Like. I know they are different and that one is less likely. But could someone explain this in a way that I can explain it to my partner? I know it is true but cannot remember anything about why and how to explain my point.
In: 6
Imagine two bags, each filled with 100 marbles.
The first bag has 70 red marbles and 30 black marbles.
The second bag has 7 red marbles and 93 black marbles.
A single 70% chance would be reaching into the first bag one time hoping for a red marble.
Ten 7% chances would be reaching your hand into the second bag, grabbing a marble, putting the marbles back and then trying again until you try 10 times.
Explain like your five ….
Imagine you had a game with 11players. Each player gets given a shuffles deck of cars each with numbers 1-100 on them.
To win the game you need to find the number 69.
10 players are allowed to choose 7 cards from their deck of 100 at random.
1 player gets to choose 70 cards from Thier deck of cards at random.
Even if the 10 players team up and agree if any one of them wins then they all win. It’s still really unlikely any one of them would win with only 7 chances to find the card I’m a deck of 100.
Even if they team up they can’t increase each others chance of winning individually. Each player still only has 7 chances to find the card in the deck of 100 card.
The guy with 70 tries in his deck of 100 though obviously has a massively increased chance to win.
If you had to bet on the team of 10 players or the Single guy, who would you choose?
The odds of flipping at least 1 head in 4 flips is about a 94% (93.75 to be exact) chance. So you might think that every flip has a 94/4 or 23.5% chance of coming up heads, but we know that’s not the case, so what’s going on?
The first flip gives us a 50% chance of giving an immediate heads. If that’s the case, we can stop after one flip, we got our heads. Only half of the time we will need a second flip.
During the second flip, half of the time, we can expect to get our heads. So half of the time we need a 2nd flip and half of half of the time (25%) we will get our heads and be done after the second flip. So far, we have a 50% chance of getting heads on the first flip and a 25% chance of getting heads on the second flip. (Remember that’s the 50% odds of getting a heads times the 50% odds that we even need to flip the 2nd coin). That’s 50+25=75% leaving 25% of the outcomes that will require a 3rd flip.
This logic continues. 25% of the time we need a 3rd flip and half of those outcomes (half of 25 is 12.5%) we get the heads and are done. So 50+25+12.5 = 87.5% of the time, we get our heads in one of the first 3 flips, only 12.5% of attempts will we need to go to the 4th flip.
Of those 12.5% of attempts that require the 4th flip, half of them will give us the heads we need (half of 12.5 is 6.25%) and the other half (the remaining 6.25%) we will fail our 4th flip as well because we flipped 4 tails in a row. What are the odds of getting 4 tails out of 4 flips? It’s 1/2⁴ because there are 2⁴ possible outcomes and only one of them is TTTT. What’s 1/2⁴? It’s 6.25% awesome, we verified our logic.
And what are the odds of getting at least one heads? 50% from the 1st flip plus 25% from the 2nd, plus 12.5% from the 3rd and plus 6.25% from the 4th gives us a grand total of 93.75% what’s more, the odds that we don’t get any heads is 6.25% and since the only possible outcomes from 4 flips is at least one heads or no heads, we should get 100% if we add up the odds of both possibilities: 93.75+6.25=100 great! More verification we did it right.
If you just take the 50+50+50+50=200% chance of each flip that is double counting. Because if the second flip comes up heads, half of the time, the first flip was already heads, so we already counted that probability once. We don’t need to count it again.
This thread is about to explain to r/futurology why their expectations of otherworldly life are probably wildly off.
