Because there’s an overlap where you can have multiple 7% chances come through.

Imagine a coin flipped twice, compared to a sure chance of 100%.

Flipping a coin twice gives you 50% chance to get 1 heads, 25% chance to get 2 heads, and 25% chance to get no heads. You still, on average, get 1 head (1*50%+2*25%), but that isn’t the same as having a 100% chance to get a head.

Think of it like raffle tickets.

In the first scenario, there are 100 raffle tickets and you own 70 of them.

In the second scenario, there are ten individual raffles. Each raffle still has 100 tickets, of which you own 7. At the end of the day, you still have 70 tickets, but there are 1000 tickets total.

Think about it like this: If something has a 70% chance of occurring then it has a 30% chance of not occurring. If something has a 7% chance of occurring it has a 93% chance of not occurring. If you have multiple trials of something the chance that it does not happen is the chance of not happening times itself for each trial. 10 trials with a 7% outcome has a 39.8% chance of happening (since 93%^10 =60.2%)

Let’s imagine that these odds apply to the odds of winning a free ice cream cone with the purchase of one at normal price.

At one ice cream shop they give you one spin of the wheel where 70% of the wheel comes up with a free extra ice cream cone and 30% you lose. It’s pretty clear they’re only two possible outcomes 70% of the time you get the free ice cream cone 30% of the time you get nothing.

Now let’s consider the second ice cream shop where they give you 10 spins of a wheel where 7% of the wheel represents winning a free ice cream cone. Here it’s possible, although unlikely, that you could actually win all 10 times. You might lose every time, you might win every time or you might get an outcome somewhere in the middle. Winning anywhere from 0 to 10 ice cream cones is very different than the other ice cream shop where you could only get 0 or 1 ice cream cone.

There are ways to calculate exactly what the odds are of getting 0, 1, 2 3 or etc ice cream cones.

That said the outcome you might be most interested in is the chance that you will get at least one ice cream cone. if we had a pie chart where each slice represented the share of times that each number of ice cream cones came up from 0 to 10, we could see how big the zero slice was, and subtract from 100% to get combined odds for winning one to 10 ice cream cones. The combined odds for getting one to ten ice cream cones can also be called the chance of getting “at least one”.

OK so let’s figure out what the odds of getting zero ice cream cones are. On the first spin the odds of not getting an ice cream cone are 93%. The odds of not getting an ice cream cone on either the first or second spin are 93% of 93%. And this continues until you multiply 93% by itself 10 times. If you do this math, you get 48% as the chance of missing on all 10 Spins.

So we can see that the odds of not getting an ice cream cone all are very different at the two different shops. At the first shop there is only a 30% chance of not getting an ice cream cone. At the second shop, there is a 48% chance of not getting an ice cream cone at all. However, at the second shop there’s a 52% chance of getting one or perhaps more ice cream cones.

Sorry this got a bit long but hopefully the first three paragraphs give you the simple explain it like I’m five answer you were looking for.

The expected number of “wins” in either case is the same: 0.7. But they’re distributed differently, so the chance of **at least one** win is different.

The reason is that your ten 7% chances includes the possibility of *multiple* wins. To get the same average number of wins, those multiple wins have to also be balanced out by a higher chance of zero wins.

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More mathematically, the chance of zero wins here is the chance of not winning at each step, multiplied together. So for your 10 7% chances, your chance of not winning is 93% (that is, 1 – 0.07), and you end up with (1 – 0.07) * (1 – 0.07) * … * (1 – 0.07) ten times. That comes out to (1 – 0.07)^10 = (0.93)^10 = about 0.48, or 48%.

It may help to think of things in smaller, easier to intuit numbers. How about one 100% chance and two 50% chances? Obviously two coin flips can come up both tails, so the two must be different, and it’s worth actually listing out every option to see why. The single 100% chance has only one option: “win”. The two 50% chances have four equally likely options: “win win”, “win lose”, “lose win”, and “lose lose”. In both cases, you average exactly one win, but in the second, the “win win” case gets balanced out by a “lose lose” one.

Flip the thing you’re looking at – instead of a 70% vs 7%x10, look at 30% vs 93%x10. In the first case, there’s a 30% chance of losing. In the second, you have a 93% chance to lose, then a 93% chance on the next one, etc. Since you’re only interested in cases where you lose every single time (because otherwise you’ll have succeeded at least once) you can multiply the chances. So what’s 93% * 93% * 93%…etc? A bit higher than 48%. So the first option has only a 30% chance of losing while the second one has a 48% chance of losing overall.

Depending on what you meant. They could be the same or different.

If you have 1 – 10 numbered on marbles and just need to pull 1-7, then each one has a 10% chance for a collective 70%

If you are talking about making attempts at something over and over you multiply the odds against each other.

(7% x 7% x 7%) = .0343% chance you hit 3 times in a row.

If you just want to win once then it’s slightly backwards, calculate the odds of failing over and over then flip it.

100% – [(100% – 7%) * (100% – 7%) etc. ]

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