More Probability for Fun

[Debater Slayer]

Suppose you have two coins and two dice. You flip both coins and then roll both dice, but only once.

What is the probability that you will end up with two heads and two sixes from your one attempt?

As usual, please use spoiler tags for your answer!

 

[Brickjectivity]

Spoiler

All of the events are independent, so there are four independent statistical events. Since there is no dependence, the probability of all four events is the multiplication of their individual probabilities. 1/2 * 1/2 * 1/6 * 1/6 which is 1/144 or about .007

 

[lewisnotmiller]

Spoiler

All of the events are independent, so there are four independent statistical events. Since there is no dependence, the probability of all four events is the multiplication of their individual probabilities. 1/2 * 1/2 * 1/6 * 1/6 which is 1/144 or about .007

Spoiler

This is where I am at as well.
I would say that the 'no dependence' comment gets a little trickier if the scenario included a head and a tail, and this would impact on the outcome.

 

[Shadow Wolf]

Suppose you have two coins and two dice. You flip both coins and then roll both dice, but only once.

What is the probability that you will end up with two heads and two sixes from your one attempt?

As usual, please use spoiler tags for your answer!

Why didn't ask this years ago when I was in college and could actually do this?

 

[Shadow Wolf]

Spoiler

All of the events are independent, so there are four independent statistical events. Since there is no dependence, the probability of all four events is the multiplication of their individual probabilities. 1/2 * 1/2 * 1/6 * 1/6 which is 1/144 or about .007

Yay! I got to be the cheater peaker!

 

[Sedim Haba]

Spoiler

 

[ChristineM]

A problem with your problem. I heardany years ago you were statistically more likely to throw a head in a coin toss. So i took a look at some research.

How random is the toss of a coin?).

 

[exchemist]

Spoiler

All of the events are independent, so there are four independent statistical events. Since there is no dependence, the probability of all four events is the multiplication of their individual probabilities. 1/2 * 1/2 * 1/6 * 1/6 which is 1/144 or about .007

Agree.

 

[SalixIncendium]

Suppose you have two coins and two dice. You flip both coins and then roll both dice, but only once.

What is the probability that you will end up with two heads and two sixes from your one attempt?

As usual, please use spoiler tags for your answer!

Spoiler

100%. You never said the heads and sixes would be facing up.

 

[Debater Slayer]

Spoiler

All of the events are independent, so there are four independent statistical events. Since there is no dependence, the probability of all four events is the multiplication of their individual probabilities. 1/2 * 1/2 * 1/6 * 1/6 which is 1/144 or about .007

Spoiler

This is correct! It doesn't matter whether we want two sixes and two heads or two tails, a three, and a five (for example). Both combinations involve independent events, so it's 1/2 * 1/2 * 1/6 * 1/6.

 

[Debater Slayer]

Spoiler

This is where I am at as well.
I would say that the 'no dependence' comment gets a little trickier if the scenario included a head and a tail, and this would impact on the outcome.

Spoiler

It wouldn't in this case, because a head and a tail in ideal conditions are equiprobable for one coin toss. The probability of two heads is 1/2 * 1/2 = 1/4, and the same goes for the probability of one head and one tail.

 

[Debater Slayer]

Spoiler

100%. You never said the heads and sixes would be facing up.

Spoiler

I like the way you thought out of the box there, but in the context of such problems, "getting a head" or "getting a six" usually implies that they're facing up.

 

[Debater Slayer]

A problem with your problem. I heardany years ago you were statistically more likely to throw a head in a coin toss. So i took a look at some research.

How random is the toss of a coin?).

In real-world conditions rather than problems like these (which typically assume perfect conditions), almost no coin toss is a perfect 50-50. This is because the coin is affected by multiple factors including (but not limited to) the force with which it is tossed each time, the condition of each face (e.g., whether one face is dirtier and heavier than the other), and even the speed of the air when each toss happens.

There are computer programs that simulate an ideal coin toss, though!

 

[lewisnotmiller]

Spoiler

It wouldn't in this case, because a head and a tail in ideal conditions are equiprobable for one coin toss. The probability of two heads is 1/2 * 1/2 = 1/4, and the same goes for the probability of one head and one tail.

Spoiler

Probability of 2 heads on 2 coins is 25%
Probability of 2 tails on 2 coins is 25%
Probability of 1 head and 1 tail on 2 coins is 50%. The first coin is basically not considered from an odds point of view because it can land heads or tails without impact. I then have a 50% chance of getting the required toss on the second coin, with the required toss being dependant on the first toss.

If you said 'first toss head, second toss tail, then the odds are the same as 2 heads'

 

[Debater Slayer]

Spoiler

Probability of 2 heads on 2 coins is 25%
Probability of 2 tails on 2 coins is 25%
Probability of 1 head and 1 tail on 2 coins is 50%. The first coin is basically not considered from an odds point of view because it can land heads or tails without impact. I then have a 50% chance of getting the required toss on the second coin, with the required toss being dependant on the first toss.

If you said 'first toss head, second toss tail, then the odds are the same as 2 heads'

Spoiler

Oh! I got confused there. Yes, you're right. The outcomes are 2² = 4, and they form a sample space S = {HH, TH, HT, TT}. Since the different outcomes can occur in two ways, that's 2/4 = 50%.

I'm thinking about another problem now for my next post where the outcome is known and then conditional probability is calculated from it.

 

[Subduction Zone]

1

Spoiler

Oops, I may not have used the spoiler correctly.

 

[Debater Slayer]

1

Spoiler

View attachment 71804

Oops, I may not have used the spoiler correctly.

That's a surefire way to start a fight in a casino.