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MeetMagicMike Inner circle Gainesville Fl 3501 Posts |
Obviously if you roll one ordinary fair die, the probability of getting a 6 (for instance) is 1/6.
It seems obvious to me that if you roll the dice twice you double your odds of getting at least one 6. But tonight I've been reading many articles saying that the odds are not 2/6. I can follow their arguments but don't find them as compelling as the simple fact that you now have two chances of success. Doesn't that double your odds? If I ask you to put a basketball through a hoop wouldn't giving you two chances double your odds of success? |
Michael Daniels Inner circle Isle of Man 1609 Posts |
You are right - the probability of getting at least one 6 on two rolls is 2/6.
However, the probability of getting EXACTLY one 6 is 2/6 minus the probability of getting two sixes = 2/6 - 1/36 = 5/36. What are the arguments for anything other than this? Mike |
MeetMagicMike Inner circle Gainesville Fl 3501 Posts |
In two rolls you could get:
1/1 1/2 1/3 1/4 1/5 1/6 2/1 2/2 2/3 2/4 2/5 2/6 3/1 3/2 3/3 3/4 3/5 3/6 4/1 4/2 4/3 4/4 4/5 4/6 5/1 5/2 5/3 5/4 5/5 5/6 6/1 6/2 6/3 6/4 6/5 6/6 That's 36 possibilities. 11 of them contain at least one 6. So the probability is apparently 11 out of 6. |
Michael Daniels Inner circle Isle of Man 1609 Posts |
Yes, my bad - brain not in gear that early in the morning.
P(A or B) = P(A) + P(B) - P(A and B) = 1/6 + 1/6 - 1/36 = 11/36 This equates to the listing you gave. By your previous logic, the probability of obtaining at least one Head when a coin is tossed twice is 0.5 + 0.5 = 1.00 (certainty) which is clearly wrong because you could get two tails. The formula provides for this: P(H or H) = P(H) + P(H) - P(H and H) = 0.5 + 0.5 - 0.25 = 0.75 Mike |
MeetMagicMike Inner circle Gainesville Fl 3501 Posts |
Right but I'm still stuck on the idea that giving you two tries at something has to double your odds.
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Michael Daniels Inner circle Isle of Man 1609 Posts |
Quote:
On Jun 4, 2014, MeetMagicMike wrote: Suppose you draw a card from a shuffled deck of 52, then replace it, shuffle, and draw again. What is the probability that at least one of the two selections is a number card? P(number card) on each draw = 40/52 = 0.769. By your reasoning, the probability of at least one number card = 0.769 + 0.769 = 1.538, which makes no sense because a probability cannot exceed 1.00. The actual probability = 0.769 + 0.769 - (0.769 * 0.769) = 0.947 This is the same as the probability that neither card drawn is a picture card, which can be calculated as follows: P(picture card) = 12/52 = 0.231. P(both are picture cards) = 0.231 * 0.231 = 0.053 Therefore probability that both are NOT picture cards (i.e., at least one is a number card) = 1 - 0.053 = 0.947 Mike |
Michael Daniels Inner circle Isle of Man 1609 Posts |
Quote:
On Jun 4, 2014, Michael Daniels wrote: I'm obviously having a bad day - I meant the same as the probability that both cards drawn are not picture cards. My calculations are correct. Mike |
wwhokie1 Special user 512 Posts |
If two tries doubled then three would triple and four would quadruple etc... Which is clearly not possible. Flip a coin and you have a 50% chance of heads. three tosses can not triple your chances, that would be 150% chance of a head.
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MeetMagicMike Inner circle Gainesville Fl 3501 Posts |
I can follow all of that but I still find the idea that having two tries at something doubles your odds very compelling.
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landmark Inner circle within a triangle 5194 Posts |
Probability can be a tough branch of mathematics even for those who have an affinity for other branches because so often the results are counter-intuitive.
If you doubled the number of 6s, you *would* double your chances of rolling a 6 in one try; but doubling the number of tries, as you can see above, does not double the probability of one try. Let's say there were *only* 6s on the die. What's the probability of rolling a 6 on one try? On two tries? On three tries? The probability remains the same i.e. 1.00, but if you were playing for $1 a win, then your pay off would be 3 times the amount for 3 rolls than it would be for one roll.
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
MeetMagicMike Inner circle Gainesville Fl 3501 Posts |
Landmark you bolstered a thought I was having.
If you look at my array above you will see that there are 36 possible outcomes and 11 of them contain at least one 6 so the probability is 11/36 But there are twelve 6's in the array. I think that corresponds to your comment about payoffs. So if I offered you $1 for every 6 that showed then giving you two tries would actually double the payoff. (But would NOT double the chances of a win) So my best explanation of why rolling the dice twice doesn't double your chances of win has to do with that time when you roll two 6's. That is the one case in which you don't benefit at all from the second dice. |
landmark Inner circle within a triangle 5194 Posts |
Yes, that's kind of another way of saying what Michael said above mathematically.
So in plain English what that formula is saying is this: If you want to know the probability of at least one of two things happening, then count the number of ways the first thing can happen, count the number of ways the second thing can happen--BUT MAKE SURE NOT TO DOUBLE COUNT, SO SUBTRACT THE NUMBER OF WAYS THEY BOTH HAPPEN TOGETHER--and then divide by the total number of possible outcomes. And the usual caveat is that each event must be independent of the other, that is, the outcome of the first event doesn't change the probability of the outcome of the second event.
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
geeta172 New user 29 Posts |
Think about it this way...the chances of you getting a head is half on a fair coin. If two tosses would double the odds...the odds of getting a head in two coin tosses would always be one...which means id never get a TT...
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Michael Daniels Inner circle Isle of Man 1609 Posts |
Or, to put it yet another way, the probability does not double because the proportion of target outcomes does not double.
I.e., tossing a coin once, the possible outcomes are H or T. The proportion of target outcomes (H) = 0.5 Tossing a coin twice (or two coins at once), the possible outcomes are HH, HT, TH, and TT. The proportion of target outcomes (at least one H) is 0.75. Mike |
TheRaven Special user 597 Posts |
Putting a ball through a hoop is not a "random" probability but a factor of skill as well as the odds of the second try being impacted by the first try -- i.e. you can make adjustments based on your first try applying learning from the first try.
Every time you flip a coin the odds of heads is 50%. If you flip 100 heads in a row, on the 101 flip, the odds of heads are still 50%. One flip does not affect the odds of the next flip. |
Michael Daniels Inner circle Isle of Man 1609 Posts |
Quote:
On Jul 11, 2014, TheRaven wrote: Theoretically you are correct on the assumption that the coin is unbiased. However, if it flipped 100 heads in a row, this would be extremely strong evidence of a biased coin. In practice, therefore, it would be more reasonable to conclude that the probability of a head on the next flip is considerably greater than 50%. Mike |
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