User talk:KalebCoberly

Monty Hall problem
Hi - I've reverted this edit because it's both unsourced and incorrect. There's a correct, full case analysis on page 138 of the Grinstead and Snell book, cited in the article. There are 12 possible situations and the player can stay or choose in each. However, they are not all equally likely so you can't just count cases and say that because you win by switching in 6 and lose by staying in 6 that the odds of winning by switching are 1/2. If you switch, the 6 cases in which you win each have probability 1/9 so your overall probability of winning by switching is 6/9 (=2/3) - while the cases in which you lose each have probability 1/18 so your overall probability of losing by switching is 6/18 (=1/3). Winning and losing flips if you don't switch (the cases where you win by switching are cases where you lose by staying, and cases where you lose by switching are cases where you win by staying).

The same tree lets you figure out what happens if the player picks door 1 and the host opens door 3 (this analysis is in the book). The probability of being in this case and the car being behind door 1 is 1/18, while the probability of being in this case and the car being behind door 2 is 1/9. Expressed as conditional probabilities given the player picks door 1 and the host opens door 3 these are 1/3 (=(1/18)/(1/18 + 1/9)) and 2/3 (=(1/9)/(1/18 + 1/9)). -- Rick Block (talk) 17:12, 15 October 2012 (UTC)