Expected value of 2 coins

modraig · January 31st, 2010, 8:41 pm

Hi Guys,I am a newbie to probs so I would really appreciate if something could give me his/her opinion on the following brainteaser.2 coins are tossed. if you get 2 H you get $0. if you get 1H and 1T (or vice versa) you get $1. If you get 2 T you play again. What is the expected value? I think is $2/3. Is this correct?Thank you

4ndre1 · January 31st, 2010, 9:50 pm

Yeah, that's correct, it is the solution of the equation x = 1/2 + 1/4x

modraig · January 31st, 2010, 11:54 pm

Thanks 4ndre1 for the quick reply. Could you please tell me what x represents?Thanks again

CrazyDave · February 1st, 2010, 4:48 pm

I think this needs another lookHalf of the time you make $1 = Expectancy of $.50Then 1/4 of the you flip again and half of the time make $1 = Expectancy of $.125(.5 * $1.00) + (.25 * .5 * $1.00) = $.625

NicolasQuant · February 1st, 2010, 5:47 pm

4ndre1 is correct yet to answer to CrazyDave you can as well see it as the sum of a series:EV = 1/2*1 + 1/4*(1/2 + 1/4*(1/2 + ... = 1/2 * (1 + 1/4 + 1/4^2 + ... = 1/2 * (1/ (1-1/4)) = 2/3

CrazyDave · February 1st, 2010, 6:41 pm

AHA!I wasn't thinking infinite series.Just one more play for TT and the game ends.Thanks

4ndre1 · February 2nd, 2010, 9:38 am

QuoteOriginally posted by: modraigThanks 4ndre1 for the quick reply. Could you please tell me what x represents?Thanks againx is the expected value of the game. half of the time you make $1 = expectancy of $.50,then 1/4 times you flip again and ... here is the mistake in CrazyDave's reasoning, youmake not $.50, but you make x, beacause you replay the same game independently ofthe previous outcome.

modraig · February 4th, 2010, 9:02 pm

Hi all,Thanks for the answers. 4ndre1 your reasoning is very clear.Thanks,