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        riddles >> general problem-solving / chatting / whatever >> Game: Coin Flips
        
(Message started by: BenVitale on Feb 20^th, 2010, 2:42pm)

Title: Game: Coin Flips
Post by BenVitale on Feb 20^th, 2010, 2:42pm

Here I post a problem w/ the solution.
I don't understand the solution. Could anyone clarify the proposed solution?

The problem:

Suppose I offer a bet on a series of coin flips. One of these bets would be bad for you. Which one?

The sequence HTT will appear before TTT. If it does, I give you $1. If not, you give me $4.
The sequence HHT will appear before TTT. If it does, I give you $1. If not, you give me $3.
The sequence THH will appear before HHT. If it does, I give you $1. If not, you give me $2.
The sequence HTH will appear before THH. If it does, I give you $1. If not, you give me $1.

The proposed solution:

Quote:

Bets are offered on a series of coin flips. Details have been added for this solution.
•The sequence HTT will appear before TTT. If it does, I give you $1. If not, you give me $4. (Good for you. Cost = .80, Expectation = .84)
•The sequence HHT will appear before TTT. If it does, I give you $1. If not, you give me $3. (Bad for you. Cost = .75, Expectation = .69)
•The sequence THH will appear before HHT. If it does, I give you $1. If not, you give me $2. (Good for you. Cost = .67, Expectation = .70)
•The sequence HTH will appear before THH. If it does, I give you $1. If not, you give me $1. (Good for you. Cost = .50, Expectation = .51)

Source: http://numb3rs.wolfram.com/515/puzzle.html

The Wolfram demo: http://demonstrations.wolfram.com/CoinFlips/

Title: Re: Game: Coin Flips
Post by rmsgrey on Feb 21^st, 2010, 5:48am

For the first one, the first time TTT appears, either it's the first three flips, or the previous flip was H - so I'll only have to pay up the one time in eight that the first three flips are all T - so a loss of $7 would be fair.

I've no idea where the .84 (a little under 6/7) comes from - the figures labelled "Cost" have the same value as the probability of my winning that would make it a fair bet at the given payouts, so if the figures labelled "Expectation" were my actual chances of winning, direct comparison would show which bets were advantageous to me. However, the "expectations" don't appear to be my actual chances of winning in an arbitrarily long series of flips.

One possibility is that the Expectation figures have been generated by listing sequences up to a given length - in which case every time TTT comes up first will have been counted, but some of the sequences just won't have continued long enough for TT to come up (except possibly as the first two flips)

Or the Expectation figures might be intended to be something other than my chance of winning - in which case your guess is as good as mine...

Title: Re: Game: Coin Flips
Post by BenVitale on Feb 23^rd, 2010, 10:40am

on 02/21/10 at 05:48:16, rmsgrey wrote:

Yeah, there's 7/8 chance of winning and 1/8 chance of losing.
In 8 throws, you have $7 gain and $4 loss.

Quote:

I've no idea where the .84 (a little under 6/7) comes from

In the link, Wolfram gives an expectation of 0.84 but I think it is 7/8

Quote:

- the figures labelled "Cost" have the same value as the probability of my winning that would make it a fair bet at the given payouts

In part #1, Wolfram gives Cost = $ 0.80

I don't know how he got that !?

Title: Re: Game: Coin Flips
Post by rmsgrey on Feb 24^th, 2010, 5:36am

on 02/23/10 at 10:40:06, BenVitale wrote:

In part #1, Wolfram gives Cost = $ 0.80

I don't know how he got that !?

If I get $1 when I win, and you get $4 when I lose, then we break-even when I win with probability 0.8 (or 4/5). I'm guessing the cost figures are meant to be the break-even probabilities for the stated payouts rather than any sort of $ amount - the figures match too well for me to buy it being a coincidence...

The "solution" given either includes at least one mistake, or requires better explanation for me to figure out what it's supposed to mean...