Strange Tales #131

Featuring: Human Torch and Thing
Release: January 12, 1965
Cover: April 1965
12 cents
Story by: Stan (Prolific) Lee
Illustrations by: Bob (Terrific) Powell
Delineation by: Dick (Specific) Ayers
Lettering by: S. (Hieroglyphic) Rosen
12 pages

Previous	#318	Next
Tales of Suspense #64, Story B	Reading order	Fantastic Four #37
Strange Tales #130, Story B	Strange Tales	Strange Tales #131, Story B
Captain America Comics #1, Story C	Prelude

Continuing through the Human Torch stories in Strange Tales while saving the Dr. Strange stories for much later.

Why is the story titled the “Bouncing Ball of Doom”? Because the Thinker’s plan involves a bouncing ball.

Huh.

How many Human Torch stories do we still have to read?

I must emphasize that the Dr. Strange story is one of the single greatest Marvel stories of all time, yet the Bouncing Ball of Doom is what gets spotlighted on the cover.

Ben and Johnny get invited to a dam opening ceremony.

This story isn’t very interesting. You know what is more interesting than this story? Math. Math is more interesting than this story.

The Mad Thinker claimed his bank robbery plan had a 99.99999% chance of success. He is saying that his schemes only fail once every 10 million or so attempts. His plan didn’t account for the possibility of superheroes passing by. In Manhattan, a city now bursting with them.

I don’t really believe his calculations, but I don’t think I even need to know information about how many Manhattan crimes are stopped by superheroes to prove he’s wrong. I just need to know that we’ve seen the Thinker fail 4 times in a row, each time while claiming a similar chance of success. Is that even possible?

He is claiming his odds of failure are 1 in 10 million. Not many things in this world are that likely. That is about certain as the sun, rising in the East.

Let’s think this through. First, let’s suppose the Thinker instead claimed a 50% chance of success for his plans. That’s just like flipping a fair coin. Heads means success for the Thinker and tails means failure. If his plans had a 50% chance of success, does it make sense he would lose 4 times in a row?

Sure. That’s just like landing tails 4 times in a row.

If we let P(F) represent the probability of failing once, then the probability of failing 4 times in a row is P(F)⁴.

If his plans were as likely to fail as succeed (like a fair coin toss), the probability of failing 4 times in a row is 6.25%, the probability of landing tails 4 times in a row. Is that a high probability? No, it doesn’t happen that often that you get 4 tails in a row. Does that mean it’s impossible to get 4 tails in a row? Not even close. 6% is well within the realm of the possible. Events of a similar likelihood happen all the time.

Suppose now the Thinker claimed a 90% chance of success. That is like flipping a weighted coin, a coin specifically designed to land on its head a lot. Suppose the coin is designed to land on its head 9/10 times, 90% of the time. Is it possible to get 4 tails in a row with such a coin?

Sure. Very unlikely, but possible.

If the Thinker’s plans had a 90% chance of success, the probability of failure would be P(F)=0.1, and the probability of 4 failures would be 0.01%, one hundredth of a percent. Is that very likely? No. Is it possible? Yes. Honestly, probably more likely than a radioactive spider bite bestowing spider powers.

It quickly becomes easier to think about dice than coins. If you want to play with it, find a ten-sided die, a D10. Pretend rolling a 1 is a failure for the Thinker and rolling a 2-10 is success. Is it possible to roll four 1’s in a row?

Possible, but very hard. If the Thinker were claiming his plan had a 90% chance of success, then that’s as likely as rolling a 2-10. That’s about as good as one could actually hope for in as complex a system as our world.

And if he tried 4 times with those odds, I am actually 99.99% confident he would have succeeded at least once. Said differently, I am 99.99% confident his plans had a success rate of less than 90%.

I decided to test it out; I have a D10 with numbers 0-9.

I think I’d have to try to roll the dice around 7000 times before I would confidently expect 4 consecutive 1s. I’m not willing to do that, but I can roll until I get bored. Remember, if we see 4 consecutive 1s, that represents 4 failures for the Mad Thinker if his plans had a 90% chance of success.

2	7	1	6	3	8	9	3
0	2	6	1	5	2	7	7
4	3	6	2	5	8	8	3
4	9	6	5	7	4	6	9
5	7	4	3	4	1	1	3
3	6	1	6	3	5	7	2
0	9	9	4	2	9	3	2
5	5	4	1	7	4	7	1
3	0	2	6	0	4	8	3
6	0	3	2	4	5	7	2

In the chart above, imagine each 1 is a failure of one of Thinker’s schemes, in the case where his chances of success are 90%. I saw two consecutive failures in this case, but not 4.

He claimed a success rate of 99.99999%. Since he has failed 4 times in a row, a possibility that only has a 0.00000000000000000001% chance of happening if his estimates were correct, I am 99.99999999999999999999% certain that he massively overestimates his chances of success.

Put differently, if his estimates were correct, his chances of having 4 consecutive failures are 1 in 10 octillion, about the odds of me winning the lottery 5 times in a row.

A caveat. I harped on his having 4 consecutive failures. I guess it’s possible there were several unseen successful heists in between the failures. We would need to think a little harder about the math if we remove the word “consecutive” everywhere, and we would need to guess at his number of successful schemes, but we’ll still find the same conclusion: that he way overestimates himself.

You see above that there are plenty of 1s spread throughout.

With a 1-in-10 chance of failure, if he pulled off 40 capers, we’d expect about 4 of them to fail. And perhaps those are the 4 that Stan chose to chronicle. But if he had a 1-in-10 million chance of failure as he claimed…

We could continue the math in this non-consecutive case, but probably enough to absorb for now. In that case, a 90% success rate was feasible, so I would want to move from the D10 to the D100 and see that a 99% success rate was not.

I am 90% confident his plans’ actual success rate is less than 44%, and he just way overestimates his own planning abilities.

For example, in this case, the Thinker’s plan involved Thing punching a ball at a very precise moment. Thing foiled the Thinker’s plan by ducking. By ducking. Not only do I think Ben ducking is a plausible occurrence, I’m beginning to think the Thinker hasn’t thought through any of his plans.

Rating: ★★☆☆☆, 35/100
Significance: ★★★☆☆

You can find this story in Marvel Masterworks: Human Torch vol. 2 or The Human Torch & The Thing: Strange Tales – The Complete Collection. Or on Kindle.

Characters:

• Human Torch/Johnny Storm
• Thing/Benjamin “Ben” Grimm
• (Mad) Thinker
• Invisible Girl
• Mr. Fantastic
• Dr. Vega

Minor characters:

• Hennessy (police officer)

Story notes:

• Dam almost floods. Dr. Vega heroically tries to stop it.
• Thinker calls himself Thinker. Reed calls him Mad Thinker.

Previous	#318	Next
Tales of Suspense #64, Story B	Reading order	Fantastic Four #37
Strange Tales #130, Story B	Strange Tales	Strange Tales #131, Story B
Captain America Comics #1, Story C	Prelude

While I couldn’t roll a die 7,000 times, I could simulate it with a computer. Here’s what I got, with the sides of my virtual die numbered 0-9. Sure enough, there is a single run of 4 1’s. See it?

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If the Thinker had successfully executed 7000 operations and we just happened to witness a short run of failures, I might believe his plans had a 90% chance of success. Will never be convinced about 99.999999 though…