Talk:1282: Monty Hall
This has absolutely nothing to do with "The Monty Hall Problem". It's strictly about the TV game show Let's Make a Deal. In the game, contestants are often given a choice of several options (Curtains, boxes, envelopes etc). Generally, one has a valuable prize (such as a car), and the others either have a lesser prize or nothing. The "nothing" prizes are often given a colorful name, such as "A pig in a poke". Colloquially, such losing prizes are known as "winning the goat". The joke here is that the contestant, having lost the car, is happy to get a goat as a pet. (In fact, the fine print of the rules make it clear that contestant do not really get such "losing" prizes) JamesCurran (talk) 15:21, 25 October 2013 (UTC)
- I believe this is correct. The focus of this comic is that he didn't win the prize, but instead got the gag prize. Most contestants on the show are bummed out, but Beret Guy is actually excited for a new pet goat! It would also be interesting to know that if you're ever on the show, if contestants actually keep the gag gift, or if it's just the same reusable goat every time. Uctriton00 (talk) 17:41, 25 October 2013 (UTC)
- Not a Monthy Python -- Monty Hall is a game host famous for Let's Make a Deal which gave birth to the Monty Hall Problem Spongebog (talk)
I don't understand "It is known that door 3 has a goat, but nothing else." What do you mean by that? At the beginning in the Monty Hall problem, a contestant knows nothing. --184.108.40.206 04:27, 25 October 2013 (UTC)
It goes like this: Player chooses door A, Monty then opens a door he knows there is a goat behind. Player is then offered a chance to switch.
If you do not switch you get a 1/3rd chance of winning because it was a 1 in 3 guess and nothing changed. But if you take into account that Monty will ALWAYS open a goat door and never a car door you can recalculate the odds. So you have a 1/3rd chance that you initially chose the car which means you will lose if you switch 1/3rd of the time, but you had a 2/3rd chance of not selecting the car initially meaning you have a 2/3rds chance if you switch at winning the car. 220.127.116.11 04:58, 25 October 2013 (UTC)
- Reminds me a bit of 1134: Goats make sense. Goats are fine. --18.104.22.168 08:39, 25 October 2013 (UTC)
- It's either A or C. Then the player should choose again between A or C. Xhfz (talk) 14:56, 25 October 2013 (UTC)
- Sequence of the events (assuming Beret Guy initially selects door A)
- Monty: Pick a door.
- Beret Guy: I choose A.
- Monty: I will open another door. It is B. (He opens it and they see a goat). Do you want to switch doors? (Meaning if he will switch from A to C.)
- Beret Guy: I choose door B.
- Beret Guy (to the goat): ...And my yard has so much grass, and I'll teach you tricks, and...
- A few minutes later, the goat from behind door C drives away in the car that was behind door A.
"In that scenario, if a goat is revealed, there is in fact an equal probability of winning by switching or keeping the initial door." I'm not sure of this, can anyone explain? It seems to be stating that simply because of the random chance of the host picking the goat door in this situation then the facts about probability change.This seems to be a very large stretch. Surely in this situation since the host is picking randomly then your probabilities of losing are increased but the moment he does randomly pick a goat door then your chances of winning by switching remain the same as they were in the original problem, 2/3. --Lackadaisical (talk) 17:29, 25 October 2013 (UTC)