The easiest was i can think of is, if you divide up a sure thing. If you have a 100% Chance it always works. If you divide it up into two chances at 50% it will sometimes fail. That means you can not just add up the chances. Or, the other way around. If you could just add up the chances of every try, you would just have to try enough times, that the chances are more than 100%. But in reality you can always fail all your tries, thus the chances are not really 100%
For independent incidents each incidents probability is multiplied, not the count of incidents. This should make sense cause it’s very easy to have greater then 100% chance of something if it was that case which doesn’t fit logic of what probability is for. For some basic math, for anything with a fixed probability that are independent the following is true. The probability to get it is X, the probability to get it N times is X^N, the average amount of times you need to play to win is 1/X (note this isn’t certain win, just expected averages, if you played a million times your mean will be this), the chance to not win is 1-X (1-X is always the probability of the not case), permutations and combinations are important if a specific case is wanted outside of all or at least one (example probability of exactly 2 heads in 3 flips, each permutations is 1/8 and 3 combinations satisfy the case, so 3/8 ). The most frequent question for probability is the at least once case, cause it only cares did you win and not how many times. This can always be shown as 1-((1-X)^N), basically the probability of not losing every attempt.
Everyone commenting so far has missed that their expected values are, in fact, identical. 1 x 0.7 = 10 x (0.07) = EV in both cases.
It depends on what OP means by “less likely”…less likely to do what, to “win” at least once? In that case, yes the (P>=1)s are indeed different as the first case is 70%, and the latter is 51.6% per the binomial distribution.
In a naive way, one can think of it as “both have the same mathematical expected value. However in return for the possibility of winning > once, multiple trials decrease the odds of winning at least once.”
I’m going to ignore your 70% analogy and change it to a 100% chance vs ten 10% chances.
Now imagine there are 10 balls in a hat. 9 are “wrong” red balls and one is a “right” green ball
In the first example (100% chance), you have ten goes at picking the green ball out of the hat. If you get one wrong, you put the ball to one side and have your next go. With 10 balls and 10 attempts, you’ll 100% get it.
In the next example, if you get it wrong (or right), you put the ball back in the bag. You could get unlucky 10 times and never pick a green. It’s therefore not 100%.
You could get really lucky and pick the green ball twice, or three times or more.
But it’s not 100% – it’s not guaranteed.
Assuming the 7% chances are independent events like a weighted coin flip, the chances are waaaayyyy smaller than 7% if you flip it 10 times and certainly minuscule compared to 70% lol! You see the problem with multiple independent events is that they multiply by a power of the chance so that if you had a 7% chance of getting heads, the chance of you getting heads 10 times is 0.07^10 which is 2.82*10^-12 or 2.82*10^-10 %. That is astronomically small. Think about a normal coin flip with 50% chance of getting heads. Well the first time you flip it’s a 50% chance of getting heads. There are 4 combinations however of coin flips in 2 flips (HH, HT, TH, TT) so the chance of getting heads twice is 25%. In three flips there are 8 combinations of events so it’s 12.5% for all 3 to be heads. You get the gist.
So clearly a 70% chance is extraordinarily more favorable than 10 7% chances unless you only require one instance of the 10 events to be that 7% probability. In that case the chance that it never happens is 0.93^10 or ~48.4% because there’s a 93% chance of said event never occurring. So the chance of your 7% chance event occurring at least once in 10 instances of the same independent event are 100-48.4 or 51.6%. Still less than 70%
The math gets a bit more complicated if the events are dependent on previous outcomes or other factors.
The expected value is the same, if you always get to play to the end:
A single 70% chance to win a dollar will, on average, net you 70 cents.
The 7% chances will, on average, net you 7 cents each, for a total of an average of 70 cents if it’s 10 completely separat attempts.
But with the single 70% chance, you can at most win $1. With the 10 chances, you could, theoretically, win $10. It’s unlikely, but possible. But there’s also a 48% chance that you’ll end up with nothing (the chance of getting nothing the first time is 93%, the chance of getting nothing the first and then second time is 93% * 93%, the chance of getting unlucky 10 times in a row is 93%^(10)).
However, if you stop the game once you win, for example because the other side only has one dollar, it’s a different game that’s much worse for you – there’s still a 48% chance of getting nothing, but no chance to get more than one dollar to make up for it.
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