June 1, 2008 – 11:52 pm
Some probability questions/puzzles:
1. I have two children. One of my children is a girl. What are the chances I have two girls?
Answer
The answer to the question as stated 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?
Answer
It’s hard (at least for me) to answer this one based purely on first intuition. 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. 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. I guess I have to put the old three door problem in here. 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?
Answer
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.
4. It’s your birthday, and a generous uncle tells you that he’ll give you in dollars whatever number comes up when you roll a die – so you will win between $1 and $6. He makes it more interesting – if you don’t like what you roll the first time, you can roll again. And if you don’t like that, you can roll one more time – for a maximum of three rolls. Note that if you choose to throw again, you can’t go back to an earlier better roll. What’s your strategy to maximize your winnings?
Answer
The trick here is to realize that on the third roll (if you get that far) you’re really stuck with whatever comes up, and that on average on your third roll you’ll win $3.50. So on your second roll, you should stick with anything 4 or higher, anything less you go on to the third roll. So, on average, your winnings on the second roll are ($6 + $5 + $4 + $3.50 + $3.50 + $3.50)/6 which equals $4.25. So then on your first roll, you should stick with a 5 or a 6, but go on to your second roll with anything less. On average, your winnings overall if you stick with this strategy will be ($6 + $5 + $4.25 + $4.25 + $4.25 + $4.25) /6 which equals $4.67.
5. You take it in turns tossing a coin with a friend – you having the first toss – and keep going until one of you tosses heads and wins. What are the chances you’re the winner?
Answer
To answer question 5 it helps to know a little about geometric progressions. The chances of me winning are the sum of my chances at each toss which is: 1/2 + 1/8 + 1/32… The chances of my friend winning are 1/4 + 1/16 + 1/64… I’m not sure this is the most rigorous way to think about this, however we know that when r<1:
and we also know that:
So if you can get your progression into one of the two forms above, you can just apply the formula on the right hand side.
Now start looking at this from the point of view of the guy starting second. His first toss has a probability of 
, his second toss 
, his third 
. So we can see that on his nth toss, his probability will be 
with n starting at 1.
We see from the above formulas that when we have n starting at 1 we need to get our progression into the form: 
.
So we need to expand as such:
Now we have c=1/4 and r=1/4 to plug in:
.
So of course, since someone has to win, the probability for the starter is 2/3.
But let's work this out. The starter's probability at toss n is 
with n starting at 0. Since n starts at 0, we need to get this into the form: 
. Expanding as above:
Now we have c=1/2 and r=1/4 to plug in:
.
7 Responses to “Probability Puzzles…”
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.
By porksauce on Jun 5, 2008
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.
By porksauce on Jun 5, 2008
you need a captcha on this site or you will get spammed
By porksauce on Jun 5, 2008
Here’s one: If you roll two dice, what is the probability the sum of the numbers you get is odd?
By porksauce on Jun 6, 2008
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?
By nolfonzo on Jun 7, 2008