All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Pick a card, any card.. (Posted on 2008-03-11)
You sit down with a well mixed deck containing A cards marked "+" and B cards marked "—". You may draw cards from this deck as long as you want, i.e., you can stop playing at any point. Each time you draw a + card you are given \$1 and each time you draw a — card you have to pay \$1. Cards are not replaced after having been drawn.

What would be a fair amount to pay for the right to play (i.e., what is the expected payoff) and under what circumstance should a player cease drawing?

 No Solution Yet Submitted by FrankM Rating: 2.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 the posted official solution is wrong | Comment 22 of 37 |

The posted official solution does have the recursion relation Leming and I have been using, but it stops when the number of minuses exceeds the number of pluses.

As Leming and I have shown, there can still be positive expected value when the negatives exceed the positives. Leming originally showed it for 2 plusses with 3 minuses, and I presented that in another form, that there was still a residual 0.20 to be contained in continuing at that level.

Of course in terminating recursion prematurely, this also affects the expected value of the other cases, where the plusses exceed or equal the minuses.

Using FrankM's formula (or recursion stopping when minuses exceed pluses) starting with A=10, B=5 has an expected value of 5.45, but if the strategy allowed for continuing to the limits stated previously, the expected value is 5.52.

As validation for, to take one example, A=8, B=10, or some other game where it has reached the point where there are 8 +'s and 10 -'s remaining, that there is an expected value of 0.36, see the simulation using this strategy at this link (my previous comment).

 Posted by Charlie on 2008-03-12 15:44:49

 Search: Search body:
Forums (0)