1. I have two children. One of my children is a girl. What are the chances I have two girls?
The answer to this question is not 50/50. After being told that one of your children is a girl, you know there are 3 options (with equal probability each): GB, BG & GG. So the answer is a 1/3 chance. Counterintuitive perhaps because people assume the question has stated that the first child is a girl – in which case it would be 50/50.
2. There is a sixgun with two bullets in consecutive chambers pointed at your head. The bad guy spins it, then pulls the trigger. It clicks on an empty chamber. He then tells you that you have a choice: he pulls the trigger again or spins it first before pulling the trigger. Which to choose?
My immediate intuition was to spin again – it would seem that if you were lucky enough for it not to fire once, you’re pushing your luck pulling the trigger again straight away. But now let’s think about it.
The probability of getting shot if you spin again is clearly 2/6, or 1/3. What if you don’t spin? The only way you’re going to get shot is if the you had landed on the one empty chamber just before the bullets – a 1/4 chance. So, perhaps counterintuitively, you’re better off just having the trigger pulled again with no spin.
3. OK, this is a really hoary old chestnut – the Monty Hall problem. You’re at a game show, and the host tells you the car is behind one of the three closed doors – and to choose a door. Before he opens your chosen door, he opens one of the two remaining doors that he tells you he knows the car is not behind, and then gives you the choice of sticking with your original pick or switching to the other unopened door. Which do you do?
It’s not 50/50, even though there’s two doors left (although the probability across both equals 1). There’s many ways to think about it, but to me the one that makes the most sense is that no matter what, the probability that you picked the door in the first place remains 1 in 3. So if you were to pick out of a million doors, the host opening 1 million minus 2 doors leaving your pick and one remaining unopened door doesn’t suddenly up your chances of having picked the door from 1 in a million to 1 in 2. In the 3 door case, if your pick still has a 1/3 chance of having the car, the other door must have a 2/3 chance – so you in effect double your chances by switching doors.
For #5, I think I have a 2/3 chance of winning. Here’s how I came up with that: The probability of my friend winning on any throw is the joint probability that I have not already rolled heads and that he’ll roll heads on that round. So on any round he has half the probability that I have of winning.
Since the probability of someone winning is 1, and I have twice as much chance of winning, I’m 2/3 likely to win. 1=x+y; x=2y; so 2y=1-y or y=1/3.
Not sure though, I’m still scratching my head a little.
I posed the sixgun problem to an intern today and he said to spin it. The gun didn’t actually fire on the second go, which unfortunately reinforced his wrong conclusion. I offered to continue the experiment but he had to rush off.
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Here’s one: If you roll two dice, what is the probability the sum of the numbers you get is odd?
porksauce – your thinking for #5 is very insightful. I just updated the blog entry with a more long-winded way of coming up with the same answer.
Regarding the question: 2 dice, probability the sum is odd. My first thought is the only way to get an odd sum is if you get an even number on one die, odd on the other. Given that half the numbers are even, half odd the chances are 50/50. True?
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