

That kinda makes sense, but i'm not sure its a full answer.

Roll1 Payoff Roll2 Payoff
1 -2.5 1 -5
2 -4
3 -3 EV -2.5
4 -2
5 -1
6 0

2 -1.5 1 -4
2 -3
3 -2 EV -1.5
4 -1
5 0
6 1

3 -0.5 1 -3
2 -2
3 -1 EV -0.5
4 0
5 1
6 2

4 0.5 1 -2
2 -1
3 0
4 1 EV 0.5
5 2
6 3

5 1.5 1 -1
2 0
3 1
4 2 EV 1.5
5 3
6 4

6 2.5 1 0
2 1
3 2
4 3 EV 2.5
5 4
6 5


If the first roll is 4 or above, then the second roll is +EV? So basically if you make any profit on the first roll, your EV of the game is zero, so that profit carries over, so when your total profit drops negative you stop playing? Would that make sense? Surely that would be a strategy that garuntees you lose money!

Something about the question just isn't sitting right, you're basing your decision on the second game on the profit from the first game. Should that profit not just go into the pool of wealth you have? Could it have been worded better?

red_fox

If by fair price you're meant to make winning as likely as losing then 3.50 is of course right. But the second game will not be influenced by the first unless you can't hit the same value again. In that case if you roll 3 or under the first round then you could play again for better odds. For example, if you got 3 the first time the expected value for the second time is 3.6 (or if you only scored 1 then the second would have an expected value of 4)

In any case the expected outcome over two games is 0 won/lost (this follows from your choice of price).

What seamus said is right too, if you've already won something then you're less likely to lose overall, it depends on your perspective, which is where the following comes in: http://en.wikipedia.org/wiki/Minimax

hivizman

I think that the second question is simply looking for an answer that points out that the outcome of each game is independent of any other game, so whatever the result of the first game shouldn't affect your decision as to whether to play the second game.

If the game has a fair price (as you've calculated, this is 3.50), then if you are willing to pay 3.50 to play the game once, you should in principle be willing to pay the same amount to play the game again, whatever the outcome of the first game. Of course, this assumes that you have enough money to enter the game a second time, and it may be worth pointing out the problem of "gambler's ruin" - because a gambler's wealth is finite, it is possible that a particularly poor run of luck will wipe you out before you can recover your losses, even though each game is fair in itself.

Fremen

If you're paying for each game each time, you shouldn't care what you previously paid, or what the previous outcomes were. This game is "memoryless", and your wealth is a martingale.

I'll take a long shot here and say you stated the question wrong, and also that you recently interviewed with a trading company. The statement of the problem that I'm familiar with is "how much would you pay for a game in which you can reroll once if you don't like the first outcome?"

In that case, your best strategy is to reroll if you roll less than the expected value of the first game, but stick if you beat the EV. That way, your EV is 3.5 if you roll a 1,2 or 3 on the first roll, 4 if you roll a 4, 5 if you roll a 5 and 6 if you roll a 6. Averaging out over these gives the total EV for the game, which is 4.25

gerry87

Fremen wrote: »

If you're paying for each game each time, you shouldn't care what you previously paid, or what the previous outcomes were. This game is "memoryless", and your wealth is a martingale.

I'll take a long shot here and say you stated the question wrong, and also that you recently interviewed with a trading company. The statement of the problem that I'm familiar with is "how much would you pay for a game in which you can reroll once if you don't like the first outcome?"

In that case, your best strategy is to reroll if you roll less than the expected value of the first game, but stick if you beat the EV. That way, your EV is 3.5 if you roll a 1,2 or 3 on the first roll, 4 if you roll a 4, 5 if you roll a 5 and 6 if you roll a 6. Averaging out over these gives the total EV for the game, which is 4.25

You're nearly right , i was talking with a recruitment guy and it was about a trading job. He was asking me some questions that could come up in interviews, but he did ask it the way I said, he breezed through an answer that didn't really make sense to me about you expecting it to be near the mean yada yada, so i was wondering if there was a chance he might have gotten the question wrong. That question makes a lot more sense, thanks!