Okito Coin Box (Brass) - 2 Euro

Trick by Tango Magic
20.00 In stock.
Free shipping Free shipping to USA
Add to wishlist

Get 200 VI Points for this purchase.

Okito Coin Box (Brass) - 2 Euro

20.00 usd

Trick by Tango Magic (20.00)

In stock.

The magician shows a coin and a box Okito. He puts the coin inside the box and the lid. He waves the box to show that the coin is there and he places it on the back of his hand, which supports a few centimeters over the table. The magician gives a flick on the box. The coin crosses the box and falls down on the table. The box is opened and is empty.

In this package you will find a finely crafted gimmicked coin set produced by Tango Magic.
Tango Magic produces the highest quality gimmicked coins in the world. Tango stands behind their products to guarantee your success. As a bonus to thank you for your purchase, Tango has included in this package a link to Tangopedia, a four-hour instructional video. This video includes basic instructions for more than 50 gimmick coin routines using a variety of our specialty coins such as Expanded Shell-Coins, Copper/Silver/Brass, Scotch and Soda, Pen Through Coin, Folding Coins, Okito Boxes, and more. From this video you will learn the basics of using your new Tango Magic products as well as gaining insight into the amazing of possibilities the Tango line of coins provides.

 

Review Okito Coin Box (Brass) - 2 Euro

 

Community questions about Okito Coin Box (Brass) - 2 Euro

Have a question about this product? It's possible others do too. Ask here and other Vanishing Inc. Magic customers will be able to respond with assistance! Alternatively, email us and we can help too.

  • Davidson asks: Some Okito boxes are a tad larger than the coins they say they fit and a US gold new dollar $1 coin is a tad larger than a €2 will a US gold new dollar coin fit this box?

    • 1. Jim answers: No. It is only for 2 Euro coins.
    Post an answer to this question
  • Davidson asks: Will a British Ha’penny coin fit into this coin box?

      Post an answer to this question
    Ask a question